Mathematics 9709/61 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The diameters, x millimetres, of a random sample of 200 discs made by a certain machine were recorded. The results are summarised below. n = 200 Σ x = 2520 Σ x2 = 31852
(a) Calculate a 95% confidence interval for the population mean diameter. [6]
(b) Jean chose 40 random samples and used each sample to calculate a 95% confidence interval for the population mean diameter.
How many of these 40 confidence intervals would be expected to include the true value of the population mean diameter? [1]
Mark scheme
1(a) Est(μ) = 2520 [= 12.6] B1 OE
200
( ) 25202 M1 Allow M1 if 200 omitted
Est(σ2) = 200 31582 − ‘12.6’2 or 1 31852 − 199
199 200 199 200
100 A1 CWO or σ = 0.7088 or 0.709
= 0.5025 or 0.503 or
199
z = 1.96 B1
‘12.6’ ± z × ‘0.5025’ ÷ 200 M1 For expression of correct form Any z but must be z
CI = 12.5 to 12.7 (3 sf) A1 CWO Must be an interval
Note:
Use of biased can score maximum B1 M1 A0 B1 M1 A0
6
1(b) 0.95 × 40 [= 38] B1 Give at early stage
1
Arvind uses an ordinary fair 6-sided die to play a game. He believes he has a system to predict the score when the die is thrown. Before each throw of the die, he writes down what he thinks the score will be. He claims that he can write the correct score more often than he would if he were just guessing.
His friend Laxmi tests his claim by asking him to write down the score before each of 15 throws of the die. Arvind writes the correct score on exactly 5 out of 15 throws.
Test Arvind’s claim at the 10% significance level. [5]
Mark scheme
2 H0: P(correct) = 1 B1 Allow p = 1
6 6
H1: P(correct) > 1 Allow p > 1
6 6
1 – (15C4 × ( 5 )11 × ( 1 )4 + 15C3 × ( 5 )12 × ( 1 )3 + 15C2 × ( 5 )13 × ( 1 )2 + 15 × M1 Expression must be seen
6 6 6 6 6 6 Allow one end error
5
( ) ×14 1 5
+( ) ) 15
6 6 6
0.0898 or 0.0897 (3 sf) A1 SC if M0 scored allow SCB1 for 0.0898 or 0.0897
0.0898 < 0.1 M1 Valid comparison
For valid comparison with 0.9 (0.9102 > 0.9 seen the
previous M1and A1 can be recovered
[Reject H0] There is evidence (at the 10% level) that Arvind can predict FTA1 Not definite, e.g. not ‘He can predict’ or ‘Claim true’
scores In context and no contradictions
5
The lengths, in centimetres, of two types of insect, A and B, are modelled by the random variables
X ∼ N 6.2, 0.36 and Y ∼ N 2.4, 0.25 respectively.
Find the probability that the length of a randomly chosen type A insect is greater than the sum of the lengths of 3 randomly chosen type B insects. [5]
Mark scheme
3 D = X – (Y1 + Y2 + Y3) OE B1 Give at early stage
E(D) = 6.2 -2.4 × 3 [= –1] OE
Var(D) = 0.36 + 3 × 0.25 [= 1.11] B1 Give at early stage
0 − ( −1) M1 No standard deviation/variance mixes
[= 0.949] Var must come from a combination attempt
‘1.11’
1 – Φ(‘0.949’) M1 Area consistent with their values
= 0.171 (3 s.f.) A1
5
The independent random variables X and Y have distributions Po 2 and B 20, 14 respectively.
(a) Find the mean and standard deviation of X − 3Y . [5]
(b) Find P Y = 15X . [3]
Mark scheme
4(a) 20 1 3 é 15 ù B1 Both
E(Y) = [= 5] , Var(Y) = 20 ´ ´ ê= ú OE, SOI
4 4 4 êë 4 úû
Var(X) = 2 B1 SOI or standard deviation = 2
E(X – 3Y) = –13 B1
Var(X – 3Y) = 2 + 9 × ‘ 15 ’ [= 35.75] M1 Correct formula using their values
4
Standard deviation of (X – 3Y) = 5.98 (3 s.f.) or 1 143 A1 CWO
2
5
4(b) (0, 0) and (1, 15) M1
e–2 × ( 3 ) 20 + e–2 × 2 × 20C15 ( 3 )5 ( 1 )15 M1
4 4 4
0.000430 (3 sf) A1 CWO (must have evidence of addition)
Allow 0.00043 or 4.3(0) × 10–4
3
Cars arrive at a fuel station at random and at a constant average rate of 13.5 per hour.
(a) Find the probability that more than 4 cars arrive during a 20-minute period. [3]
(b) Use an approximating distribution to find the probability that the number of cars that arrive during a 12-hour period is between 150 and 160 inclusive. [4]
Independently of cars, trucks arrive at the fuel station at random and at a constant average rate of
3.6 per 15-minute period.
(c) Find the probability that the total number of cars and trucks arriving at the fuel station during a
10-minute period is more than 3 and less than 7. [3]
Mark scheme
5(a) λ = 4.5 B1
2 3 4 M1 Allow one end error Allow any λ.
1 – e–4.5 (1 + 4.5 + 4.5 + 4.5 + 4.5 )
2! 3! 4! Poisson expressions must be seen
= 0.468 (3 sf) A1 If M0 awarded allow SC B1 for 0.468
3
5(b) λ = 162 B1
(X ~ Po(162) X ~ N(162, 162))
149.5 − ‘162’ and 160.5 − ‘162’ M1 One of these; allow with incorrect or no continuity
‘162’ ‘162’ correction
(= –0.982 and –0.118)
Φ(‘0.982’) – ɸ(‘0.118’) oe M1 Area consistent with their values (both standardisations
must be seen)
= 0.290 (3 sf) A1 Allow 0.29
4
5(c) λ = 13.5 + 3.6 × 2 OE or 4.65 M1 Attempt to find λ
6 3
4 5 6 M1 Allow any λ Allow one end error
e–4.65( 4.65 + 4.65 + 4.65 )
4! 5! 6! Poisson terms not be seen
0.494 (3 sf) A1 If M0 allow SC B1 for 0.494
3
A random variable X has probability density function f. The graph of f x is a straight line segment parallel to the x-axis from x = 0 to x = a, where a is a positive constant.
(a) State, in terms of a, the median of X . [1]
(b) Find P X > 34 a. [1]
1 a2 .
(c) Show that Var X = 12 [5]
(d) Given that P X < b = p, where 0 < b < 12 a, find P 13 b < X < a − 13 b in terms of p. [2]
Mark scheme
6(a) a B1
2
1
6(b) 1 B1
4
1
6(c) f(x) = 1 B1 SOI (may be seen in part (a) or part (b))
a
E(X) = a B1 SOI
2
a M1 Attempt integrate their f(x) × x2 with correct limits
1 2
0 a
x dx
3 a A1
= x = a
2
3a 0 3
a 2 – ( a )2 or a 2 – a 2 [= a 2 AG ] A1 Must see previous line and answer
3 2 3 4 12 No errors seen
5
6(d) p M1 SOI (could be on a diagram)
P(X < b ) = OR by integration: prob = 1-(2/3)(b/a)
3 3
2p A1
P( b < X < a – . b ) = 1 –
3 3 3
2
In the past, the mean time for Jenny’s morning run was 28.2 minutes. She does some extra training and she wishes to test whether her mean time has been reduced. After the training Jenny takes a random sample of 40 morning runs. She decides that if the sample mean run time is less than 27 minutes she will conclude that the training has been effective. You may assume that, after the training, Jenny’s run time has a standard deviation of 4.0 minutes.
(a) State suitable null and alternative hypotheses for Jenny’s test. [1]
(b) Find the probability that Jenny will make a Type I error. [3]
(c) Jenny found that the sample mean run time was 27.2 minutes.
Explain briefly whether it is possible for her to make a Type I error or a Type II error or both.
[2]
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(a) H0: pop mean run time = 28.2 mins B1 Allow ‘μ’. Not ‘mean journey time’
H1: pop mean run time < 28.2 mins
1
7(b) 27 − 28.2 [= –1.897] M1 For standardising
4/ 40 Must have √40
Φ(< ‘–1.897’) = 1 – Φ(‘1.897’) M1 For correct area consistent with these values
0.0289 (3 sf) A1
3
7(c) H0 is not rejected so… M1
Type II error can be made and Type I error cannot be made A1 Both needed (accept ‘only a Type II error could be
made’)
2
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Mathematics 9709/61 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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