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Cambridge AS & A Level Mathematics (9709), Paper 7, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_739709_w19_ms_739709/73/O/N/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
The random variable X has mean 2.4 and variance 3.1. (i) The random variable Y is the sum of four independent values of X . Find the mean and variance of Y . [2] (ii) The random variable Z is defined by Z = 4X − 3. Find the mean and variance of Z . [2]
Mark scheme
1(i) 9.6, 12.4 B1 B1 2 1(ii) 6.6, 49.6 B1 B1 2
Question
Cars arrive at a filling station randomly and at a constant average rate of 2.4 cars per minute. (i) Calculate the probability that fewer than 4 cars arrive in a 2-minute period. [2] (ii) Use a suitable approximating distribution to calculate the probability that at least 140 cars arrive in a 1-hour period. [4]
Mark scheme
2(i) ( λ ( = 2 × 2.4 ) = 4.8 ) M1 Any λ 4.82 4.83 e−4.8 1 + 4 + + 2 3! 0.294 (3 sf) A1 2 2(ii) ( λ ( = 60 × 2.4 ) = 144 ) M1 N and σ2=µ SOI N(‘144’, ‘144’) 139.5 − '144' M1 Allow with no continuity correction ( = −0.375) '144' φ(‘0.375’) M1 Correct area consistent with their working 0.646 (3 sf) A1 4
Question
The times, in minutes, taken by competitors to complete a puzzle have mean - and standard deviation 3. The times taken by a random sample of 10 competitors are noted and the results are given below. 25.2 26.8 18.5 25.5 30.1 28.9 27.0 26.1 26.0 24.9 (i) Stating a necessary assumption, calculate a 97% confidence interval for -. [5] (ii) Two more random samples, each of 10 competitors, are taken. Their times are used to calculate two more 97% confidence intervals for -. Find the probability that neither of these intervals contains the true value of -. [1]
Mark scheme
3(i) Assume population is normally distributed B1 x = 25.9 B1 259 Allow 10 z =2.17 B1 3 M1 Must have correct form and z. '25.9'± z × 10 23.8 to 28.0 (3 sf) A1 CWO 5 3(ii) 0.032 (=0.0009) B1 1
Question
A train company claims that 92% of trains on a particular line arrive on time. Sanjeep suspects that the true percentage is less than 92%. He chooses a random sample of 20 trains on this line and finds that exactly 16 of them arrive on time. Making an assumption that should be stated, test at the 5% significance level whether Sanjeep’s suspicion is justified. [6]
Mark scheme
4 Assume trains are independent OR probability of being on time is B1 Must be in context constant H0: P(on time)=0.92 B1 Both. H1: P(on time)<0.92 Allow ‘p’ or π 1− M1 Allow one end error ( C × 0.92 × 0.08 + C × 0.92 × 0.08 + 20 × 0.92 × 0.08 + 0.92 ) 20 17 17 3 20 18 18 2 19 20 Must have 1 – … =0.0706 (3 sf) A1 Compare with 0.05 M1 Valid comparison needed No evidence that percentage less than 92% A1FT OE No contradictions. Method using normal approximation: If the first B1B1 is earned then: 16.5 − 20 × 0.92 CV − 1.566 from , with continuity correction 20 × 0.92 × 0.08 or CV=1.978 (without continuity correction) comp z=1.645 No evidence that % decreased (1.566) or evidence that % decreased (1.978) is awarded SC2 after B marks 6
Question
(i) The random variable X has the distribution B 300, 0.01. Use a Poisson approximation to find P 2 < X < 6. [3] (ii) The random variable Y has the distribution Po ,, and P Y = 0 = P Y = 2. Find ,. [2] (iii) The random variable Z has the distribution Po 5.2 and it is given that P Z = n < P Z = n + 1. (a) Write down an inequality in n. [1] (b) Hence or otherwise find the largest possible value of n. [2]
Mark scheme
5(i) Po(3) B1 SOI 33 34 35 M1 Allow one or two extra terms (2 or 6 or both) e−3 + + 3! 4! 5! 0.493 (3 sf) A1 3 5(ii) A correct equation from P ( 0 ) = P ( 2 ) M1 λ2 leading to 1 = 2 λ = 2 or 1.41 (3 sf) A1 CWO 2 5(iii)(a) 5.2 n 5.2n +1 B1 Correct inequality leading to < n ! ( n + 1)! 1 5(iii)(b) 5.2 M1 Simplify to a stage without exponentials, powers or factorials. n + 1 < 5.2 or 1 < n +1 Largest n is 4 A1 2
Question
A random variable X has probability density function given by D k 3x − x2 0 ≤ x ≤ 3, f x = 0 otherwise. (i) Show that k = 29 . [3] (ii) Find P 1 ≤ X ≤ 2. [2] (iii) Find Var X . [5]
Mark scheme
6(i) 3 M1 Attempt to integrate f(x) and = 1 ∫ ( 3 x − x ) dx = 1 2 k 0 3 x3 3 A1 Correct integral and limits k x2 − 2 3 0 27 27 k − =1 2 3 2 A1 AG k= No errors seen 9 3 6(ii) 2 2 3 2 x3 2 2 8 3 1 M1 Attempt to integrate f(x) dx with limits 1 and 2 ∫( ) 2 3 x − x 2 dx = x − = × 6 − − + OE 91 9 2 3 1 9 3 2 3 13 A1 or 0.481 (3 sf) 27 2 6(iii) y = 3x – x2 symmetrical about x = 3 M1 3 ∫ (3x − x ) dx 2 2 3 2 Attempt 9 0 E(X)= 3 A1 2 3 M1 Attempt to integrate x2f(x) ∫ (3x − x ) dx 2 3 4 9 0 5 3 M1 Subtract their (E(X))2 from their integral x2f(x) with correct = 92 34x − x5 (= 2 × 243 = 27 ) 4 0 9 20 10 limits substituted 2 '27 ' '3' − 10 2 9 or 0.45 A1 20 5
Question
Bob is a self-employed builder. In the past his weekly income had mean $546 and standard deviation $120. Following a change in Bob’s working pattern, his mean weekly income for 40 randomly chosen weeks was $581. You should assume that the standard deviation remains unchanged at $120. (i) Test at the 2.5% significance level whether Bob’s mean weekly income has increased. [5] Bob finds his mean weekly income for another random sample of 40 weeks and carries out a similar test at the 2.5% significance level. (ii) Given that Bob’s mean weekly income is now in fact $595, find the probability of a Type II error. [5] 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
7(i) H0: Pop mean=546 B1 Both. Allow just µ, but not just ‘mean’ H1: Pop mean>546 581 − 546 M1 120 Standardising. Need 120 40 40 =1.845 allow 1.844 A1 Allow 1.84 or 1.85 AWRT 1.845<1.96 M1 OE. Or area comparison 0.0325>0.025 or large probabilities No evidence that mean weekly income has increased A1FT No contradictions. If H1: ≠, and 2.241 used, max B0M1A1M1A0 5 7(ii) a − 546 M1 120 = 1.96 Standardise to find a. Need and 546 and a value of z 120 40 40 a = 583.19 A1 Allow 583 to 3sf '583.19'− 595 M1 120 120 ( = −0.622 ) Standardise. Need and 595 40 40 φ(‘–0.622’)=1 – φ(‘0.622’) M1 Consistent area 0.267 A1 5
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Mathematics 9709/73 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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