(i)     Find an expression for \(c\) in terms of \(n\) such that \(U = c\sum\limits_{i = 1}^n {{X_i}} \) is an unbiased estimator for \(a\) .

(ii)     Determine an expression for \({\text{Var}}(U)\) in this case.

(i)     Show that \({\rm{P}}(Y \le y) = {\left( {\frac{y}{a}} \right)^{3n}},0 \le y \le a\) and deduce an expression for the probability density function of \(Y\) .

(ii)     Find \({\rm{E}}(Y)\) .

(iii)     Show that \({\rm{Var}}(Y) = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\) .

(iv)     Find an expression for \(d\) in terms of \(n\) such that \(V = dY\) is an unbiased estimator for \(a\) .

(v)     Determine an expression for \({\rm{Var}}(V)\) in this case.

Show that \(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{3n + 2}}{5}\) and hence state, with a reason, which of \(U\) or \(V\) is the more efficient estimator for \(a\) .

(i)     \({\rm{E}}(U) = c \times n \times \frac{{3a}}{4} = a \Rightarrow c = \frac{4}{{3n}}\)     M1A1

(ii)     \(Var(U) = \frac{{16}}{{9{n^2}}} \times n \times \frac{{3{a^2}}}{{80}} = \frac{{{a^2}}}{{15n}}\)     M1A1

[4 marks]

(i)     \({\rm{P}}(Y \le y) = {\rm{P}}({\text{all }}Xs \le y)\)     M1

\( = \left[ {\rm{P}} \right.{\left. {(X \le y)} \right]^n}\)    (A1)

\( = {\left( {{{\left( {\frac{y}{a}} \right)}^3}} \right)^n}\)     (A1)

Note: Only one of the two A1 marks above may be implied.

\( = {\left( {\frac{y}{a}} \right)^{3n}}\)     AG

\(g(y) = \frac{{\rm{d}}}{{{\rm{d}}y}}{\left( {\frac{y}{a}} \right)^{3n}} = \frac{{3n{y^{3n - 1}}}}{{{a^{3n}}}},(0 < y < a)\)     M1A1

(\(g(y) = 0\) otherwise)

(ii)     \({\rm{E}}(Y) = \int_0^a {\frac{{3n{y^{3n}}}}{{{a^{3n}}}}} {\rm{d}}y\)     M1

\( = \left[ {\frac{{3n{y^{3n + 1}}}}{{(3n + 1){a^{3n}}}}} \right]_0^a\)     A1

\( = \frac{{3na}}{{3n + 1}}\)     A1

(iii)     \({\rm{Var}}(Y) = \int_0^a {\frac{{3n{y^{3n + 1}}}}{{{a^{3n}}}}} {\rm{d}}y - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\)     M1

\( = \left[ {\frac{{3n{y^{3n + 2}}}}{{(3n + 2){a^{3n}}}}} \right]_0^a - {\left( {\frac{{3na}}{{3n + 1}}} \right)^2}\)     A1

\( = \frac{{3n{a^2}}}{{3n + 2}} - \frac{{9{n^2}{a^2}}}{{{{(3n + 1)}^2}}}\)     M1

\( = \frac{{3n{a^2}(9{n^2} + 6n + 1) - 9{n^2}{a^2}(3n + 2)}}{{(3n + 2)(3n + 1)}}\)     A1

\( = \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\)     AG

(iv)     \({\rm{E}}(V) = d \times \frac{{3na}}{{3n + 1}} = a \Rightarrow d = \frac{{3n + 1}}{{3n}}\)     M1A1

(v)     \(Var(V) = {\left( {\frac{{3n + 1}}{{3n}}} \right)^2} \times \frac{{3n{a^2}}}{{(3n + 2){{(3n + 1)}^2}}}\)     M1

\( = \frac{{{a^2}}}{{3n(3n + 2)}}\)     A1

[16 marks]

\(\frac{{{\rm{Var}}(U)}}{{{\rm{Var}}(V)}} = \frac{{\frac{{{a^2}}}{{15n}}}}{{\frac{{{a^2}}}{{3n(3n + 2)}}}}\)     A1

\( = \frac{{3n + 2}}{5}\)     AG

\(V\) is the more efficient estimator because \(3n + 2 > 5\) (for \(n > 1\) )     R1

[2 marks]

Find \({\text{E}}(A)\).

Show that \({\text{P}}(A = r) = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\).

Find \({\text{P}}(A \leqslant 5|A > 3)\).

\({\text{E}}(X{\text{ }}tennis) = \frac{1}{{\frac{3}{4}}} = \frac{4}{3}\)     (A1)

\({\text{E}}(X{\text{ }}cricket) = \frac{1}{{\frac{1}{4}}} = 4\)     (A1)

\({\text{E}}(A) = \frac{4}{3} \times \frac{1}{5} + 4 \times \frac{4}{5} = \frac{{52}}{{15}}\)     M1A1

\({\text{P}}(X = r|cricket) = \frac{1}{4} \cdot {\left( {\frac{3}{4}} \right)^{r - 1}}\)     (A1)

\({\text{P}}(X = r|tennis) = \frac{3}{4} \cdot {\left( {\frac{1}{4}} \right)^{r - 1}}\)     (A1)

\({\text{P}}(A = r) = {\text{P}}(X = r|cricket) \times {\text{P}}(cricket) + {\text{P}}(X = r|tennis) \times {\text{P}}(tennis)\)     (M1)

\( = \frac{1}{4} \times {\left( {\frac{3}{4}} \right)^{r - 1}} \times \frac{4}{5} + \frac{3}{4} \times {\left( {\frac{1}{4}} \right)^{r - 1}} \times \frac{1}{5}\)     A1

\( = \frac{1}{5} \times {\left( {\frac{3}{4}} \right)^{r - 1}} + \frac{3}{{20}} \times {\left( {\frac{1}{4}} \right)^{r - 1}}\)     AG

\({\text{P}}(A \leqslant 5)|(A < 3) = \frac{{{\text{P}}(A = 4{\text{ or }}5)}}{{1 - {\text{P}}(A = 1{\text{ or 2 or }}3)}}\)     M1A1A1

\({\text{P}}(A = 1) = \frac{7}{{20}}\)

\({\text{P}}(A = 2) = \frac{15}{{80}}\)

\({\text{P}}(A = 3) = \frac{39}{{320}}\)

\({\text{P}}(A = 4) = \frac{111}{{1280}}\)

\({\text{P}}(A = 5) = \frac{327}{{5120}}\)

\( \Rightarrow {\text{P}}(A > 3)|(A \leqslant 5) = \frac{{\frac{{771}}{{5120}}}}{{\frac{{1744}}{{5120}}}}\)     A1A2

Note: Award A1 for correct working for numerator and Award A2 for correct working for denominator

\( = \frac{{771}}{{1744}}\;\;\;( = 0.442)\)     A1

It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.

It was pleasing to see many correct answers to parts a) and b) with candidates correctly recognising how to work with the distribution.

Part c) caused more problems. Although a number of wholly correct solutions were seen, many candidates were unable to work meaningfully with the conditional probability.

Write down the first three terms of the infinite series for \(G(t)\), the probability generating function for \(X\).

Determine the radius of convergence of this infinite series.

By considering \(\left( {1 - \frac{t}{3}} \right)G(t)\), show that

\[G(t) = \frac{{3kt}}{{{{(3 - t)}^2}}}.\]

Hence show that \(k = \frac{4}{3}\).

Show that \(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\).

By differentiating both sides of this equation, determine the values of \(G’(1)\) and \(G’’(1)\).

Hence find \({\text{Var}}(X)\).

\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} +  \ldots \)     M1A1

[??? marks]

\(\frac{{{T_{n + 1}}}}{{{T_n}}} = \frac{{(n + 1)k{t^{n + 1}}}}{{{3^{n + 1}}}} \times \frac{{{3^n}}}{{nk{t^n}}}\)     M1A1

\( \to \frac{t}{3}{\text{ as }}n \to \infty \)     A1

for convergence, \(\left| {\frac{t}{3}} \right| < 1\) so radius of convergence \( = 3\)     A1

[??? marks]

\(G(t) = \frac{{kt}}{3} + \frac{{2k{t^2}}}{{{3^2}}} + \frac{{3k{t^3}}}{{{3^3}}} +  \ldots \)

\(\frac{t}{3}G(t) = \frac{{k{t^2}}}{{{3^2}}} + \frac{{2k{t^3}}}{{{3^3}}} +  \ldots \)

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

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

\(G(t) = \frac{{\frac{{kt}}{3}}}{{{{\left( {1 - \frac{t}{3}} \right)}^2}}} = \frac{{3kt}}{{{{(3 - t)}^2}}}\)     AG

[??? marks]

\(G(1) = 1\)     M1

so \(k = \frac{4}{3}\)     AG

[??? marks]

\(\ln G(t) = \ln 4t - \ln {(3 - t)^2}\)     M1

\(\ln G(t) = \ln 4 + \ln t - 2\ln (3 - t)\)     AG

[??? marks]

\(\frac{{G'(1)}}{{G(1)}} = \frac{1}{t} + \frac{2}{{3 - t}}\)     M1A1

putting \(t = 1\)

\(G’(1) = 2\)     A1

\(\frac{{G''(t)G(t) - {{[G'(t)]}^2}}}{{{{[G(t)]}^2}}} =  - \frac{1}{{{t^2}}} + \frac{2}{{{{(3 - t)}^2}}}\)     M1A1

putting \(t = 1\)

\(G’’(1) - 4 =  - 1 + \frac{1}{2}\)

\(G’’(1) = \frac{7}{2}\)     A1

[??? marks]

\({\text{Var}}(X) = G''(1) + G'(1) - {[G'(1)]^2} = \frac{3}{2}\)     A1

[??? marks]

Assuming the series for \({{\rm{e}}^x}\) , find the first five terms of the Maclaurin series for\[\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} {\rm{  .}}\]

(i)      Use your answer to (a) to find an approximate expression for the cumulative distributive function of \({\rm{N}}(0,1)\) .

(ii)     Hence find an approximate value for \({\rm{P}}( - 0.5 \le Z \le 0.5)\) , where \(Z \sim {\rm{N}}(0,1)\) .

State and justify an appropriate test procedure giving the null and alternate hypotheses.

What is the critical region for the sample mean if the probability of a Type I error is to be \(3.5\%\)?

If the mean weight of the bags is actually \(28\).1 kg, what would be the probability of a Type II error?

\({{\rm{e}}^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} +  \ldots \)

\({{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = 1 + \left( { - \frac{{{x^2}}}{2}} \right) + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^2}}}{{2!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^3}}}{{3!}} + \frac{{{{\left( { - \frac{{{x^2}}}{2}} \right)}^4}}}{{4!}} +  \ldots \)    M1A1

\(\frac{1}{{\sqrt {2\pi } }}{{\rm{e}}^{\frac{{ - {x^2}}}{2}}} = \frac{1}{{\sqrt {2\pi } }}\left( {1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{8} - \frac{{{x^6}}}{{48}} + \frac{{{x^8}}}{{384}}} \right)\)     A1

[3 marks]

(i)     \(\frac{1}{{\sqrt {2\pi } }}\int_0^x {1 - \frac{{{t^2}}}{2}}  + \frac{{{t^4}}}{8} - \frac{{{t^6}}}{{48}} + \frac{{{t^8}}}{{384}}{\rm{d}}t\)     M1

\( = \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}}} \right)\)     A1

\({\rm{P}}(Z \le x) = 0.5 + \frac{1}{{\sqrt {2\pi } }}\left( {x - \frac{{{x^3}}}{6} + \frac{{{x^5}}}{{40}} - \frac{{{x^7}}}{{336}} + \frac{{{x^9}}}{{3456}} -  \ldots } \right)\)     R1A1

(ii)     \({\rm{P}}( - 0.5 \le Z \le 0.5) = \frac{2}{{\sqrt {2\pi } }}\left( {0.5 - \frac{{{{0.5}^3}}}{6} + \frac{{{{0.5}^5}}}{{40}} - \frac{{{{0.5}^7}}}{{336}} + \frac{{{{0.5}^9}}}{{3456}} -  \ldots } \right)\)     M1

\( = 0.38292 = 0.383\)     A1

[6 marks]

this is a two tailed test of the sample mean \(\overline x \)

we use the central limit theorem to justify assuming that     R1

\(\overline X  \sim {\rm{N}}\left( {28,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1A1

\({{\rm{H}}_0}:\mu  = 28\)     A1

\({{\rm{H}}_1}:\mu  \ne 28\)     A1

[5 marks]

since \({\text{P(Type I error)}} = 0.035\) , critical value \(2.108\)     (M1)A1

and (\(\overline x  \le 28 - 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) or \(\overline x  \ge 28 + 2.108\sqrt {\frac{{{{0.54}^2}}}{{24}}} \) )     (M1)(A1)(A1)

\(\overline x  \le 27.7676\) or \(\overline x  \ge 28.2324\)

so \(\overline x  \le 27.8\) or \(\overline x  \ge 28.2\)     A1A1

[7 marks]

if \(\mu  = 28.1\)

\(\overline X  \sim {\rm{N}}\left( {28.1,\frac{{{{0.54}^2}}}{{24}}} \right)\)     R1

\({\text{P(Type II error)}} = {\rm{P}}(27.7676 < \overline x  < 28.2324)\)

\( = 0.884\)     A1

Note: Depending on the degree of accuracy used for the critical region the answer  for part (c) can be anywhere from \(0.8146\) to \(0.879\).

[2 marks]

The derivation of a series from a given one by substitution seems not to be well known. This made finding the required series from \(({{\rm{e}}^x})\) in part (a) to be much more difficult than it need have been. The fact that this part was worth only 3 marks was a clear hint that an easy derivation was possible.

In part (b)(i) the \(0.5\) was usually missing which meant that this part came out incorrectly.

The conditions required in part (a) were rarely stated correctly and some candidates were unable to state the hypotheses precisely. There was some confusion with "less than" and "less than or equal to".

There was some confusion with "less than" and "less than or equal to".

Levels of accuracy in the body of the question varied wildly leading to a wide range of answers to part (c).

The weights, \(X\) grams, of tomatoes may be assumed to be normally distributed with mean \(\mu \) grams and standard deviation \(\sigma \) grams. Barry weighs \(21\) tomatoes selected at random and calculates the following statistics.\[\sum {x = 1071} \) ; \(\sum {{x^2} = 54705} \]

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

  (ii)     Determine a \(95\%\) confidence interval for \(\mu \) .

The random variable \(Y\) has variance \({\sigma ^2}\) , where \({\sigma ^2} > 0\) . A random sample of \(n\) observations of \(Y\) is taken and \(S_{n - 1}^2\) denotes the unbiased estimator for \({\sigma ^2}\) .

By considering the expression

\({\rm{Var}}({S_{n - 1}}) = {\rm{E}}(S_{n - 1}^2) - {\left\{ {E\left. {({S_{n - 1}})} \right\}} \right.^2}\) ,

show that \(S_{n - 1}^{}\) is not an unbiased estimator for \(\sigma \) .

(i)     \(\overline x  = \frac{{1071}}{{21}} = 51\)     A1

\(S_{n - 1}^2 = \frac{{54705}}{{20}} - \frac{{{{1071}^2}}}{{20 \times 21}} = 4.2\)     M1A1

(ii)     degrees of freedom \( = 20\) ; \(t\)-value \( = 2.086\)     (A1)(A1)

\(95\%\) confidence limits are

\(51 \pm 2.086\sqrt {\frac{{4.2}}{{21}}} \)     (M1)(A1)

leading to \(\left[ {50.1,51.9} \right]\)     A1

[8 marks]

\({\rm{Var}}({S_{n - 1}}) > 0\)     A1

\(E(S_{n - 1}^2) = {\sigma ^2}\)     (A1)

substituting in the given equation,

\({\sigma ^2} - E(S_{n - 1}^{}) > 0\)     M1

it follows that

\(E(S_{n - 1}^{}) < \sigma \)     A1

this shows that \({S_{n - 1}}\) is not an unbiased estimator for \(\sigma \) since that would require = instead of \( < \)     R1

[5 marks]

Most candidates attempted (a) although some used the normal distribution instead of the \(t\)-distribution.

Many candidates were unable even to start (b) and many of those who did filled several pages of algebra with factors such as \(n\) / \((n - 1)\) prominent. Few candidates realised that the solution required only a few lines.

The farm manager selects a random sample of \(10\) apples and weighs them with the following results, given in grams.\[82, 98, 102, 96, 111, 95, 90, 89, 99, 101\]

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

  (ii)     Determine a \(95\%\) confidence interval for \(\mu \) .

The farm manager claims that the mean weight of apples is \(100\) grams but the buyer from the local supermarket claims that the mean is less than this. To test these claims, they select a random sample of \(100\) apples and weigh them. Their results are summarized as follows, where \(x\) is the weight of an apple in grams.\[\sum {x = 9831;\sum {{x^2} = 972578} } \]

  (i)     State suitable hypotheses for testing these claims.

  (ii)     Determine the \(p\)-value for this test.

  (iii)     At the \(1\%\) significance level, state which claim you accept and justify your answer.

(i)     from the GDC,

unbiased estimate for \(\mu  = 96.3\)     A1

unbiased estimate for \({\sigma ^2} = 8.028{ \ldots ^2} = 64.5\)     (M1)A1

(ii)     \(95\%\) confidence interval is [\(90.6\), \(102\)]     A1A1

Note: Accept \(102.0\) as the upper limit.

[5 marks]

(i)     \({H_0}:\mu  = 100;{H_1}:\mu  < 100\)     A1

(ii)     \(\overline x  = 98.31,{S_{n - 1}} = 7.8446 \ldots \)     (A1)

\(p\)-value \( = 0.0168\)     A1

(iii)     the farm manager’s claim is accepted because \(0.0168 > 0.01\)     A1R1

[5 marks]

Show that the probability generating function of X is given by

\[G\left( t \right) = \frac{{pt}}{{1 - qt}}.\]

Deduce that \({\text{E}}\left( X \right) = \frac{1}{p}\).

Two friends A and B play a ball game with the following rules.

Each player starts with zero points. Player A serves first and then the players have alternate pairs of serves so that the service order is A, B, B, A, A, … When player A serves, the probability of her scoring 1 point is \({p_A}\) and the probability of B scoring 1 point is \({q_A}\), where \({q_A} = 1 - {p_A}\).

When player B serves, the probability of her scoring 1 point is \({p_B}\) and the probability of A scoring 1 point is \({q_B}\), where \({q_B} = 1 - {p_B}\).

Show that, after the first 6 serves, the probability that each player has 3 points is

\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\).

After 6 serves the score is 3 points each. Play continues and the game ends when one player has scored two more points than the other player. Let N be the number of further serves required before the game ends. Given that \({p_A}\) = 0.7 and \({p_A}\) = 0.6 find P(N = 2).

Let \(M = \frac{1}{2}N\). Show that M has a geometric distribution and hence find the value of E(N).

\({\text{P}}\left( {X = x} \right) = p{q^{x - 1}},\,{\text{for}}\,x = 1,\,2 \ldots \)

\(G\left( t \right) = \sum\limits_{x = 1}^\infty  {{t^x}} p{q^{x - 1}}\)      M1

\( = pt\sum\limits_{x = 1}^\infty  {{{\left( {tq} \right)}^{x - 1}}} \)       A1

\( = pt\left( {1 + tq + {{\left( {tq} \right)}^2} \ldots } \right)\)      M1

\( = \frac{{pt}}{{1 - tq}}\)      AG

[3 marks]

\(G'\left( t \right) = \frac{{\left( {1 - tq} \right)p - pt\left( { - q} \right)}}{{{{\left( {1 - tq} \right)}^2}}}\)      M1A1

\({\text{E}}\left( X \right) = G'\left( 1 \right)\)      M1

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

\( = \frac{1}{p}\)        AG

[4 marks]

after 6 serves (3 serves each) we have ABBAAB

A serves   B serves

3 wins       0 losses      \({p_1} = {}^3{C_3}p_A^3q_A^0{}^3{C_0}p_B^3q_B^0\)     M1A1

2 wins       1 loss         \({p_2} = {}^3{C_2}p_A^2q_A^1{}^3{C_1}p_B^2q_B^1\)      A1

1 win         2 losses     \({p_3} = {}^3{C_1}p_A^1q_A^2{}^3{C_2}p_B^1q_B^2\)      A1

0 wins       3 losses     \({p_4} = {}^3{C_0}p_A^0q_A^3{}^3{C_3}p_B^0q_B^3\)       A1

since \({}^3{C_0} = {}^3{C_3},\,\,{}^3{C_1} = {}^3{C_2}\)

\(\sum\limits_{x = 0}^{x = 3} {{{\left( \begin{gathered}
3 \hfill \\
x \hfill \\
\end{gathered} \right)}^2}} {\left( {{p_A}} \right)^x}{\left( {{p_B}} \right)^x}{\left( {{q_A}} \right)^{3 - x}}{\left( {{q_B}} \right)^{3 - x}}\)     AG

[5 marks]

for N = 2 serves are B, A respectively

P(N = 2) = P(B wins twice) + P(A wins twice)      (M1)

= 0.6 × 0.3 + 0.4 × 0.7       A1

= 0.46       A1

[3 marks]

for \(M = \frac{1}{2}N\)

\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M}\)      M1

\({\text{P}}\left( {M = 2} \right) = {\text{P}}\left( {N = 4} \right)\)

\( = {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game does not end after}}} \\
{{\text{first two serves}}}
\end{array}} \right) \times {\text{P}}\left( {\begin{array}{*{20}{c}}
{{\text{game ends after}}} \\
{{\text{next two serves}}}
\end{array}} \right) = \left( {1 - {p_M}} \right){p_M}\)      A1

similarly \({\text{P}}\left( {M = 3} \right) = {\left( {1 - {p_M}} \right)^2}{p_M}\)      (A1)

hence \({\text{P}}\left( {M = r} \right) = {\left( {1 - {p_M}} \right)^{r - 1}}{p_M}\)       A1

hence M has a geometric distribution      AG

\({\text{P}}\left( {M = 1} \right) = {\text{P}}\left( {N = 2} \right) = {p_M} = 0.46\)      A1

hence \({\text{E}}\left( M \right) = \frac{1}{p} = \frac{1}{{0.46}} = 2.174\)

\({\text{E}}\left( N \right) = {\text{E}}\left( {2M} \right) = 2{\text{E}}\left( M \right)\)       M1

= 4.35      A1

[7 marks]

Write down the distribution of aX + bY where a, b \( \in \mathbb{R}\).

P(X₁ + Y₁ < 11).

P(3X₁ + 4Y₁ > 15).

P(X₁ + X₂ + Y₁ + Y₂ + Y₃ + Y₄ < 30).

Given that \({\bar X}\) and \({\bar Y}\) are the respective sample means, find \({\text{P}}\left( {\bar X > \bar Y} \right)\).

aX + bY ~ N\(\left( {a{\mu _1} + b{\mu _2},\,\,{a^2}\sigma _1^2 + {b^2}\sigma _2^2} \right)\)     A1A1

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

[2 marks]

X₁ + Y₁ ∼ N(5,34)     (A1)(A1)

⇒ P(X₁ + Y₁ < 11) = 0.848       A1

[3 marks]

3X₁ + 4Y₁ ∼ N(9 + 8, 9 × 9 + 16 × 25)     (A1)(M1)(A1)

Note: Award (A1) for correct expectation, (M1)(A1) for correct variance.

∼ N(17, 481)

⇒ P(3X₁ + 4Y₁ > 15) = 0.536       A1

[4 marks]

X₁ + X₂ + Y₁ + Y₂ + Y₃ + Y₄ ∼ N(6 + 8, 2 × 9 + 4 × 25)     (A1)(A1)

∼ N(14, 118)

⇒ P(X₁ + X₂ + Y₁ + Y₂ + Y₃ + Y₄ < 30) = 0.930       A1

[3 marks]

consider \(\bar X - \bar Y\)      (M1)

\({\text{E}}\left( {\bar X - \bar Y} \right) = 3 - 2 = 1\)       A1

\({\text{Var}}\left( {\bar X - \bar Y} \right) = \frac{9}{2} + \frac{{25}}{4}\left( { = 10.75} \right)\)      (M1)A1

\( \Rightarrow {\text{P}}\left( {\bar X - \bar Y > 0} \right) = 0.620\)       A1

[5 marks]

An automatic machine is used to fill bottles of water. The amount delivered, \(Y\) ml , may be assumed to be normally distributed with mean \(\mu \) ml and standard deviation \(8\) ml . Initially, the machine is adjusted so that the value of \(\mu \) is \(500\). In order to check that the value of \(\mu \) remains equal to \(500\), a random sample of 10 bottles is selected at regular intervals, and the mean amount of water, \(\overline y \) , in these bottles is calculated. The following hypotheses are set up.

\({{\rm{H}}_0}:\mu  = 500\) ; \({{\rm{H}}_1}:\mu  \ne 500\)

The critical region is defined to be \(\left( {\overline y  < 495} \right) \cup \left( {\overline y  > 505} \right)\) .

(i)     Find the significance level of this procedure.

(ii)     Some time later, the actual value of \(\mu \) is \(503\). Find the probability of a Type II error.

(i)     Under \({{\rm{H}}_0}\) , the distribution of \({\overline y }\) is N(500, 6.4) .     (A1)

Significance level \( = {\rm{P}}\overline y  < 495\) or \( > 505|{{\rm{H}}_0}\)     M2

\( = 2 \times 0.02405\)     (A1)

\( = 0.0481\)     A1 N5

Note: Using tables, answer is \(0.0478\).

(ii)     The distribution of \(\overline y \) is now N(\(503\), \(6.4\)) .     (A1)

P(Type ΙΙ error) \( = {\rm{P}}(495 < \overline y  < 505)\)     (M1)

\( = 0.785\)     A1 N3

Note: Using tables, answer is \(0.784\).

[8 marks]

Show that \(f''(x) = \frac{{ - 1}}{{1 + \sin x}}\) .

Determine the Maclaurin series for \(f(x)\) as far as the term in \({x^4}\) .

Deduce the Maclaurin series for \(\ln (1 - \sin x)\) as far as the term in \({x^4}\) .

By combining your two series, show that \(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} +  \ldots \) .

Hence, or otherwise, find \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \sec x}}{{x\sqrt x }}\) .

He makes 5 independent measurements of the concentration of a particular solution and correctly calculates the following confidence interval for \(\mu \) .

[\(22.7\) , \(26.1\)]

Determine the confidence level of this interval.

He is now given a different solution and is asked to determine a \(95\%\) confidence interval for its concentration. The confidence interval is required to have a width less than \(2\). Find the minimum number of independent measurements required.

\(f'(x) = \frac{{\cos x}}{{1 + \sin x}}\)     M1A1

\(f''(x) = \frac{{ - \sin x(1 + \sin x) - {{\cos }^2}x}}{{{{(1 + \sin x)}^2}}}\)     M1

\( = \frac{{ - \sin x - 1}}{{{{(1 + \sin x)}^2}}}\)     A1

\( = \frac{{ - 1}}{{1 + \sin x}}\)     AG

[4 marks]

\(f'''(x) = \frac{{\cos x}}{{{{(1 + \sin x)}^2}}}\)     A1

\({f^{iv}}(x) = \frac{{ - \sin x{{(1 + \sin x)}^2} - 2(1 + \sin x){{\cos }^2}x}}{{{{(1 + \sin x)}^4}}}\)     A1

\(f(0) = 0\) , \(f'(0) = 1\) , \(f''(0) =  - 1\) , \(f'''(0) = 1\) , \({f^{iv}}(0) = - 2\)     (A2)

Note: Award A1 for 2 errors and A0 for more than 2 errors.

\(\ln (1 + \sin x) = x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots \)     M1A1

[6 marks]

\(\ln (1 - \sin x) = \ln (1 + \sin ( - x)) =  - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{6} - \frac{{{x^4}}}{{12}} +  \ldots \)     M1A1

[2 marks]

Adding,     M1

\(\ln (1 - {\sin ^2}x) = \ln {\cos ^2}x\)     A1

\( = - {x^2} - \frac{{{x^4}}}{6} +  \ldots \)     A1

\(\ln \cos x = - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}} +  \ldots \)     A1

\(\ln \sec x = \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{12}} +  \ldots \)    AG

[4 marks]

\(\frac{{\ln \sec x}}{{x\sqrt x }} = \frac{{\sqrt x }}{2} + \frac{{{x^2}\sqrt x }}{{12}} +  \ldots \)     M1

Limit \( = 0\)     A1

[2 marks]

Interval width \( = 26.1 - 22.7 = 3.4\)

So \(3.4 = 2z \times \frac{{1.6}}{{\sqrt 5 }}\)     M1A1

\(z = 2.375 \ldots \)     A1

Probability \( = 0.9912\)     A1

Confidence level \( = 2 \times 0.4912 = 98.2\% \)     A1

[5 marks]

\(z\)-value \( = 1.96\)     A1

We require

\(2 \times \frac{{1.96 \times 1.6}}{{\sqrt n }} < 2\)     M1A1

Whence \(n > 9.83\)     A1

So we need \(n = 10\)     A1

Note: Accept \( = \) signs throughout.

[5 marks]

Use l’Hôpital’s rule to show that \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^3}}}{{{{\text{e}}^x}}} = 0\).

(i)     Find \({\text{E}}({X^2})\).

(ii)     Show that \({\text{Var}}(X) = 2\).

State the central limit theorem.

Find the probability that the sample mean is less than 2.3.

attempt to apply l’Hôpital’s rule     M1

\(\mathop {\lim }\limits_{x \to \infty } \frac{{3{x^2}}}{{{{\text{e}}^x}}}\)    A1

then \(\mathop {\lim }\limits_{x \to \infty } \frac{{6x}}{{{{\text{e}}^x}}}\)

then \(\mathop {\lim }\limits_{x \to \infty } \frac{6}{{{{\text{e}}^x}}}\)     A1

\( = 0\)    AG

[3 marks]

(i)     \({\text{E}}({X^2}) = \mathop {\lim }\limits_{R \to \infty } \int\limits_0^R {{x^3}{{\text{e}}^{ - x}}{\text{d}}x} \)       M1

attempt at integration by parts      M1

the integral \( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + \mathop \smallint \limits_0^R 3{x^2}{{\text{e}}^{ - x}}{\text{d}}x\)       A1A1

\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6x{{\text{e}}^{ - x}}{\text{d}}x} \)     M1

\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + \int\limits_0^R {6{{\text{e}}^{ - x}}{\text{d}}x} \)     A1

\( = [ - {x^3}{{\text{e}}^{ - x}}]_0^R + [ - 3{x^2}{{\text{e}}^{ - x}}]_0^R + [ - 6x{{\text{e}}^{ - x}}]_0^R + [ - 6{{\text{e}}^{ - x}}]_0^R\)      A1

\( = 6\) when \(R \to \infty \)       R1

(ii)     \({\text{E}}(X) = 2\)      A1

\({\text{Var}}(X) = {\text{E}}({X^2}) - {\left( {{\text{E}}(X)} \right)^2} = 6 - {2^2}\)    M1

\( = 2\)    AG

[10 marks]

if a random sample of size \(n\) is taken from any distribution \(X\), with \({\text{E}}(X) = \mu \) and \({\text{Var}}(X) = {\sigma ^2}\), then, for large n,     A1

the sample mean \(\bar X\) has approximate distribution \({\text{N}}\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{n}} \right)\)     A1

[2 marks]

\(\bar X \sim {\text{N}}\left( {2,{\text{ }}\frac{2}{{50}} = {{(0.2)}^2}} \right)\)    (A1)

\({\text{P}}(\bar X < 2.3) = \left( {{\text{P}}(Z < 1.5)} \right) = 0.933\)    A1

[2 marks]

In part (b) the infinite upper limit was rarely treated rigorously.

In answering part (c) many failed to say that the Central Limit Theorem is valid for large samples and for any initial distribution. The parameters of the distribution were often not stated.

State the distribution of \(\bar X\) , giving its mean and standard deviation.

The sample values are summarized as \(\sum {x = 3782} \) and \(\sum {{x^2} = 155341} \) where \(x\) kg is the weight of a sheep.

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

(ii)     Find a \(95\%\) confidence interval for \(\mu \).

Test, at the \(1\%\) level of significance, the null hypothesis \(\mu  = 35\) against the alternative hypothesis that \(\mu  > 35\).

\(\bar X \sim N\left( {\mu ,{\text{ }}\frac{{{\sigma ^2}}}{{100}}} \right)\)     A1A1

Note: Award A1 for \(N\), A1 for the parameters.

(i)     \(\bar x = \frac{{\sum x }}{n} = \frac{{3782}}{{100}} = 37.8\)     A1

\(s_{n - 1}^2 = \frac{{155341}}{{99}} - \frac{{{{3782}^2}}}{{9900}} = 124\)     M1A1

(ii)     \(95\% CI = 37.82 \pm 1.98\sqrt {\frac{{124.3006}}{{100}}} \)     (M1)(A1)

\( = (35.6,{\text{ }}40.0)\)     A1

METHOD 1

one tailed t-test     A1

testing \(37.82\)     A1

\(99\) degrees of freedom

reject if \(t > 2.36\)     A1

t-value being tested is \(2.5294\)     A1

since \(2.5294 > 2.36\) we reject the null hypothesis and accept the alternative hypothesis     R1

METHOD 2

one tailed t-test     (A1)

\(p = 0.00650\)     A3

since \(p{\text{ - value}} < 0.01\) we reject the null hypothesis and accept the alternative hypothesis     R1

Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.

Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.

Almost all candidates recognised the sample distribution as normal but were not always successful in stating the mean and the standard deviation. Similarly almost all candidates knew how to find an unbiased estimator for \(\mu \), but a number failed to find the correct answer for the unbiased estimator for \({\sigma ^2}\). Most candidates were successful in finding the 95% confidence interval for \(\mu \). In part c) many fully correct answers were seen but a significant number of candidates did not recognise they were working with a t-distribution.

The random variable \(X\) has the binomial distribution \({\text{B}}(n,{\text{ }}p)\), where \(n > 1\).

Show that

(a)     \(\frac{X}{n}\) is an unbiased estimator for \(p\);

(b)     \({\left( {\frac{X}{n}} \right)^2}\) is not an unbiased estimator for \({p^2}\);

(c)     \(\frac{{X(X - 1)}}{{n(n - 1)}}\) is an unbiased estimator for \({p^2}\).

(a)     \({\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}(X)\)     M1

\( = \frac{1}{n} \times np = p\)     AG

therefore unbiased     AG

[2 marks]

(b)     \({\text{E}}\left[ {{{\left( {\frac{X}{n}} \right)}^2}} \right] = \frac{1}{{{n^2}}}\left( {{\text{Var}}(X) + {{[{\text{E}}(X)]}^2}} \right)\)     M1A1

\( = \frac{1}{{{n^2}}}\left( {np(1 - p) + {n^2}{p^2}} \right)\)     A1

\( \ne {p^2}\)     A1

therefore not unbiased     AG

[4 marks]

(c)     \({\text{E}}\left[ {\left( {\frac{{X(X - 1)}}{{n(n - 1)}}} \right)} \right] = \frac{{{\text{E}}({X^2}) - {\text{E}}(X)}}{{n(n - 1)}}\)     M1

\( = \frac{{np(1 - p) + {n^2}{p^2} - np}}{{n(n - 1)}}\)     A1

\( = \frac{{n{p^2}(n - 1)}}{{n(n - 1)}}\)     A1

\( = {p^2}\)

therefore unbiased     AG

[3 marks]

The continuous random variable \(X\) takes values only in the interval [\(a\), \(b\)] and \(F\) denotes its cumulative distribution function. Using integration by parts, show that:\[E(X) = b - \int_a^b {F(x){\rm{d}}x}. \]

The continuous random variable \(Y\) has probability density function \(f\) given by:\[\begin{array}{*{20}{c}}
  {f(y) = \cos y,}&{0 \leqslant y \leqslant \frac{\pi }{2}} \\
  {f(y) = 0,}&{{\text{elsewhere}}{\text{.}}}
\end{array}\]

  (i)     Obtain an expression for the cumulative distribution function of \(Y\) , valid for \(0 \le y \le \frac{\pi }{2}\) . Use the result in (a) to determine \(E(Y)\) .

  (ii)     The random variable \(U\) is defined by \(U = {Y^n}\) , where \(n \in {\mathbb{Z}^ + }\) . Obtain an expression for the cumulative distribution function of \(U\) valid for \(0 \le u \le {\left( {\frac{\pi }{2}} \right)^n}\) .

  (iii)     The medians of \(U\) and \(Y\) are denoted respectively by \({m_u}\) and \({m_y}\) . Show that \({m_u} = m_y^n\) .

\(E(X) = \int_a^b {xf(x){\rm{d}}x} \)     M1

\( = \left[ {xF(x)} \right]_a^b - \int_a^b {F(x){\rm{d}}x} \)     A1

\( = bF(b) - aF(a) - \int_a^b {F(x){\rm{d}}x} \)     A1

\( = b - \int_a^b {F(x){\rm{d}}x} \) because \(F(a) = 0\) and \(F(b) = 1\)     A1

[4 marks]

(i)     let \(G\) denote the cumulative distribution function of \(Y\)

\(G(y) = \int_0^y {\cos t{\rm{d}}t} \)     M1

\( = \left[ {\sin t} \right]_0^y\)     (A1)

\( = \sin y\)     A1

\(E(Y) = \frac{\pi }{2} - \int_0^{\frac{\pi }{2}} {\sin y{\rm{d}}y} \)     M1

\( = \frac{\pi }{2} + \left[ {\cos y} \right]_0^{\frac{\pi }{2}}\)     A1

\( = \frac{\pi }{2} - 1\)     A1

(ii)     CDF of \(U = P(U \le u)\)     M1

\( = P({Y^n} \le u)\)     A1

\( = P({Y^{}} \le {u^{\frac{1}{n}}})\)     A1

\( = G({u^{\frac{1}{n}}})\)     (A1)

\( = \sin \left( {{u^{\frac{1}{n}}}} \right)\)     A1

(iii)     \({m_y}\) satisfies the equation \(\sin {m_y} = \frac{1}{2}\)     A1

\({m_u}\) satisfies the equation \(\sin \left( {m_u^{\frac{1}{n}}} \right) = \frac{1}{2}\)     A1

therefore \({m_y} = m_u^{\frac{1}{n}}\)     A1

\({m_u} = m_y^n\)     AG

[14 marks]

Solutions to (a) were often unconvincing. Candidates were expected to include in their solution the fact that \(F(a) = 0\) and \(F(b) = 1\) .

In (b) (i), it was not enough to state that \(G(y) = \int {\cos y{\rm{d}}y = \sin y} \) although that, fortuitously, gave the correct answer on this occasion. The correct approach was either to state that \(G(y) = \int_0^y {\cos t{\rm{d}}t}  = \sin y\) or that \(G(y) = \int {\cos y{\rm{d}}y}  = \sin x + C\) and then show that \(C = 0\) because \(F(0) = 0\) or \(F\left( {\frac{\pi }{2}} \right) = 1\) . Solutions to (b) (ii) and (iii) were often disappointing, giving the impression that many of the candidates were not familiar with dealing with cumulative distribution functions.

(i)     Find an expression for \({\rm{P}}(X > a)\) , where \(a > 0\) .

A chicken crosses a road. It is known that cars pass the chicken’s crossing route, with intervals between cars measured in seconds, according to the random variable \(X\) , with \(\lambda  = 0.03\) . The chicken, which takes \(10\) seconds to cross the road, starts to cross just as one car passes.

(ii)     Find the probability that the chicken will reach the other side of the road before the next car arrives.

Later, the chicken crosses the road again just after a car has passed.

(iii)     Show that the probability that the chicken completes both crossings is greater than \(0.5\).

A rifleman shoots at a circular target. The distance in centimetres from the centre of the target at which the bullet hits, can be modelled by \(X\) with \(\lambda = 0.4\) . The rifleman scores \(10\) points if \(X \le 1\) , \(5\) points if \(1 < X \le 5\) , \(1\) point if \(5 < X \le 10\) and no points if \(X > 10\) .

(i)     Find the expected score when one bullet is fired at the target.

A second rifleman, whose shooting can also be modelled by \(X\) , wishes to find his value of \(\lambda \) .

(ii)     Given that his expected score is \(6.5\), find his value of \(\lambda \) .

(i)     \({\rm{P}}(X > a) = \int_a^\infty  {\lambda {e^{ - \lambda x}}{\rm{d}}x} \)     M1

\(\left[ { - {e^{ - \lambda x}}} \right]_a^\infty \)     A1

\( = {e^{ - \lambda a}}\)     A1

(ii)     \({\rm{P}}(X > 10) = {e^{ - 0.3}}( = 0.74 \ldots )\)     (M1)A1

(iii)     probability of a safe double crossing \( = {e^{ - 0.6}}\) \(( = {0.74^2})\) \( = 0.55\)     A1

which is greater than \(0.5\)     AG

[6 marks]

(i)     \({\rm{P}}(X \le 1) = 0.3296 \ldots \)     (A1)

\({\rm{P}}(1 \le X \le 5) = 0.5349 \ldots \)     (A1)

\({\rm{P}}(5 \le X \le 10) = 0.1170 \ldots \)     (A1)

\({\rm{E(score)}} = 10 \times 0.3296 \ldots  + 5 \times 0.5349 \ldots  + 1 \times 0.1170 \ldots \)     M1A1 

\( = 6.09\)     A1

Note: Accept probabilities in exponential form until the final decimal answer.

(ii)      \({\rm{E(score)}}\) for X with unknown parameter can be expressed as \(10 \times (1 - {e^{ - \lambda }}) + 5 \times ({e^{ - \lambda }} - {e^{ - 5\lambda }}) + ({e^{ - 5\lambda }} - {e^{ - 10\lambda }})\)     (M1)(A1)

attempt to solve \({\rm{E(score)}} = 6.5\)     (M1) 

obtain \(\lambda  = 0.473\)     A1

[10 marks]

This question was generally well done.

This question was generally well done.

(i)     Write down the mode of \(X\) .

(ii)     Find the exact value of \(p\) if \({\rm{Var}}(X) = \frac{{28}}{9}\) .

(i)     Find the smallest value of \(n\) for which the probability of Arthur walking to school on the next \(n\) days is less than \(0.01\).

(ii)     Find the probability that Arthur cycles to school for the third time on the last of eight successive days.

(i)     the mode is \(1\)     A1

(ii)     attempt to solve \(\frac{{1 - p}}{{{p^2}}} = \frac{{28}}{9}\)     M1

obtain \(p = \frac{3}{7}\)     A1

Note: \(p = 0.429\) is awarded M1A0.

[3 marks]

(i)     require least \(n\) such that

\({0.55^n} < 0.01\)     (M1)

EITHER

listing values: \(0.55\), \(0.3025\), \(0.166\), \(0.091\), \(0.050\), \(0.028\), \(0.015\), \(0.0084\)     (M1)

obtain \(n = 8\)     A1

OR

\(n > \frac{{\ln 0.01}}{{\ln 0.55}} = 7.70 \ldots \)     (M1)

obtain \(n = 8\)     A1

(ii)     recognition of negative binomial     (M1)

\(X \sim NB(3,0.45)\)

\({\rm{P}}(X = 8) = \left( \begin{array}{l}
7\\
2
\end{array} \right) \times {0.45^3} \times {0.55^5}\)     (A1)

\( = 0.0963\)     A1

Note: If \(0.45\) and \(0.55\) are mixed up, count it as a misread – probability in that case is \(0.0645\).

[6 marks]

(a)(i) A surprising number of candidates were unaware of the definition of the mode of a distribution.

(a)(ii) Generally well done, although a few candidates gave a decimal answer.

(b) Generally well done, and it was pleasing that most were familiar with the direct use of the negative binomial distribution in (ii).