Find the value of

(i)     \({\text{A}}\);

(ii)     \({\text{B}}\);

(iii)     \({\text{C}}\).

Calculate the probability that Tomek wears

(i)     shorts and no tie;

(ii)     no tie;

(iii)     shorts given that he is not wearing a tie.

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on both days.

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on one of the days, and shorts on the other day.

(i)     \(0.7 \left( {\frac{{70}}{{100}},{\text{ }}\frac{7}{{10}},{\text{ 70% }}} \right)\)     (A1)

(ii)     \(0.2 \left( {\frac{{20}}{{100}},{\text{ }}\frac{2}{{10}},{\text{ }}\frac{1}{5},{\text{ 20% }}} \right)\)     (A1)

(iii)     \(0.85 \left( {\frac{{85}}{{100}},{\text{ }}\frac{{17}}{{20}},{\text{ 85% }}} \right)\)     (A1)

[3 marks]

(i)     \(0.7 \times 0.85\)     (M1)

Note: Award (M1) for multiplying their values from parts (a)(i) and (a)(iii).

\( = 0.595{\text{ }}\left( {\frac{{119}}{{200}},{\text{ 59.5% }}} \right)\)     (A1)(ft)(G1)

Note: Follow through from part (a).

(ii)     \(0.3 \times 0.2 + 0.7 \times 0.85\)     (M1)(M1)

Note: Award (M1) for their two products, (M1) for adding their two products.

\( = 0.655{\text{ }}\left( {\frac{{131}}{{200}},{\text{ 65.5% }}} \right)\)     (A1)(ft)(G2)

Note: Follow through from part (a).

(iii)     \(\frac{{0.595}}{{0.655}}\)     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for correct numerator, (A1)(ft) for correct denominator. Follow through from parts (b)(i) and (ii).

\( = 0.908{\text{ }}\left( {{\text{0.90839}} \ldots ,{\text{ }}\frac{{119}}{{131}},{\text{ 90,8% }}} \right)\)     (A1)(ft)(G2)

[8 marks]

\(0.3 \times 0.3\)     (M1)

\( = 0.09 \left( {\frac{9}{{100}}, 9\%} \right)\)     (A1)(G2)

[2 marks]

\(0.3 \times 0.7\)     (M1)

\(0.3 \times 0.7 \times 2\)   OR   \((0.3 \times 0.7) + (0.7 \times 0.3)\)     (M1)

Note: Award (M1) for their correct product seen, (M1) for multiplying their product by 2 or for adding their products twice.

\( = 0.42 \left( {\frac{{42}}{{100}}, \frac{{21}}{{50}}, 42\%} \right)\)     (A1)(ft)(G2)

Note: Follow through from part (a)(i).

[3 marks]

Find the probability that this person is not allergic to nuts.

Find the probability that both people chosen are not allergic to nuts.

Copy and complete the tree diagram.

Find the probability that this adult is allergic to nuts and the liquid turns blue.

Find the probability that the liquid turns blue.

Find the probability that the tested adult is allergic to nuts given that the liquid turned blue.

Estimate the number of employees, from this 38, who are allergic to nuts.

\(\frac{{34}}{{60}}{\text{ }}\left( {\frac{{17}}{{30}},{\text{ }}0.567,{\text{ }}0.566666 \ldots ,{\text{ }}56.7\% } \right)\)     (A1)(A1)

Note:     Award (A1) for correct numerator, (A1) for correct denominator.

[2 marks]

\(\frac{{34}}{{60}} \times \frac{{33}}{{59}}\)     (M1)

Note:    Award (M1) for their correct product.

\( = 0.317{\text{ }}\left( {\frac{{187}}{{590}},{\text{ }}0.316949 \ldots ,{\text{ }}31.7\% } \right)\)     (A1)(ft)(G2)

Note:    Follow through from part (a).

[2 marks]

     (A1)(A1)(A1)

Note:     Award (A1) for each correct pair of branches.

[3 marks]

\(0.006 \times 0.98\)     (M1)

Note:     Award (M1) for multiplying 0.006 by 0.98.

\( = 0.00588{\text{ }}\left( {\frac{{147}}{{25000}},{\text{ }}0.588\% } \right)\)     (A1)(G2)

[2 marks]

\(0.006 \times 0.98 + 0.994 \times 0.05{\text{ }}(0.00588 + 0.994 \times 0.05)\)     (A1)(ft)(M1)

Note:     Award (A1)(ft) for their two correct products, (M1) for adding two products.

\( = 0.0556{\text{ }}\left( {0.05558,{\text{ }}5.56\% ,{\text{ }}\frac{{2779}}{{50000}}} \right)\)     (A1)(ft)(G3)

Note:     Follow through from parts (c) and (d).

[3 marks]

\(\frac{{0.006 \times 0.98}}{{0.05558}}\)     (M1)(M1)

Note:     Award (M1) for their correct numerator, (M1) for their correct denominator.

\( = 0.106{\text{ }}\left( {0.105793 \ldots ,{\text{ }}10.6\% ,{\text{ }}\frac{{42}}{{397}}} \right)\)     (A1)(ft)(G3)

Note:     Follow through from parts (d) and (e).

[3 marks]

\(0.105793 \ldots  \times 38\)     (M1)

Note:     Award (M1) for multiplying 38 by their answer to part (f).

\( = 4.02{\text{ }}(4.02015 \ldots )\)     (A1)(ft)(G2)

Notes: Follow through from part (f). Use of 3 sf result from part (f) results in an answer of 4.03 (4.028).

[2 marks]

Draw a Venn diagram to represent this information. Label A the set that represents The Art Journal readers, B the set that represents The Beartown News readers, and C the set that represents The Currier readers.

Find the percentage of the population that does not read any of the three newspapers.

Find the percentage of the population that reads exactly one newspaper.

Find the percentage of the population that reads The Art Journal or The Beartown News but not The Currier.

A local radio station states that 83 % of the population reads either The Beartown News or The Currier.

Use your Venn diagram to decide whether the statement is true. Justify your answer.

The population of Beartown is 120 000. The local radio station claimed that 34 000 of the town’s citizens read at least two of the local newspapers.

Find the percentage error in this claim.

(A1) for three circles and a rectangle (U need not be seen)

(A1) for 5

(A1) for 3, 8 and 12

(A1) for 16, 26 and 29 OR 32, 46, 54 placed outside the circles.     (A4)

Note: Accept answers given as decimals or fractions.

[4 marks]

100 – (16 + 26 + 29) – (8 + 5 + 3 + 12)     (M1)

100 – 71 – 28

Note: Award (M1) for correct expression. Accept equivalent expressions, for example 100 – 71 – 28 or 100 – (71 + 28).

= 1     (A1)(ft)(G2)

Note: Follow through from their Venn diagram but only if working is seen.

[2 marks]

16 + 26 + 29     (M1)

Note: Award (M1) for 16, 26, 29 seen.

= 71     (A1)(ft)(G2)

Note: Follow through from their Venn diagram but only if working is seen.

[2 marks]

16 + 3 + 26     (M1)

Note: Award (M1) for their 16, 3, 26 seen.

= 45     (A1)(ft)(G2)

Note: Follow through from their Venn diagram but only if working is seen.

[2 marks]

True     (A1)(ft)

100 – (1 –16) = 83     (R1)(ft)

OR

46 + 54 – 17 = 83     (R1)(ft)

Note: Do not award (A1)(R0). Follow through from their Venn diagram.

[2 marks]

28% of 120000     (M1)

= 33600     (A1)

\({\text{%  error}} = \frac{{(34000 - 33600)}}{{33600}} \times 100\)     (M1)

Note: Award (M1) for 28 seen (may be implied by 33600 seen), award (M1) for correct substitution of their 33600 in the percentage error formula. If an error is made in calculating 33600 award a maximum of (M1)(A0)(M1)(A0), the final accuracy mark is lost.

OR

\(\frac{{34000}}{{120000}} \times 100\)     (M1)

= 28.3(28.3333…)     (A1)

\({\text{%  error}} = \frac{{(28.3333... - 28)}}{{28}} \times 100\)     (M1)

= 1.19% (1.19047...)     (A1)(ft)(G3)

Note: % sign not required. Accept 1.07 (1.0714…) with use of 28.3. 1.18 with use of 28.33 and 1.19 with use of 28.333. Award (G3) for 1.07, 1.18 or 1.19 seen without working.

[4 marks]

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and “exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and “exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and“exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and “exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and “exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

This question was accessible to the great majority of candidates. The common errors were:

• the lack of a bounding rectangle in (a);
• the lack of subtraction for the entries in the disjoint regions of the type \(A' \cap B' \cap C\) and the subsequent total exceeding 100%;
• the incorrect interpretation of “either ...or” as “exclusive or”. It is of the utmost importance to note that the ambiguity of the “or” statement will be removed and“exclusive or” signalled by the phrase “either ...or....but not both”. Otherwise, “inclusive or” must always be assumed.

A number of candidates were unable to interpret the percentage error question correctly and scored 0/4. This was somewhat disappointing.

Draw a Venn diagram to represent this information.

Find the number of families who own no pets.

Find the percentage of families that own exactly one pet.

A family is chosen at random. Find the probability that they own a cat, given that they own a bird.

     (A1)(A1)(A1)(A1)(A1)

Note: Award (A1) for rectangle (U not required), (A1) for 3 intersecting circles, (A1) for 4 in central intersection, (A1) for 16, 3, 8 and (A1) for 33, 10, 7 (ft) if subtraction is carried out, or for S(56), Q(38) and R(22) seen by the circles.

[5 marks]

100 − 81     (M1)

19     (A1)(ft)(G2)

Note: Award (M1) for subtracting their total from 100.

[2 marks]

\(33 +10 + 7\)     (M1)

Note: Award (M1) for adding their values from (a).

\(\left( {\frac{{50}}{{100}}} \right) \times 100{\text{ }}\% \)     (A1)(ft)

50 % (50)     (A1)(ft)(G3)

[3 marks]

P (own a cat given they own a bird) \( = \frac{{12}}{{22}}\left( {0.545,\frac{6}{{11}}} \right)\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for the numerator, (A1)(ft) for the denominator.

[2 marks]

Most candidates began the paper well by correctly drawing the Venn diagram and answering parts (b) and (c) correctly.

Most candidates began the paper well by correctly drawing the Venn diagram and answering parts (b) and (c) correctly.

Most candidates began the paper well by correctly drawing the Venn diagram and answering parts (b) and (c) correctly.

Conditional probability has proved difficult for many candidates; only a very small part of the candidates scored full marks for this part.

A student is chosen at random from the group. Find the probability that the student

(i)     studied Mathematics HL;

(ii)    was examined in French;

(iii)   studied Mathematics HL and was examined in French;

(iv)   did not study Mathematics SL and was not examined in English;

(v)    studied Mathematical Studies SL given that the student was examined in Spanish.

Pam believes that the Mathematics course a student chooses is independent of the language in which the student is examined.

Using your answers to parts (a) (i), (ii) and (iii) above, state whether there is any evidence for Pam’s belief. Give a reason for your answer.

Pam decides to test her belief using a Chi-squared test at the \(5\% \) level of significance.

(i)     State the null hypothesis for this test.

(ii)    Show that the expected number of Mathematical Studies SL students who took the examination in Spanish is \(41.3\), correct to 3 significant figures.

Write down

(i)     the Chi-squared calculated value;

(ii)    the number of degrees of freedom;

(iii)   the Chi-squared critical value.

State, giving a reason, whether there is sufficient evidence at the \(5\% \) level of significance that Pam’s belief is correct.

(i)     \(\frac{{100}}{{400}}{\text{ }}\left( {\frac{1}{4}{\text{, }}0.25{\text{, }}25\% } \right)\)     (A1)

(ii)    \(\frac{{90}}{{400}}{\text{ }}\left( {\frac{9}{{40}}{\text{, }}0.225{\text{, }}22.5\% } \right)\)     (A1)

(iii)   \(\frac{{20}}{{400}}{\text{ }}\left( {\frac{1}{{20}}{\text{, }}0.05{\text{, }}5\% } \right)\)     (A1)(A1)

Note: Award (A1) for numerator, (A1) for denominator.

(iv)   \(\frac{{120}}{{400}}{\text{ }}\left( {\frac{3}{{10}}{\text{, }}0.3{\text{, }}30\% } \right)\)     (A1)(A1)

Note: Award (A1) for numerator, (A1) for denominator.

(v)    \(\frac{{30}}{{110}}{\text{ }}\left( {\frac{3}{{11}}{\text{, }}0.273{\text{, }}27.3\% } \right)\) (\(0.272727 \ldots \))     (A1)(A1)

Note: Award (A1) for numerator, (A1) for denominator. Accept \(0.27\), do not accept \(0.272\), do not accept \(0.3\).

[8 marks]

\(\frac{1}{{20}} \ne \frac{1}{4} \times \frac{9}{{40}}\)     (R1)(ft)

Note: The fractions must be used as part of the reason. Follow through from (a)(i), (a)(ii) and (a)(iii).

Pam is not correct.     (A1)(ft)

Notes: Do not award (R0)(A1). Accept the events are not independent (dependent).

[2 marks]

(i)     The mathematics course and language of examination are independent.     (A1)

Notes: Accept “There is no association between Mathematics course and language”. Do not accept “not related”, “not correlated”, “not influenced”.

(ii)    \(\frac{{110}}{{400}} \times \frac{{150}}{{400}} \times 400{\text{ }}\left( { = \frac{{110 \times 150}}{{400}}} \right)\)     (M1)

 \( = 41.25\)     (A1)

 \( = 41.3\)     (AG)

Note: \(41.25\) and \(41.3\) must be seen to award final (A1).

[3 marks]

(i)     \(7.67\) (\(7.67003 \ldots \))     (G2)

Note: Accept \(7.7\), do not accept \(8\) or \(7.6\). Award (G1) if formula with all nine terms seen but their answer is not one of those above.

(ii)    \(4\)     (G1)

(iii)   \(9.488\)     (A1)(ft)

Notes: Accept \(9.49\) or \(9.5\), do not accept \(9.4\) or \(9\). Follow through from their degrees of freedom.

[4 marks]

\(7.67 < 9.488\)     (R1)

OR

\(p = 0.104 \ldots , p > 0.05\)     (R1)

Accept (Do not reject) \({H_0}\) (Pam’s belief is correct)     (A1)(ft)

Notes: Follow through from part (d). Do not award (R0)(A1).

[2 marks]

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and p-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

The simple probabilities beginning this question were successfully attempted by the great majority. Most errors in the latter parts occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table. Probability questions in this course are, in the main, contextual and the reliance of formulas is not always beneficial to the candidates. Only the best candidates realized the significance of part (b) as a link to the chi-squared test.

This was well attempted by the majority, the weakness being the sole reliance of the calculator to calculate expected value. However, there still remains confusion between critical and \(p\)-values as the basis for accepting the null hypothesis.

Write down the total number of fish in the pond.

Leanne catches a fish.

Find the probability that she

(i) catches an undersized bream;

(ii) catches either a flathead or an undersized fish or both;

(iii) does not catch an undersized whiting;

(iv) catches a whiting given that the fish was normal.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Copy and complete the probability tree diagram below.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that it is windy and Leanne catches a fish on a particular day.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that Leanne catches a fish on a particular day.

Use your answer to part (e) to calculate the probability that Leanne catches a fish on two consecutive days.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Given that Leanne catches a fish on a particular day, calculate the probability that the day was windy.

90     (A1)

[1 mark]

(i) \(\frac{3}{{90}}(0.0\bar 3,{\text{ }}0.0333,{\text{ }}0.0333...,{\text{ }}3.\bar 3\% ,{\text{ }}3.33\% )\)     (A1)(ft)

Note: For the denominator follow through from their answer in part (a).

(ii) \(\frac{{53}}{{90}}(0.5\bar 8,{\text{ }}0.588...,{\text{ }}0.589,{\text{ }}58.\bar 8\% ,{\text{ }}58.9\% )\)     (A1)(A1)(ft)(G2)

Notes: Award (A1) for the numerator. (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iii) \(\frac{{72}}{{90}}{\text{(0.8, 80}}\%)\)     (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for the numerator, (their part (a) –18) (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iv) \(\frac{{24}}{{48}}(0.5,{\text{ 50}}\% )\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[7 marks]

     (A1)(A1)(A1)

Notes: Award (A1) for each correct entry. Tree diagram must be seen for marks to be awarded.

[3 marks]

\(0.3 \times 0.1 = 0.03\left( {\frac{3}{{100}}} \right)\)     (M1)(A1)(G2)

Note: Award (M1) for correct product seen.

[2 marks]

\(0.3 \times 0.1+ 0.7\times0.65\)     (M1)(M1)

Notes: Award (M1) for \(0.7\times0.65\) (or 0.455) seen, (M1) for adding their 0.03. Follow through from their answers to parts (c) and (d).

\( = 0.485\left( {\frac{{485}}{{1000}},\frac{{97}}{{200}}} \right)\)     (A1)(ft)(G2)

Note: Follow through from their tree diagram and their answer to part (d).

[3 marks]

\(0.485 \times 0.485\)     (M1)

\(0.235\left( {\frac{{9409}}{{40000}}{\text{, }}0.235225} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (e).

[2 marks]

\(\frac{{0.03}}{{0.485}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted conditional probability formula, (A1)(ft) for their (d) as numerator and their (e) as denominator.

\(0.0619\left( {\frac{{6}}{{97}}}\text{, 0.0618556...} \right) \)     (A1)(ft)(G2)

Note: Follow through from their parts (d) and (e).

[3 marks]

(a) Most candidates found this correctly although a few wrote 180 instead of 90.

(b) This was also answered well. The main errors were putting 65/90 in part (ii) and 24/90 in part (iv).

(c) The tree diagram was completed correctly in most scripts. It appears that some candidates may have answered this on their question paper and this was not sent to the scanning centre with the answer papers.

(d) Many answered this correctly. Some added instead of multiplying.

(e) Surprisingly well answered. Again some added and multiplied in the wrong place.

(f) Most candidates added here and then divided by 2 rather than multiplying.

(g) This was badly done with very few correct answers seen.

Calculate the value of the common ratio.

Calculate the 10^th term of this sequence.

The k^th term is the first term which is greater than 2000. Find the value of k.

Write in words \((q \wedge \neg r) \Rightarrow \neg p\).

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write this statement in symbolic form.

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write the contrapositive of this statement in symbolic form.

u₁r⁴ = 324     (A1)

u₁r = 12     (A1)

r³ = 27     (M1)

r = 3     (A1)(G3)

Note: Award at most (G3) for trial and error.

[4 marks]

4 × 3⁹ = 78732 or 12 × 3⁸ = 78732     (A1)(M1)(A1)(ft)(G3)

Note: Award (A1) for u₁ = 4 if n = 9 , or u₁ = 12 if n = 8, (M1) for correctly substituted formula.

(ft) from their (a).

[3 marks]

4 × 3^k−1 > 2000     (M1)

Note: Award (M1) for correct substitution in correct formula. Accept an equation.

k > 6     (A1)

k = 7     (A1)(ft)(G2)

Notes: If second line not seen award (A2) for correct answer. (ft) from their (a).

Accept a list, must see at least 3 terms including the 6^th and 7^th.

Note: If arithmetic sequence formula is used consistently in parts (a), (b) and (c), award (A0)(A0)(M0)(A0) for (a) and (ft) for parts (b) and (c).

[3 marks]

If the number is even and the number does not end in zero, (then) the number is not a multiple of five.     (A1)(A1)(A1)

Note: Award (A1) for “if…(then)”, (A1) for “the number is even and the number does not end in zero”, (A1) for the number is not a multiple of 5.

[3 marks]

\((p \wedge \neg q) \Rightarrow \neg r\)     (A1)(A1)(A1)(A1)

(A1) for \(\Rightarrow\), (A1) for \(\wedge\), (A1) for p and \(\neg q\), (A1) for \(\neg r\)

Note: If parentheses not present award at most (A1)(A1)(A1)(A0).

[4 marks]

\(r \Rightarrow (\neg p \vee q)\)   OR   \(r \Rightarrow \neg (p \wedge \neg q)\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for reversing the order, (A1) for negating the statements on both sides.

If parentheses not present award at most (A1)(ft)(A0).

Do not penalise twice for missing parentheses in (i) and (ii).

[2 marks]

An easy ratio to find and the majority of candidates found r = 3, though many had trouble showing the appropriate method, thus losing marks.

A fairly straightforward part for most candidates.

The majority found k − 7; many without supporting work which lost them a mark. Where candidates had difficulty in this part, it was generally a case of poor algebraic skills.

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

Very poorly answered with many candidates scoring just one mark. The main error was to open the bracket and not use the “or”.

Write down the value of a.

Write down the value of b.

Use the tree diagram to find the probability that an employee encountered traffic and was late for work.

Use the tree diagram to find the probability that an employee was late for work.

Use the tree diagram to find the probability that an employee encountered traffic given that they were late for work.

Find the value of x.

Find the value of y.

Find the number of employees who, in the last year, did not travel to work by car, bicycle or public transportation.

Find \(n\left( {\left( {C \cup B} \right) \cap P'} \right)\).

a = 0.2     (A1)

[1 mark]

b = 0.85     (A1)

[1 mark]

0.25 × 0.8     (M1)

Note: Award (M1) for a correct product.

\( = 0.2\,\,\,\left( {\,\frac{1}{5},\,\,\,20\% } \right)\)     (A1)(G2)

[2 marks]

0.25 × 0.8 + 0.75 × 0.15     (A1)(ft)(M1)

Note: Award (A1)(ft) for their (0.25 × 0.8) and (0.75 × 0.15), (M1) for adding two products.

\( = 0.313\,\,\,\left( {0.3125,\,\,\,\frac{5}{{16}},\,\,\,31.3\% } \right)\)    (A1)(ft)(G3)

Note: Award the final (A1)(ft) only if answer does not exceed 1. Follow through from part (b)(i).

[3 marks]

\(\frac{{0.25 \times 0.8}}{{0.25 \times 0.8 + 0.75 \times 0.15}}\)    (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for a correct numerator (their part (b)(i)), (A1)(ft) for a correct denominator (their part (b)(ii)). Follow through from parts (b)(i) and (b)(ii).

\( = 0.64\,\,\,\left( {\frac{{16}}{{25}},\,\,64{\text{% }}} \right)\)     (A1)(ft)(G3)

Note: Award final (A1)(ft) only if answer does not exceed 1.

[3 marks]

(x =) 3     (A1)

[1 Mark]

(y =) 10     (A1)(ft)

Note: Following through from part (c)(i) but only if their x is less than or equal to 13.

[1 Mark]

54 − (10 + 3 + 4 + 2 + 6 + 8 + 13)     (M1)

Note: Award (M1) for subtracting their correct sum from 54. Follow through from their part (c).

= 8      (A1)(ft)(G2)

Note: Award (A1)(ft) only if their sum does not exceed 54. Follow through from their part (c).

[2 marks]

6 + 8 + 13     (M1)

Note: Award (M1) for summing 6, 8 and 13.

27     (A1)(G2)

[2 marks]

Write down in symbolic form the compound proposition

“If \(x\) is a factor of 60 then \(x\) is a multiple of 5 or \(x\) is not a multiple of 4.”

Write down in words the compound proposition \(\neg r \wedge (p\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \vee } q)\).

Copy the following truth table and complete the last three columns.

State why the compound proposition \(\neg r \wedge (p\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \vee } q)\) is not a logical contradiction.

A row from the truth table from part (c) is given below.

Write down one value of \(x\) that satisfies these truth values.

\(p \Rightarrow (r \vee \neg q)\)     (A1)(A1)(A1)

Note:     Award (A1) for “\(p \Rightarrow \)”.

Award (A1) for “\(r \vee \neg q\)” or “\(r \vee q\)” (or “\(\neg q \vee r\)”or “\(q \vee r\)”)

Award (A1) for “\(\neg q\)”.

Award at most (A1)(A1)(A0) if parentheses are missing for \(r \vee \neg q\).

Award (A0)(A0)(A1) for \((p \Rightarrow r) \vee \neg q\).

[3 marks]

\(x\) is not a multiple of 5 and (\(x\)) is (either) a factor of 60 or (\(x\)) is a multiple of 4, but not both     (A1)(A1)(A1)

Note:     Award (A1) for “\(x\) is not a multiple of 5”, (A1) for “(\(x\)) is a factor of 60 or (\(x\)) is a multiple of 4 but not both”, (A1) for “and” in the correct position. Accept only “but not both” in the second (A1).

Award at most (A1)(A1)(A0) for using extra statements such as “If ...then”, “if and only if” etc.

[3 marks]

     (A1)(A1)(A1)(ft)

Note:     Award (A1) for each correct column. Last column follows through from previous two.

[3 marks]

because not all the entries in the \(\neg r \wedge (p\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{ \vee } q)\) column are F     (R1)(ft)

Note:     If all entries in the last column of their truth table are T, award (R1)(ft) for an answer of “it is a tautology”. Only award (R1)(ft) if the column is identified in the justification.

[1 mark]

accept one of: 1\(\,\,\,\)OR\(\,\,\,\)2\(\,\,\,\)OR\(\,\,\,\)3\(\,\,\,\)OR\(\,\,\,\)6     (A1)

Note:     Award (A1) for any one of the above answers.

[1 mark]

Draw a Venn diagram to represent the given information using sets labelled C, M and P. Complete the diagram to include the number of families represented in each region.

Find the number of families who

(i) went to the circus only;

(ii) went to the museum and the park but not the circus;

(iii) did not go to any of the three places on the weekend.

A family is chosen at random from the group of 40 families. Find the probability that the family went to

(i) the circus;

(ii) two or more places;

(iii) the park or the circus, but not the museum;

(iv) the museum, given that they also went to the circus.

Two families are chosen at random from the group of 40 families.

Find the probability that both families went to the circus.

     (A1)(A1)(A1)(A1)

Award (A1) for 3 intersecting circles and rectangle, (A1) for 1, 3, 4 and 7, (A1) for 2, (A1) for 6 and 5.

(i) 2     (A1)(ft)

(ii) 6     (A1)(ft)

(iii) 40 − (1 + 6 + 2 + 3 + 4 + 7 + 5)     (M1)

Note: Award (M1) for subtracting all their values from 40.

= 12     (A1)(ft)(G2)

Note: Follow through from their Venn diagram for parts (i), (ii) and (iii).

(i) \(\frac{{16}}{{40}}\left( {\frac{2}{5},0.4,40\% } \right)\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

(ii) \(\frac{{20}}{{40}}\left( {\frac{1}{2},0.5,50\% } \right)\)     (A1)(ft) (A1) (G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

(iii) \(\frac{6}{{40}}\left( {\frac{3}{{20}},0.15,15\% } \right)\)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

(iv) \(\frac{{11}}{{16}}\left( {0.6875,68.75\% } \right)\)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram. Answer must be less than 1 otherwise award (A0)(A0). Award (A0)(A0) if answer is given as incorrect reduced fraction without working.

\(\frac{{16}}{{40}} \times \frac{{15}}{{39}}\)     (A1)(A1)(ft)

Note: Award (A1) for multiplication of their probabilities, (A1)(ft) for their correct probabilities.

\(\frac{{240}}{{1560}}\left( {\frac{2}{{13}},0.153846...,15.4\% } \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (c)(i). Answer must be less than 1 otherwise award at most (A1)(A1)(A0)(ft).

Draw a Venn diagram to show this information.

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

Find the number of tourists that saw Cheetah and Rhino but not Leopard.

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

Calculate the probability that a tourist chosen at random from the group

(i)     saw Leopard;

(ii)     saw only one of the three types of animal;

(iii)     saw only Leopard, given that he saw only one of the three types of animal.

There were 25 tourists in the group and every tourist saw at least one of the three types of animal.

If a tourist chosen at random from the group saw Leopard, find the probability that he also saw Cheetah.

     (A1)(A1)(A1)(A1)

Note: Award (A1) for rectangle and three labelled intersecting circles (the rectangle need not be labelled), (A1) for 5, (A1) for 2 and 1, (A1) for 4, 3 and 9.

[4 marks]

\(25 - (5 + 2 + 1 + 4 + 3 + 9)\)     (M1)

Notes: Award (M1) for their \(5 + 2 + 1 + 4 + 3 + 9\) seen even if total is greater than \(25\).

     Do not award (A1)(ft) if their total is greater than \(25\).

\( = 1\)     (A1)(ft)(G2)

[2 marks]

(i)   \(\frac{{12}}{{25}}{\text{ }}(0.48,{\text{ }}48\% )\)     (A1)(ft)(A1)(G2)

Notes: Award (A1)(ft) for numerator, (A1) for denominator.

     Follow through from Venn diagram.

(ii)     \(\frac{{16}}{{25}}{\text{ }}(0.64,{\text{ }}64\% )\)     (A1)(A1)(G2)

Notes: Award (A1) for numerator, (A1) for denominator.

     There is no follow through; all information is given.

(iii)     \(\frac{4}{{16}}{\text{ }}(0.25,{\text{ }}25\% )\))     (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

     Follow through from part (c)(ii) only.

[6 marks]

\(\frac{6}{{12}}{\text{ }}(0.5,{\text{ }}50\% )\)     (A1)(A1)(ft)(G2)

Notes: Award (A1) for numerator, (A1)(ft) for denominator.

     Follow through from Venn diagram.

[2 marks]

Write the following argument in symbolic form.

“If the land has been purchased and the building permit has been obtained, then the land can be used for residential purposes.”

In your answer booklet, copy and complete a truth table for the argument in part (a).

Begin your truth table as follows.

Use your truth table to determine whether the argument in part (a) is valid.

Give a reason for your decision.

Write down the inverse of the argument in part (a)

(i)     in symbolic form;

(ii)     in words.

\((p \wedge q) \Rightarrow r\)     (A1)(A1)(A1)

Notes: Award (A1) for conjunction seen, award (A1) for implication seen, award (A1) for correct simple propositions in correct order (the parentheses are required). Accept \(r \Leftarrow (p \wedge q)\).

     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for each correct column, follow through to the final column from their \((p \wedge q)\) column. For the second (A1)(ft) to be awarded there must be an implication in part (a).

Follow through from part (a).

The argument is not valid since not all entries in the final column are T.     (A1)(ft)(R1)

Notes: Do not award (A1)(ft)(R0). Follow through from part (b).

Accept “The argument is not valid since \((p \wedge q) \Rightarrow r\) is not a tautology”.

(i)     \(\neg (p \wedge q) \Rightarrow \neg r\)     (A1)(ft)(A1)(ft)

OR

\((\neg p \vee \neg q) \Rightarrow \neg r\)     (A1)(ft)(A1)(ft)

Notes: Award (A1)(ft) for the negation of their antecedent and the negation of their consequent, (A1)(ft) for their fully correct answer.

Follow through from part (a). Accept \(\neg r \Leftarrow \neg (p \wedge q)\) or \(\neg r \Leftarrow (\neg p \vee \neg q)\). Follow through from part (a).

(ii)     if it is not the case that the land has been purchased and the building permit has been obtained then the land can not be used for residential purposes.     (A1)(A1)(ft)

OR

if (either) the land has not been purchased or the building permit has not been obtained then the land can not be used for residential purposes.     (A1)(A1)(ft)

Notes: Award (A1) for “if… then…” seen, (A1)(ft) for correct statements in correct order. Follow through from part (d)(i).

Forming the statement in part (a) was attainable by the great majority, although the lack of parentheses was a common fault.

The truth table in part (b) saw less success and it was clear that some centres simply had not prepared their candidates in this area of the course.

Where the truth table was correctly constructed many candidates were not aware of the conditions required for an argument to be valid and in part (d) the converse and the inverse were often confused.

Where the truth table was correctly constructed many candidates were not aware of the conditions required for an argument to be valid and in part (d) the converse and the inverse were often confused.

Draw a Venn diagram, using sets labelled \(E\), \(S\) and \(A\), to show this information.

Calculate the value of \(x\).

Explain, in words, the meaning of \((E \cup S) \cap A'\).

Write down \(n\left( {(E \cup S \cup A)'} \right)\).

Find the probability that a woman selected at random from the group had visited Europe.

Find the probability that a woman selected at random from the group had visited Europe, given that she had visited Asia.

Two women from the group are selected at random.

Find the probability that both women selected had visited South America.

     (A1)(A1)(A1)(A1)(A1)

Notes: Award (A1) for rectangle and three labelled intersecting circles.

     Award (A1) for \(7\) in correct place.

     Award (A1) for \(28\), \(22\) and \(16\) in the correct places.

     Award (A1) for \(15\), \(x\) and \(2x\) in the correct places.

     Award (A1) for \(20\) in the correct place.

     Accept \(4\) and \(8\) instead of \(x\) and \(2x\).

     Do not penalize if \(U\) is omitted from the diagram.

[5 marks]

\(3x = 120 - (20 + 28 + 15 + 22 + 7 + 16)\)     (M1)

Note: Award (M1) for setting up a correct equation involving \(x\), the \(120\) and values from their diagram.

\(x = 4\)     (A1)(ft)(G2)

Note: Follow through from part (a). For the follow through to be awarded \(x\) must be a positive integer.

[2 marks]

(Women who had visited) Europe or South America and (but had) not (visited) Asia     (A1)(A1)

Notes: Award (A1) for “(visited) Europe or South America” (or both).

     Award (A1) for “and (but) had not visited Asia”.

     \(E\)(urope) union \(S\)(outh America) intersected with not \(A\)(sia) earns no marks, (A0).

[2 marks]

\(20\)     (A1)

Note: Award (A0) for the embedded answer of \(n(20)\).

[1 mark]

\(\frac{{58}}{{120}}{\text{ }}\left( {\frac{{29}}{{60}},{\text{ 0.483, 48.3% }}} \right){\text{ (0.48333}} \ldots {\text{)}}\)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, follow through from their value of \(x\), or their diagram, (A1) for denominator.

[2 marks]

\(\frac{{15}}{{35}}{\text{ }}\left( {\frac{3}{7},{\text{ 0.429, 42.9% }}} \right){\text{ (0.428571}} \ldots {\text{)}}\)     (A1)(ft)(A1)(ft)(G2)

Note: Award (A1)(ft) for numerator, (A1)(ft) for denominator, follow through from their value of \(x\) or their diagram. 

[2 marks]

\(\frac{{48}}{{120}} \times \frac{{47}}{{119}}\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for two correct fractions, follow through from their denominator in part (e), follow through the numerator from their answer to part (b) or from their diagram, (M1) for multiplication of their two fractions.

\( = \frac{{2256}}{{14\,280}}\left( {\frac{{94}}{{595}},{\text{ 0.158, 15,8% }}} \right){\text{ (0.157983}} \ldots {\text{)}}\)     (A1)(ft)(G2)

Notes: Award (A1)(M1)(A1) for correct fractions, correctly multiplied together with an answer of \(0.16\).

     Award (A0)(M1)(A0) for \(\frac{{48}}{{120}} \times \frac{{48}}{{120}} = 0.16\).

     Award (G1) for an answer of \(0.16\) with no working seen.

[3 marks]

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Candidates seemed to be well-drilled in the technique of creating Venn diagrams and using the data from their diagrams to solve problems in probability and this question was well answered. Except for the odd mistake in determining the value of x in part (b), many candidates scored full marks on the first two parts of the question. Indeed, those who calculated an incorrect value of x were able to recover many of the marks in the remainder of the question with the use of follow through marks. ‘Explain in words…’ required candidates to answer part (c) in the context of the question so ‘E union S intersection not A’ earned no marks. Of those candidates who did answer in context, many scored 1 mark for ‘had not visited Asia’ but a significant number used ‘and’ rather than ‘or’ and consequently were not awarded the other mark for expressing \(E \cup S\) in words. Whilst many correct answers of 20 were seen for part (d), a significant number of candidates wrote down the incorrect value of 113 which presumably was arrived at by evaluating \(n((E \cap S \cap A)')\) rather than the actual demand of the question. Having a Venn diagram seemed to be a good aid for parts (e) and (f) and much good work was seen in these two parts. However, in part (g), a significant number of candidates either chose a “with replacement” method or simply did not know what to do with the probabilities once they were found. As a consequence, this part of the question proved to be quite a discriminator.

Draw a Venn diagram to represent the given information, using sets labelled \(B\), \(C\) and \(H\).

Show that \(x = 3\).

Write down the value of \(n(B \cap C)\).

Find the probability that this person

(i)     went on at most one trip;

(ii)     went on the coach trip, given that this person also went on both the helicopter trip and the boat trip.

     (A5)

Notes:     Award (A1) for rectangle and three labelled intersecting circles (U need not be seen),

(A1) for 3 in the correct region,

(A1) for 8 in the correct region,

(A1) for 5, 13 and 16 in the correct regions,

(A1) for \(x\), \(2x\) and \(4x\) in the correct regions.

[5 marks]

\(8 + 13 + 16 + 3 + 5 + x + 2x + 4x = 66\)    (M1)

Note:     Award (M1) for either a completely correct equation or adding all the terms from their diagram in part (a) and equating to 66.

Award (M0)(A0) if their equation has no \(x\).

\(7x = 66 - 45\) OR \(7x + 45 = 66\)     (A1)

Note:     Award (A1) for adding their like terms correctly, but only when the solution to their equation is equal to 3 and is consistent with their original equation.

\(x = 3\)    (AG)

Note:     The conclusion \(x = 3\) must be seen for the (A1) to be awarded.

[2 marks]

15     (A1)(ft)

Note:     Follow through from part (a). The answer must be an integer.

[1 mark]

(i)     \(\frac{{42}}{{66}}{\text{ }}\left( {\frac{7}{{11}},{\text{ }}0.636,{\text{ }}63.6\% } \right)\)     (A1)(ft)(A1)(G2)

Note:     Award (A1)(ft) for numerator, (A1) for denominator. Follow through from their Venn diagram.

(ii)     \(\frac{3}{9}{\text{ }}\left( {\frac{1}{3},{\text{ }}0.333,{\text{ }}33.3\% } \right)\)     (A1)(A1)(ft)(G2)

Note:     Award (A1) for numerator, (A1)(ft) for denominator. Follow through from their Venn diagram.

[4 marks]

Write down the value of

(i)     \(p\) ;

(ii)     \(q\) ;

(iii)     \(r\).

(i)     Find the probability that Mike reaches Trap B.

(ii)     Find the probability that Mike reaches Trap A.

(iii)     Find the probability that Mike escapes from the maze.

Sonya, a lab assistant, counts the number of paths that lead to traps or escape doors. She believes that the probability that Mike will be trapped is greater than the probability that he will escape.

State whether Sonya is correct. Give a mathematical justification for your conclusion.

During the first trial Mike escapes.

Given that Mike escaped, find the probability that he went directly from S to Escape Door 3.

(i)     \(\frac{1}{3}\;\;\;(0.333333…,{\text{ }}33.3333...\% )\)     (A1)

(ii)     \(\frac{1}{2}\;\;\;(0.5,{\text{ }}50\% )\)     (A1)

(iii)     \(\frac{1}{4}\;\;\;(0.25,{\text{ }}25\% )\)     (A1)

(i)     \(\frac{1}{3} \times \frac{1}{4}\)     (M1)

\( = \frac{1}{{12}}\;\;\;({\text{0.0833333..., 8.33333...% )}}\)     (A1)(G2)

(ii)     \(\frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3} \times \frac{1}{4}\)     (A1)(ft)(M1)

Note: Award (A1)(ft) for their three correct products seen, (M1) for addition of their products.

\( = \frac{1}{3}\;\;\;(0.333333...,{\text{ }}33.3333...\% )\)     (A1)(ft)(G2)

Note: Follow through from their parts (a)(i) and (a)(iii).

(iii)     \(1 - \frac{1}{{12}} - \frac{1}{3}\)     (M1)

Note: Follow through from parts (b)(i) and (b)(ii).

OR

\(\frac{1}{3} \times \frac{1}{2} + \frac{1}{3} \times \frac{1}{4} + \frac{1}{3}\)     (M1)

Note: Follow through from parts (a)(i) and (a)(ii).

\( = \frac{7}{{12}}\;\;\;(0.583333...,{\text{ }}58.3333...\% )\)     (A1)(ft)(G2)

Sonya is not correct.     (A1)(ft)

The probability that Mike escapes is \(\frac{7}{{12}}\), which is

greater than \(\frac{5}{{12}}{\text{ }}\left( {{\text{or greater than }}\frac{1}{2}} \right)\).     (R1)(ft)

Notes: Do not award (A1)(R0).

Follow through from their answers to part (b).

\(\frac{{\frac{1}{3}}}{{\frac{7}{{12}}}}\)     (A1)(A1)(ft)

Note: Award (A1) for correct numerator, (A1) for correct denominator.

\( = \frac{4}{7}\;\;\;\left( {\frac{{12}}{{21}},{\text{ 0.571428..., 57.1428...% }}} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (b)(iii).

Write down a formula for \(A\), the surface area to be coated.

Express this volume in \({\text{c}}{{\text{m}}^3}\).

Write down, in terms of \(r\) and \(h\), an equation for the volume of this water container.

Show that \(A = \pi {r^2}\frac{{1\,000\,000}}{r}\).

Show that \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\).

Find \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).

Using your answer to part (e), find the value of \(r\) which minimizes \(A\).

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

\((A = ){\text{ }}\pi {r^2} + 2\pi rh\)    (A1)(A1)

Note:     Award (A1) for either \(\pi {r^2}\) OR \(2\pi rh\) seen. Award (A1) for two correct terms added together.

[2 marks]

\(500\,000\)    (A1)

Notes:     Units not required.

[1 mark]

\(500\,000 = \pi {r^2}h\)    (A1)(ft)

Notes:     Award (A1)(ft) for \(\pi {r^2}h\) equating to their part (b).

Do not accept unless \(V = \pi {r^2}h\) is explicitly defined as their part (b).

[1 mark]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(A = \pi {r^2} + 2\pi r\left( {\frac{{500\,000}}{{\pi {r^2}}}} \right)\)    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their \({\frac{{500\,000}}{{\pi {r^2}}}}\) seen.

Award (M1) for correctly substituting only \({\frac{{500\,000}}{{\pi {r^2}}}}\) into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to \(\pi rh = \frac{{500\,000}}{r}\) and substituting for \(\pi rh\) in expression for \(A\).

\(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\)    (AG)

Notes:     The conclusion, \(A = \pi {r^2} + \frac{{1\,000\,000}}{r}\), must be consistent with their working seen for the (A1) to be awarded.

Accept \({10^6}\) as equivalent to \({1\,000\,000}\).

[2 marks]

\(2\pi r - \frac{{{\text{1}}\,{\text{000}}\,{\text{000}}}}{{{r^2}}}\)    (A1)(A1)(A1)

Note:     Award (A1) for \(2\pi r\), (A1) for \(\frac{1}{{{r^2}}}\) or \({r^{ - 2}}\), (A1) for \( - {\text{1}}\,{\text{000}}\,{\text{000}}\).

[3 marks]

\(2\pi r - \frac{{1\,000\,000}}{{{r^2}}} = 0\)    (M1)

Note:     Award (M1) for equating their part (e) to zero.

\({r^3} = \frac{{1\,000\,000}}{{2\pi }}\) OR \(r = \sqrt[3]{{\frac{{1\,000\,000}}{{2\pi }}}}\)     (M1)

Note:     Award (M1) for isolating \(r\).

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

\((r = ){\text{ }}54.2{\text{ }}({\text{cm}}){\text{ }}(54.1926 \ldots )\)    (A1)(ft)(G2)

[3 marks]

\(\pi {(54.1926 \ldots )^2} + \frac{{1\,000\,000}}{{(54.1926 \ldots )}}\)    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

\( = 27\,700{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}(27\,679.0 \ldots )\)    (A1)(ft)(G2)

[2 marks]

\(\frac{{27\,679.0 \ldots }}{{2000}}\)    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

\( = 13.8395 \ldots \)    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their \(13.8395 \ldots \) to the next integer.

[3 marks]

Represent this information on a Venn Diagram.

There were \(28\) students who used a bus to travel to school. Calculate the number of students
(i)     who travelled by car and by bus but did not walk;
(ii)    who travelled by car.

Tomoko used a bus to travel to school yesterday.

Find the probability that she also walked.

Two students are chosen at random from all \(50\) students.

Find the probability that
(i)     both students walked;
(ii)    only one of the students walked.

     (A4)
Note: Award (A1) for rectangle and three labelled intersecting circles, (A1) for \(3\), (A1) for \(5\) and \(10\), (A1) for \(7\) and \(12\).

(i)     \(28 - (10 + 3 + 7) = 8\)     (M1)(A1)(ft)(G2)

Note: Follow through from their Venn diagram.

(ii)    \(5 + 3 + 8 + 12 = 28\)     (M1)(A1)(ft)(G2)

Note: Follow through from part (b)(i) and their Venn diagram.

\({\text{P(}}\left. {{\text{walk}}} \right|{\text{bus}}) = \frac{{13}}{{28}}\) \((0.464{\text{, }}46.4\% )\) (\(0.464285 \ldots \))     (A1)(A1)(ft)(G2)

Note: Award (A1)(ft) for the numerator, (A1) for denominator.

(i)     \(\frac{{23}}{{50}} \times \frac{{22}}{{49}}\)     (A1)(M1)(M1)

Note: Award (A1) for \(23\) seen, (M1) for non replacement, (M1) for multiplying their fractions.

\( = \frac{{506}}{{2450}}\) \((0.207{\text{, }}20.7\% )\) (\(0.206530 \ldots \))     (A1)(G3)

(ii)    \(\frac{{23}}{{50}} \times \frac{{27}}{{49}} + \frac{{27}}{{50}} \times \frac{{23}}{{49}}\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for two products, (M1) for adding two products. Do not penalise in (ii) for consistent use of with replacement.

\( = \frac{{1242}}{{2450}}\) \((0.507{\text{, }}50.7\% )\) (\(0.509638 \ldots \))     (A1)(ft)(G2)

Copy the tree diagram below and fill in all the probabilities, where NL represents not late, to represent this information.

Find the probability that Geraldine travels by bus and is late.

Find the probability that Geraldine is late.

Find the probability that Geraldine travelled by train, given that she is late.

Sketch the curve of the function \(f (x) = x^3 − 2x^2 + x − 3\) for values of \(x\) from −2 to 4, giving the intercepts with both axes.

On the same diagram, sketch the line \(y = 7 − 2x\) and find the coordinates of the point of intersection of the line with the curve.

Find the value of the gradient of the curve where \(x = 1.7\) .

Award (A1) for 0.5 at B, (A1) for 0.3 at T, then (A1) for each correct pair. Accept fractions or percentages.     (A5)

[5 marks]

0.06 (accept \(0.5 \times 0.12\) or 6%)     (A1)(ft)

[1 mark]

for a relevant two-factor product, either \(C \times L\) or \(T \times L\)     (M1)

for summing three two-factor products     (M1)

\((0.2 \times 0.05 + 0.06 + 0.3 \times 0.08)\)

0.094     (A1)(ft)(G2)

[3 marks]

\(\frac{{0.3 \times 0.08}}{{0.094}}\)     (M1)(A1)(ft)

award (M1) for substituted conditional probability formula seen, (A1)(ft) for correct substitution

= 0.255     (A1)(ft)(G2)

[3 marks]

     (G3)

[3 marks]

line drawn with –ve gradient and +ve y-intercept     (G1)

(2.45, 2.11)     (G1)(G1)

[3 marks]

\(f ' (1.7) = 3(1.7)^2 - 4(1.7) + 1\)     (M1)

award (M1) for substituting in their \(f' (x)\)

2.87     (A1)(G2)

[2 marks]

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

This should have been an easy first question but, even so, there were some candidates who were unable to fill in the tree diagram correctly let alone evaluate any probabilities. The majority of candidates were confident with answering parts (a), (b) and (c) but the conditional probability question was not well answered with few candidates managing to recognise that it was a conditional type.

The curve sketching and straight line were well drawn but not all candidates indicated the intersection points with the axes. In finding the line / curve intersection some candidates did not use the intersection function on the GDC. Few candidates managed the last part. Many just chose two sets of coordinates and used the gradient formula.

Draw a Venn diagram to represent the information above.

Write down the number of students who study Music but not Economics.

There are 22 Economics students in total.

(i) Calculate the number of students who study Economics only.

(ii) Find the number of students who study none of these three subjects.

A student is chosen at random from the 100 that were asked above.

Find the probability that this student

(i) studies Economics;

(ii) studies Music and Chemistry but not Economics;

(iii) does not study either Music or Economics;

(iv) does not study Music given that the student does not study Economics.

(A1) for rectangle and three labelled circles (U need not be seen)

(A1) for 10 in the correct region

(A1) for 2, 7 and 5 in the correct regions

(A1) for 6 and 11 in the correct regions     (A4)

16     (A1)(ft)

Note: Follow through from their Venn diagram.

(i) \(10 + 7 + 2\)     (M1)

Note: Award (M1) for summing their 10, 7 and 2.

\(22 - 19\)

\(= 3\)     (A1)(ft)(G2)

Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram irrespective of whether working seen. Award a maximum of (M1)(A0) for a negative answer.

(ii) \(22 + 11+ 5 + 6\)     (M1)

Note: Award (M1) for summing 22, and their 11, 5 and 6.

\(100 - 44\)

\(= 56\)     (A1)(ft)(G2)

Note: Follow through from their diagram. Award (M1)(A1)(ft) for answers consistent with their diagram and the use of 22 irrespective of whether working seen. If negative values are used or implied award (M0)(A0).

(i) \(\frac{{22}}{{100}}\left( {\frac{{11}}{{50}},0.22,22\% } \right)\)     (A1)(G1)

(ii) \(\frac{{5}}{{100}}\left( {\frac{{1}}{{20}},0.05,5\% } \right)\)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for their 5 in numerator, (A1) for denominator.

   Follow through from their diagram.

(iii) \(\frac{{62}}{{100}}\left( {\frac{{31}}{{50}},0.62,62\% } \right)\)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for \(100 - (22 + 11 + {\text{their }}5)\), (A1) for denominator.

   Follow through from their diagram.

(iv) \(\frac{{62}}{{78}}\left( {\frac{{31}}{{39}},0.795,79.5\% } \right)\) (0.794871...)     (A1)(ft)(A1)(G2)

Note: Award (A1)(ft) for numerator, (A1) for denominator. Follow

through from part (d)(iii) for numerator.

This question divided the candidates into two parts: those who knew how to interpret the information in a manner the led to a consistent Venn diagram and those who did not. The use of the word “only” is crucial in this regard.

Follow through to the probability part of the question was contingent on the use of the given \(n(E) = 22\) ; given information should be used in subsequent parts. As ever, conditional probability proves a trial for many.

It is recommended that candidates write probabilities as unsimplified fractions as this increase their chances of gaining follow through from previous parts.

This question divided the candidates into two parts: those who knew how to interpret the information in a manner the led to a consistent Venn diagram and those who did not. The use of the word “only” is crucial in this regard.

Follow through to the probability part of the question was contingent on the use of the given \(n(E) = 22\) ; given information should be used in subsequent parts. As ever, conditional probability proves a trial for many.

It is recommended that candidates write probabilities as unsimplified fractions as this increase their chances of gaining follow through from previous parts.

This question divided the candidates into two parts: those who knew how to interpret the information in a manner the led to a consistent Venn diagram and those who did not. The use of the word “only” is crucial in this regard.

Follow through to the probability part of the question was contingent on the use of the given \(n(E) = 22\) ; given information should be used in subsequent parts. As ever, conditional probability proves a trial for many.

It is recommended that candidates write probabilities as unsimplified fractions as this increase their chances of gaining follow through from previous parts.

This question divided the candidates into two parts: those who knew how to interpret the information in a manner the led to a consistent Venn diagram and those who did not. The use of the word “only” is crucial in this regard.

Follow through to the probability part of the question was contingent on the use of the given \(n(E) = 22\) ; given information should be used in subsequent parts. As ever, conditional probability proves a trial for many.

It is recommended that candidates write probabilities as unsimplified fractions as this increase their chances of gaining follow through from previous parts.

Jorge applies a \({\chi ^2}\) test at the \(5\% \) level to investigate whether wearing a seat belt is associated with the time a driver has had their licence.

(i)     Write down the null hypothesis, \({{\text{H}}_0}\).

(ii)    Write down the number of degrees of freedom.

(iii)   Show that the expected number of drivers that wear a seat belt and have had their driving licence for more than \(15\) years is \(22\), correct to the nearest whole number.

(iv)   Write down the \({\chi ^2}\) test statistic for this data.

(v)    Does Jorge accept \({{\text{H}}_0}\) ? Give a reason for your answer.

Consider the \(200\) drivers surveyed. One driver is chosen at random. Calculate the probability that

(i)     this driver wears a seat belt;

(ii)    the driver does not wear a seat belt, given that the driver has held a licence for more than \(15\) years.

Two drivers are chosen at random. Calculate the probability that

(i)     both wear a seat belt.

(ii)    at least one wears a seat belt.

(i)    \({{\text{H}}_0} = \) wearing of a seat belt and the time a driver has held a licence are independent.     (A1)

Note: For independent accept 'not associated' but do not accept 'not related' or 'not correlated'

(ii)    \(2\)     (A1)

(iii)   \(\frac{{98 \times 45}}{{200}} = 22.05 = 22\) (correct to the nearest whole number)     (M1)(A1)(AG)

Note: (M1) for correct formula and (A1) for correct substitution. Unrounded answer must be seen for the (A1) to be awarded.

(iv)   \({\chi ^2} = 8.12\)     (G2)

Note: For unrounded answer award (G1)(G0)(AP). If formula used award (M1) for correct substituted formula with correct substitution (6 terms) (A1) for correct answer.

(v)   “Does not accept \({{\text{H}}_0}\)”     (A1)(ft)

\(p{\text{-}}value < 0.05\)     (R1)(ft)

Note: Allow “Reject \({{\text{H}}_0}\)” or equivalent. Follow through from their \({\chi ^2}\) statistic. Award (R1)(ft) for comparing the appropriate values. The (A1)(ft) can be awarded only if the conclusion is valid according to the comparison given. If no reason given or if reason is wrong the two marks are lost.

[8 marks]

(i)     \(\frac{{98}}{{200}}( = 0.49{\text{, }}49\% )\)     (A1)(A1)(G2)

Note: (A1) for numerator, (A1) for denominator.

(ii)    \(\frac{{15}}{{45}}( = 0.333{\text{, }}33.3\% )\)     (A1)(A1)(G2)

Note: (A1) for numerator, (A1) for denominator.

[4 marks]

(i)     \(\frac{{98}}{{200}} \times \frac{{97}}{{199}} = 0.239{\text{ }}(23.9\% )\)     (A1)(M1)(A1)(G3)

Note: (A1) for correct probabilities seen, (M1) for multiplying two probabilities, (A1) for correct answer.

(ii)     \(1 - \frac{{102}}{{200}} \times \frac{{101}}{{199}} = 0.741{\text{ }}(74.1\% )\)     (M1)(M1)(A1)(ft)(G2)

Note: (M1) for showing the product, (M1) for using the probability of the complement, (A1) for correct answer. Follow through for consistent use of with replacement.

OR

\(\frac{{98}}{{200}} \times \frac{{97}}{{199}} + \frac{{98}}{{200}} \times \frac{{102}}{{199}} + \frac{{102}}{{200}} \times \frac{{98}}{{199}} = 0.741{\text{ }}(74.1\% )\)     (M1)(M1)(A1)(ft)(G2)

Note: (M1) for adding three products of fractions (or equivalent), (M1) for using the correct fractions, (A1) for correct answer. Follow through for consistent use of with replacement.

[6 marks]

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.

The first part of the question was relatively well done. The null hypothesis and the degrees of freedom were well answered by the majority of the students. In the show that question some students used the GDC to find the expected values table and highlighted the correct value \(22.05\). This procedure gained no mark; the expected value formula was expected to be used here. Also those who did use the formula were expected to show the unrounded value \(22.05\) to gain full marks in this part question. Many lost the answer mark for not doing so. GDC was used by most of the students to find the chi-squared test though some students attempted to find this value by hand which made them waste time. Correct values were compared when deciding whether to accept or not the null hypothesis and follow through marks were awarded from their degrees of freedom and chi-squared test when incorrect.

The second part was not as successful as the first one. Simple probability was well answered. Not all the students changed the denominator to \(45\) for the second probability showing their weaknesses in conditional probability. It would have been useful for the students to use a tree diagram to help them solve the last part of this question but very few did so. Some of those students that reached the last part of the question forgot to add one of the three terms. Very few used the probability of the complement.

Draw a Venn diagram to show this information. Use T to represent the set of students who have a television, C the set of students who have a computer and I the set of students who have an iPhone.

Write down the number of students that

(i) have a computer only;

(ii) have an iPhone and a computer but no television.

Write down \(n[T \cap (C \cup I)']\).

Calculate the number of students who have none of the three.

Two students are chosen at random from the 450 students. Calculate the probability that

(i) neither student has an iPhone;

(ii) only one of the students has an iPhone.

The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.

In the first 10 months they collect 13 050 dollars.

(i) Write down a second equation in x and y to represent this information.

(ii) Write down the value of x and of y .

The students are asked to collect money for charity. In the first month, the students collect x dollars and the students collect y dollars in each subsequent month. In the first 6 months, they collect 7650 dollars. This can be represented by the equation x + 5y = 7650.

In the first 10 months they collect 13 050 dollars.

Calculate the number of months that it will take the students to collect 49 500 dollars.

     (A1)(A1)(A1)(A1)

Notes: Award (A1) for labelled sets T, C, and I included inside an enclosed universal set. (Label U is not essential.) Award (A1) for central entry 40. (A1) for 20, 30 and 35 in the other intersecting regions. (A1) for 60, 110 and 115 or T(150), C(205), I(220).

[4 marks]

In parts (b), (c) and (d) follow through from their diagram.

(i) 110     (A1)(ft)

(ii) 35     (A1)(ft)

[2 marks]

In parts (b), (c) and (d) follow through from their diagram.

60     (A1)(ft)

[2 marks]

In parts (b), (c) and (d) follow through from their diagram.

450 − (60 + 20 + 40 + 30 + 115 + 35 + 110)     (M1)

Note: Award (M1) for subtracting all their values from 450.

= 40     (A1)(ft)(G2)

[2 marks]

(i) \(\frac{{230}}{{450}} \times \frac{{229}}{{449}}\)     (A1)(M1)

Note: Award (A1) for correct fractions, (M1) for multiplying their fractions.

\(\frac{{52670}}{{202050}}\left( {\frac{{5267}}{{20205}},{\text{ 0}}{\text{.261, 26}}{\text{.1% }}} \right)(0.26067...)\)     (A1)(G2)

Note: Follow through from their Venn diagram in part (a).

(ii) \(\frac{{220}}{{450}} \times \frac{{230}}{{449}} + \frac{{230}}{{450}} \times \frac{{220}}{{449}}\)     (A1)(A1)

Note: Award (A1) for addition of their products, (A1) for two correct products.

OR

\(\frac{{230}}{{450}} \times \frac{{220}}{{449}} \times 2\)     (A1)(A1)

Notes: Award (A1) for their product of two fractions multiplied by 2, (A1) for correct product of two fractions multiplied by 2. Award (A0)(A0) if correct product is seen not multiplied by 2.

\(\frac{{2024}}{{4041}}(0.501,{\text{ 50}}{\text{.1% )(0}}{\text{.50086}}...{\text{)}}\)     (A1)(G2)

Note: Follow through from their Venn diagram in part (a) and/or their 230 used in part (e)(i).

Note: For consistent use of replacement in parts (i) and (ii) award at most (A0)(M1)(A0) in part (i) and (A1)(ft)(A1)(A1)(ft) in part (ii).

[6 marks]

(i) x + 9y = 13050     (A1)

(ii) x = 900     (A1)(ft)

y = 1350     (A1)(ft)

Notes: Follow through from their equation in (f)(i). Do not award (A1)(ft) if answer is negative. Award (M1)(A0) for an attempt at solving simultaneous equations algebraically but incorrect answer obtained.

[3 marks]

49500 = 900 + 1350n     (A1)(ft)

Notes: Award (A1)(ft) for setting up correct equation. Follow through from candidate’s part (f).

n = 36     (A1)(ft)

The total number of months is 37.     (A1)(ft)(G2)

Note: Award (G1) for 36 seen as final answer with no working. The value of n must be a positive integer for the last two (A1)(ft) to be awarded.

OR

49500 = 900 + 1350(n − 1)     (A2)(ft)

Notes: Award (A2)(ft) for setting up correct equation. Follow through from candidate’s part (f).

n = 37     (A1)(ft)(G2)

Note: The value of n must be a positive integer for the last (A1)(ft) to be awarded.

[3 marks]

The question was moderately well answered. The majority of candidates answered part (a) and at least parts of (b), and (d).

The question was moderately well answered. The majority of candidates answered part (a) and at least parts of (b), and (d).

The question was moderately well answered. Part (c) proved to be difficult, as it required understanding and interpreting set notation.

The question was moderately well answered. The majority of candidates answered part (a) and at least parts of (b), and (d).

The question was moderately well answered. Part (e) was rarely answered in its entirety.

The question was moderately well answered. Part (f) was answered by many candidates, but most of them offered a partial answer to part (g); a typical response was 36 instead of 37.

The question was moderately well answered. Part (f) was answered by many candidates, but most of them offered a partial answer to part (g); a typical response was 36 instead of 37.

Find the total number of fish in the net.

Find (i) the modal length interval,

(ii) the interval containing the median length,

(iii) an estimate of the mean length.

(i) Write down an estimate for the standard deviation of the lengths.

(ii) How many fish (if any) have length greater than three standard deviations above the mean?

The fishing company must pay a fine if more than 10% of the catch have lengths less than 40cm.

Do a calculation to decide whether the company is fined.

A sample of 15 of the fish was weighed. The weight, W was plotted against length, L as shown below.

Exactly two of the following statements about the plot could be correct. Identify the two correct statements.

Note: You do not need to enter data in a GDC or to calculate r exactly.

(i) The value of r, the correlation coefficient, is approximately 0.871.

(ii) There is an exact linear relation between W and L.

(iii) The line of regression of W on L has equation W = 0.012L + 0.008 .

(iv) There is negative correlation between the length and weight.

(v) The value of r, the correlation coefficient, is approximately 0.998.

(vi) The line of regression of W on L has equation W = 63.5L + 16.5.

Total = 2 + 3 + 5 + 7 + 11 + 5 + 6 + 9 + 2 + 1     (M1)

(M1) is for a sum of frequencies.

= 51     (A1)(G2)

[2 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(i) modal interval is 60 – 70

Award (A0) for 65     (A1)

(ii) median is length of fish no. 26,     (M1)(A1)

also 60 – 70     (G2)

Can award (A1)(ft) or (G2)(ft) for 65 if (A0) was awarded for 65 in part (i).

(iii) mean is \(\frac{{2 \times 25 + 3 \times 35 + 5 \times 45 + 7 \times 55 + ...}}{{51}}\)     (M1)

(UP) = 69.5 cm (3sf)     (A1)(ft)(G1)

Note: (M1) is for a sum of (frequencies multiplied by midpoint values) divided by candidate’s answer from part (a). Accept mid-points 25.5, 35.5 etc or 24.5, 34.5 etc, leading to answers 70.0 or 69.0 (3sf) respectively. Answers of 69.0, 69.5 or 70.0 (3sf) with no working can be awarded (G1).

[5 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) (i) standard deviation is 21.8 cm     (G1)

For any other answer without working, award (G0). If working is present then (G0)(AP) is possible.

(ii) \(69.5 + 3 \times 21.8 = 134.9 > 120\)     (M1)

no fish     (A1)(ft)(G1)

For ‘no fish’ without working, award (G1) regardless of answer to (c)(i). Follow through from (c)(i) only if method is shown.

[3 marks]

5 fish are less than 40 cm in length,     (M1)

Award (M1) for any of \(\frac{5}{51}\), \(\frac{46}{51}\), 0.098 or 9.8%, 0.902, 90.2% or 5.1 seen.

hence no fine.     (A1)(ft)

Note: There is no G mark here and (M0)(A1) is never allowed. The follow-through is from answer in part (a).

[2 marks]

(i) and (iii) are correct.     (A1)(A1)

[2 marks]

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

a) b), c) There was much confusion about how to present the intervals. Often the mid-point only was seen. (eg. 65 instead of 60-70). Understanding of mode, median and mean was usually good but too many candidates wasted time calculating standard deviation by hand and got it wrong. In c(ii) 'greater than three' caused no problems but 'above the mean' was often ignored.

d) This was often well done, even if earlier parts were poorly done.

e) Rather mixed performance here. It was hard to identify any consistency in the errors made.

Too much time was spent on this question. It was only worth two marks and candidates ought to have realised that it relied on a general pictorial understanding of the concepts, possibly supplemented by a little elementary arithmetic only, to compare (iii) and (vi). With good understanding, many of the options could be ruled out in a few seconds.

Write down the value of a.

Write down the value of b.

Find the probability that both biscuits are plain.

Copy and complete the tree diagram below.

Find the probability that he chooses a chocolate biscuit.

Find the probability that he chooses a biscuit from the red tin given that it is a chocolate biscuit.

a = 0 \(\left( {\frac{0}{4}} \right)\)     (A1)

[1 mark]

\(b = \frac{3}{4}(0.75,{\text{ }}75\% )\)     (A2)(G2)

[2 marks]

\(\frac{4}{5} \times \frac{3}{4}\)     (M1)(A1)

\(\frac{{12}}{{20}}\left( {\frac{3}{5},{\text{ }}0.6,{\text{ 60% }}} \right)\)     (A1)(ft)(G2)

Note: Award (M1) for multiplying two probabilities, (A1) for using their probabilities, (A1) for answer.

[3 marks]

     (A1)(A1)(A1)

Note: Award (A1) for each pair.

[3 marks]

\(\frac{1}{2} \times \frac{3}{{10}} + \frac{1}{2} \times \frac{1}{5}\)     (M1)(M1)

\( = \frac{5}{{20}}\left( {\frac{1}{4},{\text{ }}0.25,{\text{ 25% }}} \right)\)     (A1)(ft)(G2)

Note: Award (M1) for two products seen with numbers from the problem, (M1) for adding two products. Follow through from their tree diagram.

[3 marks]

\(\frac{{\frac{1}{2} \times \frac{3}{{10}}}}{{\frac{1}{4}}}\)     (M1)(A1)

\( = \frac{3}{5}{\text{ }}\left( {0.6,{\text{ 60 }}\% } \right)\)     (A1)(ft)(G2)

Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.

Follow through from their part (b) and part (c) (i).

[3 marks]

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

This question was well handled by most of the candidates except for (c)(ii) in which they had to find a conditional probability. Some candidates did not copy the second tree diagram in the answer sheets and instead wrote their answers in the exam booklet thus losing the 3 marks allocated to part (b).

Use the table to find the probability that

(i) a television will be bought by a female;

(ii) a television with a screen size of 32 < S ≤ 46 will be bought;

(iii) a television with a screen size of 32 < S ≤ 46 will be bought by a female;

(iv) a television with a screen size greater than 46 inches will be bought, given that it is bought by a male.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Write down the null hypothesis.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Show that the expected frequency for females who bought a screen size of 32 < S ≤ 46, is 79, correct to the nearest integer.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Write down the number of degrees of freedom.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Write down the \({\chi ^2}\) calculated value.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Write down the critical value for this test.

The manager of the store wants to determine whether the screen size is independent of gender. A Chi-squared test is performed at the 1 % significance level.

Determine if the null hypothesis should be accepted. Give a reason for your answer.

(i) \(\frac{{220}}{{500}}\left( {\frac{{11}}{{25}},{\text{ 0}}{\text{.44, 44}}\% } \right)\)     (A1)(G1)

(ii) \(\frac{{180}}{{500}}\left( {\frac{{9}}{{25}},{\text{ 0}}{\text{.36, 36}}\% } \right)\)     (A1)(G1)

(iii) \(\frac{{40}}{{500}}\left( {\frac{{2}}{{25}},{\text{ 0}}{\text{.08, 8}}\% } \right)\)     (A1)(A1)(G2)

(iv) \(\frac{{55}}{{500}}\left( {\frac{{11}}{{56}},{\text{ 0}}{\text{.196, 19.6}}\% } \right)\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator. Award (A0)(A0) if answers are given as incorrect reduced fractions without working.

[6 marks]

“The size of the television screen is independent of gender.”     (A1)

Note: Accept “not associated”, do not accept “not correlated”.

[1 mark]

\(\frac{{180}}{{500}} \times \frac{{220}}{{500}} \times 500\) OR \(\frac{{180 \times 220}}{{500}}\)     (M1)

= 79.2     (A1)

= 79     (AG)

Note: Both the unrounded and the given answer must be seen for the final (A1) to be awarded.

[2 marks]

3     (A1)

[1 mark]

\(\chi _{calc}^2\) = 104(103.957...)     (G2)

Note: Award (M1) if an attempt at using the formula is seen but incorrect answer obtained.

[2 marks]

11.345     (A1)(ft)

Notes: Follow through from their degrees of freedom.

[1 mark]

\(\chi _{calc}^2\) > \(\chi _{crit}^2\) OR p < 0.01     (R1)

Do not accept H₀.     (A1)(ft) 

Note: Do not award (R0)(A1)(ft). Follow through from their parts (d), (e) and (f).

[2 marks]

Part (a) was generally well answered by most of the students, except for part (a)(iv) which called for conditional probability.

Most students correctly stated the null hypothesis in part (b), and answered parts (d), (e), (f) and (g).

In some responses to part (c) it seemed that the difference between calculation of the expected value and showing that the value is 79 was not clear to the candidates. It is important that teachers explain to their students that in a “show that” question they are expected to demonstrate the mathematical reasoning through which the given answer is obtained.

Most students correctly stated the null hypothesis in part (b), and answered parts (d), (e), (f) and (g).

Most students correctly stated the null hypothesis in part (b), and answered parts (d), (e), (f) and (g).

Most students correctly stated the null hypothesis in part (b), and answered parts (d), (e), (f) and (g).

Most students correctly stated the null hypothesis in part (b), and answered parts (d), (e), (f) and (g).

Represent this information on a Venn diagram.

Calculate the number of students who own none of the items mentioned above.

If a student is chosen at random, write down the probability that the student owns a digital camera only.

If two students are chosen at random, calculate the probability that they both own a cell phone only.

If a student owns an iPod, write down the probability that the student also owns a digital camera.

Copy and complete the probability tree diagram below.

Calculate the probability that

(i)     Kate goes to the cinema and is not late;

(ii)    the person who goes to the cinema arrives home late.

     (A1)(A1)(A1)(A1)(ft)

Note: (A1) for rectangle with 3 intersecting circles, (A1) for \(1\), (A1) for \(5\), \(3\), \(2\), (A1)(ft) for \(9\), \(7\), \(20\) if subtraction is carried out, or \(18\), \(15\), \(26\) seen by the letters D, I and C.

[4 marks]

\(50 - 47\)     (M1)

Note: (M1) for subtracting their value from \(50\).

\( = 3\)     (A1)(ft)(G2)

[2 marks]

\(\frac{9}{{50}}\)     (A1)(ft)

[1 mark]

\(\frac{{20}}{{50}} \times \frac{{19}}{{49}}\)     (A1)(ft)(M1)

\( = \frac{{38}}{{245}}{\text{ }}\left( {\frac{{380}}{{2450}}{\text{, }}0.155{\text{, }}15.5\% } \right)\)     (A1)(ft)(G2)

Notes: (A1)(ft) for correct fractions from their Venn diagram
(M1) for multiplying their fractions
(A1)(ft) for correct answer.

[3 marks]

\(\frac{6}{{15}}{\text{ }}\left( {\frac{2}{5}{\text{, }}0.4{\text{, }}40\% } \right)\)     (A1)(ft)(A1)(ft)

Note: (A1)(ft) for numerator, (A1)(ft) denominator.

[2 marks]

     (A1)(A1)(A1)

Note: (A1) for \(0.8\), (A1) for \(0.7\), (A1) for \(0.6\) and \(0.4\).

[3 marks]

(i)     \(0.2 \times 0.7 = 0.14\)     (M1)(A1)(ft)(G2)

Note: (M1) for multiplying correct numbers.

[2 marks]

(ii)    \(0.2 \times 0.3 + 0.8 \times 0.6\)     (M1)(M1)

\( = 0.54\)     (A1)(ft)(G3)

Note: (M1) for each correct product (use candidate’s tree), (A1)(ft) for answer.

[3 marks]

The Venn diagram was well drawn on the whole although some of the candidates missed out the Universal box and others filled in the intersections wrongly but still gained ft marks for the remaining parts of the question.

Well answered.

Well answered.

Few correct answers. Either candidates added instead of multiplying or they used replacement and so the fractions given were the same.

Again few correct answers. Candidates wrote the answer out of \(50\) instead of \(15\).

The tree diagram was well done on the whole. It appears as if some candidates may have completed this on the exam paper and this was not included with their papers. However, the question did state clearly “Copy and complete …”

(i)     This part was well done by those candidates who remembered to multiply instead of add.

(ii)    Many candidates just wrote down “Claire” for this answer. Others wrongly multiplied or added \(0.3\) with \(0.6\).

Draw a Venn Diagram, using sets labelled \(A\) , \(S\) and \(P\) , that shows this information.

There were 48 customers of Pete’s Eats that day. Calculate the number of people who were customers of all three cafés.

There were 50 customers of Sarah’s Snackbar that day. Calculate the total number of people who were customers of Alan’s Diner.

Write down the number of customers of Alan’s Diner that were also customers of Pete’s Eats.

Find \(n[(S \cup P) \cap A']\).

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”;

One of the survey forms was chosen at random, find the probability that the form showed “Satisfied” and was completed at Sarah’s Snackbar;

One of the survey forms was chosen at random, find the probability that the form showed “Dissatisfied”, given that it was completed at Alan’s Diner.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the null hypothesis, \({{\text{H}}_0}\) , for the \({\chi ^2}\) test.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Write down the number of degrees of freedom for the test.

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

Using your graphic display calculator, find \({\chi ^2}_{calc}\) .

A \({\chi ^2}\) test at the \(5\% \) significance level was carried out to determine whether there was any difference in the level of customer satisfaction in each of the cafés.

State, giving a reason, the conclusion to the test.

(A1) for rectangle and three labelled intersecting circles
(A1) for \(15\), \(27\) and \(17\)
(A1) for \(10\) and \(8\)     (A3)

[3 marks]

\(48 - (8 +10 +17)\) or equivalent     (M1)

\( = 13\)     (A1)(ft)(G2)

[2 marks]

\(50 - (27 +10 +13)\)     (M1)

Note: Award (M1) for working seen.

\( = 0\)     (A1)
number of elements in A \(= 36\)     (A1)(ft)(G3)

Note: Follow through from (b).

[3 marks]

\(21\)     (A1)(ft)

Note: Follow through from (b) even if no working seen.

[1 mark]

\(54\)     (M1)(A1)(ft)(G2)

Note: Award (M1) for \(17\), \(10\), \(27\) seen. Follow through from (a).

[2 marks]

\(\frac{{40}}{{120}}{\text{ }}\left( {\frac{1}{3}{\text{, }}0.333{\text{, }}33.3\% } \right)\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[2 marks]

\(\frac{{34}}{{120}}{\text{ }}\left( {\frac{{17}}{{60}}{\text{, }}0.283{\text{, }}28.3\% } \right)\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[2 marks]

\(\frac{8}{{28}}{\text{ }}\left( {\frac{2}{7}{\text{, }}0.286{\text{, }}28.6\% } \right)\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[2 marks]

customer satisfaction is independent of café     (A1)

Note: Accept “customer satisfaction is not associated with the café”.

[1 mark]

\(2\)     (A1)

[1 mark]

\(0.754\)     (G2)

Note: Award (G1)(G1)(AP) for \(0.75\) or for correct answer incorrectly rounded to 3 s.f. or more, (G0) for \(0.7\).

[2 marks]

since \({\chi ^2}_{calc} < {\chi ^2}_{crit}5.991 accept (or Do not reject) H₀     (R1)(A1)(ft)

Note: Follow through from their value in (e).

OR

Accept (or Do not reject) H₀ as \(p\)-value \((0.686) > 0.05\)     (R1)(A1)(ft)

Notes: Do not award (A1)(R0). Award the (R1) for comparison of appropriate values.

[2 marks]

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

Part A

This part was successfully attempted by the great majority. A common mistake was the failure to intersect all three sets.

A surprising number seemed unfamiliar with set notation in (e) and thus did not attempt this part.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Part B

The work on probability also proved accessible to the great majority with a large number of candidates attaining full marks. Most errors occurred due to candidates trying to use the algebraic form of laws of probability, rather than by interpreting the contingency table.

The chi-squared test was well done by the great majority, however, it was clear that a number of centres do not teach this subject, since there were a number of scripts which either were left blank or showed no understanding in the responses seen.

Copy and complete the Venn diagram on your answer paper.

Write down the number of students who study Mathematics but not Further Mathematics.

Write down the total number of students in the school.

Write down \(n({\text{B}} \cup {\text{C}})\).

Write the following compound statement in symbolic form.

“It is snowing and the roads are not open.”

Write the following compound statement in words.

\((\neg p \wedge q) \Rightarrow r\)

An incomplete truth table for the compound proposition \((\neg p \wedge q) \Rightarrow r\) is given below.

Copy and complete the truth table on your answer paper.

     (A1)(A1)(A1)

Note: Award (A1) for each correct number in the correct position.

[3 marks]

28     (A1)(ft)

Note: 20 + their 8.

[1 mark]

59     (A1)(ft)

[1 mark]

10 + 12 + 20 + 6     (M1)

Note: Award (M1) for use of the correct regions.

= 48     (A1)(ft)(G2)

OR

59 − 8 − 3     (M1)

= 48     (A1)(ft)

[2 marks]

\(p \wedge \neg q\)     (A1)(A1)

Note: Award (A1) for \(\wedge\), (A1) for both statements in the correct order.

[2 marks]

If it is not snowing and the roads are open (then) we will go skiing.     (A1)(A1)(A1)

Note: Award (A1) for “if…(then)”, (A1) for “not snowing and the roads are open”, (A1) for “we will go skiing”.

[3 marks]

     (A1)(A1)(ft)(A1)(ft)

Note: Award (A1) for each correct column.

[3 marks]

This part was successfully attempted by the great majority. The less familiar form of the Venn diagram seemed not to cause too many problems, although a common mistake was the failure to add the 20 in set A in part (b). A surprising number seemed unfamiliar with set notation in (d) and thus were not able to attempt this part.

This part was successfully attempted by the great majority. The less familiar form of the Venn diagram seemed not to cause too many problems, although a common mistake was the failure to add the 20 in set A in part (b). A surprising number seemed unfamiliar with set notation in (d) and thus were not able to attempt this part.

This part was successfully attempted by the great majority. The less familiar form of the Venn diagram seemed not to cause too many problems, although a common mistake was the failure to add the 20 in set A in part (b). A surprising number seemed unfamiliar with set notation in (d) and thus were not able to attempt this part.

This part was successfully attempted by the great majority. The less familiar form of the Venn diagram seemed not to cause too many problems, although a common mistake was the failure to add the 20 in set A in part (b). A surprising number seemed unfamiliar with set notation in (d) and thus were not able to attempt this part.

The work on logic also proved accessible to the great majority with a large number of candidates attaining full marks. The most common errors were the omission of the “If” in the conditional statement in (b) and the inability to follow the implication in the truth table in (c).

The work on logic also proved accessible to the great majority with a large number of candidates attaining full marks. The most common errors were the omission of the “If” in the conditional statement in (b) and the inability to follow the implication in the truth table in (c).

The work on logic also proved accessible to the great majority with a large number of candidates attaining full marks. The most common errors were the omission of the “If” in the conditional statement in (b) and the inability to follow the implication in the truth table in (c).

Represent this information on a Venn diagram.

Find the number of students who had none of the three choices for breakfast.

Write down the percentage of students who had fruit for breakfast.

Describe in words what the students in the set \(X \cap Y'\) had for breakfast.

Find the probability that a student had at least two of the three choices for breakfast.

Two students are chosen at random. Find the probability that both students had all three choices for breakfast.

Write down the null hypothesis, H₀, for this test.

Write down the number of degrees of freedom for this test.

Write down the critical value for this test.

Show that the expected number of females that have more than 5 meals per day is 13, correct to the nearest integer.

Use your graphic display calculator to find the \(\chi _{calc}^2\) for this data.

Decide whether H₀ must be accepted. Justify your answer.

(A1) for rectangle and three intersecting circles

(A1) for 10, (A1) for 8, 13 and 9, (A1) for 12, 15 and 17     (A4)

[4 marks]

100 – (9 +12 +13 +15 +10 +17 + 8) =16     (M1)(A1)(ft)(G2) 

Note: Follow through from their diagram.

[2 marks]

\(\frac{{51}}{{100}}(0.51)\)     (A1)(ft)

= 51%     (A1)(ft)(G2)

Note: Follow through from their diagram.

[2 marks]

Note: The following statements are correct. Please note that the connectives are important. It is not the same (had cereal) and (not bread) and (had cereal) or (not bread). The parentheses are not needed but are there to facilitate the understanding of the propositions.

(had cereal) and (did not have bread)

(had cereal only) or (had cereal and fruit only)

(had either cereal or (fruit and cereal)) and (did not have bread)     (A1)(A1)

Notes: If the statements are correct but the connectives are wrong then award at most (A1)(A0). For the statement (had only cereal) and (cereal and fruit) award (A1)(A0). For the statement had cereal and fruit award (A0)(A0).

[2 marks]

\(\frac{{54}}{{100}}(0.54,{\text{ 54 % }})\)     (A1)(ft)(A1)(ft)(G2)

Note: Award (A1)(ft) for numerator, follow through from their diagram, (A1)(ft) for denominator. Follow through from total or denominator used in part (c).

[2 marks]

\(\frac{{10}}{{100}} \times \frac{9}{{99}} = \frac{1}{{110}}(0.00909,{\text{ 0}}{\text{.909 % }})\)     (A1)(ft)(M1)(A1)(ft)(G2)

Notes: Award (A1)(ft) for their correct fractions, (M1) for multiplying two fractions, (A1)(ft) for their correct answer. Answer 0.009 with no working receives no marks. Follow through from denominator in parts (c) and (e) and from their diagram.

[3 marks]

H₀ : The (average) number of meals per day a student has and gender are independent     (A1)

Note: For “independent” accept “not associated” but do not accept “not related” or “not correlated”.

[1 mark]

2     (A1)

[1 mark]

5.99 (accept 5.991)     (A1)(ft)

Note: Follow through from their part (b).

[1 mark]

\(\frac{{28 \times 45}}{{100}} = 12.6 = 13\) or \(\frac{{28}}{{100}} \times \frac{{25}}{{100}} \times 100 = 12.6 = 13\)     (M1)(A1)(AG)

Notes: Award (M1) for correct formula and (A1) for correct substitution. Unrounded answer must be seen for the (A1) to be awarded.

[2 marks]

0.0321      (G2)

Note: For 0.032 award (G1)(G1)(AP). For 0.03 with no working award (G0).

[2 marks]

0.0321 < 5.99 or 0.984 > 0.05     (R1)

accept H₀     (A1)(ft)

Note: If reason is incorrect both marks are lost, do not award (R0)(A1).

[2 marks]

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

This question was in general well done. Candidates began the paper well by drawing the Venn diagram correctly. Some students omitted the rectangle (universal set) around the three circles. There were quite a few errors in (c) as some students forgot to convert their answers to percentages. Also describing in words what the students in \(X \cap Y'\) had for breakfast seemed to be difficult for the majority of the candidates. Some misread what Y was and even more missed the complement sign. However, the main problem in answering this question seemed to be the lack of knowledge in the relationship between set theory and logic (use of "and" and "or"). Combining probabilities caused problems to many. Common wrong answers were \(\frac{{10}}{{100}}\), \(\frac{{10}}{{100}} \times \frac{{10}}{{100}}\) or \(\frac{{10}}{{100}} + \frac{9}{{99}}\).

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

In general this part question was well answered. The major concerns of the examining team were the following:

• In (f) many students wrote down the expected values table (from the GDC) and highlighted the correct expected value, 12.6. As this is a "show that" question the use of the GDC is not expected and therefore no marks are awarded for this working. Instead it is expected the use of the formula for the expected value with the correct substitutions.
• In (e) surprisingly many candidates found the \(\chi _{calc}^2\) through the use of the formula. Unfortunately this led to some incorrect answers and also to a bad use of time. The question clearly says "use your graphic display calculator" and it is worth 2 marks therefore a student should not spend more than 2 minutes to answer this part question. Time management is essential in this type of examinations and the IB rule is one minute – one mark.

Copy and complete the tree diagram to represent this information.

Find the probability that Lisa cooks dinner and they do not have pasta.

Find the probability that they do not have pasta.

Given that they do not have pasta, find the probability that Lisa cooked dinner.

Write down \(n(R \cap S \cap B)\).

Find \(n(R')\).

Describe which groups the pupils in the set \(S \cap B\) like.

Use set notation to describe the group of pupils who like the Rockers and the Bluers but do not like the Salseros.

There are 33 pupils in the class.

Find x.

There are 33 pupils in the class.

Find the number of pupils who like the Rockers.

Note: Award (A1) for each correct pair.     (A3)

[3 marks]

\(0.7 \times 0.88 = 0.616{\text{   }}\left( {\frac{{77}}{{125}},{\text{ }}61.6{\text{ }}\% } \right)\)    (M1)(A1)(ft)(G2)

Note: Award (M1) for multiplying the correct probabilities.

[2 marks]

\(0.3 \times 0.25 + 0.7 \times 0.88\)     (M1)(M1)

Notes: Award (M1) for a relevant two-factor product, could be \(S \times NP\) OR \(L \times NP\).

Award (M1) for summing 2 two-factor products.

\({\text{P}} = 0.691{\text{   }}\left( {\frac{{691}}{{1000}},{\text{ }}69.1{\text{ }}\% } \right)\)     (A1)(ft)(G2)

Notes: (ft) from their answer to (b).

[3 marks]

\(\frac{{0.616}}{{0.691}}\)     (M1)(A1)

Note: Award (M1) for substituted conditional probability formula, (A1) for correct substitution.

\({\text{P}} = 0.891{\text{   }}\left( {\frac{{616}}{{691}},{\text{ }}89.1{\text{ }}\% } \right)\)     (A1)(ft)(G2)

[3 marks]

3     (A1)

[1 mark]

For 5, 4, 7 (0) seen with no extra values     (A1)

16     (A1)(G2)

[2 marks]

They like (both) the Salseros (S) and they like the Bluers (B)     (A1)(A1)

Note: Award (A1) for “and”, (A1) for the correct groups.

[2 marks]

\(R \cap B \cap S'\)     (A1)(A1)

Note: Award (A1) for \(R \cap B\), (A1) for \( \cap S'\)

[2 marks]

\(21+ 3x = 33\)     (M1)

\(x = 4\)     (A1)(G2)

[2 marks]

17     (A1)(ft)

[1 mark]

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

The tree diagram was quite well answered by many students, but sometimes it was missing on many papers. It seemed they had it on their examination paper because the subsequent questions were answered accurately. Conditional probability was of great difficulty to many candidates.

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

This question was well handled although part (d) proved too difficult for many candidates and demonstrated, overall, a poor level of understanding of basic set notation. Students generally had the algebraic skills required to solve for x in part (e)(ii).

Write down the null hypothesis, H₀, for this test.

Find the expected value of female footballers.

Write down the number of degrees of freedom.

Write down the critical value of \(\chi ^2\), at the 5 % level of significance.

Use your graphic display calculator to determine the \(\chi _{calc}^2\) value.

Determine whether H₀ should be accepted. Justify your answer.

One student is chosen at random from the 120 students.

Find the probability that this student

(i) is male;

(ii) plays tennis.

Two students are chosen at random from the 120 students.

Find the probability that

(i) both play football;

(ii) neither play basketball.

H₀ : Gender and choice of afterschool sport are independent.     (A1)

Note: Accept “not associated”, do not accept “not related”, “not correlated”, or “not linked”. Accept “the relation between  gender and sport is independent”.

[1 mark]

\(\frac{{85}}{{120}} \times \frac{{48}}{{120}} \times 120\left( {\frac{{85 \times 48}}{{120}}} \right)\)     (M1)

Note: Award (M1) for correct expression.

= 34     (A1)(G2)

[2 marks]

2     (A1)

[1 mark]

5.99 (5.991)     (A1)(ft)

Note: Follow through from part (c).

[1 mark]

2.42 (2.42094…)     (G2)

[2 marks]

Since 2.42 < 5.99 therefore accept (do not reject) H₀     (R1)(A1)(ft)

Note: The numerical values need not be seen, but must be consistent with their parts (d) and (e).

OR

p-value 0.298 > 0.05 therefore accept (do not reject) H₀     (R1)(A1)

Note: p-value comparison may not be used as part of a follow through solution. Do not award (A1)(R0). Follow through from parts (c), (d) and (e).

[2 marks]