Mathematics 9709/62 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
In a game, a ball is thrown and lands in one of 4 slots, labelled A, B, C and D. Raju wishes to test whether the probability that the ball will land in slot A is 41 .
(a) State suitable null and alternative hypotheses for Raju’s test. [1]
The ball is thrown 100 times and it lands in slot A 15 times.
(b) Use a suitable approximating distribution to carry out the test at the 2% significance level. [5]
Mark scheme
1(a) 1 B1 or H0: μ = 25
H0: p = or H1: μ ≠ 25
4
1
H1: p ≠
4
1
1(b) 75 B1 SOI.
N 25,
4 75
Allow B1 for N 25, or N(0.25,0.001875) SOI.
4
15.5 M1 Standardise with their N(25,…)
− 0.25 Allow with no or wrong continuity correction.
15.5 − 25 100
± or
75 0.25 × 0.75
4 100
± –2.194 (2.19) A1
–2.326 < –2.194 or 0.0141 > 0.01 or 0.9859 < 0.99 M1 For valid comparison (accept 2.326 to 2.329)
1 A1 FT OE must be in context and not definite, e.g. not ‘Claim untrue’.
No evidence to reject that the probability is No contradictions.
4
FT their z ; dependent on two–tailed test (one-tailed test can
score B1 M1 A1 M1 A0)
SC for use of Binomial B(100,0.25) P = 0.0111 for B1 and
then comparison with 0.01 and correct conclusion for B1,
maximum 2 out of 5 marks.
5
The random variable X has the distribution B 400, 0.01.
(a) Find Var 4X + 2. [3]
(b) (i) State an appropriate approximating distribution for X , giving the values of any parameters.
Justify your choice of approximating distribution. [2]
(ii) Use your approximating distribution to find P 2 ≤ X ≤ 5. [2]
Mark scheme
2(a) Var(X) = 400 × 0.01 × 0.99 (= 3.96) M1
Var(4X + 2) = 16 × Var(X) M1 For 16 × their Var(X)
63.36 A1 Accept 63.4
3
2(b)(i) Po(4) B1
n = 400 > 50 and either np = 4 < 5 or p=0.01 < 0.1 B1 Must quote values 400 and 4 or clearly see n=400 and np=4
(or p=0.01) in working
2
2(b)(ii) 42 43 44 45 M1 FT their ‘4’
e–-4 + + + Allow one end error
2! 3! 4! 5! FT from (b)(i) Use of Normal allow M1 for attempt at
standardising (with correct continuity correction) using their
N(4,3.96) and attempt at probability.
FT from (b)(i) Use of Binomial allow M1 for attempt at
P(2,3,4,5) Binomial terms clearly seen and added
0.694 (3 sf) A1 CWO
SCB1 only for unsupported answer of 0.694
2
f x
p
x
O 1 p
The random variable X takes values in the range 1 ≤ x ≤ p, where p is a constant. The graph of the probability density function of X is shown in the diagram.
(a) Show that p = 2. [2]
(b) Find E X . [5]
Mark scheme
3(a) 1 M1 For area =1
p(p – 1) = 1
2 1
For verification methods accept × 2 × 1 = 1 or
2
1 1
× 2 × ( p − 1) = 1 or × 1× p = 1 as indication that area=1
2 2
p=2 A1 AG - Convincing method and answer.
Must see quadratic rearranged to =0 and no errors seen.
N.B. Accept convincing verification methods (e.g. statement
such as ‘assume p = 2’ or ‘if p = 2’ or ‘using p = 2’ or showing
1
by clear substitution that p = 2 fits p(p – 1) = 1 with clear
2
conclusion)
2
3(b) Gradient = 2 M1 Award for attempting equation of line y=mx+c with m = 2, -2,
equation of line is y = 2x + c 1 1
line passes through (1, 0), hence c = –2 or − and numerical c (c≠0)
2 2
y = 2x – 2 A1 May be seen in (a)
M1 can be implied by correct answer
2 M1 For attempting ʃxf(x)dx.
2 ( x 2 − x)dx
1
Ignore limits, FT their equation.
x3 x 2 2 A1 FT Correct integration FT their f(x)and correct limits
2 −
3 2 1
5 A1
or 1.67 (3 sf)
3
5
Wendy’s journey to work consists of three parts: walking to the train station, riding on the train and then walking to the office. The times, in minutes, for the three parts of her journey are independent and have the distributions N 15.0, 1.12 , N 32.0, 3.52 and N 8.6, 1.22 respectively.
(a) Find the mean and variance of the total time for Wendy’s journey. [2]
If Wendy’s journey takes more than 60 minutes, she is late for work.
(b) Find the probability that, on a randomly chosen day, Wendy will be late for work. [3]
(c) Find the probability that the mean of Wendy’s journey times over 15 randomly chosen days will be less than 54.5 minutes. [3]
Mark scheme
4(a) Mean = 15.0+32.0+8.6 [= 55.6] B1 Allow unsimplified
Var = 1.12+3.52+1.22 [= 14.9] B1 Allow unsimplified
2
4(b) 60 − "55.6" M1 FT their 55.6 and 14.9
[= 1.140] Ignore continuity correction
"14.9"
1 – ɸ("1.140") M1 For correct probability area consistent with their working
0.127 (3 sf) A1 CWO
3
4(c) 54.5 − "55.6" 817.5 − 834 M1 FT their 55.6 and 14.9
or [= –1.104] No mixed methods
"14.9" 223.5
15
1 – ɸ("1.104") M1 For correct probability area consistent with their working
0.135 (3 sf) A1 As final answer
3
The time, in minutes, spent by customers at a particular gym has the distribution N -, 38.2. In the past the value of - has been 42.4. Following the installation of some new equipment the management wishes to test whether the value of - has changed.
(a) State what is meant by a Type I error in this context. [1]
(b) The mean time for a sample of 20 customers is found to be 45.6 minutes.
Test at the 2.5% significance level whether the value of - has changed. [5]
Mark scheme
5(a) Conclude that (population) mean time has changed (or is not 42.4) B1 OE. In context.
although μ has not changed (or is still 42.4)
1
5(b) H0: population mean (or μ) = 42.4 B1 Not just ‘mean’. (could be seen in (a))
H1: population mean (or μ) ≠ 42.4
45.6 − 42.4 M1 For standardising (must have 20 )
±
38.2 ÷ 20
± 2.315 A1
2.240 < ‘2.315’ M1 For valid comparison (accept 2.241)
or P(z > 2.315) = 0.0103 < 0.0125 oe
There is evidence that μ or mean time has changed A1 FT FT their z
In context, not definite. No contradictions.
Note: Accept correct alternative methods
SC: One tail test no FT. Can score B0 M1 A1 M1 (comparison
with 1.96) A0 (maximum 3 out of 5)
5
The heights, h centimetres, of a random sample of 100 fully grown animals of a certain species were measured. The results are summarised below. n = 100 Σ h = 7570 Σ h2 = 588 050
(a) Find unbiased estimates of the population mean and variance. [3]
(b) Calculate a 99% confidence interval for the mean height of animals of this species. [3]
Four random samples were taken and a 99% confidence interval for the population mean, -, was found from each sample.
(c) Find the probability that all four of these confidence intervals contain the true value of -. [2]
Mark scheme
6(a) 7570 B1
est (μ) = (= 75.7)
100
2
est(σ ) =
100 h
Σ 2
− '75.7 '2 or
1
588050 −
75702
M1 Attempted
99 100
99 100 (Note: Biased variance (150.01) scores M0 )
100 588050
= − '75.7 '2 [= 151.525]
99 100
= 152 (3 sf) A1 15001
Or
99
3
6(b) '151.525' M1 For expression of correct form. Must be a z value. Condone
‘75.7’ ± z just + or just -.
100
z = 2.576 B1 Accept 2.574 to 2.579
72.5 to 78.9 A1 FT FT biased variance only
Must be an interval
3
6(c) 0.994 B1
0.961 (3 sf) B1
2
Customers arrive at a particular shop at random times. It has been found that the mean number of customers who arrive during a 5-minute interval is 2.1.
(a) Find the probability that exactly 4 customers arrive during a 10-minute interval. [2]
(b) Find the probability that at least 4 customers arrive during a 20-minute interval. [2]
(c) Use a suitable approximating distribution to find the probability that fewer than 40 customers arrive during a 2-hour interval. [4]
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) 4.24 M1 P(4), allow any λ
e −4.2 ×
4!
0.194 (3 sf) A1 As final answer.
SC Unsupported correct answer scores B1 only.
2
7(b) 8.42 8.43 M1 Allow M1 with incorrect λ. Accept one end error.
1 − e −8.4 1 + 8.4 + +
2 3!
0.968 (3 sf) A1 As final answer.
SC Unsupported correct answer scores B1 only.
2
7(c) N(50.4, 50.4) M1 SOI
39.5 − 50.4 M1 Allow wrong or no continuity correction.
[= –1.535] Must have √
50.4
Φ(‘–1.535’) = 1 – Φ(‘1.535’) M1 For correct probability area consistent with their working.
0.0624 (3 sf) or 0.0623 A1
4
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Mathematics 9709/62 for May/June 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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