Mathematics 9709/62 February/March 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics 2, February/March 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Anita carried out a survey of 140 randomly selected students at her college. She found that 49 of these students watched a TV programme called Bunch.
(a) Calculate an approximate 98% confidence interval for the proportion, p, of students at Anita’s college who watch Bunch. [3]
Carlos says that the confidence interval found in (a) is not useful because it is too wide.
(b) Without calculation, explain briefly how Carlos can use the results of Anita’s survey to find a narrower confidence interval for p. [1]
Mark scheme
1(a) 140
49
= 0.35
0.35(1− 0.35) M1 Use of formula of correct form, ft their 49 , any z (not a
0.35 ± z 140
140
probability).
z = 2.326 B1 Accept 2.326 to 2.329 .
Confidence interval = 0.256 to 0.444 (3 sf) A1 Must be an interval.
3
1(b) Find a smaller percentage confidence interval/ lower level of confidence B1 ISW if 2 reasons given. Just saying ‘use smaller z’ oe
B0.
Accept a correct example e.g. 90% (even if not qualified
with statement).
1
The number of orders arriving at a shop during an 8-hour working day is modelled by the random variable X with distribution Po 25.2.
(a) State two assumptions that are required for the Poisson model to be valid in this context. [2]
(b) (i) Find the probability that the number of orders that arrive in a randomly chosen 3-hour period is between 3 and 5 inclusive. [3]
(ii) Find the probability that, in two randomly chosen 1-hour periods, exactly 1 order will arrive in one of the 1-hour periods, and at least 2 orders will arrive in the other 1-hour period. [4]
(c) The shop can only deal with a maximum of 120 orders during any 36-hour period.
Use a suitable approximating distribution to find the probability that, in a randomly chosen
36-hour period, there will be too many orders for the shop to deal with. [4]
Mark scheme
2(a) Orders arrive at constant mean rate (must say mean or rate) Must be in context (accept 25.2 as context).
Orders arrive at random
Orders arrive independently B1 Any one reason correctly stated.
Orders arrive singly
B1 A second reason correctly stated.
SC B1: both correct, not in context.
2
2(b)(i) λ = 83 × 25.2 [= 9.45] B1
e–’9.45’( 9.45
3
+ 9.45
4
+ 9.45
5
) or e–‘9.45’ (140.65 + 332.29 + 628.03) or 0.01107 + M1 Allow any λ. Allow end errors.
3! 4! 5!
Expression must be seen.
0.02615 + 0.04942
= 0.0866 (3 sf) A1 If M0 allow SC B1 for 0.0866 no working seen.
3
2(b)(ii) e–3.15 ×3.15 or (1 – e–3.15(1 + 3.15)) or 0.135 or 0.822 (3 sf) B1
e–3.15 × 3.15 × (1 – e–3.15(1 + 3.15) ) M1 M1 for product of two Poisson probabilities P(1) × (1–
P(0,1)) (no end errors accepted). Accept any λ.
× 2 or 0.111 × 2 M1 M1 for their product of two Poisson probabilities
(accept end errors) × 2. Accept any λ
0.222 (3 sf) A1
4
2(c) N(113.4, 113.4) B1 SOI
120.5−113.4 M1 Standardise with their values. Allow wrong or no cc.
113.4
[= 0.667]
Must have √.
1 – ɸ(their ‘0.667’) M1 For probability area consistent with their values.
= 0.252 (3 sf) A1
4
f x
x
0 0.6 1.2 1.5 3
The diagram shows the graph of the probability density function, f, of a random variable X that takes values between x = 0 and x = 3 only. The graph is symmetrical about the line x = 1.5.
(a) It is given that P X < 0.6 = a and P 0.6 < X < 1.2 = b.
Find P 0.6 < X < 1.8 in terms of a and b. [2]
(b) It is now given that the equation of the probability density function of X is
T 2 kx 3 − x2 0 ≤ x ≤ 3, f x =
0 otherwise, where k is a constant.
(i) Show that k = 10
81
. [3]
(ii) Find Var X . [3]
Mark scheme
3(a) 1 – 2(a + b) or 1 – 2a or 0.5 − a – b or 1-(a+b) or a+a+b M1 OE. Seen or implied – may be on the diagram
(or for correct un-simplified final expression).
P(0.6 < X < 1.8) = 1 – 2a – b A1 Accept 1 – (2a + b).
2
3(b)(i) 3 M1 Attempt integrate f(x) ignore limits and ‘= 1’.
k (9 x 2 − 6 x3 + x 4 )dx = 1
0
3 A1 Correct integration seen, correct limits.
k 93x − 64x + x5
3 4 5
=1
0
81
k × 10 = 1, k = 10 A1 AG. Convincingly obtained. No errors seen.
81
(Must see integration).
3
3(b)(ii) 3 M1 Attempt integrate x2f(x) between 0 and 3
(9 x − 6 x + x )dx
10 4 5 6
81 condone missing k.
0 Must see integration or correct answer of 18/7 seen or
10 9 x5 3 implied.
81 5 − x + 7 0 [= 7 or 2.57...]
6 x7 18
18
7 – ‘1.5’2 M1 Their integral of x2f(x) – 1.52 (or their mean2).
= 289 or 0.321 A1
3
2
The number of accidents per 3-month period on a certain road has the distribution Po ,. In the past the value of , has been 5.7. Following some changes to the road, the council carries out a hypothesis test to determine whether the value of , has decreased. If there are fewer than 3 accidents in a randomly chosen 3-month period, the council will conclude that the value of , has decreased.
(a) Find the probability of a Type I error. [2]
(b) Find the probability of a Type II error if the mean number of accidents per 3-month period is now actually 0.9. [3]
Mark scheme
4(a) e-5.7(1 + 5.7 + 5.72! ) or e-5.7(1 + 5.7 + 16.245) or 0.003346 + 0.01907 + 0.05436 M1 Allow one end error.
Must see this expression.
= 0.0768 (3 sf) A1 SC B1 for unsupported answer of 0.0768 .
2
2
4(b) e –0.9(1 + 0.9 + 0.92! ) M1 Attempted; allow one end error (must see expression).
2
= 1 – e –0.9(1 + 0.9 + 0.92! )= 1 – e –0.9(1 + 0.9 + 0.405) = 1 – (0.4066 + 3659 + A1 Correct expression P(X ⩾ 3) no end errors (must see
expression).
0.1647)
= 0.0629 (3 sf) A1 SC B2 for unsupported answer of 0.0629 .
3
The masses, in grams, of large and small packets of Maxwheat cereal have the independent distributions
N 410.0, 3.62 and N 206.0, 3.72 respectively.
(a) Find the probability that a randomly chosen large packet has a mass that is more than double the mass of a randomly chosen small packet. [5]
The packets are placed in boxes. The boxes are identical in appearance. 60% of the boxes contain exactly 10 randomly chosen large packets. 40% of the boxes contain exactly 20 randomly chosen small packets.
(b) Find the probability that a randomly chosen box contains packets with a total mass of more than
4080 grams. [6]
Mark scheme
5(a) D = L – 2S E(D) = 410 – 2(206) = –2 B1 SOI. OE using 2S–L.
Var(D) = 3.62 + 4 × 3.72 [= 67.72] B1 SOI
0 −( −2) M1 For standardising using their values.
'67.72'
[= 0.243]
1 – ɸ(their ‘0.243’) M1 For probability area consistent with their values.
= 0.404 (3 sf) A1 As final answer.
5
5(b) B1 One of N(4100, 129.6) or N(4120, 273.8) USED
TL ~ N(4100, 10×3.62) TS ~ N(4120, 20×3.72)
(unchanged) in a standardising equation.
4080−4100 4080−4120
M1 Standardising with either their N(4100, 129.6) or
(= –1.757) (= –2.417) N(4120, 273.8) or their N(…,…) (could be from a
'129.6' '273.8'
combination).
M1 One area consistent with their working (could be from a
1 – ɸ(‘–1.757’) = ɸ(1.757) 1 – ɸ(‘–2.417’) = ɸ(2.417)
combination). Do not ISW.
A1 Both of these correct. Do not ISW.
= 0.9605 or 0.961 = 0.9921 or 0.9922 or 0.992
0.6 × ‘their 0.9605’ + 0.4 × ‘their 0.9921’ M1 Must be using probabilities.
= 0.973 (3 sf) A1
6
Last year, the mean time taken by students at a school to complete a certain test was 25 minutes.
Akash believes that the mean time taken by this year’s students was less than 25 minutes. In order to test this belief, he takes a large random sample of this year’s students and he notes the time taken by each student. He carries out a test, at the 2.5% significance level, for the population mean time,
- minutes. Akash uses the null hypothesis H0 : - = 25.
(a) Give a reason why Akash should use a one-tailed test. [1]
Akash finds that the value of the test statistic is z = −2.02.
(b) Explain what conclusion he should draw. [2]
In a different one-tailed hypothesis test the z-value was found to be 2.14.
(c) Given that this value would lead to a rejection of the null hypothesis at the !% significance level, find the set of possible values of !. [3]
The population mean time taken by students at another school to complete a test last year was m minutes. Sorin carries out a one-tailed test to determine whether the population mean this year is less than m, using a random sample of 100 students. He assumes that the population standard deviation of the times is 3.9 minutes. The sample mean is 24.8 minutes, and this result just leads to the rejection of the null hypothesis at the 5% significance level.
(d) Find the value of m. [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
6(a) He is expecting a decrease (in μ) B1 OE
1
6(b) −2.02 < −1.96 M1 For valid comparison.
Allow 2.02 > 1.96 or 0.0217 < 0.025 or 0.9783 > 0.975
(Reject H0 ) A1 OE (such as evidence to support Akash’s belief), in
There is evidence to suggest that this year’s (mean) time is less than 25 context, not definite.
No contradictions.
2
6(c) 1 – ɸ(2.14) [= 0.0162] M1
1.62 A1 Allow 1.62% or 1.6 or 1.6%.
α ⩾ 1.62 (3 sf) A1ft FT their 1.62 .
Allow α ⩾ 1.62% or 1.6 or 1.6%. Condone >.
3
6(d) 24.8 − m M1 For standardising.
3.9 ÷10
24.8 − m
= −1.645 M1 Equate their standardised value to −1.645 (signs must be
3.9 ÷10
consistent).
m = 25.4 (3 sf) A1
3
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Mathematics 9709/62 for February/March 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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