Find the probability that a randomly chosen male bird weighs between 4.75 kg and 4.85 kg.

Find the probability that the weight of a randomly chosen male bird is more than twice the weight of a randomly chosen female bird.

Two randomly chosen male birds and three randomly chosen female birds are placed on a weighing machine that has a weight limit of 18 kg. Find the probability that the total weight of these five birds is greater than the weight limit.

Note: In question 1, accept answers that round correctly to 2 significant figures.

P(4.75 < X < 4.85) = 0.197      A1

[1 mark]

Note: In question 1, accept answers that round correctly to 2 significant figures.

consider the random variable X − 2Y     (M1)

E(X − 2Y) =  − 0.6     (A1)

Var(X − 2Y) = Var(X) + 4Var(Y)     (M1)

= 0.13     (A1)

X − 2Y ∼ N(−0.6, 0.13)

P(X − 2Y > 0)     (M1)

= 0.0480     A1

[6 marks]

Note: In question 1, accept answers that round correctly to 2 significant figures.

let W = X₁ + X₂ + Y₁ + Y₂ + Y₃ be the total weight

E(W) = 17.7     (A1)

Var(W) = 2Var(X) + 3Var(Y) = 0.1475     (M1)(A1)

W ∼ N(17.7, 0.1475)

P(W > 18) = 0.217     A1

[4 marks]

State suitable hypotheses to investigate whether or not \(U\), \(V\) are independent.

Find the least value of \(|r|\) for which the test concludes that \(\rho  \ne 0\).

\({{\text{H}}_0}:\rho  = 0;{\text{ }}{{\text{H}}_1}:\rho  \ne 0\)     A1A1

[2 marks]

\(\nu  = 10\)     (A1)

\({t_{0.005}} = 3.16927 \ldots \)     (M1)(A1)

we reject \({{\text{H}}_0}:\rho  = 0\) if \(\left| t \right| > 3.16927 \ldots \)     (R1)

attempting to solve \(\left| r \right|\sqrt {\frac{{10}}{{1 - {r^2}}}}  > 3.16927 \ldots \) for \(\left| r \right|\)     M1

Note:     Allow = instead of >.

(least value of \(\left| r \right|\) is) 0.708 (3 sf)     A1

Note:     Award A1M1A0R1M1A0 to candidates who use a one-tailed test. Award A0M1A0R1M1A0 to candidates who use an incorrect number of degrees of freedom or both a one-tailed test and incorrect degrees of freedom.

Note: Possible errors are

10 DF 1-tail, \(t = 2.763 \ldots \), least value \( = \) 0.658

11 DF 2-tail, \(t = 3.105 \ldots \), least value \( = \) 0.684

11 DF 1-tail, \(t = 2.718 \ldots \), least value \( = \) 0.634.

[6 marks]

(i)     Find \({\text{E}}({X_1})\).

(ii)     Hence obtain an unbiased estimator for \(p\).

The die is rolled a second time, and the score \({X_2}\) is noted.

(i)     Show that \(k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)\) is also an unbiased estimator for \(p\) for all values of \(k \in \mathbb{R}\).

(ii)     Find the value for \(k\), which maximizes the efficiency of this estimator.

let \(X\) denote the score on the die

(i)     \({\text{P}}(X = x) = \left\{ {\begin{array}{*{20}{c}} {\frac{{1 - p}}{5},}&{x = 1,{\text{ 2}},{\text{ 3}},{\text{ 4}},{\text{ 5}}} \\ {p,}&{x = 6} \end{array}} \right.\)     (M1)

\(E({X_1}) = (1 + 2 + 3 + 4 + 5)\frac{{1 - p}}{5} + 6p\)     M1

\( = 3 + 3p\)     A1

(ii)     so an unbiased estimator for \(p\) would be \(\frac{{{X_1} - 3}}{3}\)     A1

[4 marks]

(i)     \(E\left( {k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)} \right)\)     M1

\( = kE({X_1} - 3) + \left( {\frac{1}{3} - k} \right)E({X_2} - 3)\)     M1

\( = k(3p) + \left( {\frac{1}{3} - k} \right)(3p)\)     A1

any correct expression involving just \(k\) and \(p\)

\( = p\)     AG

hence \(k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)\) is an unbiased estimator of \(p\)

(ii)     \({\text{Var}}\left( {k({X_1} - 3) + \left( {\frac{1}{3} - k} \right)({X_2} - 3)} \right)\)     M1

\( = {k^2}{\text{Var}}({X_1} - 3) + {\left( {\frac{1}{3} - k} \right)^2}{\text{Var}}({X_2} - 3)\)     A1

\( = \left( {{k^2} + {{\left( {\frac{1}{3} - k} \right)}^2}} \right){\sigma ^2}\) (where \({\sigma ^2}\) denotes \({\text{Var}}(X)\))

valid attempt to minimise the variance     M1

\(k = \frac{1}{6}\)      A1

Note:     Accept an argument which states that the most efficient estimator is the one having equal coefficients of \({X_1}\) and \({X_2}\).

[7 marks]

Total [11 marks]

One hundred men from that country, selected at random, had their heights measured.

The mean of this sample was \(185\) cm. Calculate a \(95\% \) confidence interval for the mean height of the population.

A second random sample of size \(n\) is taken from the same population. Find the minimum value of \(n\) needed for the width of a \(95\% \) confidence interval to be less than \(3\) cm.

valid attempt to use \(\bar x \pm z\frac{\sigma }{{\sqrt n }}\)     (M1)

\([182,{\text{ }}188]\)     A1A1

Note:     Accept answers that round to the correct \(3\) sf.

[3 marks]

\(1.96 \times \frac{{15.0}}{{\sqrt n }} < 1.5\)     M1A1

\(n > {\left( {\frac{{15.0}}{{1.5}} \times 1.96} \right)^2}\)     (M1)

Note:     Award M1 for attempting to solve the inequality.

Note:     Allow the use of \( = \).

minimum value \(n = 385\)     A1

[4 marks]

Total [7 marks]

Determine the product moment correlation coefficient for these data.

Perform a two-tailed test, at the \(5\% \) level of significance, of the hypothesis that strength is independent of moisture content.

If the moisture content of a beam is found to be \(9.5\), use the appropriate regression line to estimate the strength of the beam.

\(r =  - 0.762\)     (M1)A1

Note:     Accept answers that round to \( - 0.76\).

[2 marks]

\({H_0}:\) Moisture content and strength are independent or \(\rho  = 0\)

\({H_1}:\) Moisture content and strength are not independent or \(\rho  \ne 0\)     A1

EITHER

test statistic is \(-3.33\)     A1

critical value is \(( \pm ){\text{ }}2.306\)     A1

since \( - 3.33 <  - 2.306\) or \(3.33 > 2.306\),     R1

reject \({H_0}\;\;\;\)(or equivalent)     A1

OR

\(p\)-value is \(0.0104\)     A2

as \(0.0104 < 0.05\),     R1

reject \({H_0}\;\;\;\)(or equivalent)     A1

Note:     The R1 and A1 can be awarded as follow through from their test statistic or \(p\)-value.

[5 marks]

\(x = {\text{strength}}\)

\(y = {\text{moisture content}}\)

\(x =  - 0.629y + 28.1\)     (M1)(A1)

if \(y = 9.5\) so \(x = 22.1\)     (M1)A1

Note:     Only accept answers that round to \(22.1\).

Note:     Award M1A1M0A0 for the other regression line \(y = 30.1 - 0.924x\).

[4 marks]

Total [11 marks]

Anna cycles to her new school. She records the times taken for the first ten days with the following results (in minutes).

12.4 13.7 12.5 13.4 13.8 12.3 14.0 12.8 12.6 13.5

Assume that these times are a random sample from the \({\text{N}}(\mu ,{\text{ }}{\sigma ^2})\) distribution.

(a)     Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\).

(b)     Calculate a 95 % confidence interval for \(\mu \).

(c)     Before Anna calculated the confidence interval she thought that the value of \(\mu \) would be 12.5. In order to check this, she sets up the null hypothesis \({{\text{H}}_0}:\mu  = 12.5\).

(i)     Use the above data to calculate the value of an appropriate test statistic. Find the corresponding p-value using a two-tailed test.

(ii)     Interpret your p-value at the 1 % level of significance, justifying your conclusion.

(a)     estimate of \(\mu = 13.1\)     A1

estimate of \({\sigma ^2} = 0.416\)     A1

[2 marks]

(b)     using a GDC (or otherwise), the 95% confidence interval is     (M1)

[12.6, 13.6]     A1A1

Note: Accept open or closed intervals.

[3 marks]

(c)     (i)     \(t = \frac{{13.1 - 12.5}}{{0.6446 \ldots /\sqrt {10} }} = 2.94\)     (M1)A1

\(v = 9\)     (A1)

p-value \( = 2 \times {\text{P}}(T > 2.9433 \ldots )\)     (M1)

\( = 0.0164\,\,\,\,\,\)(accept 0.0165)     A1

(ii)     we accept the null hypothesis (the mean travel time is 12.5 minutes)     A1

because 0.0164 (or 0.0165) > 0.01     R1

Note: Allow follow through on their p-value.

[7 marks]

Total [12 marks]

This was well answered by many candidates. In (a), some candidates chose the wrong standard deviation from their calculator and often failed to square their result to obtain the unbiased variance estimate. Candidates should realise that it is the smaller of the two values (ie the one obtained by dividing by (n – 1)) that is required. The most common error was to use the normal distribution instead of the t-distribution. The signpost towards the t-distribution is the fact that the variance had to be estimated in (a). Accuracy penalties were often given for failure to round the confidence limits, the t-statistic or the p-value to three significant figures.

Show that \({\text{P}}(X \leqslant n) = 1 - {(1 - p)^n},{\text{ }}n \in {\mathbb{Z}^ + }\) .

Deduce an expression for \({\text{P}}(m < X \leqslant n)\,,{\text{ }}m\,,{\text{ }}n \in {\mathbb{Z}^ + }\) and m < n .

Given that p = 0.2, find the least value of n for which \({\text{P}}(1 < X \leqslant n) > 0.5\,,{\text{ }}n \in {\mathbb{Z}^ + }\) .

\({\text{P}}(X \leqslant n) = \sum\limits_{{\text{i}} = 1}^n {{\text{P}}(X = {\text{i}}) = \sum\limits_{{\text{i}} = 1}^n {p{q^{{\text{i}} - 1}}} } \)     M1A1

\( = p\frac{{1 - {q^n}}}{{1 - q}}\)     A1

\( = 1 - {(1 - p)^n}\)     AG

[3 marks]

\({(1 - p)^m} - {(1 - p)^n}\)     A1

[1 mark]

attempt to solve \(0.8 - {(0.8)^n} > 0.5\)     M1

obtain n = 6     A1

[2 marks]

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

In part (a) some candidates thought that the geometric distribution was continuous, so attempted to integrate the pdf! Others, less seriously, got the end points of the summation wrong.

In part (b) It was very disappointing that may candidates, who got an incorrect answer to part (a), persisted with their incorrect answer into this part.

For these data find the product moment correlation coefficient, \(r\).

(i)     State the distribution of the test statistic (including any parameters).

(ii)     Find the \(p\)-value for the test.

(iii)     State the conclusion, in the context of the question, with the word “correlation” in your answer. Justify your answer.

Using a suitable regression line, find an estimate for his score on Exam 2, giving your answer to the nearest integer.

(i)     State the distribution of your test statistic (including any parameters).

(ii)     Find the \(p\)-value.

(iii)     State the conclusion, justifying the answer.

\(r = 0.804\)    A2

Note: Accept any number that rounds to 0.80.

[2 marks]

(i)     \(t\) distribution with 8 degrees of freedom     A1A1

(ii)     \(p{\text{ - value}} = 0.00254\)     A2

Notes: Accept any number that rounds to 0.0025.

Award A1 for 2-tail test giving an answer that rounds to 0.0051.

(iii)     \(p{\text{ - value}} < 0.01\), so conclude that there is positive correlation     R1A1

Notes: Only award the A1 if the R1 is awarded.

Do not accept just “reject \({H_0}\)” or “accept \({H_1}\)”.

The words “positive correlation” must be seen.

[6 marks]

regression line of \(y\) (Exam 2 mark) on \(x\) (Exam 1 mark) is     (M1)

\(y = 0.59407 \ldots x + 21.387 \ldots \)    (A1)

\(x = 11\) gives \(y = 28\) (to nearest integer)     A1

[3 marks]

(i)     applying the \(t\) test to the differences

\(t\) distribution with 9 degrees of freedom     A1A1

(ii)     \(p{\text{ - value}} = 0.239\)     A2

Notes: Accept any number that rounds to 0.24.

Award A1 if subtraction done the wrong way round giving \(p{\text{ - value}} = 0.109\).

(iii)     \(p{\text{ - value}} > 0.05\), so accept \({H_0}\) or \({u_d} = 6\)     R1A1

[6 marks]

Show that \(U = a{\bar X_1} + (1 - a){\bar X_2},{\text{ }}a \in \mathbb{R}\), is an unbiased estimator of \(\mu \).

Show that \({\text{Var}}(U) = {a^2}\frac{{{\sigma ^2}}}{{{n_1}}} + {(1 - a)^2}\frac{{{\sigma ^2}}}{{{n_2}}}\).

Find, in terms of \({n_1}\) and \({n_2}\), an expression for \(a\) which gives the most efficient estimator of this form.

Hence find an expression for the most efficient estimator and interpret the result.

\({\text{E}}(U) = E(a{\bar X_1} + (1 - a){\bar X_2}) = a{\text{E}}({\bar X_1}) + (1 - a){\text{E}}({\bar X_2})\)     (M1)

\({\text{E}}({\bar X_1}) = \mu \) and \({\text{E}}({\bar X_2}) = \mu \)

\({\text{E}}(U) = a\mu  + (1 - a)\mu \) (or equivalent)      A1

\( = \mu \)     A1

hence \(U\) is an unbiased estimator of \(\mu \)     AG

[3 marks]

\({\text{Var}}(U) = {\text{Var}}(a{\bar X_1} + (1 - a){\bar X_2})\)

\( = {a^2}{\text{Var}}({\bar X_1}) + {(1 - a)^2}{\text{Var}}({\bar X_2})\)     M1

stating that \({\text{Var}}({\bar X_1}) = \frac{{{\sigma ^2}}}{{{n_1}}}\) and \({\text{Var}}({\bar X_2}) = \frac{{{\sigma ^2}}}{{{n_2}}}\)     A1

\( \Rightarrow {\text{Var}}(U) = {a^2}\frac{{{\sigma ^2}}}{{{n_1}}} + {(1 - a)^2}\frac{{{\sigma ^2}}}{{{n_2}}}\)     AG

Note:     Line 3 or equivalent must be seen somewhere.

[2 marks]

let \({\text{Var}}(U) = V\)

EITHER

\(\frac{{{\text{d}}V}}{{{\text{d}}a}} = 2a\frac{{{\sigma ^2}}}{{{n_1}}} - 2(1 - a)\frac{{{\sigma ^2}}}{{{n_2}}}\)     M1

attempting to solve \(\frac{{{\text{d}}V}}{{{\text{d}}a}} = 0\) for \(a\)     R1

Note:     Award M1 for obtaining \(a\) in terms of \({n_1},{\text{ }}{n_2}\) and \(\sigma \).

OR

forming a quadratic in \(a\)

\(V = \left( {\frac{{{\sigma ^2}}}{{{n_1}}} + \frac{{{\sigma ^2}}}{{{n_2}}}} \right){a^2} - 2\frac{{{\sigma ^2}}}{{{n_2}}}a + \frac{{{\sigma ^2}}}{{{n_2}}}\)     M1

attempting to find the axis of symmetry of V     R1

THEN

\(a = \frac{{\frac{{2{\sigma ^2}}}{{{n_2}}}}}{{2{\sigma ^2}\left( {\frac{1}{{{n_1}}} + \frac{1}{{{n_2}}}} \right)}}\)     (A1)

\(a = \frac{{{n_1}}}{{{n_1} + {n_2}}}\)     A1 

[4 marks]

substituting \(a\) into \(U\)     (M1)

\(U = \frac{{{n_1}{{\bar X}_1} + {n_2}{{\bar X}_2}}}{{{n_1} + {n_2}}}\)     A1

Note:     Do not FT an incorrect \(a\) for A1, the M1 may however be awarded.

this is an expression for the mean of the combined samples

OR this is a weighted mean of the two sample means     R1

[3 marks]

A teacher has forgotten his computer password. He knows that it is either six of the letter J followed by two of the letter R (i.e. JJJJJJRR) or three of the letter J followed by four of the letter R (i.e. JJJRRRR). The computer is able to tell him at random just two of the letters in his password.

The teacher decides to use the following rule to attempt to find his password.

If the computer gives him a J and a J, he will accept the null hypothesis that his password is JJJJJJRR.

Otherwise he will accept the alternative hypothesis that his password is JJJRRRR.

(a)     Define a Type I error.

(b)     Find the probability that the teacher makes a Type I error.

(c)     Define a Type II error.

(d)     Find the probability that the teacher makes a Type II error.

(a)     a Type I error is when \({{\text{H}}_0}\) is rejected, when \({{\text{H}}_0}\) is actually true     A1

[1 mark]

(b)     \({\text{P(}}{{\text{H}}_0}{\text{ rejected}}|{{\text{H}}_0}{\text{ true)}} = {\text{P(at least one R}}|{\text{6 J and 2 R)}}\)     M1

EITHER

\({\text{P(no R}}|{{\text{H}}_0}{\text{ true)}} = \frac{6}{8} \times \frac{5}{7} = \frac{{15}}{{28}}\)     (A1)

OR

let X count the number of R’s given by the computer under \({{\text{H}}_0},{\text{ }}X \sim {\text{Hyp(}}2,{\text{ }}2,{\text{ }}8)\)

\({\text{P}}(X = 0) = \frac{{\left( {\begin{array}{*{20}{c}}
  2 \\
  0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  6 \\
  2
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  8 \\
  2
\end{array}} \right)}} = \frac{{15}}{{28}}\)     (A1)

THEN

\({\text{P(at least one R}}|{{\text{H}}_0}{\text{ true)}} = 1 - \frac{{15}}{{28}}\)     (M1)

\({\text{P(Type I error)}} = \frac{{13}}{{28}}\,\,\,\,\,( = 0.464)\)     A1

[4 marks]

(c)     a Type II error is when \({{\text{H}}_0}\) is accepted, when \({{\text{H}}_0}\) is actually false     A1

[1 mark]

(d)     \({\text{P(}}{{\text{H}}_0}{\text{ accepted}}|{{\text{H}}_0}{\text{ false)}} = {\text{P(2 J}}|{\text{3 J and 4 R)}}\)     M1

EITHER

\({\text{P(2 J}}|{{\text{H}}_0}{\text{ false)}} = \frac{3}{7} \times \frac{2}{6} = \frac{1}{7}\)     (A1)

OR

let Y count the number of R’s given by the computer.

\({{\text{H}}_0}\) false implies \(Y \sim {\text{Hyp(}}2,{\text{ }}4,{\text{ }}7)\)

\({\text{P}}(Y = 0) = \frac{{\left( {\begin{array}{*{20}{c}}
  4 \\
  0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  3 \\
  2
\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}
  7 \\
  2
\end{array}} \right)}} = \frac{1}{7}\)     (A1)
THEN

\({\text{P}}({\text{Type II error)}} = \frac{1}{7}( = 0.143)\)     A1

[3 marks]

Total [9 marks]

Poorer candidates just gained the 2 marks for saying what a Type I and Type II error were and could not then apply the definitions to obtain the conditional probabilities required. It was clear from some crossings out that even the 2 definition continue to cause confusion. Good, clear-thinking candidates were able to do the question correctly.

The random variable X has the negative binomial distribution NB(3, p) .

Let \(f(x)\) denote the probability that X takes the value x .

(i)     Write down an expression for \(f(x)\) , and show that

\[\ln f(x) = 3\ln \left( {\frac{p}{{1 - p}}} \right) + \ln (x - 1) + \ln (x - 2) + x\ln (1 - p) - \ln 2{\text{ .}}\]

(ii)     State the domain of f .

(iii)     The domain of f is extended to \(]2,{\text{ }}\infty [\) . Show that

\(\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln (1 - p){\text{ .}}\)

Jo has a biased coin which has a probability of 0.35 of showing heads when tossed. She tosses this coin successively and the \({3^{{\text{rd}}}}\) head occurs on the \({Y^{{\text{th}}}}\) toss. Use the result in part (a)(iii) to find the most likely value of Y .

(i)     \(f(x) = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}{(1 - p)^{x - 3}}\)     M1A1

Note: Award M1A0 for \(f(x) = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}{q^{x - 3}}\)

taking logs,     M1

\(\ln f(x) = \left( {\ln \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  2
\end{array}} \right){p^3}(1 - p){}^{x - 3}} \right)\)

\( = \ln \left( {\frac{{(x - 1)(x - 2)}}{2} \times {p^3}{{(1 - p)}^{x - 3}}} \right)\)     A1

Note: Award A1 for simplifying binomial coefficient, seen anywhere.

\( = \ln \left( {\frac{{(x - 1)(x - 2)}}{2} \times {p^3}\frac{{{{(1 - p)}^x}}}{{{{(1 - p)}^3}}}} \right)\)     A1

Note: Award A1 for correctly splitting \({{{(1 - p)}^{x - 3}}}\) , seen anywhere.

\( = 3\ln \left( {\frac{p}{{1 - p}}} \right) + \ln (x - 1) + \ln (x - 2) + x\ln (1 - p) - \ln 2\)     AG

(ii)     the domain is {3, 4, 5, …}     A1

Note: Do not accept \(x \geqslant 3\)

(iii)     differentiating with respect to x ,     M1

\(\frac{{f'(x)}}{{f(x)}} = \frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln (1 - p)\)     AG 

[7 marks]

setting \(f'(x) = 0\) and putting p = 0.35 ,

\(\frac{1}{{x - 1}} + \frac{1}{{x - 2}} + \ln 0.65 = 0\)     M1A1

solving, x = 6.195…     A1

we need to check x = 6 and 7

f (6) = 0.1177… and f (7) = 0.1148…     A1

the most likely value of Y is 6     A1

Note: Award the final A1 for the correct conclusion even if the previous A1 was not awarded.

[5 marks]

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being \(x \geqslant 3\) . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

In general, candidates were able to start this question, but very few wholly correct answers were seen. Most candidates were able to write down the probability function but the process of taking logs was often unconvincing. The vast majority of candidates gave an incorrect domain for f, the most common error being \(x \geqslant 3\) . Most candidates failed to realise that the solution to (b) was to be found by setting the right-hand side of the given equation equal to zero. Many of the candidates who obtained the correct answer, 6.195…, then rounded this to 6 without realising that both 6 and 7 should be checked to see which gave the larger probability.

If the dice is fair, write down the distribution of X , including the value of any parameter(s).

Write down E(X ) for the distribution in part (a).

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Write down the distribution of Y , including the value of any parameter(s).

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find the value of y such that \({\text{P}}(Y = y) = \frac{1}{{36}}\).

Before Jenny’s Dad can start, he has to throw two “sixes” using a fair, ordinary six-sided dice. Let the random variable Y denote the total number of times Jenny’s Dad has to throw the dice until he obtains his second “six”.

Find \({\text{P}}(Y \leqslant 6)\) .

\(X \sim {\text{Geo}}\left( {\frac{1}{6}} \right){\text{ or NB}}\left( {1,\frac{1}{6}} \right)\)     A1

[1 mark]

\({\text{E}}(X) = 6\)     A1

[1 mark]

Y is \({\text{NB}}\left( {2,\frac{1}{6}} \right)\)     A1

[1 mark]

\({\text{P}}(Y = y) = \frac{1}{{36}}{\text{ gives }}y = 2\)     A1

(as all other probabilities would have a factor of 5 in the numerator)

[1 mark]

\({\text{P}}(Y \leqslant 6) = {\left( {\frac{1}{6}} \right)^2} + 2\left( {\frac{5}{6}} \right){\left( {\frac{1}{6}} \right)^2} + 3{\left( {\frac{5}{6}} \right)^2}{\left( {\frac{1}{6}} \right)^2} + 4{\left( {\frac{5}{6}} \right)^3}{\left( {\frac{1}{6}} \right)^2} + 5{\left( {\frac{5}{6}} \right)^4}{\left( {\frac{1}{6}} \right)^2}\)     (M1)

\( = 0.263\)     A1

[2 marks]

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

This was well answered as the last question should be the most difficult. It seemed accessible to many candidates, if they realised what the distributions were. The goodness of fit test was well used in (c) with hardly any candidates mistakenly combining cells. Part (e) was made more complicated than it needed to be with calculator solutions when a bit of pure maths would have sufficed. Part (f) caused some problems but good candidates did not have too much difficulty.

(a)     Find the probability that the weight of a randomly chosen apple is more than double the weight of a randomly chosen pear.

(b)     A shopper buys 3 apples and 4 pears. Find the probability that the total weight is greater than 1000 grams.

(a)     Let X, Y (grams) denote respectively the weights of a randomly chosen apple, pear.

Then

\(X - 2Y{\text{ is N}}(200 - 2 \times 120,{\text{ }}{15^2} + 4 \times {10^2}),\)     (M1)(A1)(A1)

i.e. \({\text{N}}( - 40,{\text{ }}{25^2})\)     A1

We require

\({\text{P}}(X > 2Y) = {\text{P}}(X - 2Y > 0)\)     (M1)(A1)

\( = 0.0548\)     A2

[8 marks]

(b)     Let \(T = {X_1} + {X_2} + {X_3} + {Y_1} + {Y_2} + {Y_3} + {Y_4}\) (grams) denote the total weight.

Then

\(T{\text{ is N}}(3 \times 200 + 4 \times 120,{\text{ }}3 \times {15^2} + 4 \times {10^2}),\)     (M1)(A1)(A1)

i.e. \({\text{N(1080, 1075)}}\)     A1

\({\text{P}}(T > 1000) = 0.993\)     A2

[6 marks]

Total [14 marks]

The response to this question was disappointing. Many candidates are unable to differentiate between quantities such as \(3X{\text{ and }}{X_1} + {X_2} + {X_3}\) . While this has no effect on the mean, there is a significant difference between the variances of these two random variables.

Find the cumulative distribution function \(F(t)\), for \(0 \leqslant t \leqslant 2\).

Sketch the graph of \(F(t)\) for \(0 \leqslant t \leqslant 2\), clearly indicating the coordinates of the endpoints.

Given that \(P(T < a) = 0.75\), find the value of \(a\).

\(F(t) = \int_0^t {\left( {x - \frac{{{x^3}}}{4}} \right){\text{d}}x{\text{ }}\left( { = \int_0^t {\frac{{x(4 - {x^2})}}{4}{\text{d}}x} } \right)} \)     M1

\( = \left[ {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{{16}}} \right]_0^t{\text{ }}\left( { = \left[ {\frac{{{x^2}(8 - {x^2})}}{{16}}} \right]_0^t} \right){\text{ }}\left( { = \left[ {\frac{{ - 4 - {x^2}{)^2}}}{{16}}} \right]_0^t} \right)\)     A1

\( = \frac{{{t^2}}}{2} - \frac{{{t^4}}}{{16}}{\text{ }}\left( { = \frac{{{t^2}(8 - {t^2})}}{{16}}} \right){\text{ }}\left( { = 1 - \frac{{{{(4 - {t^2})}^2}}}{{16}}} \right)\)     A1

Note:     Condone integration involving \(t\) only.

Note:     Award M1A0A0 for integration without limits eg, \(\int {\frac{{t(4 - {t^2})}}{4}{\text{d}}t = \frac{{{t^2}}}{2} - \frac{{{t^4}}}{{16}}} \) or equivalent.

Note:     But allow integration \( + \) \(C\) then showing \(C = 0\) or even integration without \(C\) if \(F(0) = 0\) or \(F(2) = 1\) is confirmed.

[3 marks]

correct shape including correct concavity     A1

clearly indicating starts at origin and ends at \((2,{\text{ }}1)\)     A1

Note:     Condone the absence of \((0,{\text{ }}0)\).

Note:     Accept 2 on the \(x\)-axis and 1 on the \(y\)-axis correctly placed.

[2 marks]

attempt to solve \(\frac{{{a^2}}}{2} - \frac{{{a^4}}}{{16}} = 0.75\) (or equivalent) for \(a\)     (M1)

\(a = 1.41{\text{ }}( = \sqrt 2 )\)     A1

Note:     Accept any answer that rounds to 1.4.

[2 marks]

Given that \(U\) is unbiased, show that \(2a + 6b = 1\).

Show that \({\text{Var}}(U) = (39{b^2} - 12b + 1){\sigma ^2}\).

Hence find the value of \(a\) and the value of \(b\) which give the best unbiased estimator of this form, giving your answers as fractions.

Hence find the variance of this best unbiased estimator.

\({\text{E}}(U) = a\left( {{\text{E}}({X_1}) + {\text{E}}({X_2})} \right) + b\left( {{\text{E}}({Y_1}) + {\text{E}}({Y_2}) + {\text{E}}({Y_3})} \right)\)     (M1)

\( = 2a\mu  + 6b\mu \)     A1

(for an unbiased estimator,) \({\text{E}}(U) = \mu \)     R1

giving \(2a + 6b = 1\)     AG

Note:     Condone omission of E on LHS.

[3 marks]

\({\text{Var}}(U) = {a^2}\left( {{\text{Var}}({X_1}) + {\text{Var}}({X_2})} \right) + {b^2}\left( {{\text{Var}}({Y_1}) + {\text{Var}}({Y_2}) + {\text{Var}}({Y_3})} \right)\)     (M1)

\( = 4{a^2}{\sigma ^2} + 3{b^2}{\sigma ^2}\)     A1

\( = 4{\left( {\frac{{1 - 6b}}{2}} \right)^2}{\sigma ^2} + 3{b^2}{\sigma ^2}\)     A1

\( = (39{b^2} - 12b + 1){\sigma ^2}\)     AG

[3 marks]

the best unbiased estimator (of this form) will be found by minimising \({\text{Var}}(U)\)     (R1)

For example, \(\frac{{\text{d}}}{{{\text{d}}b}}\left( {{\text{Var}}(U)} \right) = (78b - 12){\sigma ^2}\)     (A1)

for a minimum, \(b = \frac{{12}}{{78}}\,\,\,\left( { = \frac{2}{{13}}} \right)\) so that \(a = \frac{3}{{78}}\,\,\,\left( { = \frac{1}{{26}}} \right)\)     A1

[3 marks]

\({\text{Var}}U = \left( {39{{\left( {\frac{2}{{13}}} \right)}^2} - 12\left( {\frac{2}{{13}}} \right) + 1} \right){\sigma ^2}\)

\( = \frac{{{\sigma ^2}}}{{13}}\,\,\,(0.0769{\sigma ^2})\)     A1

[1 mark]

The random variable X represents the height of a wave on a particular surf beach.

It is known that X is normally distributed with unknown mean \(\mu \) (metres) and known variance \({\sigma ^2} = \frac{1}{4}{\text{ (metre}}{{\text{s}}^2}{\text{)}}\) . Sally wishes to test the claim made in a surf guide that \(\mu  = 3\) against the alternative that \(\mu  < 3\) . She measures the heights of 36 waves and calculates their sample mean \({\bar x}\) . She uses this value to test the claim at the 5 % level.

(i)     Find a simple inequality, of the form \(\bar x < A\) , where A is a number to be determined to 4 significant figures, so that Sally will reject the null hypothesis, that \(\mu  = 3\) , if and only if this inequality is satisfied.

(ii)     Define a Type I error.

(iii)     Define a Type II error.

(iv)     Write down the probability that Sally makes a Type I error.

(v)     The true value of \(\mu \) is 2.75. Calculate the probability that Sally makes a Type II error.

The random variable Y represents the height of a wave on another surf beach. It is known that Y is normally distributed with unknown mean \(\mu \) (metres) and unknown variance \({\sigma ^2}{\text{ (metre}}{{\text{s}}^2}{\text{)}}\) . David wishes to test the claim made in a surf guide that \(\mu = 3\) against the alternative that \(\mu < 3\) . He is also going to perform this test at the 5 % level. He measures the heights of 36 waves and finds that the sample mean, \(\bar y = 2.860\) and the unbiased estimate of the population variance, \(s_{n - 1}^2 = 0.25\).

(i)     State the name of the test that David should perform.

(ii)     State the conclusion of David’s test, justifying your answer by giving the p-value.

(iii)     Using David’s results, calculate the 90 % confidence interval for \(\mu \) , giving your answers to 4 significant figures.

(i)     \({H_0}:\mu = 3,{\text{ }}{H_1}:\mu < 3\)

1 tailed z test as \({\sigma ^2}\) is known

under \({H_0}{\text{, }}X \sim {\text{N}}\left( {3,\frac{1}{4}} \right){\text{ so }}\bar X \sim {\text{N}}\left( {3,\frac{{\frac{1}{4}}}{{36}}} \right) = N\left( {3,\frac{1}{{144}}} \right)\)     (M1)

\(z = \frac{{\bar x - 3}}{{\frac{1}{{12}}}}{\text{ is N(0, 1)}}\)     (A1)

\({\text{P}}(z < - 1.64485...) = 0.05\)     (A1)

so inequality is given by \(\frac{{\bar x - 3}}{{\frac{1}{{12}}}} < - 1.64485...{\text{ giving }}\bar x < 2.8629…\)     M1

\(\bar x < 2.863{\text{ (4sf)}}\)     A1

Note: Candidates can get directly to the answer from \({\text{N}}\left( {3,\frac{1}{{144}}} \right)\) they do not have to go via z is N(0, 1) . However they must give some explanation of what they have done; they cannot just write the answer down.

(ii)     a Type I error is accepting \({H_1}\) when \({H_0}\) is true     A1

(iii)     a Type II error is accepting \({H_0}\) when \({H_1}\) is true     A1

(iv)     0.05     A1

Note: Accept anything that rounds to 0.050 if they do the conditional calculation.

(v)     \(\bar X \sim {\text{N}}\left( {2.75,\frac{1}{{144}}} \right)\)     (M1)

\({\text{P}}(\bar x > 2.8629...) = 0.0877{\text{ (3sf)}}\)     (M1)A1

Note: Accept any answer between 0.0875 and 0.0877 inclusive.

Note: Accept anything that rounded is between 0.087and 0.089 if there is evidence that the candidate has used tables.

[11 marks]

(i)     t-test     A1

(ii)     \({{\text{H}}_0}:\mu = 3,{\text{ }}{{\text{H}}_1}:\mu < 3\)

1 tailed t test as \({\sigma ^2}\) is unknown

\(t = \frac{{\bar y - 3}}{{\frac{1}{{12}}}}\) has the t-distribution with \(v = 35\)     (M1)

the p-value is 0.0509…     A2

this is \( > 0.05\)     R1

so we accept that the mean wave height is 3     R1

Note: Allow “Accept \({{\text{H}}_0}\) ” provided \({{\text{H}}_0}\) has been stated.

Note: Accept FT on the p-value for the R1s.

(iii)     \(2.719 < \mu < 3.001{\text{ (4 sf)}}\)     A1A1

Note: \(2.860 \pm 1.6896... \times \frac{{\frac{1}{2}}}{6}\) would gain M1.

Note: Award A1A0 if answer are only given to 3sf.

[8 marks]

(a) There were many reasonable answers. In (i) not all candidates explained their method so that they could gain good partial marks even if they had the wrong final answer. A common mistake was to give an answer above 3. It was pleasing that almost all candidates had (ii) and (iii) correct, as this had caused problems in the past. In (iv) it was amusing to see a few candidates work out 5% using conditional probability rather than just write down the answer as asked.

(b) It was pleasing that almost all candidates realised that it was a t-test rather than a z-test.

There was good understanding on how to use the calculator in parts (ii) and (iii). The correct confidence interval to the desired accuracy was not always given.

The most common mistake in question 3 was forgetting to take into account the variance of the sample mean.

(a) There were many reasonable answers. In (i) not all candidates explained their method so that they could gain good partial marks even if they had the wrong final answer. A common mistake was to give an answer above 3. It was pleasing that almost all candidates had (ii) and (iii) correct, as this had caused problems in the past. In (iv) it was amusing to see a few candidates work out 5% using conditional probability rather than just write down the answer as asked.

(b) It was pleasing that almost all candidates realised that it was a t-test rather than a z-test.

There was good understanding on how to use the calculator in parts (ii) and (iii). The correct confidence interval to the desired accuracy was not always given.

The most common mistake in question 3 was forgetting to take into account the variance of the sample mean.

Find \(F(u)\), the cumulative distribution function of \(U\), for \(u = 1,{\text{ }}2,{\text{ }}3 \ldots \)

Hence, or otherwise, find the value of \(P(U > 20)\).

Prove that the probability generating function of \(U\) is given by \({G_u}(t) = \frac{t}{{4 - 3t}}\).

Given that \({U_i} \sim {\text{Geo}}\left( {\frac{1}{4}} \right),{\text{ }}i = 1,{\text{ }}2,{\text{ }}3\), and that \(V = {U_1} + {U_2} + {U_3}\), find

(i)     \({\text{E}}(V)\);

(ii)     \({\text{Var}}(V)\);

(iii)     \({G_v}(t)\), the probability generating function of \(V\).

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 - 3t)}^3}}}\).

By differentiating \({G_w}(t)\), find \({\text{E}}(W)\).

A third random variable \(W\), has probability generating function \({G_w}(t) = \frac{1}{{{{(4 - 3t)}^3}}}\).

Prove that \(V = W + 3\).

METHOD 1

\({\text{P}}(U = u) = \frac{1}{4}{\left( {\frac{3}{4}} \right)^{u - 1}}\)     (M1)

\(F(u) = {\text{P}}(U \le u) = \sum\limits_{r = 1}^u {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r - 1}}\;\;\;} \)(or equivalent)

\( = \frac{{\frac{1}{4}\left( {1 - {{\left( {\frac{3}{4}} \right)}^u}} \right)}}{{1 - \frac{3}{4}}}\)     (M1)

\( = 1 - {\left( {\frac{3}{4}} \right)^u}\)     A1

METHOD 2

\({\text{P}}(U \le u) = 1 - {\text{P}}(U > u)\)     (M1)

\({\text{P}}(U > u) = \) probability of \(u\) consecutive failures     (M1)

\({\text{P}}(U \le u) = 1 - {\left( {\frac{3}{4}} \right)^u}\)     A1

[3 marks]

\({\text{P}}(U > 20) = 1 - {\text{P}}(U \le 20)\)     (M1)

\( = {\left( {\frac{3}{4}} \right)^{20}}\;\;\;( = 0.00317)\)     A1

[2 marks]

\({G_U}(t) = \sum\limits_{r = 1}^\infty  {\frac{1}{4}{{\left( {\frac{3}{4}} \right)}^{r - 1}}{t^r}\;\;\;} \)(or equivalent)     M1A1

\( = \sum\limits_{r = 1}^\infty  {\frac{1}{3}{{\left( {\frac{3}{4}t} \right)}^r}} \)     (M1)

\( = \frac{{\frac{1}{3}\left( {\frac{3}{4}t} \right)}}{{1 - \frac{3}{4}t}}\;\;\;\left( { = \frac{{\frac{1}{4}t}}{{1 - \frac{3}{4}t}}} \right)\)     A1

\( = \frac{t}{{4 - 3t}}\)     AG

[4 marks]

(i)     \(E(U) = \frac{1}{{\frac{1}{4}}} = 4\)     (A1)

\(E({U_1} + {U_2} + {U_3}{\text{)}} = 4 + 4 + 4 = 12\)     A1

(ii)     \({\text{Var}}(U) = \frac{{\frac{3}{4}}}{{{{\left( {\frac{1}{4}} \right)}^2}}}=12\)     A1

\({\text{Var(}}{U_1} + {U_2} + {U_3}) = 12 + 12 + 12 = 36\)     A1

(iii)     \({G_v}(t) = {\left( {{G_U}(t)} \right)^3}\)     (M1)

\( = {\left( {\frac{t}{{4 - 3t}}} \right)^3}\)     A1

[6 marks]

\({G_W}^\prime (t) =  - 3{(4 - 3t)^{ - 4}}( - 3)\;\;\;\left( { = \frac{9}{{{{(4 - 3t)}^4}}}} \right)\)     (M1)(A1)

\(E(W) = {G_W}^\prime (1) = 9\)     (M1)A1

Note:     Allow the use of the calculator to perform the differentiation.

[4 marks]

EITHER

probability generating function of the constant 3 is \({t^3}\)     A1

OR

\({G_{W - 3}}(t) = E({t^{W + 3}}) = E({t^W})E({t^3})\)     A1

THEN

\(W + 3\) has generating function \({G_{W + 3}} = \frac{1}{{{{(4 - 3t)}^3}}} \times {t^3} = {G_V}(t)\)     M1

as the generating functions are the same \(V = W + 3\)     R1AG

[3 marks]

Total [22 marks]

Determine \(P(0.25 \leqslant X \leqslant 0.75)\);

Determine the median of \(X\).

Show that the probability density function \(f\) of \(X\) is given, for \(0 \leqslant x \leqslant 1\), by

\[f(x) = (x + 1){{\text{e}}^{x - 1}}.\]

Hence determine the mean and the variance of \(X\).

State the central limit theorem. 

A random sample of 100 observations is obtained from the distribution of \(X\). If \(\bar X\) denotes the sample mean, use the central limit theorem to find an approximate value of \(P(\bar X > 0.65)\). Give your answer correct to two decimal places.

\(P(0.25 \leqslant X \leqslant 0.75) = F(0.75) - F(0.25)\)     (M1)

\( = 0.466\)     A1

Note:     Accept any answer that rounds correctly to 0.466.

[2 marks]

the median \(m\) satisfies \(F(m) = 0.5\)     (M1)

\(m = 0.685\)     A1

Note:     Accept any answer that rounds correctly to 0.685.

[2 marks]

\(f(x) = F’(x)\)     (M1)

\( = {{\text{e}}^{x - 1}} + x{{\text{e}}^{x - 1}}\)     A1

\( = (x + 1){{\text{e}}^{x - 1}}\)     AG

[2 marks]

\(\mu  = \int\limits_0^1 {x\left( {x + 1} \right){{\text{e}}^{x - 1}}{\text{d}}x} \)    (M1)

\( = 0.632\,\,\,\left( {1 - \frac{1}{{\text{e}}}} \right)\)     A1

Note:     Accept any answer that rounds correctly to 0.632.

\({\sigma ^2} = \int\limits_0^1 {x\left( {x + 1} \right){{\text{e}}^{x - 1}}{\text{d}}x}  - 0.632{ \ldots ^2}\)     (M1)

\( = 0.0719\,\,\,\left( {\frac{6}{{\text{e}}} - 2 - \frac{1}{{{{\text{e}}^2}}}} \right)\)     A1

Note:     Accept any answer that rounds correctly to 0.072.

[4 marks]

the central limit theorem states that the mean of a large sample from any distribution (with a finite variance) is approximately normally distributed     A1

[1 mark]

\(\bar X\) is approximately \(N(0.632 \ldots ,{\text{ }}0.000719 \ldots )\)     (M1)(A1)

\(P(\bar X > 0.65) = 0.25\) (2 dps required)     A1

[3 marks]

Given that, on a randomly chosen day, the probability that he completes the crossword in less than \(a\) minutes is equal to 0.8, find the value of \(a\).

Find the probability that the total time taken for him to complete five randomly chosen crosswords exceeds 120 minutes.

Find the probability that, on a randomly chosen day, the time taken by Beatrice to complete the crossword is more than twice the time taken by Adam to complete the crossword. Assume that these two times are independent.

\(z = 0.841 \ldots \)    (A1)

\(a = \mu  + z\sigma \)    (M1)

\( = 26.2\)    A1

[3 marks]

let \(T\) denote the total time taken to complete 5 crosswords.

\(T\) is \({\text{N}}(110,{\text{ }}125)\)     (A1)(A1)

Note:     A1 for the mean and A1 for the variance.

\({\text{P}}(T > 120) = 0.186\)    A1

[3 marks]

consider the random variable \(U = Y - 2X\)     (M1)

\({\text{E}}(U) =  - 4\)    A1

\({\text{Var}}(U) = {\text{Var}}(Y) + 4{\text{Var}}(X)\)    (M1)

\( = 136\)    A1

\({\text{P}}(Y > 2X) = {\text{P}}(U > 0)\)    (M1)

\( = 0.366\)    A1

[6 marks]

Part (a) was very well answered with only a very few weak candidates using 0.8 instead of 0.841...

Part (b) was well answered with only a few candidates calculating the variance incorrectly.

Part (c) was again well answered. The most common errors, not often seen, were writing the variance of \(Y - 2X\) as either \({\text{Var}}(Y) + 2{\text{Var}}(X)\) or \({\text{Var}}(Y) - 2(or{\text{ }}4){\text{Var}}(X)\).

State suitable hypotheses to investigate whether or not \(X\), \(Y\) are independent.

(i)     Determine the \(p\)-value.

(ii)     State your conclusion at the 5% significance level.

Explain why the equation of the regression line of \(y\) on \(x\) should not be used to predict the value of \(y\) corresponding to \(x = {x_0}\), where \({x_0}\) lies within the range of values of \(x\) in the sample.

\({H_0}:{\text{ }}\rho  = 0;{\text{ }}{H_1}:{\text{ }}\rho  \ne 0\)    A1A1

[2 marks]

(i)     \(t = 0.486 \times \sqrt {\frac{{10 - 2}}{{1 - {{0.486}^2}}}} \)     (M1)

\( = 1.572 \ldots \)    (A1)

degrees of freedom \( = 8\)     (A1)

\({\text{P}}(T > 1.5728 \ldots )\)    (M1)

\( = 0.0772\)    (A1)

\(p{\text{ - value }} = {\text{ }}0.154\)    A1

Note:     Do not follow through for the final A1 if their \({H_1}\) is one-sided.

(ii)     accept \({H_0}\) or equivalent statement involving \({H_0}\) or \({H_1}\) (at the 5% significance level)     R1

Note:     Follow through the candidate’s \(p\)-value.

[7 marks]

EITHER

because the above analysis suggests that \(X\), \(Y\) are independent     R1

OR

the value of \(r\) suggests that \(X\) and \(Y\) are weakly correlated     R1

[1 mark]

Part (a) was well answered with only a few candidates using inappropriate symbols, for example \(r\) or \(\mu \). Also, only very few candidates failed to realise that the wording of the question indicated that a two-tailed test was required.

The test in (b) was generally well carried out and the \(p\)-value found correctly. The most common errors were using incorrect degrees of freedom and evaluating a one-tailed \(p\)-value instead of a two-tailed \(p\)-value.

In (c), many realised that the earlier work meant that the regression line should not be used because the variables had been found to be independent. Incorrect reasons, however, were not uncommon, for example the suggestions that either the regression line of \(x\) on \(y\) should be used or that there were insufficient data.

Find

(i)     \({\text{E}}(X)\);

(ii)     \({\text{Var}}(X)\).

Explain why a normal distribution can be used to give an approximate model for \(X\).

Use this model to find the values of \(A\) and \(B\) such that \({\text{P}}(A < X < B) = 0.9\), where \(A\) and \(B\) are symmetrical about the mean of \(X\).

Calculate the probability that he makes a Type II error.

(i)     \({\text{mean}} = 119 \times 2 = 238\)     A1

(ii)     \({\text{variance}} = 119 \times \frac{1}{9} = \frac{{119}}{9}{\text{ }}( = 13.2)\)     (M1)A1

Note: If 120 is used instead of 119 award A0(M1)A0 for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes \((234,{\text{ }}246)\). In (d) the first 2 A1 marks are for \(0.3633 \ldots \) and \(0.0174 \ldots \) so the final answer will round to 0.017.

[3 marks]

justified by the Central Limit Theorem     R1

since \(n\) is large     A1

Note: Accept \(n > 30\).

[2 marks]

\(X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)\)

\(Z = \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} \sim N(0,{\text{ }}1)\)     (M1)(A1)

\({\text{P}}(Z < q) = 0.95 \Rightarrow q = 1.644 \ldots \)    (A1)

so \({\text{P}}( - 1.644 \ldots  < Z < 1.644 \ldots ) = 0.9\)     (R1)

\({\text{P}}( - 1.644 \ldots  < \frac{{X - 238}}{{\frac{{\sqrt {119} }}{3}}} < 1.644 \ldots ) = 0.9\)    (M1)

interval is \(232 < X < 244{\text{ }}({\text{3sf}}){\text{ }}(A = 232,{\text{ }}B = 244)\)     A1A1

Notes: Accept the use of inverse normal applied to the distribution of \(X\).

Alternative is to use the GDC to find a pretend \(Z\) confidence interval for a mean and then convert by multiplying by 119.

Either \(A\) or \(B\) correct implies the five implied marks.

Accept any numbers that round to these 3sf numbers.

[7 marks]

under \({{\text{H}}_1},{\text{ }}X \sim N\left( {238,{\text{ }}\frac{{119}}{9}} \right)\)     (M1)

\({\text{P}}(236 \leqslant X \leqslant 240) = 0.41769 \ldots \)    (A1)

probability that all 4 values of \(X\) lie in this interval is

\({(0.41769 \ldots )^4} = 0.030439 \ldots \)     (M1)(A1)

so probability of a Type II error is 0.0304 (3sf)     A1

Note: Accept any answer that rounds to 0.030.

[5 marks]

State suitable hypotheses to investigate whether or not a negative linear association exists between X and Y.

Determine the p-value.

State your conclusion at the 1 % significance level.

Show that Cov(U, V) = E(UV) − E(U)E(V).

Hence show that if U, V are independent random variables then the population product moment correlation coefficient, ρ, is zero.

H₀: ρ = 0; H₁: ρ < 0       A1

[1 mark]

\(t =  - 0.708\sqrt {\frac{{11 - 2}}{{1 - {{\left( { - 0.708} \right)}^2}}}} \,\, = \,\,\left( { - 3.0075 \ldots } \right)\)       (M1)

degrees of freedom = 9        (A1)

P(T < −3.0075...) = 0.00739       A1

Note: Accept any answer that rounds to 0.0074.

[3 marks]

reject H₀ or equivalent statement       R1

Note: Apply follow through on the candidate’s p-value.

[1 mark]

Cov(U, V) + E((U − E(U))(V − E(V)))

= E(UV − E(U)V − E(V)U + E(U)E(V))       M1

= E(UV) − E(E(U)V) − E(E(V)U) + E(E(U)E(V))       (A1)

= E(UV) − E(U)E(V) − E(V)E(U) + E(U)E(V)       A1

Cov(U, V) = E(UV) − E(U)E(V)       AG

[3 marks]

E(UV) = E(U)E(V) (independent random variables)       R1

⇒Cov(U, V) = E(U)E(V) − E(U)E(V) = 0      A1

hence, ρ = \(\frac{{{\text{Cov}}\left( {U,\,V} \right)}}{{\sqrt {{\text{Var}}\left( U \right)\,{\text{Var}}\left( V \right)} }} = 0\)     A1AG

Note: Accept the statement that Cov(U,V) is the numerator of the formula for ρ.

Note: Only award the first A1 if the R1 is awarded.

[3 marks]

State suitable hypotheses for a two-tailed test.

Find the critical region for testing \({\bar b}\) at the 5 % significance level.

Find the probability of making a Type II error.

Another model of smartphone whose battery life may be assumed to be normally distributed with mean μ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of [10.2, 11.4] for μ.

Calculate the confidence level of this interval.

Note: In question 3, accept answers that round correctly to 2 significant figures.

\({{\text{H}}_0}\,{\text{:}}\,\mu  = 9.5{\text{;}}\,\,{{\text{H}}_1}\,{\text{:}}\,\mu  \ne 9.5\)     A1

[1 mark]

Note: In question 3, accept answers that round correctly to 2 significant figures.

the critical values are \(9.5 \pm 1.95996 \ldots  \times \frac{{0.4}}{{\sqrt {20} }}\)     (M1)(A1)

i.e. 9.3247…, 9.6753…

the critical region is \({\bar b}\) < 9.32, \({\bar b}\) > 9.68     A1A1

Note: Award A1 for correct inequalities, A1 for correct values.

Note: Award M0 if t-distribution used, note that t(19)_97.5 = 2.093 …

[4 marks]

Note: In question 3, accept answers that round correctly to 2 significant figures.

\(\bar B \sim {\text{N}}\left( {9.8,\,{{\left( {\frac{{0.4}}{{\sqrt {20} }}} \right)}^2}} \right)\)     (A1)

\({\text{P}}\left( {9.3247 \ldots  < \bar B < 9.6753 \ldots } \right)\)     (M1)

=0.0816     A1

Note: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899.

[3 marks]

Note: In question 3, accept answers that round correctly to 2 significant figures.

METHOD 1

\(X \sim {\text{N}}\left( {{\text{10}}{\text{.8,}}\,\frac{{{{1.2}^2}}}{6}} \right)\)     (M1)(A1)

P(10.2 < X < 11.4) = 0.7793…     (A1)

confidence level is 77.9%    A1

Note: Accept 78%.

METHOD 2

\(11.4 - 10.2 = 2z \times \frac{{1.2}}{{\sqrt 6 }}\)      (M1)

\(z = 1.224 \ldots \)     (A1)

P(−1.224… < Z < 1.224…) = 0.7793…      (A1)

confidence level is 77.9%      A1

Note: Accept 78%.

[4 marks]

The random variable X has the negative binomial distribution NB(5, p), where p < 0.5, and \({\text{P}}(X = 10) = 0.05\). By first finding the value of p, find the value of \({\text{P}}(X = 11)\).

\({\text{P}}(X = 10) = \left( {\begin{array}{*{20}{c}}
  9 \\
  4
\end{array}} \right){p^5}{(1 - p)^5}\)     (= 0.05)     (M1)A1A1

Note: First A1 is for the binomial coefficient. Second A1 is for the rest.

solving by any method, \(p = 0.297 \ldots \)     A4

Notes: Award A2 for anything which rounds to 0.703.

Do not apply any AP at this stage.

\({\text{P}}(X = 10) = \left( {\begin{array}{*{20}{c}}
  {10} \\
  4
\end{array}} \right) \times {(0.297...)^5} \times {(1 - 0.297...)^6}\)     (M1)A1

= 0.0586     A1

Note: Allow follow through for incorrect p-values.

[10 marks]

Questions on these discrete distributions have not been generally well answered in the past and it was pleasing to note that many candidates submitted a reasonably good solution to this question. In (b), the determination of the value of p was often successful using a variety of methods including solving the equation \(p(1 - p) = {(0.000396{\text{ }} \ldots )^{1/5}}\), graph plotting or using SOLVER on the GDC or even expanding the equation into a \({10^{{\text{th}}}}\) degree polynomial and solving that. Solutions to this particular question exceeded expectations.

Find the probability that the weight of a randomly selected male is more than twice the weight of a randomly selected female.

Two males and five females stand together on a weighing machine. Find the probability that their total weight is less than 175 kg.

we are given that \(M \sim {\text{N(30, 9)}}\) and \(F \sim {\text{N(20, 6.25)}}\)

let \(X = M - 2F;{\text{ }}X \sim {\text{N}}\)(\( - 10\), \(34\))     M1A1A1

we require \({\text{P}}(X > 0)\)     (M1)

= 0.0432     A1

[5 marks]

let \(Y = {M_1} + {M_2} + {F_1} + {F_2} + {F_3} + {F_4} + {F_5};{\text{ }}Y \sim {\text{N(160, 49.25)}}\)     M1A1A1

we require \({\text{P}}(Y < 175) = 0.984\)     A1

[4 marks]

State suitable hypotheses for this investigation.

State, in context, what conclusion should be drawn from this \(p\)-value.

The teacher then asks Anne for the values of the \(t\)-statistic and the product moment correlation coefficient \(r\) produced by the calculator but she has deleted these. Starting with the \(p\)-value, calculate these values of \(t\) and \(r\).

\({H_0}:\rho  = 0;{\text{ }}{H_1}:\rho  > 0\)     A1

Note:     Do not accept \(r\) in place of \(\rho \).

[1 mark]

insufficient evidence to conclude that there is a (positive) association between marks in these two subjects (or equivalent statement in context)     A1

[1 mark]

degrees of freedom \( = 10\)     (A1)

required value of \(t = {\text{inverse }}t(0.823)\)     (M1)

\( = 0.972\)     A1

attempt to solve \(t = r\sqrt {\frac{{n - 2}}{{1 - {r^2}}}} \)     (M1)

\(r = 0.294\)     A1

Note:     Accept any \(r\) value that rounds to 0.29.

Note:     Follow through their \(t\) value to determine \(r\).

[5 marks]

(a)     The heating in a residential school is to be increased on the third frosty day during the term. If the probability that a day will be frosty is 0.09, what is the probability that the heating is increased on the \({25^{{\text{th}}}}\) day of the term?

(b)     On which day is the heating most likely to be increased?

(a)     the distribution is NB(3, 0.09)     (M1)(A1)

the probability is \(\left( {\begin{array}{*{20}{c}}
  {24} \\
  2
\end{array}} \right){0.91^{22}} \times {0.09^3} = 0.0253\)     (M1)(A1)A1

[5 marks]

(b)     P(Heating increased on \({n^{{\text{th}}}}\) day)

\(\left( {\begin{array}{*{20}{c}}
  {n - 1} \\
  2
\end{array}} \right){0.91^{n - 3}} \times {0.09^3}\)     (M1)(A1)(A1)

by trial and error n = 23 gives the maximum probability     (M1)A3

(neighbouring values: 0.02551 (n = 22) ; 0.02554 (n = 23) ; 0.02545 (n = 24) )

[7 marks]

Total [12 marks]

Most candidates understood the context of this question, and the negative binomial distribution was usually applied, albeit occasionally with incorrect parameters. Good solutions were seen to part(b), using lists in their GDC or trial and error.

If \(X\) and \(Y\) are two random variables such that \({\text{E}}(X) = {\mu _X}\) and \({\text{E}}(Y) = {\mu _Y}\) then \({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X - {\mu _X})(Y - {\mu _Y})} \right)\).

Prove that if \(X\) and \(Y\) are independent then \({\text{Cov}}(X,{\text{ }}Y) = 0\).

In a particular company, it is claimed that the distance travelled by employees to work is independent of their salary. To test this, 20 randomly selected employees are asked about the distance they travel to work and the size of their salaries. It is found that the product moment correlation coefficient, \(r\), for the sample is \( - 0.35\).

You may assume that both salary and distance travelled to work follow normal distributions.

Perform a one-tailed test at the \(5\% \) significance level to test whether or not the distance travelled to work and the salaries of the employees are independent.

METHOD 1

\({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X - {\mu _X})(Y - {\mu _Y})} \right)\)

\( = {\text{E}}(XY - X{\mu _Y} - Y{\mu _X} + {\mu _X}{\mu _Y})\)     (M1)

\( = {\text{E}}(XY) - {\mu _Y}{\text{E}}(X) - {\mu _X}{\text{E}}(Y) + {\mu _X}{\mu _Y}\)

\( = {\text{E}}(XY) - {\mu _X}{\mu _Y}\)     A1

as \(X\) and \(Y\) are independent \({\text{E}}(XY) = {\mu _X}{\mu _Y}\)     R1

\({\text{Cov}}(X,{\text{ }}Y) = 0\)     AG

METHOD 2

\({\text{Cov}}(X,{\text{ }}Y) = {\text{E}}\left( {(X - {\mu _x})(Y - {\mu _y})} \right)\)

\( = {\text{E}}(X - {\mu _x}){\text{E}}(Y - {\mu _y})\)     (M1)

since \(X,Y\) are independent     R1

\( = ({\mu _x} - {\mu _x})({\mu _y} - {\mu _y})\)     A1

\( = 0\)     AG

[3 marks]

\({H_0}:\rho  = 0\;\;\;{H_1}:\rho  < 0\)     A1

Note:     The hypotheses must be expressed in terms of \(\rho \).

test statistic \({t_{test}} =  - 0.35\sqrt {\frac{{20 - 2}}{{1 - {{( - 0.35)}^2}}}} \)     (M1)(A1)

\( =  - 1.585 \ldots \)     (A1)

\({\text{degrees of freedom}} = 18\)     (A1)

EITHER

\(p{\text{ - value}} = 0.0652\)     A1

this is greater than \(0.05\)     M1

OR

\({t_{5\% }}(18) =  - 1.73\)     A1

this is less than \( - {\text{1.59}}\)     M1

THEN

hence accept \({H_0}\) or reject \({H_1}\) or equivalent or contextual equivalent     R1

Note:     Allow follow through for the final R1 mark.

[8 marks]

Total [11 marks]

Solutions to (a) were often disappointing with few candidates gaining full marks, a common error being failure to state that

\(E(XY) = E(X)E(Y)\) or \({\text{E}}\left( {(X - {\mu _x})(Y - {\mu _y})} \right) = {\text{E}}(X - {\mu _x}){\text{E}}(Y - {\mu _y})\) in the case of independence.

In (b), the hypotheses were sometimes given incorrectly. Some candidates gave \({H_1}\) as \(\rho  \ne 0\), not seeing that a one-tailed test was required. A more serious error was giving the hypotheses as \({H_0}:r = 0,{\text{ }}{H_1}:r < 0\) which shows a complete misunderstanding of the situation. Subsequent parts of the question were well answered in general.

Find an expression for \({u_n}\) in terms of \(n\).

For every prime number \(p > 3\), show that \(p|{u_{p - 1}}\).

the auxiliary equation is \({\lambda ^2} - 5\lambda  + 6 = 0\)     M1

\( \Rightarrow \lambda  = 2,{\text{ }}3\)     (A1)

the general solution is \({u_n} = A \times {2^n} + B \times {3^n}\)     A1

imposing initial conditions (substituting \(n = 0,{\text{ }}1\))     M1

\(A + B = 0\) and \(2A + 3B = 1\)     A1

the solution is \(A =  - 1,{\text{ }}B = 1\)

so that \({u_n} = {3^n} - {2^n}\)     A1

[6 marks]

\({u_{p - 1}} = {3^{p - 1}} - {2^{p - 1}}\)

\(p > 3\), therefore 3 or 2 are not divisible by \(p\)     R1

hence by FLT, \({3^{p - 1}} \equiv 1 \equiv {2^{p - 1}}(\bmod p)\) for \(p > 3\)     M1A1

\({u_{p - 1}} \equiv 0(\bmod p)\)     A1

\(p|{u_{p - 1}}\) for every prime number \(p > 3\)     AG

[4 marks]

State suitable hypotheses.

The marks obtained by the 12 students who sat both papers are given in the following table. 

(i)     Determine the product moment correlation coefficient for these data and state its p-value.

(ii)     Interpret your p-value in the context of the problem.

George obtained a mark of 63 on Paper 1 but was unable to sit Paper 2 because of illness. Predict the mark that he would have obtained on Paper 2.

Another class of 16 students sat examinations in Physics and Chemistry and the product moment correlation coefficient between the marks in these two subjects was calculated to be 0.524. Using a 1 % significance level, determine whether or not this value suggests a positive association between marks in Physics and marks in Chemistry.

\({{\text{H}}_0}:\rho  = 0;{\text{ }}{{\text{H}}_1}:\rho  > 0\)     A1

[1 mark]

(i)     correlation coefficient = 0.905     A2

p-value \( = 2.61 \times {10^{ - 5}}\)     A2

(ii)     very strong evidence to indicate a positive association between marks in Mechanics and marks in Statistics     R1 

[5 marks]

the regression line of y on x is \(y = 8.71 + 0.789x\)     (M1)A1

George’s estimated mark on Paper 2 \( = 8.71 + 0.789 \times 63\)     (M1)

= 58     A1

[4 marks]

\(t = r\sqrt {\frac{{n - 2}}{{1 - {r^2}}}}  = 2.3019 \ldots \)     M1A1

degrees of freedom = 14     (A1)

p-value \( = 0.0186 \ldots \)     A1

at the 1 % significance level, this does not indicate a positive association between the marks in Physics and Chemistry     R1

[5 marks]

Calculate the significance level of the test.

Given that the value of \(\mu \) is actually 2.5, determine the probability of a Type II error.

under \({H_0}\) , \(S{\text{ is Po}}(30)\)     (A1)

EITHER

\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \)     A1

\({\text{P}}(S \geqslant 38) = {\text{0.089012}} \ldots \)     A1

significance level = 0.080569… + 0.089012…     (M1)

= 0.170     A1

OR

\({\text{P}}(S \leqslant 22) = {\text{0.080569}} \ldots \)     A1

\({\text{P}}(S \leqslant 37) = {\text{0.910987}} \ldots \)     A1

significance level = 1 – (0.910987…) + 0.089012…     (M1)

= 0.170     A1

Note: Accept 17 % or 0.17.

Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.144 without a continuity correction and 0.171 with a continuity correction. The first (A1) is independent.

[5 marks]

S is now Po(25)     (A1)

P (Type II error) = P (accept \({H_0}|\mu = 2.5\))     (M1)

\( = {\text{P}}\left( {23 \leqslant S \leqslant 37|S{\text{ is Po}}(25)} \right)\)     (M1)

Note: Only one of the above M1 marks can be implied.

= 0.990789… – 0.317533…     (A1)

= 0.673     A1

Note: Award 2 marks out of the final 4 marks for correct use of the Central Limit Theorem, giving 0.647 without a continuity correction and 0.685 with a continuity correction. The first (A1) is independent.

[5 marks]

Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.

Solutions to this question were often disappointing with many candidates not knowing what had to be done. Even those candidates who knew what to do sometimes made errors in evaluating the probabilities, often by misinterpreting the inequality signs. Candidates who used the Central Limit Theorem to evaluate the probabilities were given only partial credit on the grounds that the answers obtained were approximate and not exact.

The following table gives the average yield of olives per tree, in kg, and the rainfall, in cm, for nine separate regions of Greece. You may assume that these data are a random sample from a bivariate normal distribution, with correlation coefficient \(\rho \).

A scientist wishes to use these data to determine whether there is a positive correlation between rainfall and yield.

(a)     State suitable hypotheses.

(b)     Determine the product moment correlation coefficient for these data.

(c)     Determine the associated p-value and comment on this value in the context of the question.

(d)     Find the equation of the regression line of y on x.

(e)     Hence, estimate the yield per tree in a tenth region where the rainfall was 19 cm.

(f)     Determine the angle between the regression line of y on x and that of x on y . Give your answer to the nearest degree.

(a)     \({H_0}:\rho  = 0\)     A1

\({H_1}:\rho  > 0\)     A1

[2 marks]

(b)     0.853     A2

Note:     Accept any answer that rounds to 0.85.

[2 marks]

(c)     p-value = 0.00173 (1-tailed)     A1

Note:     Accept any answer that rounds to 0.0017.

     Accept any answer that rounds to 0.0035 obtained from 2-tailed test.

strong evidence to reject the hypothesis that there is no correlation between rainfall and yield or to accept the hypothesis that there is correlation between rainfall and yield     R1

Note:     Follow through the p-value for the conclusion.

[2 marks]

(d)     \(y = 1.78x + 40.5\)     A1A1

Note:     Accept numerical coefficients that round to 1.8 and 41.

[2 marks]

(e)     \(y = 1.77 \ldots (19) + 14.5 \ldots \)     M1

74.3     A1

Note:     Accept any answer that rounds to 74 or 75.

[2 marks]

(f)     the gradient of the regression line y on x is 1.78 or equivalent     A1

the regression line of x on y is \(x = 0.409y - 12.2\)     (A1)

the gradient of the regression line x on y is \(\frac{1}{{0.409}}{\text{ }}( = 2.44)\)     (M1)A1

calculate \(\arctan (2.44) - \arctan (1.78)\)     (M1)

angle between regression lines is 7 degrees     A1

Note:     Accept any answer which rounds to ±7 degrees.

[6 marks]

Total [16 marks]

State suitable hypotheses to test the inspector’s claim.

Find unbiased estimates of \(\mu \) and \({\sigma ^2}\).

Carry out an appropriate test and state the \(p\)-value obtained.

Using a 10% significance level and justifying your answer, state your conclusion in context.

\({H_0}:\mu = 7,{\text{ }}{H_1}:\mu < 7\)     A1

[1 mark]

\(\bar x = \frac{{83.64}}{{12}} = 6.97\)     A1

\(s_{n - 1}^2 = \frac{{583.05}}{{11}} - \frac{{{\text{ }}{{83.64}^2}}}{{132}} = 0.0072\)     (M1)A1

[3 marks]

\(t = \frac{{6.97 - 7}}{{\sqrt {\frac{{0.0072}}{{12}}} }} = - 1.22(474 \ldots )\)     (M1)(A1)

\({\text{degrees of freedom}} = 11\)     (A1)

\(p{\text{ - value}} = 0.123\)     A1

Note:     Accept any answer that rounds correctly to 0.12.

[4 marks]

because \(p > 0.1\)     R1

the inspector’s claim is not supported (at the 10% level)

(or equivalent in context)     A1

Note:     Only award the A1 if the R1 has been awarded

[2 marks]

Find the cumulative distribution function, \(F(x)\), of X.

Find the probability that the lifetime of a particular battery is more than twice the mean.

Find the median of X in terms of \(\lambda \).

Find the probability that the lifetime of a particular battery lies between the median and the mean.

\(\int {\lambda {{\text{e}}^{ - \lambda t}}{\text{d}}t = - {{\text{e}}^{ - \lambda t}}{\text{ }}( + c)} \)     A1

\( \Rightarrow F(x) = \left[ { - {{\text{e}}^{ - \lambda t}}} \right]_0^x\)     (M1)

\( = 1 - {{\text{e}}^{ - \lambda t}}{\text{ }}(x \geqslant 0)\)     A1

[3 marks]

\(1 - F\left( {\frac{2}{\lambda }} \right)\)     M1

\( = {{\text{e}}^{ - 2}}\,\,\,\,\,( = 0.135)\)     A1

[2 marks]

\(F(m) = \frac{1}{2}\)     (M1)

\( \Rightarrow {{\text{e}}^{ - \lambda m}} = \frac{1}{2}\)     A1

\( \Rightarrow - \lambda m = \ln \frac{1}{2}\)

\( \Rightarrow m = \frac{1}{\lambda }\ln 2\)     A1

[3 marks]

\(F\left( {\frac{1}{\lambda }} \right) - F\left( {\frac{{\ln 2}}{\lambda }} \right)\)     M1

\( = \frac{1}{2} - {{\text{e}}^{ - 1}}\,\,\,\,\,( = 0.132)\)     A1

[2 marks]

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.

Determine unbiased estimates for \(\mu \) and \({\sigma ^2}\).

Use a two-tailed test to determine the \(p\)-value for the above results.

Interpret your \(p\)-value at the 5% level of significance, justifying your conclusion.

UE of \(\mu \) is \(8.01{\text{ }}( = 8.0125)\)     A1

UE of \({\sigma ^2}\) is 0.404     (M1)A1

Note:     Accept answers that round correctly to 2 sf.

Note:      Condone incorrect notation, ie, \(\mu \) instead of UE of \(\mu \) and \({\sigma ^2}\) instead of UE of \({\sigma ^2}\).

Note:     M0 for squaring \(0.594 \ldots \) giving 0.354, M1A0 for failing to square \(0.635 \ldots \)

[3 marks]

attempting to use the \(t\)-test     (M1)

\(p\)-value is 0.0566     A2

Note:     Accept any answer that rounds correctly to 2 sf.

[3 marks]

\(0.0566 > 0.05\)     R1

we accept the null hypothesis (mean pumpkin weight is 7.5 kg)     A1

Note:     Apply follow through on the candidate’s \(p\)-value.

Note:     Do not award A1 if R1 is not awarded.

[2 marks]

(i)     Find \({\text{P}}(X = 6)\).

(ii)     Find \({\text{P}}(X = 6|5 \leqslant X \leqslant 8)\).

\(\bar X\) denotes the sample mean of \(n > 1\) independent observations from \(X\).

(i)     Write down \({\text{E}}(\bar X)\) and \({\text{Var}}(\bar X)\).

(ii)     Hence, give a reason why \(\bar X\) is not a Poisson distribution.

A random sample of \(40\) observations is taken from the distribution for \(X\).

(i)     Find \({\text{P}}(7.1 < \bar X < 8.5)\).

(ii)     Given that \({\text{P}}\left( {\left| {\bar X - 8} \right| \leqslant k} \right) = 0.95\), find the value of \(k\).

(i)     \({\text{P}}(X = 6) = 0.122\)     (M1)A1

(ii)     \({\text{P}}(X = 6|5 \leqslant X \leqslant 8) = \frac{{{\text{P}}(X = 6)}}{{{\text{P}}(5 \leqslant X \leqslant 8)}} = \frac{{0.122 \ldots }}{{0.592 \ldots  - 0.0996 \ldots }}\)     (M1)(A1)

\( = 0.248\)     A1

[5 marks]

(i)     \({\text{E}}(\bar X) = 8\)     A1

\({\text{Var}}(\bar X) = \frac{8}{n}\)     A1

(ii)     \({\text{E}}(\bar X) \ne {\text{Var}}(\bar X)\)   \({\text{(for }}n > 1)\)     R1

Note:     Only award the R1 if the two expressions in (b)(i) are different.

[3 marks]

(i)     EITHER

\(\bar X \sim {\text{N(8, 0.2)}}\)     (M1)A1

Note:     M1 for normality, A1 for parameters.

\({\text{P}}(7.1 < \bar X < 8.5) = 0.846\)     A1

OR

The expression is equivalent to

\({\text{P}}(283 \leqslant \sum {X \leqslant 339)} \) where \(\sum X \) is \({\text{Po(320)}}\)     M1A1

\( = 0.840\)     A1

Note:     Accept 284, 340 instead of 283, 339

     Accept any answer that rounds correctly to 0.84 or 0.85.

(ii)     EITHER

\(k = 1.96\frac{\sigma }{{\sqrt n }}\) or \(1.96{\text{ std}}(\bar X)\)     (M1)(A1)

\(k = 0.877\) or \(1.96\sqrt {0.2} \)     A1

OR

The expression is equivalent to

\(P(320 - 40k \leqslant \sum {X \leqslant 320 + 40k) = 0.95} \)     (M1)

\(k = 0.875\)     A2

Note:     Accept any answer that rounds to 0.87 or 0.88.

     Award M1A0 if modulus sign ignored and answer obtained rounds to 0.74 or 0.75

[6 marks]

Find the probability that a randomly chosen Supermug tea bag contains more than 3.9 g of tea.

Find the probability that, of two randomly chosen Megamug tea bags, one contains more than 5.4 g of tea and one contains less than 5.4 g of tea.

Find the probability that five randomly chosen Supermug tea bags contain a total of less than 20.5 g of tea.

Find the probability that the total weight of tea in seven randomly chosen Supermug tea bags is more than the total weight in five randomly chosen Megamug tea bags.

let S be the weight of tea in a random Supermug tea bag

\(S \sim {\text{N(4.2, 0.1}}{{\text{5}}^2})\)

\({\text{P}}(S > 3.9) = 0.977\)     (M1)A1

[2 marks]

let M be the weight of tea in a random Megamug tea bag

\(M \sim {\text{N(5.6, 0.1}}{{\text{7}}^2})\)

\({\text{P}}(M > 5.4) = 0.880 \ldots \)     (A1)

\({\text{P}}(M < 5.4) = 1 - 0.880 \ldots = 0.119 \ldots \)     (A1)

required probability \( = 2 \times 0.880 \ldots \times 0.119 \ldots = 0.211\)     M1A1

[4 marks]

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} < 20.5)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} = A\)     (M1)

\({\text{E}}(A) = 5{\text{E}}(S)\)

= 21     A1

\({\text{Var}}(A) = 5{\text{Var}}(S)\)

= 0.1125     A1

\(A \sim {\text{N(21, 0.1125}})\)

\({\text{P}}(A < 20.5) = 0.0680\)     A1

[4 marks]

\({\text{P}}({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} - ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) > 0)\)

let \({S_1} + {S_2} + {S_3} + {S_4} + {S_5} + {S_6} + {S_7} - ({M_1} + {M_2} + {M_3} + {M_4} + {M_5}) = B\)     (M1)

\({\text{E}}(B) = 7{\text{E}}(S) - 5{\text{E}}(M)\)

= 1.4     A1

Note: Above A1 is independent of first M1.

\({\text{Var}}(B) = 7{\text{Var}}(S) + 5{\text{Var}}(M)\)     (M1)

= 0.302     A1

\({\text{P}}(B > 0) = 0.995\)     A1

[5 marks]

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

For most candidates this was a reasonable start to the paper with many candidates gaining close to full marks. The most common error was in (b) where, surprisingly, many candidates did not realise the need to multiply the product of the two probabilities by 2 to gain the final answer. Weaker candidates often found problems in understanding how to correctly find the variance in both (c) and (d).

Show that the probability generating function for \(X\) is given by \(G(t) = \frac{P}{{1 - q{t^2}}}\).

Hence determine \({\text{E}}(X)\) in terms of \(p\) and \(q\).

The random variable \(Y\) is given by \(Y = 2X + 1\). Find the probability generating function for \(Y\).

\(G(t) = \sum {P(X = x){t^x}} \)     (M1)

\( = p + pq{t^2} + p{q^2}{t^4} +  \ldots \)

(summing \(GP\)) \({u_1} = p,{\text{ }}r = q{t^2}\)     A1

\( = \frac{p}{{1 - q{t^2}}}\)     AG

[2 marks]

\(G’(t) =  - \frac{p}{{{{(1 - q{t^2})}^2}}} \times  - 2qt\)     M1A1

\({\text{E}}(X) = G’(1)\)     (M1)

\( = \frac{{2pq}}{{{{(1 - q)}^2}}}\,\,\,\left( { = \frac{{2q}}{p}} \right)\)     A1

[4 marks]

METHOD 1

\({\text{PGF of }}Y = \sum {P(Y = y){t^y}} \)     (M1)

\( = pt + pq{t^5} + p{q^2}{t^9} +  \ldots \)     A1

\( = \frac{{pt}}{{1 - q{t^4}}}\)     A1

METHOD 2

\({\text{PGF of }}Y = {\text{E}}({t^Y})\)     (M1)

\( = {\text{E}}({t^{2X + 1}})\)

\( = {\text{E}}\left( {{{({t^2})}^X}} \right) \times {\text{E}}(t)\)     A1

\( = \frac{{pt}}{{1 - q{t^4}}}\)     A1

[3 marks]

Find expressions for \({G’_X}(t)\) and \({G’’_X}(t)\).

Hence show that \({\text{Var}}(X) = \lambda \).

By considering the probability generating function, \({G_{X + Y}}(t)\), of \(X + Y\), show that \(X + Y\) follows a Poisson distribution with mean \(\lambda  + \mu \).

Show that \({\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 - \frac{\lambda }{{\lambda + \mu }}} \right)^{n - x}}\), where \(n\), \(x\) are non-negative integers and \(n \geqslant x\).

Identify the probability distribution given in part (c)(i) and state its parameters.

\({G’_X}(t) = \lambda {{\text{e}}^{\lambda (t - 1)}}\)     A1

\({G’’_X}(t) = {\lambda ^2}{{\text{e}}^{\lambda (t - 1)}}\)     A1

[2 marks]

\({\text{Var}}(X) = {G''_X}(1) + {G'_X}(1) - {\left( {{{G'}_X}(1)} \right)^2}\)     (M1)

\({G’_X}(1) = \lambda \) and \({G’’_X}(1) = {\lambda ^2}\)     (A1)

\({\text{Var}}(X) = {\lambda ^2} + \lambda  - {\lambda ^2}\)     A1

\( = \lambda \)     AG

[3 marks]

\({G_{X + Y}}(t) = {{\text{e}}^{\lambda (t - 1)}} \times {{\text{e}}^{\mu (t - 1)}}\)     M1

Note:     The M1 is for knowing to multiply pgfs.

\( = {{\text{e}}^{(\lambda  + \mu )(t - 1)}}\)     A1

which is the pgf for a Poisson distribution with mean \(\lambda  + \mu \)     R1AG

Note:     Line 3 identifying the Poisson pgf must be seen.

[3 marks]

\({\text{P}}(X = x|X + Y = n) = \frac{{{\text{P}}(X = x \cap Y = n - x)}}{{{\text{P}}(X + Y = n)}}\)     (M1)

\( = \left( {\frac{{{{\text{e}}^{ - \lambda }}{\lambda ^x}}}{{x!}}} \right)\left( {\frac{{{{\text{e}}^{ - \mu }}{\mu ^{n - x}}}}{{(n - x)!}}} \right)\left( {\frac{{n!}}{{{{\text{e}}^{ - (\lambda  + \mu )}}{{(\lambda  + \mu )}^n}}}} \right)\) (or equivalent)     M1A1

\( = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right)\frac{{{\lambda ^x}{\mu ^{n - x}}}}{{{{(\lambda + \mu )}^n}}}\)     A1

\( = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {\frac{\mu }{{\lambda + \mu }}} \right)^{n - x}}\)     A1

leading to \({\text{P}}(X = x|X + Y = n) = \left( {\begin{array}{*{20}{c}} n \\ x \end{array}} \right){\left( {\frac{\lambda }{{\lambda + \mu }}} \right)^x}{\left( {1 - \frac{\lambda }{{\lambda + \mu }}} \right)^{n - x}}\)     AG

[5 marks]

\({\text{B}}\left( {n,{\text{ }}\frac{\lambda }{{\lambda  + \mu }}} \right)\)     A1A1

Note:     Award A1 for stating binomial and A1 for stating correct parameters.

[2 marks]

State the distribution of X.

Show that the probability generating function, \(G\left( t \right)\), for X is given by \(G\left( t \right) = \frac{t}{{4 - 3t}}\).

Find \(G'\left( t \right)\).

Determine the mean number of throws required to obtain a 1.

X is geometric (or negative binomial)      A1

[1 mark]

\(G\left( t \right) = \frac{1}{4}t + \frac{1}{4}\left( {\frac{3}{4}} \right){t^2} + \frac{1}{4}{\left( {\frac{3}{4}} \right)^2}{t^3} +  \ldots \)     M1A1

recognition of GP \(\left( {{u_1} = \frac{1}{4}t,\,\,r = \frac{3}{4}t} \right)\)     (M1)

\( = \frac{{\frac{1}{4}t}}{{1 - \frac{3}{4}t}}\)     A1

leading to \(G\left( t \right) = \frac{t}{{4 - 3t}}\)     AG

[4 marks]

attempt to use product or quotient rule      M1

\(G'\left( t \right) = \frac{4}{{{{\left( {4 - 3t} \right)}^2}}}\)     A1

[2 marks]

4      A1

Note: Award A1FT to a candidate that correctly calculates the value of \(G'\left( 1 \right)\) from their \(G'\left( t \right)\).

[1 mark]

(i)     State the distribution of \(X\), including its parameters.

(ii)     Calculate \({\text{E}}(X)\).

(iii)     Calculate \({\text{P}}(X = 5)\).

(i)     Calculate \({\text{E}}(Y)\).

(ii)     Calculate \({\text{P}}(Y = 5)\).

(i)     \({\text{NB}}\left( {2,{\text{ }}\frac{1}{7}} \right)\)     A1A1A1

Note: The final A1 mark can be awarded for knowing that \(p = \frac{1}{7}\) independent of the other two marks.

(ii)     \({\text{E}}(X) = \frac{r}{p} = 14\)     A1

(iii)     \(\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){\left( {\frac{6}{7}} \right)^3}{\left( {\frac{1}{7}} \right)^2} = 0.0514\)     (M1)A1

Note: Accept any number that rounds to this 3sf number.

[6 marks]

(i)     \(Y = {Y_1} + {Y_2}\) (number up to1st + number up to 2nd)     (M1)

\({Y_1} \sim Geo\left( {\frac{1}{7}} \right),{\text{ }}{Y_2} \sim Geo\left( {\frac{3}{7}} \right)\)    (A1)

Notes: The above (A1) is independent of the (M1).

Could have \({\text{NB }}(1,{\text{ }}p)\), instead of \(Geo(p)\).

\({\text{E}}(Y) = \frac{1}{{\left( {\frac{1}{7}} \right)}} + \frac{1}{{\left( {\frac{3}{7}} \right)}} = 7 + \frac{7}{3} = 9\frac{1}{3}{\text{ (9.33)}}\)    M1A1

(ii)     \(Y = {Y_1} + {Y_2} = 5\) happens when     (M1)

\({Y_1} = 1,{\text{ }}{Y_2} = 4\) or \({Y_1} = 2,{\text{ }}{Y_2} = 3\) or \({Y_1} = 3,{\text{ }}{Y_2} = 2\) or \({Y_1} = 4,{\text{ }}{Y_2} = 1\)     (A1)

so probability is \(\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{4}{7}\frac{3}{7} + \frac{6}{7}\frac{6}{7}\frac{6}{7}\frac{1}{7}\frac{3}{7}\)     (M1)(A1)

\( = 0.0928{\text{ }}\left( {\frac{{1560}}{{16807}}} \right)\)    A1

Note: Accept any answer that rounds to 0.093.

[9 marks]

Write down an expression for \(J(t)\) in terms of \(G(t)\) and \(H(t)\).

By differentiating \(J(t)\), prove that

(i)     \({\text{E}}(Z) = {\text{E}}(X) + {\text{E}}(Y)\);

(ii)     \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\).

\(J(t) = G(t)H(t)\)    A1

[1 mark]

(i)     \(J'(t) = G'(t)H(t) + G(t)H'(t)\)     M1A1

\(J'(1) = G'(1)H(1) + G(1)H'(1)\)    M1

\(J'(1) = G'(1) + H'(1)\)    A1

so \(E(Z) = E(X) + E(Y)\)     AG

(ii)     \(J''(t) = G''(t)H(t) + G'(t)H'(t) + G'(t)H'(t) + G(t)H''(t)\)     M1A1

\(J''(1) = G''(1)H(1) + 2G'(1)H'(1) + G(1)H''(1)\)

\( = G''(1) + 2G'(1)H'(1) + H''(1)\)    A1

\({\text{Var}}(Z) = J''(1) + J'(1) - {\left( {J'(1)} \right)^2}\)    M1

\( = G''(1) + 2G'(1)H'(1) + H''(1) + G'(1) + H'(1) - {\left( {G'(1) + H'(1)} \right)^2}\)    A1

\( = G''(1) + G'(1) - {\left( {G'(1)} \right)^2} + H''(1) + H'(1) - {\left( {H'(1)} \right)^2}\)    A1

so \({\text{Var}}(Z) = {\text{Var}}(X) + {\text{Var}}(Y)\)     AG

Note: If addition is wrongly used instead of multiplication in (a) it is inappropriate to give FT apart from the second M marks in each part, as the working is too simple.

[10 marks]

Find the mean and the variance of

(i)     \({X_1} + {X_2}\) ;

(ii)     \(3{X_1}\);

(iii)     \({X_1} + {X_2} - {X_3}\) ;

(iv)     \(\bar X = \frac{{({X_1} + {X_2} + ... + {X_n})}}{n}\).

Find \({\text{E}}(X_1^2)\) in terms of \(\mu \) and \(\sigma \) .

(i)     \(2\mu ,{\text{ }}2{\sigma ^2}\)     A1A1

(ii)     \(3\mu ,{\text{ }}9{\sigma ^2}\)     A1A1

(iii)     \(\mu ,{\text{ }}3{\sigma ^2}\)     A1A1

(iv)     \(\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}\)     A1A1

Note: If candidate clearly and correctly gives the standard deviations rather than the variances, give A1 for 2 or 3 standard deviations and A1A1 for 4 standard deviations.

[8 marks]

\({\text{Var}}({X_1}) = {\text{E}}(X_1^2) - {\left( {{\text{E}}({X_1})} \right)^2}\)     (M1)

\({\sigma ^2} = {\text{E}}(X_1^2) - {\mu ^2}\)     (A1)

\({\text{E}}(X_1^2) = {\sigma ^2} + {\mu ^2}\)     A1

[3 marks]

This was very well answered indeed with very many candidates gaining full marks including, pleasingly, part (b). Candidates who could not do question 2, struggled on the whole paper.

This was very well answered indeed with very many candidates gaining full marks including, pleasingly, part (b). Candidates who could not do question 2, struggled on the whole paper.

(a)     Having thrown the die once, she lets \({X_2}\) denote the number of additional throws required to obtain a different number from the one obtained on the first throw. State the distribution of \({X_2}\) and hence find \({\text{E}}({X_2})\) .

(b)     She then lets \({X_3}\) denote the number of additional throws required to obtain a different number from the two numbers already obtained. State the distribution of \({X_3}\) and hence find \({\text{E}}({X_3})\) .

(c)     By continuing the process, show that the expected number of tosses needed to obtain all six possible scores is 14.7.

(a)     \({X_2}\) is a geometric random variable     A1

with \(p = \frac{5}{6}.\)     A1

Therefore \({\text{E}}({X_2}) = \frac{6}{5}.\)     A1

[3 marks]

(b)     \({X_3}\) is a geometric random variable with \(p = \frac{4}{6}.\)     A1

Therefore \({\text{E}}({X_3}) = \frac{6}{4}.\)     A1

[2 marks]

(c)     \({\text{E}}({X_4}) = \frac{6}{3},{\text{ E}}({X_5}) = \frac{6}{2},{\text{ E}}({X_6}) = \frac{6}{1}\)     A1A1A1

\({\text{E}}({X_1}) = 1\,\,\,\,\,{\text{(or }}{X_1} = 1)\)     A1

Expected number of tosses \(\sum\limits_{n = 1}^6 {{\text{E}}({X_n})} \)     M1

\( = 14.7\)     AG

[5 marks]

Total [10 marks]

Many candidates were unable even to start this question although those who did often made substantial progress.

A coin was tossed 200 times and 115 of these tosses resulted in ‘heads’. Use a two-tailed test with significance level 1 % to investigate whether or not the coin is biased.

The number of’ ‘heads’ X is B(200, p)     (M1)

\({{\text{H}}_0}:p = 0.5;{\text{ }}{{\text{H}}_1}:p \ne 0.5\)     A1A1

Note: Award A1A0 for the statement “ \({{\text{H}}_0}:\) coin is fair; \({{\text{H}}_1}:\) coin is biased”.

EITHER

\({\text{P}}(\left. {X \geqslant 115} \right|{{\text{H}}_0}) = 0.0200\)     (M1)(A1)

p-value = 0.0400     A1

This is greater than 0.01.     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

(Using a proportion test on a GDC) p-value = 0.0339     N3

This is greater than 0.01.     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

Under \({{\text{H}}_0}X\) is approximately N(100, 50)     (M1)

\(z = \frac{{115 - 100}}{{\sqrt {50} }} = 2.12\)     (M1)A1

(Accept 2.05 with continuity correction)

This is less than 2.58     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

OR

99 % confidence limits for p are \(\frac{{115}}{{200}} \pm 2.576\sqrt {\frac{{115}}{{200}} \times \frac{{85}}{{200}} \times \frac{1}{{200}}} \)     (M1)A1

giving [0.485, 0.665]     A1

This interval contains 0.5     R1

There is insufficient evidence to conclude that the coin is biased (or the coin is not biased).     R1

[8 marks]

This question was well answered in general with several correct methods seen. The most popular method was to use a GDC to carry out a proportion test which is equivalent to using a normal approximation. Relatively few candidates calculated an exact p-value using the binomial distribution. Candidates who found a 95% confidence interval for p, the probability of obtaining a head, and noted that this contained 0.5 were given full credit.

The random variable Y is such that \({\text{E}}(2Y + 3) = 6{\text{ and Var}}(2 - 3Y) = 11\).

Calculate

(i)     E(Y) ;

(ii)     \({\text{Var}}(Y)\) ;

(iii)     \({\text{E}}({Y^2})\) .

Independent random variables R and S are such that

\[R \sim {\text{N}}(5,{\text{ 1}}){\text{ and }}S \sim {\text{N(8, 2).}}\]

The random variable V is defined by V = 3S – 4R.

Calculate P(V > 5).

(i)     \({\text{E}}(2Y + 3) = 6\)

\(2{\text{E}}(Y) + 3 = 6\)     M1

\({\text{E}}(Y) = \frac{3}{2}\)     A1

(ii)     \({\text{Var}}(2 - 3Y) = 11\)

\({\text{Var}}( - 3Y) = 11\)     (M1)

\(9{\text{Var}}(Y) = 11\)

\({\text{Var}}(Y) = \frac{{11}}{9}\)     A1

(iii)     \({\text{E}}({Y^2}) = {\text{Var}}(Y) + {\left[ {{\text{E}}(Y)} \right]^2}\)     M1

\( = \frac{{11}}{9} + \frac{9}{4}\)

\( = \frac{{125}}{{36}}\)     A1     N0

[6 marks]

 E(V) = E(3S – 4R)

 = 3E(S) – 4E(R)     M1

 = 24 – 20 = 4     A1

 Var(3S – 4R) = 9Var(S) + 16Var(R) , since R and S are independent random variables     M1

 =18 + 16 = 34     A1

 \(V \sim {\text{N}}(4,{\text{ 34}})\)

 \({\text{P}}(V > 5) = 0.432\)     A2     N0

 [6 marks]

E(Y) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. V was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 - P(V \leqslant 5)\).

E(Y) was calculated correctly but many could not go further to find \(Var(Y){\text{ and }}E({Y^2})Var(2)\) was often taken to be 2. V was often taken to be discrete leading to calculations such as \(P(V > 5) = 1 - P(V \leqslant 5)\).

Determine unbiased estimates for the mean and variance of the distribution.

In spite of these results the baker insists that his claim is correct.

Stating appropriate hypotheses, test the baker’s claim at the 10 % level of significance.

unbiased estimate of the mean: 795 (grams)     A1

unbiased estimate of the variance: 108 \((gram{s^2})\)     (M1)A1

[3 marks]

null hypothesis \({H_0}:\mu  = 800\)     A1

alternative hypothesis \({H_1}:\mu  < 800\)     A1

using 1-tailed t-test     (M1)

EITHER

p = 0.0812...     A3

OR

with 9 degrees of freedom     (A1)

\({t_{calc}} = \frac{{\sqrt {10} (795 - 800)}}{{\sqrt {108} }} = - 1.521\)     A1

\({t_{crit}} = - 1.383\)     A1 

Note: Accept 2sf intermediate results.

THEN

so the baker’s claim is rejected     R1 

Note: Accept “reject \({H_0}\) ” provided \({H_0}\) has been correctly stated.

Note: FT for the final R1.

[7 marks]

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

A successful question for many candidates. A few candidates did not read the question and adopted a 2-tailed test.

Let \(X\) and \(Y\) be independent random variables with \(X \sim {P_o}{\text{ (3)}}\) and \(Y \sim {P_o}{\text{ (2)}}\).

Let \(S = 2X + 3Y\).

(a)     Find the mean and variance of \(S\).

(b)     Hence state with a reason whether or not \(S\) follows a Poisson distribution.

Let \(T = X + Y\).

(c)     Find \({\text{P}}(T = 3)\).

(d)     Show that \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \).

(e)     Hence show that \(T\) follows a Poisson distribution with mean 5.

Find the probability of a Type II error if the leaves are in fact from a plant of species B.

(a)     \({\text{E}}(S) = 2{\text{E}}(X) + 3{\text{E}}(Y) = 6 + 6 = 12\)     A1

\({\text{Var}}(S) = 4{\text{Var}}(X) + 9{\text{Var}}(Y) = 12 + 18 = 30\)     A1

[2 marks]

(b)     \(S\) does not have a Poisson distribution     A1

because \({\text{Var}}(S) \ne {\text{E}}(S)\)     R1

Note:     Follow through their \({\text{E}}(S)\) and \({\text{Var}}(S)\) if different.

[2 marks]

(c)     EITHER

\({\text{P}}(T = 3) = {\text{P}}\left( {(X,{\text{ }}Y) = (3,{\text{ }}0)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (2,{\text{ }}1)} \right) + \)

\( + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}2)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}3)} \right)\)     (M1)

\( = {\text{P}}(X = 3){\text{P}}(Y = 0) + {\text{P}}(X = 2){\text{P}}(Y = 1) + \)

\( + {\text{P}}(X = 1){\text{P}}(Y = 2) + {\text{P}}(X = 0){\text{P}}(Y = 3)\)     (M1)

\( = \frac{{125{e^{ - 5}}}}{6}{\text{ }}( = 0.140)\)     A2

Note:     Accept answers which round to 0.14.

OR

\(T\) is \({{\text{P}}_o}(2 + 3) = {{\text{P}}_o}(5)\)     (M1)(A1)

\({\text{P}}(T = 3) = \frac{{125{e^{ - 5}}}}{6}{\text{ }}( = 0.140)\)     A2

Note:     Accept answers which round to 0.14.

[4 marks]

(d)     \({\text{P}}(T = t) = {\text{P}}\left( {(X,{\text{ }}Y) = (0,{\text{ }}t)} \right) + {\text{P}}\left( {(X,{\text{ }}Y) = (1,{\text{ }}t - 1)} \right) +  \ldots {\text{P}}\left( {(X,{\text{ }}Y) = (t,{\text{ }}0)} \right)\)     (M1)

\( = {\text{P}}(X = 0){\text{P}}(Y = t) + {\text{P}}(X = 1){\text{P}}(Y = t - 1) +  \ldots  + {\text{P}}(X = t){\text{P}}(Y = 0)\)     A1

\( = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \)     AG

[2 marks]

(e)     \({\text{P}}(T = t) = \sum\limits_{r = 0}^t {{\text{P}}(X = r){\text{P}}(Y = t - r)} \)

\( = \sum\limits_{r = 0}^t {\frac{{{e^{ - 3}}{3^r}}}{{r!}} \times \frac{{{e^{ - 2}}{2^{t - r}}}}{{(t - r)!}}} \)     M1A1

\( = \frac{{{e^{ - 5}}}}{{t!}}\sum\limits_{r = 0}^t {\frac{{t!}}{{r!(t - r)!}} \times {3^r}{2^{t - r}}} \)     M1

\( = \frac{{{e^{ - 5}}}}{{t!}}{(3 + 2)^t}\)     A1

\(\left( { = \frac{{{e^{ - 5}}{5^t}}}{{t!}}} \right)\)

hence \(T\) follows a Poisson distribution with mean 5     AG

[4 marks]

type II error probability \( = {\text{P}}(\bar X > 4.70654 \ldots |\bar X{\text{ is }}N\left( {4.6,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\)     (M1)

\( = 0.361\)     A1

Parts (a) and (b) were well answered by most candidates. The most common error in (a) was to calculate \(E(2X + 3Y)\) correctly as 12 and then state that, because the sum is Poisson, the variance is also 12. Many of these candidates then stated in (b) that the sum is Poisson because the mean and variance are equal, without apparently realising the circularity of their argument. Although (c) was intended as a possible hint for solving (d) and (e), many candidates simply noted that \(X + Y\) is \({{\text{P}}_o}{\text{(5)}}\) which led immediately to the correct answer. Some candidates tended to merge (d) and (e), often unsuccessfully, while very few candidates completed (e) correctly where the need to insert \(t!\) in the numerator and denominator was not usually spotted.

Alan and Brian are athletes specializing in the long jump. When Alan jumps, the length of his jump is a normally distributed random variable with mean 5.2 metres and standard deviation 0.1 metres. When Brian jumps, the length of his jump is a normally distributed random variable with mean 5.1 metres and standard deviation 0.12 metres. For both athletes, the length of a jump is independent of the lengths of all other jumps. During a training session, Alan makes four jumps and Brian makes three jumps. Calculate the probability that the mean length of Alan’s four jumps is less than the mean length of Brian’s three jumps.

Colin joins the squad and the coach wants to know the mean length, \(\mu \) metres, of his jumps. Colin makes six jumps resulting in the following lengths in metres.

5.21, 5.30, 5.22, 5.19, 5.28, 5.18

(i)     Calculate an unbiased estimate of both the mean \(\mu \) and the variance of the lengths of his jumps.

(ii)     Assuming that the lengths of these jumps are independent and normally distributed, calculate a 90 % confidence interval for \(\mu \) .

let \(\bar A,{\text{ }}\bar B\) denote the means of Alan’s and Brian’s jumps

attempting to find the distributions of \(\bar A,{\text{ }}\bar B\)     (M1)

\(\bar A{\text{ is N}}\left( {5.2,\frac{{{{0.1}^2}}}{4}} \right)\)     A1

\(\bar B{\text{ is N}}\left( {5.1,\frac{{{{0.12}^2}}}{3}} \right)\)     A1

attempting to find the distribution of \(\bar A - \bar B\)     (M1)

\(\bar A - \bar B{\text{ is N}}\left( {5.2 - 5.1,\frac{{{{0.1}^2}}}{4} + \frac{{{{0.12}^2}}}{3}} \right)\)     (A1)(A1)

i.e. \({\text{N}}(0.1,{\text{ }}0.0073)\)     A1

\({\text{P}}(\bar A < \bar B) = {\text{P}}(\bar A - \bar B < 0)\)     M1

\( = 0.121\)     A1

[9 marks]

(i)     \(\sum {x = 31.38,{\text{ }}\sum {{x^2} = 164.1294} } \)

\(\bar x = \frac{{31.38}}{6} = 5.23\)     (M1)A1

EITHER

\(s_{n - 1}^2 = \frac{{164.1294}}{5} - \frac{{{{31.38}^2}}}{{5 \times 6}} = 0.00240\)     (M1)(A1)A1

OR

\({s_{n - 1}} = 0.04899 \Rightarrow s_{n - 1}^2 = 0.00240\)     (M1)(A1)A1

Note: Accept the exact answer 0.0024 without an arithmetic penalty.

(ii)     using the t-distribution with DF = 5     (A1)

critical value of t = 2.015     A1

90 % confidence limits are \(5.23 \pm 2.015\sqrt {\frac{{0.0024}}{6}} \)     M1A1

giving [5.19, 5.27]     A1     N5

[10 marks]

In (a), it was disappointing to note that many candidates failed to realise that the question was concerned with the mean lengths of the jumps and worked instead with the sums of the lengths.

Most candidates obtained correct estimates in (b)(i), usually directly from the GDC. In (b)(ii), however, some candidates found a z-interval instead of a t-interval.

Find the probability that

(i)     he hits the target exactly 4 times in his first 8 shots;

(ii)     he hits the target for the \({4^{{\text{th}}}}\) time with his \({8^{{\text{th}}}}\) shot.

Ben hits the target for the \({10^{{\text{th}}}}\) time with his \({X^{{\text{th}}}}\) shot.

(i)     Determine the expected value of the random variable X.

(ii)     Write down an expression for \({\text{P}}(X = x)\) and show that

\[\frac{{{\text{P}}(X = x)}}{{{\text{P}}(X = x - 1)}} = \frac{{3(x - 1)}}{{5(x - 10)}}.\]

(iii)     Hence, or otherwise, find the most likely value of X.

(i)     the number of hits, \(X \sim {\text{B(8, 0.4)}}\)     (A1)

\(P(X = 4) = \left( {\begin{array}{*{20}{c}}
  8 \\
  4
\end{array}} \right) \times {0.4^4} \times {0.6^4}\)     (M1)

= 0.232     A1

Note: Accept any answer that rounds to 0.23.

(ii)     let the \({4^{{\text{th}}}}\) hit occur on the \({Y^{{\text{th}}}}\) shot so that \(Y \sim {\text{NB(4, 0.4)}}\)     (A1)

\(P(Y = 8) = \left( {\begin{array}{*{20}{c}}
  7 \\
  3
\end{array}} \right) \times {0.4^4} \times {0.6^4}\)     (M1)

= 0.116     A1

Note: Accept any answer that rounds to 0.12.

[6 marks]

(i)     \(X \sim {\text{NB(10, 0.4)}}\)     (M1)

\({\text{E}}(X) = \frac{{10}}{{0.4}} = 25\)     A1

(ii)     let \({{\text{P}}_x}\) denote \({\text{P}}(X = x)\)

\({P_x} = \left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  9
\end{array}} \right) \times {0.4^{10}} \times {0.6^{x - 10}}\)     A1

\(\frac{{{P_x}}}{{{P_{x - 1}}}} = \frac{{\left( {\begin{array}{*{20}{c}}
  {x - 1} \\
  9
\end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x - 10}}}}{{\left( {\begin{array}{*{20}{c}}
  {x - 2} \\
  9
\end{array}} \right) \times {{0.4}^{10}} \times {{0.6}^{x - 11}}}}\)     M1A1

\( = \frac{{(x - 1)!}}{{9!(x - 10)!}} \times \frac{{9!(x - 11)! \times 0.6}}{{(x - 2)!}}\)     A1

Note: Award A1 for correct evaluation of combinatorial terms.

\( = \frac{{3(x - 1)}}{{5(x - 10)}}\)     AG

(iii)     \({{\text{P}}_x} > {{\text{P}}_{x - 1}}\) as long as

\(3x - 3 > 5x - 50\)     (M1)

i.e. \(x < 23.5\)     (A1)

the most likely value is 23     A1

Note: Allow solutions based on creating a table of values of \({{\text{P}}_x}\).

[9 marks]

Part (a) was well answered in general although some candidates were unable to distinguish between the binomial and negative binomial distributions.

In (b)(ii), most candidates knew what to do but algebraic errors were not uncommon. Candidates often used equal instead of inequality signs and this was accepted if it led to \(x = 23.5\). The difficulty for these candidates was whether to choose \(23\) or \(24\) for the final answer and some made the wrong choice. Some candidates failed to see the relevance of the result in (b)(ii) to finding the most likely value of \(X\) and chose an ‘otherwise’ method, usually by creating a table of probabilities and selecting the largest.

The mean weight of a certain breed of bird is believed to be 2.5 kg. In order to test this belief, it is planned to determine the weights \({x_1}{\text{ , }}{x_2}{\text{ , }}{x_3}{\text{ , }} \ldots {\text{, }}{x_{16}}\) (in kg) of sixteen of these birds and then to calculate the sample mean \({\bar x}\) . You may assume that these weights are a random sample from a normal distribution with standard deviation 0.1 kg.

(a)     State suitable hypotheses for a two-tailed test.

(b)     Find the critical region for \({\bar x}\) having a significance level of 5 %.

(c)     Given that the mean weight of birds of this breed is actually 2.6 kg, find the probability of making a Type II error.

(a)     \({H_0}:\mu = 2.5\)     A1

\({H_1}:\mu \ne 2.5\)     A1

[2 marks]

(b)     the critical values are \(2.5 \pm 1.96 \times \frac{{0.1}}{{\sqrt {16} }}\) ,     (M1)(A1)(A1)

i.e. 2.45, 2.55     (A1)

the critical region is \(\bar x < 2.45 \cup \bar x > 2.55\)     A1A1

Note: Accept \( \leqslant ,{\text{ }} \geqslant \) .

[6 marks]

(c)     \({\bar X}\) is now \({\text{N}}(2.6,{\text{ }}{0.025^2})\)     A1

a Type II error is accepting \({H_0}\) when \({H_1}\) is true     (R1)

thus we require

\({\text{P}}(2.45 < \bar X < 2.55)\)     M1A1

\( = 0.0228\,\,\,\,\,\)(Accept 0.0227)     A1

Note: If critical values of 2.451 and 2.549 are used, accept 0.0207.

[5 marks]

Total [13 marks]

In (a), some candidates incorrectly gave the hypotheses in terms of \({\bar x}\) instead of \(\mu \). In (b), many candidates found the correct critical values but then some gave the critical region as \(2.45 < \bar x < 2.55\) instead of \(\bar x < 2.45 \cup \bar x > 2.55\) Many candidates gave the critical values correct to four significant figures and therefore were given an arithmetic penalty. In (c), many candidates correctly defined a Type II error but were unable to calculate the corresponding probability.

The apple trees in a large orchard have, for several years, suffered from a disease for which the outward sign is a red discolouration on some leaves.

The fruit grower knows that the mean number of discoloured leaves per tree is 42.3. The fruit grower suspects that the disease is caused by an infection from a nearby group of cedar trees. He cuts down the cedar trees and, the following year, counts the number of discoloured leaves on a random sample of seven apple trees. The results are given in the table below.

(a)     From these data calculate an unbiased estimate of the population variance.

(b)     Stating null and alternative hypotheses, carry out an appropriate test at the 10 % level to justify the cutting down of the cedar trees.

(a)     \(n = 7,{\text{ sample mean }} = 35\)     (A1)

\(s_{n - 1}^2 = \frac{{\sum {{{(x - 35)}^2}} }}{6} = 322\)     (M1)A1

[3 marks]

(b)     null hypothesis \({{\text{H}}_0}:\mu = 42.3\)     A1

alternative hypothesis \({{\text{H}}_1}:\mu < 42.3\)     A1

using one-sided t-test

\(\left| {{t_{{\text{calc}}}}} \right| = \sqrt 7 \frac{{42.3 - 35}}{{\sqrt {322} }} = 1.076\)     (M1)(A1)

with 6 degrees of freedom , \({t_{{\text{crit}}}} = 1.440 > 1.076\)

\({\text{(or }}p{\text{-value }} = 0.162 > 0.1)\)     A1

we conclude that there is no justification for cutting down the cedar trees     R1     N0

Note: FT on their t or p-value.

[6 marks]

Total [9 marks]

This question was generally well attempted as an example of the t-test. Very few used the Z statistic, and many found p-values.

Determine \({\text{E}}(X)\) and show that \({\text{Var}}(X) = 6\theta - 16{\theta ^2}\).

In order to estimate \(\theta \), a random sample of n observations is obtained from the distribution of X .

(i)     Given that \({\bar X}\) denotes the mean of this sample, show that

\[{{\hat \theta }_1} = \frac{{3 - \bar X}}{4}\]

is an unbiased estimator for \(\theta \) and write down an expression for the variance of \({{\hat \theta }_1}\) in terms of n and \(\theta \).

(ii)     Let Y denote the number of observations that are equal to 1 in the sample. Show that Y has the binomial distribution \({\text{B}}(n,{\text{ }}\theta )\) and deduce that \({{\hat \theta }_2} = \frac{Y}{n}\) is another unbiased estimator for \(\theta \). Obtain an expression for the variance of \({{\hat \theta }_2}\).

(iii)     Show that \({\text{Var}}({{\hat \theta }_1}) < {\text{Var}}({{\hat \theta }_2})\) and state, with a reason, which is the more efficient estimator, \({{\hat \theta }_1}\) or \({{\hat \theta }_2}\).

\({\text{E}}(X) = 1 \times \theta + 2 \times 2\theta + 3(1 - 3\theta ) = 3 - 4\theta \)     M1A1

\({\text{Var}}(X) = 1 \times \theta + 4 \times 2\theta + 9(1 - 3\theta ) - {(3 - 4\theta )^2}\)     M1A1

\( = 6\theta - 16{\theta ^2}\)     AG

[4 marks]

(i)     \({\text{E}}({\hat \theta _1}) = \frac{{3 - {\text{E}}(\bar X)}}{4} = \frac{{3 - (3 - 4\theta )}}{4} = \theta \)     M1A1

so \({\hat \theta _1}\) is an unbiased estimator of \(\theta \)     AG

\({\text{Var}}({{\hat \theta }_1}) = \frac{{6\theta  - 16{\theta ^2}}}{{16n}}\)     A1

(ii)     each of the n observed values has a probability \(\theta \) of having the value 1     R1

so \(Y \sim {\text{B}}(n,{\text{ }}\theta )\)     AG

\({\text{E}}({{\hat \theta }_2}) = \frac{{{\text{E}}(Y)}}{n} = \frac{{n\theta }}{n} = \theta \)     A1

\({\text{Var}}({{\hat \theta }_2}) = \frac{{n\theta (1 - \theta )}}{{{n^2}}} = \frac{{\theta (1 - \theta )}}{n}\)     M1A1

(iii)     \({\text{Var}}({{\hat \theta }_1}) - {\text{Var}}({{\hat \theta }_2}) = \frac{{6\theta  - 16{\theta ^2} - 16\theta  + 16{\theta ^2}}}{{16n}}\)     M1

\( = \frac{{ - 10\theta }}{{16n}} < 0\)     A1

\({{\hat \theta }_1}\) is the more efficient estimator since it has the smaller variance     R1

[10 marks]

(a)     Consider the random variable \(X\) for which \({\text{E}}(X) = a\lambda  + b\), where \(a\) and \(b\)are constants and \(\lambda \) is a parameter.

Show that \(\frac{{X - b}}{a}\) is an unbiased estimator for \(\lambda \).

(b)     The continuous random variable Y has probability density function

\(f(y) = \left\{ \begin{array}{r}{\textstyle{2 \over 9}}(3 + y - \lambda ),\\0,\end{array} \right.\begin{array}{*{20}{l}}{{\rm{ for}}\, \lambda  - 3 \le y \le \lambda }\\{{\rm{ otherwise}}}\end{array}\)

where \(\lambda \) is a parameter.

          (i)     Verify that \(f(y)\) is a probability density function for all values of \(\lambda \).

          (ii)     Determine \({\text{E}}(Y)\).

          (iii)     Write down an unbiased estimator for \(\lambda \).

(a)     \({\text{E}}\left( {\frac{{X - b}}{a}} \right) = \frac{{a\lambda  + b - b}}{a}\)     M1A1

\( = \lambda \)     A1

(Therefore \(\frac{{X - b}}{a}\) is an unbiased estimator for \(\lambda \))     AG

[3 marks]

(b)     (i)     \(f(y) \geqslant 0\)     R1

Note:     Only award R1 if this statement is made explicitly.

          recognition or showing that integral of f is 1 (seen anywhere) R1

          EITHER

          \(\int_{\lambda  - 3}^\lambda  {\frac{2}{9}(3 + y - \lambda ){\text{d}}y} \)     M1

          \( = \frac{2}{9}\left[ {(3 - \lambda )y + \frac{1}{2}{y^2}} \right]_{\lambda  - 3}^\lambda \)     A1

          \( = \frac{2}{9}\left( {\lambda (3 - \lambda ) + \frac{1}{2}{\lambda ^2} - (3 - \lambda )(\lambda  - 3) - \frac{1}{2}{{(\lambda  - 3)}^2}} \right)\) or equivalent     A1

          \( = 1\)

          OR

          the graph of the probability density is a triangle with base length 3 and height \(\frac{2}{3}\)     M1A1

          its area is therefore \(\frac{1}{2} \times 3 \times \frac{2}{3}\)     A1

          \( = 1\)

          (ii)     \({\text{E}}(Y) = \int_{\lambda  - 3}^\lambda  {\frac{2}{9}y(3 + y - \lambda ){\text{d}}y} \)     M1

          \( = \frac{2}{9}\left[ {(3 - \lambda )\frac{1}{2}{y^2} + \frac{1}{3}{y^3}} \right]_{\lambda  - 3}^\lambda \)     A1

          \( = \frac{2}{9}\left( {(3 - \lambda )\frac{1}{2}\left( {{\lambda ^2} - {{(\lambda  - 3)}^2}} \right) + \frac{1}{3}\left( {{\lambda ^3} - {{(\lambda  - 3)}^3}} \right)} \right)\)     M1

          \( = \lambda  - 1\)     A1A1

Note:     Award 3 marks for noting that the mean is \(\frac{2}{3}{\text{rds}}\) the way along the base and then A1A1 for \(\lambda  - 1\).

Note:     Award A1 for \(\lambda \) and A1 for –1.

          (iii)     unbiased estimator: \(Y + 1\)     A1

Note:     Accept \(\bar Y + 1\).

     Follow through their \({\text{E}}(Y)\) if linear.

[11 marks]

Total [14 marks]

Find unbiased estimates of \(\mu \) and \({\sigma ^2}\).

Determine a 95 % confidence interval for \(\mu \).

Given the hypotheses

\[{{\text{H}}_0}:\mu  = 15;{\text{ }}{{\text{H}}_1}:\mu  \ne 15,\]

find the p-value of the above results and state your conclusion at the 1 % significance level.

\(\bar x = 14\)     A1

\(s_{n - 1}^2 = \frac{{3977.57}}{{19}} - \frac{{{{280}^2}}}{{380}}\)     (M1)

\( = 3.03\)     A1

[3 marks]

Note: Accept any notation for these estimates including \(\mu \) and \({\sigma ^2}\).

Note: Award M0A0 for division by 20.

the 95% confidence limits are

\(\bar x \pm t\sqrt {\frac{{s_{n - 1}^2}}{n}} \)     (M1)

Note: Award M0 for use of z.

ie, \(14 \pm 2.093\sqrt {\frac{{3.03}}{{20}}} \)     (A1)

Note:FT their mean and variance from (a).

giving [13.2, 14.8]     A1

Note: Accept any answers which round to 13.2 and 14.8.

[3 marks]

Use of t-statistic \(\left( { = \frac{{14 - 15}}{{\sqrt {\frac{{3.03}}{{20}}} }}} \right)\)     (M1)

Note:FT their mean and variance from (a).

Note: Award M0 for use of z.

Note: Accept \(\frac{{15 - 14}}{{\sqrt {\frac{{3.03}}{{20}}} }}\).

\( = - 2.569 \ldots \)     (A1)

Note: Accept \(2.569 \ldots \)

\(p{\text{ - value}} = 0.009392 \ldots  \times 2 = 0.0188\)     A1

Note: Accept any answer that rounds to 0.019.

Note: Award (M1)(A1)A0 for any answer that rounds to 0.0094.

insufficient evidence to reject \({{\text{H}}_0}\) (or equivalent, eg accept \({{\text{H}}_0}\) or reject \({{\text{H}}_1}\))     R1

Note:FT on their p-value.

[4 marks]

In (a), most candidates estimated the mean correctly although many candidates failed to obtain a correct unbiased estimate for the variance. The most common error was to divide \(\sum {{x^2}} \) by \(20\) instead of \(19\). For some candidates, this was not a costly error since we followed through their variance into (b) and (c).

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

In (b) and (c), since the variance was estimated, the confidence interval and test should have been carried out using the t-distribution. It was extremely disappointing to note that many candidates found a Z-interval and used a Z-test and no marks were awarded for doing this. Candidates should be aware that having to estimate the variance is a signpost pointing towards the t-distribution.

Find the probability that a randomly chosen orange weighs more than 200 grams.

Five of these oranges are selected at random to be put into a bag. Find the probability that the combined weight of the five oranges is less than 1 kilogram.

The farm also produces lemons whose weights may be assumed to be normally distributed with mean 75 grams and standard deviation 3 grams. Find the probability that the weight of a randomly chosen orange is more than three times the weight of a randomly chosen lemon.

\(z = \frac{{200 - 205}}{{10}} = - 0.5\)     (M1)

probability = 0.691 (accept 0.692)     A1

Note: Award M1A0 for 0.309 or 0.308

[2 marks]

let X be the total weight of the 5 oranges

then \({\text{E}}(X) = 5 \times 205 = 1025\)     (A1)

\({\text{Var}}(X) = 5 \times 100 = 500\)     (M1)(A1)

\({\text{P}}(X < 1000) = 0.132\)     A1

[4 marks]

let Y = B – 3C where B is the weight of a random orange and C the weight of a random lemon     (M1)

\({\text{E}}(Y) = 205 - 3 \times 75 = - 20\)     (A1)

\({\text{Var}}(Y) = 100 + 9 \times 9 = 181\)     (M1)(A1)

\({\text{P}}(Y > 0) = 0.0686\)     A1

[5 marks]

Note: Award A1 for 0.0681 obtained from tables

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .

As might be expected, (a) was well answered by many candidates, although those who gave 0.6915 straight from tables were given an arithmetic penalty. Parts (b) and (c), however, were not so well answered with errors in calculating the variances being the most common source of incorrect solutions. In particular, some candidates are still uncertain about the difference between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) .

(i)     Determine an expression for \(F(x)\), valid for \(1 \leqslant x \leqslant 2\), where F denotes the cumulative distribution function of X.

(ii)     Hence, or otherwise, determine the median of X.

(i)     State the central limit theorem.

(ii)     A random sample of 150 observations is taken from the distribution of X and \(\bar X\) denotes the sample mean. Use the central limit theorem to find, approximately, the probability that \(\bar X\) is greater than 1.6.

(i)     \(F(x) = \int_1^x {\frac{{3{u^2} + 2u}}{{10}}{\text{d}}u} \)     (M1)

\( = \left[ {\frac{{{u^3} + {u^2}}}{{10}}} \right]_1^x\)     A1

Note: Do not penalise missing or wrong limits at this stage.

Accept the use of x in the integrand.

\( = \frac{{{x^3} + {x^2} - 2}}{{10}}\)     A1

(ii)     the median m satisfies the equation \(F(m) = \frac{1}{2}\) so     (M1)

\({m^3} + {m^2} - 7 = 0\)     (A1)

Note: Do not FT from an incorrect \(F(x)\).

\(m = 1.63\)     A1

Note: Accept any answer that rounds to 1.6.

[6 marks]

(i)     the mean of a large sample from any distribution is approximately 

normal     A1

Note: This is the minimum acceptable explanation.

(ii)     we require the mean \(\mu \) and variance \({\sigma ^2}\) of X

\(\mu  = \int_1^2 {\left( {\frac{{3{x^3} + 2{x^2}}}{{10}}} \right){\text{d}}x} \)     (M1)

\( = \frac{{191}}{{120}}{\text{ }}(1.591666 \ldots )\)     A1

\({\sigma ^2} = \int_1^2 {\left( {\frac{{3{x^4} + 2{x^3}}}{{10}}} \right){\text{d}}x - {\mu ^2}} \)     (M1)

\( = 0.07659722 \ldots \)     A1

the central limit theorem states that

\(\bar X \approx N\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right),\) i.e. \(N(1.591666 \ldots ,{\text{ }}0.0005106481 \ldots )\)     M1A1

\({\text{P}}(\bar X > 1.6) = 0.356\)     A1

Note: Accept any answer that rounds to 0.36.

[8 marks]

Solutions to (a)(i) were disappointing in general, suggesting that many candidates are unfamiliar with the concept of the cumulative distribution function. Many candidates knew that it was something to do with the integral of the probability density function but some thought it was \(\int\limits_1^2 {f(x){\text{d}}x} \) which they then evaluated as \(1\) while others thought it was just \(\int {f(x){\text{d}}x}  = \frac{{\left( {{x^2} + {x^3}} \right)}}{{10}}\) which is not, in general, a valid method. However, most candidates solved (a)(ii) correctly, usually by integrating the probability density function from \(1\) to \(m\).

In (b)(i), the statement of the central limit theorem was often quite dreadful. The term ‘sample mean’ was often not mentioned and a common misconception appears to be that the actual distribution rather than the sample mean tends to normality as the sample size increases. Solutions to (b)(ii) often failed to go beyond finding the mean and variance of \(X\) . In calculating the variance, some candidates rounded the mean from \(1.5916666..\) to \(1.59\) which resulted in an incorrect value for the variance. It is important to note that calculating a variance usually involves a small difference of two large numbers so that full accuracy must be maintained.

(a)     A random variable, X , has probability density function defined by

\[f(x) = \left\{ {\begin{array}{*{20}{l}}
  {100,}&{{\text{for }} - 0.005 \leqslant x < 0.005} \\
  {0,}&{{\text{otherwise}}{\text{.}}}
\end{array}} \right.\]

Determine E(X) and Var(X) .

(b)     When a real number is rounded to two decimal places, an error is made.

Show that this error can be modelled by the random variable X .

(c)     A list contains 20 real numbers, each of which has been given to two decimal places. The numbers are then added together.

(i)     Write down bounds for the resulting error in this sum.

(ii)     Using the central limit theorem, estimate to two decimal places the probability that the absolute value of the error exceeds 0.01.

(iii)     State clearly any assumptions you have made in your calculation.

(a)     f(x)is even (symmetrical about the origin)     (M1)

\({\text{E}}(X) = 0\)     A1

\({\text{Var}}(X) = {\text{E}}({X^2}) = \int_{ - 0.005}^{0.005} {100{x^2}{\text{d}}x} \)     (M1)(A1)

\( = 8.33 \times {10^{ - 6}}\left( {{\text{accept }}0.83 \times {{10}^{ - 5}}{\text{ or }}\frac{1}{{120\,000}}} \right)\)     A1

[5 marks]

(b)     rounding errors to 2 decimal places are uniformly distributed     R1

and lie within the interval \( - 0.005 \leqslant x < 0.005.\)     R1

this defines X     AG

[2 marks]

(c)     (i)     using the symbol y to denote the error in the sum of 20 real numbers each rounded to 2 decimal places

\( - 0.1 \leqslant y( = 20 \times x) < 0.1\)     A1

(ii)     \(Y \approx {\text{N}}(20 \times 0,{\text{ }}20 \times 8.3 \times {10^{ - 6}}) = {\text{N}}(0,{\text{ }}0.00016)\)     (M1)(A1)

\({\text{P}}\left( {\left| Y \right| > 0.01} \right) = 2\left( {1 - {\text{P}}(Y < 0.01)} \right)\)     (M1)(A1)

\( = 2\left( {1 - {\text{P}}\left( {Z < \frac{{0.01}}{{0.0129}}} \right)} \right)\)

\( = 0.44\) to 2 decimal places     A1     N4

(iii)     it is assumed that the errors in rounding the 20 numbers are independent     R1

and, by the central limit theorem, the sum of the errors can be modelled approximately by a normal distribution     R1

[8 marks]

Total [15 marks]

This was the only question on the paper with a conceptually ‘hard’ final part. Part(a) was generally well done, either by integration or by use of the standard formulae for a uniform distribution. Many candidates were not able to provide convincing reasoning in parts (b) and (c)(iii). Part(c)(ii), the application of the Central Limit Theorem was only very rarely tackled competently.

Write down the value of \({\text{P}}(X = 1)\) and show that \({\text{P}}(X = 2) = \frac{1}{6}\).

Show that the probability generating function for X is given by

\[G(t) = \frac{{2t + {t^2}}}{{6 - 3{t^2}}}.\]

Hence determine \({\text{E}}(X)\).

\({\text{P}}(X = 1) = \frac{1}{3}\)     A1

\({\text{P}}(X = 2) = \frac{2}{3} \times \frac{1}{4}\)     A1

\(= \frac{1}{6}\)     AG

[2 marks]

\(G(t) = \frac{1}{3}t + \frac{2}{3} \times \frac{1}{4}{t^2} + \frac{2}{3} \times \frac{3}{4} \times \frac{1}{3}{t^3} + \frac{2}{3} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{4}{t^4} +  \ldots \)     M1A1

\( = \frac{1}{3}t\left( {1 + \frac{1}{2}{t^2} +  \ldots } \right) + \frac{1}{6}{t^2}\left( {1 + \frac{1}{2}{t^2} +  \ldots } \right)\)     M1A1

\( = \frac{{\frac{t}{3}}}{{1 - \frac{{{t^2}}}{2}}} + \frac{{\frac{{{t^2}}}{6}}}{{1 - \frac{{{t^2}}}{2}}}\)     A1A1

\( = \frac{{2t + {t^2}}}{{6 - 3{t^2}}}\)     AG

[6 marks]

\(G'(t) = \frac{{(2 + 2t)(6 - 3{t^2}) + 6t(2t + {t^2})}}{{{{(6 - 3{t^2})}^2}}}\)     M1A1

\({\text{E}}(X) = G'(1) = \frac{{10}}{3}\)     M1A1

[4 marks]

(i)     Prove that \({\text{E}}(X) = \lambda \).

(ii)     Prove that \({\text{Var}}(X) = \lambda \).

\(Y\) is a random variable, independent of \(X\), that also follows a Poisson distribution with mean \(\lambda \).

If \(S = 2X - Y\) find

(i)     \({\text{E}}(S)\);

(ii)     \({\text{Var}}(S)\).

Let \(T = \frac{Y}{2} + \frac{Y}{2}\).

(i)     Show that \(T\) is an unbiased estimator for \(\lambda \).

(ii)     Show that \(T\) is a more efficient unbiased estimator of \(\lambda \) than \(S\).

Could either \(S\) or \(T\) model a Poisson distribution? Justify your answer.

By consideration of the probability generating function, \({G_{X + Y}}(t)\), of \(X + Y\), prove that \(X + Y\) follows a Poisson distribution with mean \(2\lambda \).

Find

(i)     \({G_{X + Y}}(1)\);

(ii)     \({G_{X + Y}}( - 1)\).

Hence find the probability that \(X + Y\) is an even number.

(i)     \(G'(t) = \lambda {e^{\lambda (t - 1)}}\)     A1

\({\text{E}}(X) = G'(1)\)     M1

\( = \lambda \)     AG

(ii)     \(G''(t) = {\lambda ^2}{e^{\lambda (t - 1)}}\)     M1

\( \Rightarrow G''(1) = {\lambda ^2}\)     (A1)

\({\text{Var}}(X) = G''(1) + G'(1) - {\left( {G'(1)} \right)^2}\)     (M1)

\( = {\lambda ^2} + \lambda  - {\lambda ^2}\)     A1

\( = \lambda \)     AG

[6 marks]

(i)     \({\text{E}}(S) = 2\lambda  - \lambda  = \lambda \)     A1

(ii)     \({\text{Var}}(S) = 4\lambda  + \lambda  = 5\lambda \)     (A1)A1

Note:     First A1 can be awarded for either \(4\lambda \) or \(\lambda \).

[3 marks]

(i)     \({\text{E}}(T) = \frac{\lambda }{2} + \frac{\lambda }{2} = \lambda \;\;\;\)(so \(T\) is an unbiased estimator)     A1

(ii)     \({\text{Var}}(T) = \frac{1}{4}\lambda  + \frac{1}{4}\lambda  = \frac{1}{2}\lambda \)     A1

this is less than \({\text{Var}}(S)\), therefore \(T\) is the more efficient estimator     R1AG

Note:     Follow through their variances from (b)(ii) and (c)(ii).

[3 marks]

no, mean does not equal the variance     R1

[1 mark]

\({G_{X + Y}}(t) = {e^{\lambda (t - 1)}} \times {e^{\lambda (t - 1)}} = {e^{2\lambda (t - 1)}}\)     M1A1

which is the probability generating function for a Poisson with a mean of \(2\lambda \)     R1AG

[3 marks]

(i)     \({G_{X + Y}}(1) = 1\)     A1

(ii)     \({G_{X + Y}}( - 1) = {e^{ - 4\lambda }}\)     A1

[2 marks]

\({G_{X + Y}}(1) = p(0) + p(1) + p(2) + p(3) \ldots \)

\({G_{X + Y}}( - 1) = p(0) - p(1) + p(2) - p(3) \ldots \)

so \({\text{2P(even)}} = {G_{X + Y}}(1) + {G_{X + Y}}( - 1)\)     (M1)(A1)

\({\text{P(even)}} = \frac{1}{2}(1 + {e^{ - 4\lambda }})\)     A1

[3 marks]

Total [21 marks]

Solutions to the different parts of this question proved to be extremely variable in quality with some parts well answered by the majority of the candidates and other parts accessible to only a few candidates. Part (a) was well answered in general although the presentation was sometimes poor with some candidates doing the differentiation of \(G(t)\) and the substitution of \(t = 1\) simultaneously.

Part (b) was well answered in general, the most common error being to state that \({\text{Var}}(2X - Y) = {\text{Var}}(2X) - {\text{Var}}(Y)\).

Parts (c) and (d) were well answered by the majority of candidates.

Parts (c) and (d) were well answered by the majority of candidates.

Solutions to (e), however, were extremely disappointing with few candidates giving correct solutions. A common incorrect solution was the following:

\(\;\;\;{G_{X + Y}}(t) = {G_X}(t){G_Y}(t)\)

Differentiating,

\(\;\;\;{G'_{X + Y}}(t) = {G'_X}(t){G_Y}(t) + {G_X}(t){G'_Y}(t)\)

\(\;\;\;{\text{E}}(X + Y) = {G'_{X + Y}}(1) = {\text{E}}(X) \times 1 + {\text{E}}(Y) \times 1 = 2\lambda \)

This is correct mathematics but it does not show that \(X + Y\) is Poisson and it was given no credit. Even the majority of candidates who showed that \({G_{X + Y}}(t) = {{\text{e}}^{2\lambda (t - 1)}}\) failed to state that this result proved that \(X + Y\) is Poisson and they usually differentiated this function to show that \({\text{E}}(X + Y) = 2\lambda \).

In (f), most candidates stated that \({G_{X + Y}}(1) = 1\) even if they were unable to determine \({G_{X + Y}}(t)\) but many candidates were unable to evaluate \({G_{X + Y}}( - 1)\). Very few correct solutions were seen to (g) even if the candidates correctly evaluated \({G_{X + Y}}(1)\) and \({G_{X + Y}}( - 1)\).

Explain what can be done with this data to decrease the probability of making a type I error.

(i)     State the meaning of a type II error.

(ii)     Write down how to proceed if it is required to decrease the probability of making both a type I and type II error.

reduce the significance level (or equivalent statement)     R2

[2 marks]

(i)     accepting \({{\text{H}}_0}\) (or failing to reject \({{\text{H}}_0}\)) when it is false (or equivalent)     A1

(ii)     increase the number of trials     A1

[2 marks]

It was disappointing to see that some candidates wrote incorrect hypotheses, eg ‘\({{\text{H}}_0}\): Data are binomial; \({{\text{H}}_1}\): Data are not binomial’ without specifying any parameters. Part (b) caused unexpected problems for many candidates who misunderstood the question and gave ‘increase the number of trials’ as their answer.

Show that \(P = \frac{X}{n}\) is an unbiased estimator of \(p\).

Show that \({\text{E}}\left( U \right) = \left( {n - 1} \right)p\left( {1 - p} \right)\).

Hence write down an unbiased estimator of Var(X).

\({\text{E}}\left( P \right) = {\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}\left( X \right)\)      M1

\( = \frac{1}{n}\left( {np} \right) = p\)      A1

so P is an unbiased estimator of \(p\)     AG

[2 marks]

\({\text{E}}\left( {nP\left( {1 - P} \right)} \right) = {\text{E}}\left( {n\left( {\frac{X}{n}} \right)\left( {1 - \frac{X}{n}} \right)} \right)\)

\( = {\text{E}}\left( X \right) = \frac{1}{n}{\text{E}}\left( {{X^2}} \right)\)      M1A1

use of \({\text{E}}\left( {{X^2}} \right) = {\text{Var}}\left( X \right) + {\left( {{\text{E}}\left( X \right)} \right)^2}\)      M1

Note: Allow candidates to work with P rather than X for the above 3 marks.

\( = np - \frac{1}{n}\left( {np\left( {1 - p} \right) + {{\left( {np} \right)}^2}} \right)\)       A1

\( = np - p\left( {1 - p} \right) - n{p^2}\)

\( = np\left( {1 - p} \right) - p\left( {1 - p} \right)\)      A1

Note: Award A1 for the factor of \(\left( {1 - p} \right)\).

\( = \left( {n - 1} \right)p\left( {1 - p} \right)\)     AG

[5 marks]

an unbiased estimator is \(\frac{{{n^2}P\left( {1 - P} \right)}}{{n - 1}}\left( { = \frac{{nU}}{{n - 1}}} \right)\)       A1

[1 mark]

Two students are selected at random from a large school with equal numbers of boys and girls. The boys’ heights are normally distributed with mean \(178\) cm and standard deviation \(5.2\) cm, and the girls’ heights are normally distributed with mean \(169\) cm and standard deviation \(5.4\) cm.

Calculate the probability that the taller of the two students selected is a boy.

let \(X\) denote boys’ height and \(Y\) denote girls’ height

if \(BB,{\text{ P(taller is boy)}} = 1\)     (A1)

if \(GG,{\text{ P(taller is boy)}} = 0\)     (A1)

if \(BG\) or \(GB\):

consider \(X - Y\)     (M1)

\(E(X - Y) = 178 - 169 = 9\)     A1

\({\text{Var}}(X - Y) = {5.2^2} + {5.4^2}\;\;\;( = 56.2)\)     (M1)A1

\({\text{P}}(X - Y > 0) = 0.885\)     A1

answer is \(\frac{1}{4} \times 1 + \frac{1}{2} \times 0.885 = 0.693\)     (M1)A1

[9 marks]

A hospital specializes in treating overweight patients. These patients have weights that are independently, normally distributed with mean 200 kg and standard deviation 15 kg. The elevator in the hospital will break if the total weight of people inside it exceeds 1150 kg. Six patients enter the elevator.

Find the probability that the elevator breaks.

let \(W = \sum\limits_{i = 1}^6 {{w_i}} \)     (M1)

\({w_i}{\text{ is N}}(200,{\text{ 1}}{{\text{5}}^2})\)

\({\text{E}}(W) = \sum\limits_{i = 1}^6 {{\text{E}}({w_i}) = 6 \times 200 = 1200} \)     A1

\({\text{Var}}(W) = \sum\limits_{i = 1}^6 {{\text{Var}}({w_i}) = 6 \times {{15}^2} = 1350} \)     A2

\(W{\text{ is N}}(1200,{\text{ 1350}})\)     (M1)

\({\text{P}}(W > 1150) = 0.913\) by GDC     A1A1

Note: Using 6 times the mean or a lower bound for the mean are acceptable methods.

[7 marks]

Candidates will often be asked to solve these problems that test if they can distinguish between a number of individuals and a number of copies. The wording of the question was designed to make the difference clear. If candidates wrote \({w_1} +  \ldots  + {w_6}\) in (a) and 12w in (b), they usually went on to gain full marks.

Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).

(i)     State suitable hypotheses to test the claim that extra tuition improves examination marks.

(ii)     Calculate the \(p\)-value of the sample.

(iii)     Determine whether or not the above claim is supported at the \(5\% \) significance level.

unbiased estimate of \(\mu \) is \(2.36(36 \ldots )\;\;\;(26/11)\)     (M1)A1

unbiased estimate of \({\sigma ^2}\) is \(33.65(45 \ldots ) = ({5.801^2})\;\;\;(1851/55)\)     (M1)A1

Note:     Accept any answer that rounds correctly to \(3\) significant figures.

Note:     Award M1A0 for any unbiased estimate of \({\sigma ^2}\) that rounds to \(5.80\).

[4 marks]

(i)     \({{\text{H}}_0}:\mu  = 0;{\text{ }}{{\text{H}}_1}:\mu  > 0\)     A1A1

Note:     Award A1A0 if an inappropriate symbol is used for the mean, eg, \(r\), \({\rm{\bar d}}\).

(ii)     attempt to use t-test     (M1)

\(t = 1.35\)     (A1)

\({\text{DF}} = 10\)     (A1)

\(p\)-value \( = 0.103\)     A1

Note:     Accept any answer that rounds correctly to \(3\) significant figures.

(iii)     \(0.103 > 0.05\)     A1

there is insufficient evidence at the \(5\% \) level to support the claim (that extra tuition improves examination marks)

OR

the claim (that extra tuition improves examination marks) is not supported at the \(5\% \) level (or equivalent statement)     R1

Note:     Follow through the candidate’s \(p\)-value.

Note:     Do not award R1 for Accept \({{\text{H}}_0}\) or Reject \({{\text{H}}_1}\).

[8 marks]

Total [12 marks]

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the p-value. Instead, many candidates found the p-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the \(p\)-value. Instead, many candidates found the \(p\)-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

Find the critical region for this test.

It is now known that in the area in which the plant was found \(90\% \) of all the plants are of species \(A\) and \(10\% \) are of species \(B\).

Find the probability that \(\bar X\) will fall within the critical region of the test.

If, having done the test, the sample mean is found to lie within the critical region, find the probability that the leaves came from a plant of species \(A\).

\(\bar X \sim N\left( {5.2,{\text{ }}\frac{{{{1.2}^2}}}{{16}}} \right)\)     (M1)

critical value is \(5.2 - 1.64485 \ldots  \times \frac{{1.2}}{4} = 4.70654 \ldots \)     (A1)

critical region is \(] - \infty ,{\text{ }}4.71]\)     A1

Note:     Allow follow through for the final A1 from their critical value.

Note:     Follow through previous values in (b), (c) and (d).

[3 marks]

\(0.9 \times 0.05 + 0.1 \times (1 - 0.361 \ldots ) = 0.108875997 \ldots  = 0.109\)     M1A1

Note:     Award M1 for a weighted average of probabilities with weights \(0.1,0.9\).

[2 marks]

attempt to use conditional probability formula     M1

\(\frac{{0.9 \times 0.05}}{{0.108875997 \ldots }}\)     (A1)

\( = 0.41334 \ldots  = 0.413\)     A1

[3 marks]

Total [10 marks]

Solutions to this question were generally disappointing.

In (a), the standard error of the mean was often taken to be \(\sigma (1.2)\) instead of \(\frac{\sigma }{{\sqrt n }}(0.3)\) and the solution sometimes ended with the critical value without the critical region being given.

In (c), the question was often misunderstood with candidates finding the weighted mean of the two means, ie \(0.9 \times 5.2 + 0.1 \times 4.6 = 5.14\) instead of the weighted mean of two probabilities.

Without having the solution to (c), part (d) was inaccessible to most of the candidates so that very few correct solutions were seen.

A factory makes wine glasses. The manager claims that on average 2 % of the glasses are imperfect. A random sample of 200 glasses is taken and 8 of these are found to be imperfect.

Test the manager’s claim at a 1 % level of significance using a one-tailed test.

Let X denote the number of imperfect glasses in the sample     (M1)

For recognising binomial or proportion or Poisson     A1

(\(X \sim {\text{B}}(200,{\text{ }}p)\) where p-value is the probability of a glass being imperfect)

Let \({{\text{H}}_0}:p{\text{-value}} = 0.02{\text{ and }}{{\text{H}}_1}:p{\text{-value}} > 0.02\)     A1A1

EITHER

p-value = 0.0493     A2

Using the binomial distribution \(p{\text{-value}} = 0.0493 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0511     A2

Using the Poisson approximation to the binomial distribution since \(p{\text{-value}} = 0.0511 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

OR

p-value = 0.0217     A2

Using the one proportion z-test since \(p{\text{-value}} = 0.0217 > 0.01{\text{ we accept }}{{\text{H}}_0}\)     R1

Note: Use of critical values is acceptable.

[7 marks]

Many candidates used a t-test on this question. This was possibly because the sample was large enough to approximate normality of a proportion. The need to use a one-tailed test was often missed. When using the z-test of proportions p = 0.04 was often used instead of p = 0.02 . Not many candidates used the binomial distribution.

Determine unbiased estimates of the mean and variance of the loss in weight achieved over the month by people using this diet.

(i)     State suitable hypotheses for testing whether or not this diet causes a mean loss in weight.

(ii)     Determine the value of a suitable statistic for testing your hypotheses.

(iii)     Find the 1 % critical value for your statistic and state your conclusion.

the weight losses are

2.2\(\,\,\,\,\,\)3.5\(\,\,\,\,\,\)4.3\(\,\,\,\,\,\)–0.5\(\,\,\,\,\,\)4.2\(\,\,\,\,\,\)–0.2\(\,\,\,\,\,\)2.5\(\,\,\,\,\,\)2.7\(\,\,\,\,\,\)0.1\(\,\,\,\,\,\)–0.7     (M1)(A1)

\(\sum {x = 18.1} \), \(\sum {{x^2} = 67.55} \)

UE of mean = 1.81     A1

UE of variance \( = \frac{{67.55}}{9} - \frac{{{{18.1}^2}}}{{90}} = 3.87\)     (M1)A1

Note: Accept weight losses as positive or negative. Accept unbiased estimate of mean as positive or negative.

Note: Award M1A0 for 1.97 as UE of variance.

[5 marks]

(i)     \({H_0}:{\mu _d} = 0\) versus \({H_1}:{\mu _d} > 0\)     A1

Note: Accept any symbol for \({\mu _d}\)

(ii)     using t test     (M1)

\(t = \frac{{1.81}}{{\sqrt {\frac{{3.87}}{{10}}} }} = 2.91\)     A1

(iii)     DF = 9     (A1)

Note: Award this (A1) if the p-value is given as 0.00864

1% critical value = 2.82     A1

accept \({H_1}\)     R1

Note: Allow FT on final R1.

[6 marks]

In (a), most candidates gave a correct estimate for the mean but the variance estimate was often incorrect. Some candidates who use their GDC seem to be unable to obtain the unbiased variance estimate from the numbers on the screen. The way to proceed, of course, is to realise that the larger of the two ‘standard deviations’ on offer is the square root of the unbiased estimate so that its square gives the required result. In (b), most candidates realised that the t-distribution should be used although many were awarded an arithmetic penalty for giving either t = 2.911 or the critical value = 2.821. Some candidates who used the p-value method to reach a conclusion lost a mark by omitting to give the critical value. Many candidates found part (c) difficult and although they were able to obtain t = 2.49…, they were then unable to continue to obtain the confidence interval.

In (a), most candidates gave a correct estimate for the mean but the variance estimate was often incorrect. Some candidates who use their GDC seem to be unable to obtain the unbiased variance estimate from the numbers on the screen. The way to proceed, of course, is to realise that the larger of the two ‘standard deviations’ on offer is the square root of the unbiased estimate so that its square gives the required result. In (b), most candidates realised that the t-distribution should be used although many were awarded an arithmetic penalty for giving either t = 2.911 or the critical value = 2.821. Some candidates who used the p-value method to reach a conclusion lost a mark by omitting to give the critical value. Many candidates found part (c) difficult and although they were able to obtain t = 2.49…, they were then unable to continue to obtain the confidence interval.

State hypotheses to enable the quality control manager to test the mean weight using a two-tailed test.

(i)     Calculate unbiased estimates of the mean and the variance of the weights of the bricks being produced.

(ii)     Assuming that the weights of the bricks are normally distributed, determine the \(p\)-value of the above results and state the conclusion in context using a 5% significance level.

The owner is more familiar with using confidence intervals. Determine a 95% confidence interval for the mean weight of bricks produced on that particular day.

\({H_0}:{\text{ }}\mu  = 2.2;{\text{ }}{H_1}:{\text{ }}\mu  \ne 2.2\)    A1A1

[2 marks]

(i)     UE of mean \( = \frac{{42.0}}{{20}}{\text{ = }}2.1\)     A1

UE of variance \( = \frac{{89.2}}{{19}} - \frac{{20 \times {{2.1}^2}}}{{19}} = 0.0526{\text{ }}\left( {\frac{1}{{19}}} \right)\)     (M1)A1