Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 6 · Variant 1

1 When Shona goes to college she either catches the bus with probability 0.8 or she cycles with probability 0.2. If she catches the bus, the probability that she is late is 0.4. If she cycles, the probability that she is late is x. The probability that Shona is not late for college on a randomly chosen day is 0.63. Find the value of x. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 0.8 0.6 0.2 1 0.63 x × + − = 0.6 or 0.4 and C = 0.63 or 0.37 0.2 0.05 x = M1 Correct unsimplified equation 0.25 x = A1 Alternative method for question 1 0.8 0.4 0.2 1 0.63 x × + = − M1 Equation of form 0.8 × A + 0.2 × B = C, A,B involving x and 0.6 or 0.4 and C = 0.63 or 0.37 0.2 0.05 x = M1 Correct unsimplified equation 0.25 x = A1 3

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2 Annan has designed a new logo for a sportswear company. A survey of a large number of customers found that 42% of customers rated the logo as good. (i) A random sample of 10 customers is chosen. Find the probability that fewer than 8 of them rate the logo as good. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) On another occasion, a random sample of n customers of the company is chosen. Find the smallest value of n for which the probability that at least one person rates the logo as good is greater than 0.995. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) 10Capa(1 – p)b 0 < p < 1 any p, 0 ⩽ a,b ⩽ 10 A1 Correct unsimplified expression 0.983 A1 3 2(ii) 1 – P(0) > 0.995 0.58n < 0.005 M1 Equation or inequality involving 0.58n or 0.42n and 0.995 or 0.005 log0.005 log0.58 n > n > 9.727 M1 Attempt to solve using logs or Trial and Error. May be implied by their answer (rounded or truncated) n = 10 A1 CAO 3

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3 The mean and standard deviation of 20 values of x are 60 and 4 respectively. (i) Find the values of Σx and Σx2. 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Another 10 values of x are such that their sum is 550 and the sum of their squares is 40 500. (ii) Find the mean and standard deviation of all these 30 values of x. 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Mark scheme: 3(i) 60 20 1200 x B1 2 2 2 60 4 20 x ∑ − = M1 Correct variance formula used, condone = 4 2 3616 20 72320 x ∑ = × = A1 Exact value 3 Question Answer Marks Guidance 3(ii) x ∑ = 1200 + 550 = 1750 2 72320 40500 112800 x ∑ = + = M1 Summing both values of x ∑ and 2 x ∑ Mean = 1 750 30 their = 58.3 B1FT FT their 1750 (not 550 or 1200)/their(20+10), accept unsimplified Variance = ( ) 2 1 12820 1 750 357.89 30 30 their their   − =     M1 substitute their Σx and Σx2 into correct variance formula s.d. = 18.9 A1 4

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5 Ransha measured the lengths, in centimetres, of 160 palm leaves. His results are illustrated in the cumulative frequency graph below. 180 160 140 120 100 frequency 80 60 Cumulative 40 20 0 0 5 10 15 20 25 30 Length in centimetres (i) Estimate how many leaves have a length between 14 and 24 centimetres. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) 10% of the leaves have a length of L centimetres or more. Estimate the value of L. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Estimate the median and the interquartile range of the lengths. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Sharim measured the lengths, in centimetres, of 160 palm leaves of a different type. He drew a box-and-whisker plot for the data, as shown on the grid below. 0 5 10 15 20 25 30 Length in centimetres (iv) Compare the central tendency and the spread of the two sets of data. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) 1 5(ii) 90% of 160 = 144 M1 144 seen, may be marked on graph (L =) 22 A1 2 5(iii) Median = 15.6 UQ = 18.8, LQ = 12.7 B1 15.5 < median < 15.8 IQR = 18.8 – 12.7 M1 18.5 < UQ < 19 – 12.5 < LQ < 13 6.1 A1 6.0 ⩽ IQR ⩽ 6.2 3 5(iv) The Median higher for Ransha (1st set of data) B1 Any correct comparison of central tendency, must mention median IQR lower for Ransha (1st set of data) B1 Any correct comparison of spread, must refer to IQR 2

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6 (i) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Four letters are selected from the 12 letters of the word STEEPLECHASE. (iii) Find the number of different selections if the four letters include exactly one S. 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Mark scheme: 6(i) 9! 2! 181440 = B1 Exact value 1 6(ii) Total no of ways = 12! 9 979 200 2!4! = (A) B1 Accept unevaluated With Ss together = 11! 1663 200 4! = (B) B1 Accept unevaluated With Ss not together = (B) – (A) M1 Correct or 12! 8!, , integers 1 m n m n − > or their identified total – their identified Ss together 8 316 000 A1 Exact value Alternative method for question 6(ii) _ T _ E _ E _ P _ L _ E _ C _ H _ A _ E _ B1 10! × k in numerator k integer ⩾ 1 10! 11 10 4! 2! × × B1 4! × k in numerator k integer ⩾ 1 1 0! 4! their their × 11C2 or 11P2 M1 OE 8 316 000 A1 Exact value 4 Question Answer Marks Guidance 6(iii) S E E E : 1 M1 6Cx seen alone or times K > 1 S E E _ : 6C1 = 6 S E _ _ : 6C2 = 15 S _ _ _ : 6C3 = 20 B1 6C3 or 6C2 or 6C1 alone Add 3 or 4 correct scenarios M1 No extras Total = 42 A1 4

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7 The shortest time recorded by an athlete in a 400 m race is called their personal best (PB). The PBs of the athletes in a large athletics club are normally distributed with mean 49.2 seconds and standard deviation 2.8 seconds. (i) Find the probability that a randomly chosen athlete from this club has a PB between 46 and 53 seconds. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) It is found that 92% of athletes from this club have PBs of more than t seconds. Find the value of t. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Three athletes from the club are chosen at random. (iii) Find the probability that exactly 2 have PBs of less than 46 seconds. 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Mark scheme: 7(i) P(46 < X < 53) = P 46 49.2 53 49.2 2.8 2.8 Z − −   < <     continuity correction, σ2 or √σ P( 1.143 1.357) Z − < < A1 Both standardisations correct unsimplified ( ) ( ) Φ 1.357 Φ 1.143 1 + − = 0.9126 + 0.8735 – 1 M1 Correct final area 0.786 A1 Final answer 4 Question Answer Marks Guidance 7(ii) 49.2 1.406 2.8 t − = − B1 ±1.406 seen M1 An equation using ± standardisation formula with a z-value, condone σ2 or √σ 45.3 A1 3 7(iii) P(X < 46) = 0.1265 M1 Calculated or ft from (i) P(2PB < 46) = ( ) 2 3 1 0.1265 0.1265 − M1 3(1-p)p2, 0<p<1 0.0419 A1 3

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