Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 5 · Variant 2

1 A fair six-sided die, with faces marked 1, 2, 3, 4, 5, 6, is thrown repeatedly until a 4 is obtained. (a) Find the probability that obtaining a 4 requires fewer than 6 throws. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ On another occasion, the die is thrown 10 times. (b) Find the probability that a 4 is obtained at least 3 times. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) 5 5 1 6   −    or 2 3 4 1 5 1 5 1 5 1 5 1 6 6 6 6 6 6 6 6 6       + × + × + × + ×             or p + pq + pq2+pq3+ pq4 (+ pq5) 0 < p < 1, p + q = 1, 0·598, 4651 7776 A1 2 1(b) (1 – P(0, 1, 2)) 1 – 10 5 6        +10C1 9 1 5 6 6   +       10C2 2 8 1 5 6 6              M1 10Cx ( ) 10 1 , − − x x p p 0 < p < 1, any p, x ≠ 0,10 1 – (0·1615056 + 0·3230111 + 0·290710) A1 Correct expression, accept unsimplified, condone omission of final bracket 0·225 A1 0·2247 < p ≤ 0·225, WWW 3

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3 Pia runs 2 km every day and her times in minutes are normally distributed with mean 10.1 and standard deviation 1.3. (a) Find the probability that on a randomly chosen day Pia takes longer than 11.3 minutes to run 2 km. 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(b) On 75% of days, Pia takes longer than t minutes to run 2 km. Find the value of t. 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(c) On how many days in a period of 90 days would you expect Pia to take between 8.9 and 11.3 minutes to run 2 km? 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Mark scheme: 3(a) P(X > 11.3) = P( 11.3 10.1 1.3 − > z ) = P( 0 9231) > ⋅ z M1 Using ± standardisation formula, no 2 or σ σ , continuity correction 1 – 0.822 M1 Appropriate area Φ , from standardisation formula P(z>…) in final solution 0·178 A1 0.1779… 3 3(b) 0 674 = −⋅ z B1 ±0.674 seen (critical value) 10.1 0 674 1.3 − = −⋅ t M1 An equation using ±standardisation formula with a z-value, condone 2 or σ σ , continuity correction. t = 9·22 A1 AWRT. Only dependent on M1 3 Question Answer Marks Guidance 3(c) P(8.9 < X < 11.3) = 1 – 2 × their 3(a) ≡ 2(1 – their 3(a)) – 1 ≡ 2(0·5 – their 3(a)) =0.644 B1 FT FT from their 3(a) < 0·5 or correct, accept unevaluated probability OE Number of days = 90 × 0·644 = 57·96 M1 90 × their p seen, 0 < p < 1 So 57 (days) A1 FT Accept 57 or 58, not 57·0 or 58·0, no approximation/rounding stated FT must be an integer value Alternative method for question 3(c) P 8 9 10 1 11.3 10.1 1 3 1 3 ⋅− ⋅ −   < <   ⋅ ⋅   z ( ) ( ) ( ) Φ 0 9231 1 Φ 0 9231 = ⋅ − − ⋅ oe = 0·822 – (1 – 0·822) = 0·644 B1 Accept unevaluated probability Number of days = 90 × 0·644 = 57·96 M1 90 × their p seen, 0 < p < 1 So 57 (days) A1 FT Accept 57 or 58, not 57·0 or 58·0, no approximation/rounding stated FT must be an integer value 3

More questions on The normal distribution

5 The following table gives the weekly snowfall, in centimetres, for 11 weeks in 2018 at two ski resorts, Dados and Linva. Dados 6 8 12 15 10 36 42 28 10 22 16 Linva 2 11 15 16 0 32 36 40 10 12 9 (a) Represent the information in a back-to-back stem-and-leaf diagram. [4] (b) Find the median and the interquartile range for the weekly snowfall in Dados. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) The median, lower quartile and upper quartile of the weekly snowfall for Linva are 12, 9 and 32 cm respectively. Use this information and your answers to part (b) to compare the central tendency and the spread of the weekly snowfall in Dados and Linva. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Dados Linva 8 6 0 0 2 9 6 5 2 0 0 1 0 1 2 5 6 8 2 2 6 3 2 6 2 4 0 KEY 6| 3| 2 means 36 cm (snow) in Dados and 32 cm (snow) in Linva B1 Correct stem can be upside down, ignore extra values B1 Correct Dados labelled, leaves in order and lined up vertically (less than midway to next column), no commas etc, no extra terms B1 Correct Linva on opposite side of stem labelled, leaves in order and lined up vertically (less than midway to next column), no commas etc, no extra terms B1 Correct single key for their diagram, need both resorts identified and ‘cm’ stated at least once here or in leaf headings or title. SC If 2 separate diagrams drawn, SCB1 if both keys meet these criteria B0B1B0SCB1 max. 4 5(b) Median or Q2 = 15 (cm) B1 Correct UQ or Q3 = 28 cm, LQ or Q1 = 10 cm IQR = 28 – 10 M1 22 ⩽ UQ ⩽ 36 – 8 ⩽ LQ ⩽ 10 18 (cm) A1 WWW 3 5(c) On average the snowfall in Davos is higher B1 FT FT from their 5(b) values for Dados. Statement comparing central tendency in context The amount of snowfall in Linva varies more than in Davos B1 FT Statement comparing spread in context Note: simply stating and comparing the values is not sufficient. 2

More questions on Representation of data

6 Mr and Mrs Ahmed with their two children, and Mr and Mrs Baker with their three children, are visiting an activity centre together. They will divide into groups for some of the activities. (a) In how many ways can the 9 people be divided into a group of 6 and a group of 3? [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 5 of the 9 people are selected at random for a particular activity. (b) Find the probability that this group of 5 people contains all 3 of the Baker children. 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All 9 people stand in a line. (c) Find the number of different arrangements in which Mr Ahmed is not standing next to Mr Baker. 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(d) Find the number of different arrangements in which there is exactly one person between Mr Ahmed and Mr Baker. 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Mark scheme: 6(a) Condone 9C6 + 3C3, 9P6 × 3P3 84 A1 Accept unevaluated. 2 6(b) Number with 3 Baker children = 6C2 or 15 B1 Correct seen anywhere, not multiplied or added Total no of selections = 9C5 or 126 Probability = number of selections with 3 Baker children total number of selections M1 Seen as denominator of fraction 15 126 , 0·119 A1 OE, e.g. 5 42 Alternative method for question 6(b) 5 3 3 2 1 6 5 9 8 7 6 5    × × × × ×       C B1 5C3 (OE) or 10 seen anywhere, multiplied by fractions only, not added M1 3 2 1 6 5 9 8 7 6 5    × × × × ×       k , 1 ⩽ k, k integer 15 126 , 0·119 A1 OE, e.g. 5 42 3 Question Answer Marks Guidance 6(c) [Total no of arrangements = 9!] [Arrangements with men together = 8! × 2] Not together: 9! – M1 9! – k or 362880 – k, k an integer<362 880 8! × 2 B1 8! × 2(!) or 80 640 seen anywhere 282 240 A1 Exact value Alternative method for question 6(c) 7! × 8 × 7 B1 7! × k, k positive integer > 1 M1 m × 8 × 7, m × 8P2, m × 8C2 m positive integer > 1 282 240 A1 Exact value 3 6(d) 7! × 2 × 7 M1 7! × k, k positive integer > 1 If 7! not seen, condone 7 × 6 × 5 × 4 × 3 × 2 × (1) × k or 7 × 6! × k only M1 m × 2 × 7, m positive integer > 1 70 560 A1 3

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