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Cambridge AS & A Level Mathematics (9709), Paper 7, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s19_qp_739709_s19_ms_739709/73/M/J/19
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
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 M1 Recognisable value of z 0.6 ± z 0.4100 z = 2.326 B1 2.326 to 2.329 0.486 to 0.714 (3 sf) A1 Must be an interval 3
Question
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 Σ x2 = 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 50 ( 4361 − x 2 ) = 9.62 M1 or ( 4361 − ( Σx )2 ) = 9.62 BOD regarding symbols used 49 50 49 50×49 x 2 = 4361 49 = 77.7924 - 9.62× 50 A1 (Σx)2 = 4361 × 50 – 9.62 × 50 × 49 = 194481or Σx =441 ( Σx ) or ( x ) must be 50 correctly identified x = 8.82 (3 sf) A1 SC use of ‘biased’ leading to 8.81 B1 3
Question
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 3(iii) AB AC AD BC BD CD B1 Allocate as follows: B1 or similar 1: AB; 2: AC; 3: AD; 4: BC; 5: BD 6: CD 2
Question
A factory supplies boxes of children’s bricks. Each box contains 10 randomly chosen large bricks and 20 randomly chosen small bricks. The masses, in grams, of large and small bricks have the distributions N 60, 1.2 and N 30, 0.7 respectively. The mass of an empty box is 8 g. Find the probability that the total weight of a box and its contents is less than 1200 g. [5]
Mark scheme
4 Total ~ N(1208, .....) B1 Var(total) (= 10×1.2 + 20×0.7 (+ 0)) = 26 B1 May be implied by next line ± 1200−"1208" (= –1.569) M1 FT their mean and var of total mass, e.g. allow 1200 and 11.24 "26" (from 10 × 1.22 + 20 × 0.72) 1 – Φ(“1.569”) M1 Correct area consistent with their working = 0.0583 (3 sf) A1 5
Question
The amount of money, in dollars, spent by a customer on one visit to a certain shop is modelled by the distribution N -, 1.94. In the past, the value of - has been found to be 20.00, but following a rearrangement in the shop, the manager suspects that the value of - has changed. He takes a random sample of 6 customers and notes how much they each spend, in dollars. The results are as follows. 17.60 23.50 17.30 22.00 31.00 15.50 The manager carries out a hypothesis test using a significance level of !%. The test does not support his suspicion. Find the largest possible value of !. [6]
Mark scheme
5 H0: Pop mean = 20 B1 H1: Pop mean ≠ 20 Accept µ Σx (= 126.9 = 21.15) M1 Attempted 6 6 or 126.9 and 11.64 attempted '21.15'−20 M1 Must have 6 1.94 6 120 − 126.9 or no mixed method 11.64 = 2.022 A1 2(1 - ɸ(‘2.022’) ) 2 (1 – '0.9784)' = 0.0432) ( M1 FT 2 × 1 −′ .9784' ) α = 4.32 (3 sf) A1 FT Allow 4.3 or 4, if correct working seen, or clearly implied, as far as 0.0216 FT their z, no error seen One-tail test scores maximum 3/6 6
Question
A function f is defined by t 3x2 0 ≤ x ≤ a, f x = a3 0 otherwise, where a is a constant. (i) Show that f is a probability density function for all positive values of a. [3] The random variable X has probability density function f and the median of X is 2. (ii) Show that a = 2.52, correct to 3 significant figures. [3] (iii) Find E X . [3]
Mark scheme
6(i) a M1 ∫ x dx 3 2 3 a3 Attempt to integrate f(x) with limits 0 and a (condone missing ) 0 a3 3 a (= 33 x3 ) a 0 3 A1 3a3 = 3a3 − 0 or better seen 3a 3a3 = 1 Hence f is pdf for all a A1 Answer = 1 and comment 3 6(ii) 2 M1 ∫ x dx = 0.5 3 2 3 a3 Attempt to integrate f(x)=0.5, limits 0 and 2 oe, condone missing 0 a3 3 x3 2 = 0.5 a3 3 0 3 × 8 = 0.5 oe A1 23 3 a3 3 − 0 or better, condone missing 3 3 a a3 = 16 or a = 3 16 A1 Convincingly obtained Note: Attempt to verify 2.52, M1 as stated except not equated to 0.5.A1 as ( = 2.52 AG) stated, A1 for evaluation to 0.499..apprx 0.5 3 6(iii) 2.52 a M1 3 16 ∫0 x3dx 3 or 16 ∫0 x3dx Attempt integ xf(x), correct limits, condone missing 3 a3 3 x 4 2.52 3 x 4 a = 16 or 16 4 0 4 0 3 × 40.317 = 16 A1 2.524 3 4 − 0 or better, condone missing 3 4 a = 1.89 (3 sf) A1 3
Question
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) Use of Po(2.8) M1 May be implied 2 1 – e-2.8(1 + 2.8 + 2.82 ) ) 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 × 5.86 M1 Any λ 6! = 0.16(0) (3 sf) A1 3 7(iii) Use of N(58, 58) M1 May be implied or N(58, 55.38) 50.5 − '58' M1 Standardised with their values, allow wrong or incorrect cc (= -0.985) '58' Φ('0.985') M1 Correct area consistent with their working or Φ ("1.008 ) = 0.838 (3 sf) A1 or 0.843 4
Question
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 41 . 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 53 , calculate the probability of a Type II error. [3] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
8(i) H0: p = 14 B1 H1: p > 14 10 C6 ( 14 )6 ( 43 )4 + 10C7 ( 14 )7 ( 34 )3 + 10C8 ( 14 )8 ( 43 ) 2 + M1 Correct terms, allow one term incorrect or omitted or extra 10 ( 14 )9 ( 43 ) + ( 14 )10 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 > 14 A1 FT No contradictions Use of two-tail test can score BOM1A1M1(comparison with 0.005) A0 5 8(ii) 10 C7 ( 14 )7 ( 34 )3 + 10C8 ( 14 )8 ( 43 ) 2 +10 ( 14 )9 ( 43 ) + ( 14 )10 M1 Their P(X . 6) − 10C ( 0.25)6 ( 0.75)4 6 P(Type I) = 0.00351 (3 sf) A1 Accept 0.00348 to 0.00351 2 8(iii) C.R is X ⩾ 7 M1 May be implied P(Type II) = 1 – P(X ⩾ 7 | p = 53 ) = 1– (10C7 ( 53 )7 ( 52 )3 + 10C8 ( 53 )8 ( 52 ) 2 +10 ( 53 )9 ( 52 ) + ( 53 )10 ) M1 Accept 1 – P(X ⩾ 8 | p = 3 ) or 1 – P(X ⩾ 6 | p = 3 ) 5 5 = 0.618 A1 3
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Mathematics 9709/73 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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