Cambridge A Level Mathematics 9709 — 2019 May/June Paper 7 · Variant 3

1 A coin is thrown 100 times and it shows heads 60 times. Calculate an approximate 98% confidence interval for the probability, p, that the coin shows heads on any throw. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 0.6 ± z 0.4 0.6 100 × M1 z = 2.326 B1 2.326 to 2.329 0.486 to 0.714 (3 sf) A1 Must be an interval 3

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2 The length of worms is denoted by X cm. The lengths of a random sample of 50 worms were measured. Some of the results were lost, but the following results are available. ³ Σx2 = 4361 ³ An unbiased estimate of the population variance of X is 9.62. Calculate the mean length of the 50 worms. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 49 50 ( ) x − = 9.62 M1 or 2 ( ) 4361 49 50 49 ( ) Σ × − x = 9.62 BOD regarding symbols used 2 x = 4361 50 - 9.62× 49 50 = 77.7924 A1 (Σx)2 = 4361 × 50 – 9.62 × 50 × 49 = 194481or Σx =441 ( ) Σx or ( ) x must be correctly identified x = 8.82 (3 sf) A1 SC use of ‘biased’ leading to 8.81 B1 3

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3 Luis has to choose one person at random from four people, A, B, C and D. He throws a fair six-sided die. If the score is 1, he will choose A. If the score is 2 he will choose B. If the score is 3, he will choose C. If the score is 4 or more he will choose D. (i) Explain why the choice made by this method is not random. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Describe how Luis could use a single throw of the die to make a random choice. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ On another day, Luis has to choose two people at random from the same four people, A, B, C and D. (iii) List the possible choices of two people and hence describe how Luis could use a single throw of the die to make this random choice. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) D more likely to be chosen B1 oe, e.g. P(D) > P(A) e.g. P(A)=P(B)=P(C)=1/6 P(D)=1/2 no contradictions 1 3(ii) Reject scores of 5 or 6 B1 or other correct: choose D when the score is 4 1 Question Answer Marks Guidance 3(iii) AB AC AD BC BD CD B1 Allocate as follows: 1: AB; 2: AC; 3: AD; 4: BC; 5: BD 6: CD B1 or similar 2

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7 Each day at a certain doctor’s surgery there are 70 appointments available in the morning and 60 in the afternoon. All the appointments are filled every day. The probability that any patient misses a particular morning appointment is 0.04, and the probability that any patient misses a particular afternoon appointment is 0.05. All missed appointments are independent of each other. Use suitable approximating distributions to answer the following. (i) Find the probability that on a randomly chosen morning there are at least 3 missed appointments. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the probability that on a randomly chosen day there are a total of exactly 6 missed appointments. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the probability that in a randomly chosen 10-day period there are more than 50 missed appointments. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) M1 1 – e-2.8(1 + 2.8 + 2 2.8 2 ) ) M1 Any λ allowing one end error = 0.531or 0.53(0) (3 sf) A1 SC Binomial 0.534 B1 3 7(ii) Use of Po(5.8) M1 May be implied e-5.8 × 6 5.8 6! M1 Any λ = 0.16(0) (3 sf) A1 3 Question Answer Marks Guidance 7(iii) Use of N(58, 58) M1 May be implied or N(58, 55.38) 50.5 '58' '58' − (= -0.985) M1 Standardised with their values, allow wrong or incorrect cc Φ('0.985') M1 Correct area consistent with their working or ( ) Φ "1.008 = 0.838 (3 sf) A1 or 0.843 4

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8 The four sides of a spinner are A, B, C, D. The spinner is supposed to be fair, but Sonam suspects that the spinner is biased so that the probability, p, that it will land on side A is greater than 14. He spins the spinner 10 times and finds that it lands on side A 6 times. (i) Test Sonam’s suspicion using a 1% significance level. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Later Sonam carries out a similar test at the 1% significance level, using another 10 spins of the spinner. (ii) Calculate the probability of a Type I error. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Assuming that the value of p is actually 5,3 calculate the probability of a Type II error. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(i) H0: p = 1 4 H1: p > 1 4 B1 10C6 6 4 3 1 4 4 ( ) ( ) + 10C7 7 3 3 1 4 4 ( ) ( ) + 10C8 8 2 3 1 4 4 ( ) ( ) + 10 9 3 1 4 4 ( ) ( ) + 10 1 4( ) M1 Correct terms, allow one term incorrect or omitted or extra or summing all correct terms from 0 to 5 allow one term incorrect or omitted or extra = 0.0197 A1 or 0.9803 comp '0.0197' with 0.01 M1 Valid comparison with 0.01 or valid comparison with 0.99 No evidence to conclude p > 1 4 A1 FT No contradictions Use of two-tail test can score BOM1A1M1(comparison with 0.005) A0 5 8(ii) 10C7 7 3 3 1 4 4 ( ) ( ) + 10C8 8 2 3 1 4 4 ( ) ( ) +10 9 3 1 4 4 ( ) ( ) + 10 1 4( ) M1 Their P(X ( ) ( ) 6 4 10 6 6) 0.25 0.75 − . C P(Type I) = 0.00351 (3 sf) A1 Accept 0.00348 to 0.00351 2 8(iii) C.R is X ⩾ 7 P(Type II) = 1 – P(X ⩾ 7 | p = 3 5 ) = M1 May be implied 1– (10C7 7 3 3 2 5 5 ( ) ( ) + 10C8 8 2 3 2 5 5 ( ) ( ) +10 9 3 2 5 5 ( ) ( ) + 10 3 5( ) ) M1 Accept 1 – P(X ⩾ 8 | p = 3 5 ) or 1 – P(X ⩾ 6 | p = 3 5 ) = 0.618 A1 3

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