Mathematics 9709/62 February/March 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics 2, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The lengths, in millimetres, of a random sample of 12 rods made by a certain machine are as follows.
200 201 198 202 200 199 199 201 197 202 200 199
(a) Find unbiased estimates of the population mean and variance. [3]
(b) Give a statistical reason why these estimates may not be reliable. [1]
Mark scheme
1(a) 1199 2398 B1 Accept in any form
Est (μ) = or 199.833 or 200 or [mm]
6 12
12 479226 '1199'2 1 '2398'2 M1 Use of their values in correct formula (may be implied)
Est (σ2) = − or '479226'−
11 12 6 11 6
= 2.33 (3 sf) [mm2] A1 7
Accept
3
3
1(b) Small sample B1 Accept not ‘not representative’ unless qualified.
1
Harry has a five-sided spinner with sectors coloured blue, green, red, yellow and black. Harry thinks the spinner may be biased. He plans to carry out a hypothesis test with the following hypotheses.
H0 : P the spinner lands on blue = 15
H1 : P the spinner lands on blue ≠ 15
Harry spins the spinner 300 times. It lands on blue on 45 spins.
Use a suitable approximation to carry out Harry’s test at the 5% significance level. [5]
Mark scheme
2 B(300, 15 ) → N(60,48) B1 SOI
45.5 − 60 M1 Condone with wrong or no continuity correction
48
= –2.093 A1
‘2.093’ > 1.96 M1 Valid comparison
Note: ϕ('–2.093') (= 0.0182),
0.0182< 0.025
[Evidence to reject H0] There is evidence that P(landing on blue) ≠ 15 A1 FT Allow ‘There is evidence that the spinner is biased.’
In context, not definite, no contradictions
Condone critical values method (critical value 46.42
M1 A1 and 45.5 < ‘46.42’ M1 for valid comparison
A1 for correct conclusion)
SC: 0.0182 unsupported:
0.0182 < 0.025
And there is evidence that the spinner is biased.
In context, not definite B1 only
5
A random sample of 500 households in a certain town was chosen. Using this sample, a confidence interval for the proportion, p, of all households in that town that owned two or more cars was found to be 0.355 < p < 0.445.
Find the confidence level of this confidence interval. Give your answer correct to the nearest integer.
[5]
Mark scheme
3 est(p) = 0.4 B1
'0.4'× (1 − '0.4') M1 OE
'0.4'+ z [ = 0.445] Use of their 0.4 in a correct expression
500
'0.4'× (1 − '0.4') A1 Condone 2.053 and 2.05
z = 0.045 ÷ = 2.054
500
0.98 – (1 – 0.98) M1
96% confidence A1 CWO, must be integer
5
In the past the time, in minutes, taken by students to complete a certain challenge had mean 25.5 and standard deviation 5.2. A new challenge is devised and it is expected that students will take, on average, less than 25.5 minutes to complete this challenge. A random sample of 40 students is chosen and their mean time for the new challenge is found to be 23.7 minutes.
(a) Assuming that the standard deviation of the time for the new challenge is 5.2 minutes, test at the 1% significance level whether the population mean time for the new challenge is less than
25.5 minutes. [5]
(b) State, with a reason, whether it is possible that a Type I error was made in the test in part (a).
[1]
Mark scheme
4(a) H0: μ = 25.5 B1
H1: μ < 25.5
23.7 − 25.5 M1 Must have √40
5.2 ÷ 40
= –2.189 A1
'2.189' < 2.326 M1 For valid comparison
For two-tailed test: allow compare 2.576 if H1: μ ≠
25.5
[Accept H0] No evidence that mean time has decreased A1 FT In context, not definite, no contradictions
FT their 2.189 but no FT for two-tailed test
N.B. Use of two-tailed test can score max B0 M1 A1
M1 A0
Condone use of critical value method
(23.59 M1 A1 and 23.7 > 23.59 M1 A1 correct
conclusion or 25.612 M1 A1 and 25.5 < 25.612 M1 A1
with correct conclusion)
5
4(b) No, because H0 was not rejected B1 FT FT their conclusion in (a)
1
The heights of buildings in a large city are normally distributed with mean 18.3 m and standard deviation 2.5 m.
(a) Find the probability that the total height of 5 randomly chosen buildings in the city is more than 95 m. [4]
(b) Find the probability that the difference between the heights of two randomly chosen buildings in the city is less than 1 m. [5]
Mark scheme
5(a) Mean = 5×18.3 and Variance = 5×2.52 [= N(91.5, 31.25)] B1 SOI
95 − '91.5 M1 FT their mean and variance
[= 0.626]
'31.25'
1 – Φ(‘0.626’) M1 For finding area consistent with their values
0.266 (3 sf) A1
4
5(b) E(D) = 0 B1 Or E(D ‒ 1) = ‒1
Var(D) = 2.52 × 2 [= 12.5] B1
1− 0 −1 − 0 M1 FT their E and Var
[= 0.283] or [= –0.283]
'12.5' '12.5'
Φ(‘0.283’) – (1 – ɸ(0.283)) [= 0.6115 – 0.3885] M1 For finding area consistent with their values
0.223 (3 sf) A1
5
In a game a ball is rolled down a slope and along a track until it stops. The distance, in metres, travelled by the ball is modelled by the random variable X with probability density function
D
−k x − 1 x − 3 1 ≤ x ≤ 3, f x =
0 otherwise, where k is a constant.
(a) Without calculation, explain why E X = 2. [1]
(b) Show that k = 34 . [3]
(c) Find Var X . [3]
One turn consists of rolling the ball 3 times and noting the largest value of X obtained. If this largest value is greater than 2.5, the player scores a point.
(d) Find the probability that on a particular turn the player scores a point. [4]
Mark scheme
6(a) Quadratic curve, hence symmetrical B1 OE. Allow sketch and ‘symmetrical’ or just ‘curve
symmetrical’
1
6(b) 3 M1 Attempt to integrate f(x) and ‘= 1’. Ignore limits at this
1 ( x − 4 x + 3)dx = 1
2
–k
stage
x3 3 A1 Fully correct expression (correct integration and limits)
–k − 2 x 2 + 3 x
3 1
4 4 A1 AG, OE. Correctly substitute limits and ‘= 1’ and
−k × 0 − = 1 or k × = 1 correctly obtain result with no errors seen.
3 3
3
k = 4
3
6(c) 3 3 4 M1 Attempt to integrate x2 f(x) from 1 to 3
− (
4 1
)
x − 4 x3 + 3x 2 dx
ù3 A1 Correct integration and limits
3 éê x5 4 3ú
- ´ê - x + x ú
4 ë5 û1
é 3 28 21ù
ê= ´ = ú
êë 4 5 5 úû
21 2 A1
5 − 2 = 0.2
3
6(d) 3
3 M1 OE. Attempt to integrate f(x), from 2.5 to 3 (or 1 to
-
4ò
( x2 - 4 x + 3)dx 2.5)
2.5
ù3 A1
3 éê x3 5
= - ´ ê - 2 x + 3 x úú =
2
or 0.15625
4 ë3 û 2.5 32
æ 5 ö
3 M1 OE. FT their 325 .
1 - çç1 - ' '÷÷÷
çè 32 ø
= 0.399 (3 sf) A1
4
(a) Two ponds, A and B, each contain a large number of fish. It is known that 2.4% of fish in pond A are carp and 1.8% of fish in pond B are carp. Random samples of 50 fish from pond A and 60 fish from pond B are selected.
Use appropriate Poisson approximations to find the following probabilities.
(i) The samples contain at least 2 carp from pond A and at least 2 carp from pond B. [3]
(ii) The samples contain at least 4 carp altogether. [3]
(b) The random variables X and Y have the distributions Po , and Po - respectively. It is given that
³ P X = 0 = P Y = 02 ,
³ P X = 2 = kP Y = 12 , where k is a non-zero constant.
Find the value of k. [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)(i) 0.024 × 50 [= 1.2] and 0.018 × 60 [= 1.08] B1
(1 – e–1.2(1 + 1.2)) × (1 – e–1.08(1 + 1.08)) M1 For (1 – e–λ(1+λ)) × (1 – e–μ(1+μ)) any λ, μ (λ ≠ μ)
Allow one end error on either or both terms
= 0.0991 (3 sf) A1 Unsupported answer scores maximum SC B1 B1
SC Use of binomial 0.0994 scores B1 only
3
7(a)(ii) λ = 0.024 × 50 + 0.018 × 60 M1 or their 1.2 + 1.08 (NB 0.024+0.018 is M0)
æ 2.282 2.283 ö÷ M1 any λ and allow one end error
1 - e-2.28 ´ççç1 + 2.28 + + ÷÷
çè 2! 3! ÷ø
= 0.197 (3 sf) A1 Unsupported answer scores maximum SC B2
3
7(b) 2 M1
e-λ = éê e-μ ùú = e-2 μ
ë û
λ2 2 M1
e-λ ´ = k éê e-μ ´ μ ùú
2 ë û
e–2μ × 2μ2 = k × e–2μ × μ2 M1 OE. Use of λ = 2μ to find equation in μ and k only (or λ
and k only)
k=2 A1
4
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/62, February/March 2022 (9709_m22_ms_62), and the question paper (9709_m22_qp_62), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/62 for February/March 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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