Mathematics 9709/61 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A bus station has exactly four entrances. In the morning the numbers of passengers arriving at these entrances during a 10-second period have the independent distributions Po(0.4), Po(0.1), Po(0.2) and
Po(0.5).
Find the probability that the total number of passengers arriving at the four entrances to the bus station during a randomly chosen 1-minute period in the morning is more than 3. [3]
Mark scheme
1 λ = 7.2 B1
2 3
P(X > 3) = 1 − e-7.2(1 + 7.2 + 7.22! + 7.23! ) M1 Allow any λ.
Allow one end error.
or 1 − e-7.2(1 + 7.2 + 25.92 + 62.21)
Must see expression. Allow fully correct sigma notation.
or 1 − (0.0007466 + 0.005375 + 0.01935 + 0.04644)
= 0.928 (3sf) A1 SC 0.928 with no working seen scores B1 B1.
3
The random variable X has the distribution N `31.2, 10.4 2j . Two independent random values of X, denoted by X 1 and X 2 , are chosen.
Find P (X 1 2 3X2) . [5]
Mark scheme
2 E(X1 − 3X2) = 31.2 − 3×31.2 [ = −62.4] B1 OE E(3X2 – X1) = +62.4
Var(X1 − 3X2) = 10.42 + 32 × 10.42 [= 1081.6] B1
0 − (' − 62.4')
[= 1.897] M1 Standardising (with attempt at E and Var, not just using 31.2 and
'1081.6'
10.4).
1 − Φ(‘1.897’) M1 For area consistent with their working.
= 0.0289 (3 sf) A1
5
The time taken in minutes for a certain daily train journey has a normal distribution with standard deviation 5.8 . For a random sample of 20 days the journey times were noted and the mean journey time was found to be 81.5 minutes.
(a) Calculate a 98% confidence interval for the population mean journey time. [3]
A student was asked for the meaning of this confidence interval. The student replied as follows.
‘The times for 98% of these journeys are likely to be within the confidence interval.’
(b) Explain briefly whether this statement is true or not. [1]
Two independent 98% confidence intervals are found.
(c) Given that at least one of these intervals contains the population mean, find the probability that both intervals contain the population mean. [2]
Mark scheme
3(a) 81.5 ± z × 5.8 M1 For a correct expression (accept if only one side of the interval
20
calculated).
Any z (must be a z).
z = 2.326 B1
78.5 to 84.5 (3sf) A1 Must be an interval.
3
3(b) Not true. B1 OE
C. I. is for mean time, not individual times. Both comments needed.
1
3(c) 0.982 M1 Attempt P(both contain μ ) with numerator attempt 0.982 and
1− 0.022 P(at least one contains μ )
denominator attempt involving 0.02.
Must see their quotient.
= 0.961 (3sf) A1 NB: [0.982 = ] 0.9604 scores M0 A0.
2
(a) A random sample of 8 boxes of cereal from a certain supplier was taken. Each box was weighed and the masses in grams were as follows.
261 249 259 252 255 256 258 254
Find unbiased estimates of the population mean and variance. [3]
(b) The supplier claims that the mean mass of boxes of cereal is 253 g. A quality control officer suspects that the mean mass is actually more than 253 g. In order to test this claim, he weighs a random sample of 100 boxes of cereal and finds that the total mass is 25 360 g.
(i) Given that the population standard deviation of the masses is 3.5 g, test at the 5% significance level whether the population mean mass is more than 253 g. [5]
An employee says, ‘This test is invalid because it uses the normal distribution, but we do not know whether the masses of the boxes are normally distributed.’
(ii) Explain briefly whether this statement is true or not. [1]
Mark scheme
4(a) 2044 B1 Accept 3sf if nothing better seen.
Est(μ) = [=255.5]
8
8 522348 1 ‘2044’2 M1 Attempt to find Σx2 and substitute in correct formula. May be
2
Est(σ ) = − "255.5"2 or (‘522348’ – ) implied by correct answer.
7 8 7 8
Biased 13.25 scores M0.
106 A1 OE
= 15.1 (3 sf) or
7
3
4(b)(i) H0: μ = 253 B1 Allow ‘Population mean’ but not just ‘mean’.
H1: μ > 253
100 − 253
25360 M1 Standardising must have 100.
3.5 ÷ 100
= 1.714 A1
1.714 > 1.645 or 0.0432 < 0.05 M1 OE
[Reject H0 ] There is sufficient evidence (at 5% level) to suggest A1FT OE
[mean] mass is greater than 253 FT their ‘1.714’ in context, not definite, no contradictions.
Accept critical value method of 253.57 < 253.60 or 253.02 > 253.
Use of a two-tailed test scores B0 M1 A1 M1 A0 (comp with 0.025
1.96
).
5
4(b)(ii) Not true. Large sample, [so sample mean is approx normally B1 OE
distributed]. Allow ‘Not true. Large sample’
or ‘Not true. n is large’
or ‘Not true. CLT used’.
1
Sales of cell phones at a certain shop occur singly, randomly and independently.
(a) State one further condition that must be satisfied for the number of sales in a certain time period to be well modelled by a Poisson distribution. [1]
The average number of sales per hour is 1.2 .
Assume now that a Poisson distribution is a suitable model.
(b) Find the probability that the number of sales during a randomly chosen 12-hour period will be more than 12 and less than 16. [3]
(c) Use a suitable approximating distribution to find the probability that the number of sales during a randomly chosen 1-month period (140 hours) will be less than 150. [4]
Mark scheme
5(a) Constant average rate B1 OE
Accept constant rate. Allow without context.
1
5(b) λ = 14.4 B1
13
e −14.4 ( 14.4 + 14.4
14
+ 14.4 )
15
M1 Poisson P(13, 14, 15). Expression must be seen.
13! 14! 15!
-14.4 Allow one end error; allow any λ. Allow fully correct sigma
or e (183837 + 189089 + 181526)
notation.
or (0.102469 + 0.105396 + 0.101181)
= 0.309 (3sf) A1 SC: 0.309 with no working scores B1 B1.
3
5(c) N(140×1.2, 140×1.2) or N(168, 168) B1 Stated or implied.
149.5 − 168 M1 Standardising using their mean and variance. Allow with wrong or
[= −1.427] no continuity correction.
168
Φ(“−1.427”) = 1 − Φ(“1.427”) M1 For area consistent with their working.
= 0.0768 or 0.0767 (3sf) A1
4
f(x)
a
O a x
The diagram shows the graph of the probability density function, f , of a random variable X . The graph is a quarter circle entirely in the first quadrant with centre (0, 0) and radius a, where a is a positive constant. Elsewhere f (x) = 0 .
2
(a) Show that a = . [2]
r
4
(b) Show that f (x) = - x2 . [2]
r
8
(c) Show that E (X ) = . [4]
3 r3
Mark scheme
6(a) 1
π a2 = 1 M1 OE
4
Attempt to set area = 1.
a= 2 A1 AG
π
Correct equation and correctly rearranged to a = …
No errors seen.
2
6(b)
( ) M1 Or x2 + y2 = a2.
2
x2 + y2 = 2
π
[y2 = π4 − x2 ] Must see at least one intermediate step
2 A1 AG
π −x
4
f(x) =
Convincingly rearranged to reach given answer.
No errors seen.
2
6(c) 2
π B1 Correct expression for E(X)= ∫xf(x) dx with correct limits (accept
0
x π
− x 2 dx
4 limits 0 and a).
2 M1 Integrate xf(x) with any limits or none.
− 13 (
4 − x 2 32 π
π 0) Must reach expression of form:
( −x ) .
3
4 2 2
any constant × π
A1 Wholly correct integration and limits.
8 A1 AG
Correctly obtained with no errors seen.
3 π3
4
Every July, as part of a research project, Rita collects data about sightings of a particular kind of bird.
Each day in July she notes whether she sees this kind of bird or not, and she records the number X of days on which she sees it. She models the distribution of X by B (31, p), where p is the probability of seeing this kind of bird on a randomly chosen day in July.
Data from previous years suggests that p = 0.3, but in 2022 Rita suspected that the value of p had been reduced. She decided to carry out a hypothesis test.
In July 2022, she saw this kind of bird on 4 days.
(a) Use the binomial distribution to test at the 5% significance level whether Rita’s suspicion is justified. [5]
In July 2023, she noted the value of X and carried out another test at the 5% significance level using the same hypotheses.
(b) Calculate the probability of a Type I error. [2]
Rita models the number of sightings, Y , per year of a different, very rare, kind of bird by the distribution
B (365, 0.01).
(c) (i) Use a suitable approximating distribution to find P (Y = 4). [3]
(ii) Justify your approximating distribution in this context. [1]
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: p = 0.3 B1
H1: p < 0.3
B(31, 0.3), P(X ⩽ 4) = M1 No end errors.
0.731 + 31×0.730×0.3 + 31C2×0.729×0.32 + 31C3×0.728×0.33 +
31
C4×0.727×0.34
= 0.00001577 + 0.0002096 + 0.0013475 + 0.0055826 + 0.016748
= 0.0239 (3sf) A1 SC 0.0239 with no working scores B1.
‘0.0239’ < 0.05 M1 Valid comparison.
[reject H0] A1FT In context.
‘There is sufficient evidence (at 5% level) to support Rita’s Not definite. No contradictions.
suspicion’, FT their 0.0239.
or ‘There is sufficient evidence to suggest the probability of seeing
this type of bird has decreased’
5
7(b) P(X < 5) = [‘0.0239’ + 31C5×0.726×0.35] = 0.0627 [which is > 0.05] B1FT Attempt P(X ⩽ 5).
Only FT if > 0.05.
Only FT their 0.0239 if P(X ⩽ 4) attempted in (a); arithmetic error
only.
P(Type I error) = ‘0.0239’ B1FT Only FT their 0.0239 if P(X ⩽ 4) attempted in (a); arithmetic error
only and their 0.0239 < 0.05.
2
7(c)(i) [λ=] 3.65 B1 Stated or implied.
e−3.65 × 3.65
4
M1 Must see expression. Any λ.
4!
= 0.192 (3sf) A1 SC: Use of Binomial. 0.193 scores B1.
SC: 0.192 with no working scores B1 B1.
3
7(c)(ii) n = 365 > 50 B1 Explicit. Both needed.
np = 3.65 < 5 or p = 0.01 < 0.1 Note: and ‘n large, p small’ is insufficient.
1
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Mathematics 9709/61 for May/June 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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