Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 5 · Variant 3

1 The times taken to swim 100 metres by members of a large swimming club have a normal distribution with mean 62 seconds and standard deviation 5 seconds. (a) Find the probability that a randomly chosen member of the club takes between 56 and 66 seconds to swim 100 metres. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) 13% of the members of the club take more than t minutes to swim 100 metres. Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) P(56 < X < 66) = P 56 62 66 62 5 5 z − −   < <     = P( 1.2 0.8) − < < z 2 or σ σ , allow continuity correction ( ) ( ) Φ 0.8 Φ 1.2 1 + − = 0.7881 + 0.8849 – 1 M1 Appropriate area Φ , from standardisation formula in final solution 0.673 A1 3 1(b) 1.127 = z B1 ±(1.126 – 1.127) seen, 4 sf or more 60 62 1.127 5 − = t 60t = 5.635+62=67.635 M1 z-value = ( ) 60 62 5 − ± t condone z-value = ( ) 62 5 − ± t no continuity correction, condone 2 or σ σ t = 1.13 A1 CAO 3

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2 An ordinary fair die is thrown until a 6 is obtained. (a) Find the probability that obtaining a 6 takes more than 8 throws. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Two ordinary fair dice are thrown together until a pair of 6s is obtained. The number of throws taken is denoted by the random variable X. (b) Find the expected value of X. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that obtaining a pair of 6s takes either 10 or 11 throws. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 8 5 6       0.233 A1 2 2(b) 36 B1 1 2(c) P(X =10) + P(X=11) = 9 10 35 1 35 1 36 36 36 36     +         M1 OE, unsimplified expression in form 9 10 + p q p q , p + q = 1, no × 0.0425 A1 2

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3 A committee of 6 people is to be chosen from 9 women and 5 men. (a) Find the number of ways in which the 6 people can be chosen if there must be more women than men on the committee. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The 9 women and 5 men include a sister and brother. (b) Find the number of ways in which the committee can be chosen if the sister and brother cannot both be on the committee. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Scenarios: 6W 0M 9C6 = 84 5W 1M 9C5 × 5C1 = 126 × 5 = 630 4W 2M 9C4 × 5C2 = 126 × 10 = 1260 M1 Correct number of ways for either 5 or 4 women, accept unsimplified M1 Summing the number of ways for 2 or 3 correct scenarios (can be unsimplified), no incorrect scenarios. Total = 1974 A1 3 3(b) Total number of ways = 14C6 (3003) Number with sister and brother = 12C4 (495) Number required = 14C6 – M1 14C6 – a value 12C4 = 3003 – 495 M1 12Cx or nC4 seen on its own or subtracted from their total, x ⩽ 6, n ⩽ 13 2508 A1 Alternative method for question 3(b) Number of ways with neither = 12C6 = 924 M1 12C6 + a value Number of ways with either brother or sister (not both) = 12C5 × 2 (= 792 × 2) = 1584 M1 12Cx × 2 or nC5 × 2 seen on its own or added to their number of ways with neither, x ⩽ 5, n ⩽ 12 Number required = 924 + 1584 = 2508 A1 3

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4 The 13 00 train from Jahor to Keman runs every day. The probability that the train arrives late in Keman is 0.35. (a) For a random sample of 7 days, find the probability that the train arrives late on fewer than 3 days. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A random sample of 142 days is taken. (b) Use an approximation to find the probability that the train arrives late on more than 40 days. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 7 0.65 + 7C1 6 1 0.65 0.35 + 7C2 5 2 0.65 0.35 M1 Binomial term of form 7Cx ( ) 7 1 , − − x x p p 0 < p < 1, any p, x ≠ 0, 7 0.049022 + 0.184776 + 0.29848 A1 Correct unsimplified answer 0.532 A1 3 4(b) Mean = 142 0.35 49.7 × = Variance = 142 0.35 0.65 32.305 × × = B1 Correct unsimplified np and npq (condone σ = 5.684 evaluated) P(X > 40) = P( 40.5 49.7) 32.305 − > z M1 Substituting their µ and σ (no 2 or σ σ ) into ±standardisation formula with a numerical value for '40.5' P( 1.619) > − z M1 Using either 40.5 or 39.5 within a ±standardisation formula M1 Appropriate area Φ , from standardisation formula P(z >…) in final solution, must be probability 0.947 A1 Correct final answer 5

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5 The 8 letters in the word RESERVED are arranged in a random order. (a) Find the probability that the arrangement has V as the first letter and E as the last letter. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that the arrangement has both Rs together given that all three Es are together. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Total number of ways = 8! 3!2! (= 3360) B1 Correct unsimplified expression for total number of ways Number of ways with V and E in correct positions = 6! 2! 2! × (= 180) B1 6! 2! 2! × alone or as numerator in an attempt to find the number of ways with V and E in correct positions. No , × ± Probability = 180 3360 3 56   =     or 0.0536 B1 FT Final answer from their 6! 2! 2! × divided by their total number of ways Alternative method for question 5(a) 1 3 8 7 × M1 8 7 × a b seen, no other terms (correct denominators) M1 1 3 × c d seen, no other terms (correct numerators) 3 56 or 0.0536 A1 3 Question Answer Marks Guidance 5(b) Rs together and Es together: 5! (120) B1 Alone or as numerator of probability to represent the number of ways with Rs and Es together, no ×, +, – Es together: ( ) 6! 360 2! = B1 Alone or as denominator of probability to represent the number of ways with Es together, no ×, + or – Probability = 5! 6! 2! M1 5! 6! 2! their their seen 1 3 A1 OE Alternative method for question 5(b) P(Rs together and Es together): 5! 1 total number of ways 28 their   =     B1 P(Es together): 6! 3 2! 28 total number of ways their   =     B1 Alone or as numerator of probability to represent the P(Rs and Es together), no ×, +, – Probability = 1 28 3 28 M1 Alone or as denominator of probability to represent the P(Es together), no ×, + or – 1 3 A1 OE, 1 28 3 28 their their seen 4

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7 A particular piece of music was played by 91 pianists and for each pianist, the number of incorrect notes was recorded. The results are summarised in the table. Number of incorrect notes 1 −5 6 −10 11 −20 21 −40 41 −70 Frequency 10 5 26 32 18 (a) Draw a histogram to represent this information. [5] (b) State which class interval contains the lower quartile and which class interval contains the upper quartile. Hence find the greatest possible value of the interquartile range. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Calculate an estimate for the mean number of incorrect notes. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Class widths: 5, 5, 10, 20, 30 Frequency density: 2, 1, 2.6, 1.6, 0.6 M1 At least 3 class widths correct and used in a calculation M1 At least 3 correct frequency densities unsimplified – FT their class widths A1 All correct heights on a histogram using a linear vertical scale from zero – no FT B1 Correct upper bar ends (5.5, 10.5, 20.5, 40.5, 70.5) and 4 correct lower bar ends of 5.5, 10.5, 20.5, 40.5. Condone 0 or 1. B1 Linear scales with at least 3 values indicated on each axis, vertical scale from 0, axes labelled ‘fd’ and ‘no. of (incorrect) notes’, or better. 5 7(b) LQ: 11 – 20 UQ: 21 – 40 B1 Both UQ and LQ correct Greatest IQR = 40 – 11 = 29 B1 FT Subtract lower end of their LQ interval from upper end of their UQ interval 2 Question Answer Marks Guidance 7(c) Midpoints: 3 8 15.5 30.5 55.5 M1 At least 4 midpoints correct and used Mean = 3 10 8 5 15.5 26 30.5 32 55.5 18 91 × + × + × + × + × = 30 40 403 976 999 91 + + + + = 2448 91 M1 Correct formula with their midpoints (not upper boundary, lower boundary, class width, frequency density, frequency or cumulative frequency) 82 26.9, 26 91 A1 Accept 26 or 27 3

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