Mathematics 9709/63 October/November 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The heights, in metres, of a random sample of 10 mature trees of a certain variety are given below.
5.9 6.5 6.7 5.9 6.9 6.0 6.4 6.2 5.8 5.8
Find unbiased estimates of the population mean and variance of the heights of all mature trees of this variety. [3]
Mark scheme
1 62.1 B1 OE
= 6.21
10
[Σx2 = 387.05] M1 Can be implied.
Accept alternative methods (e.g. working mean of 6).
10 their '387.05' 2 Biased 0.1409 M0.
− ( their '6.21')
9 10
1 their '387.05' ( their '6.21')
2
or −
9 10 10
1409 A1
= 0.157 (3 sf) or
9000
3
A spinner has five sectors, each printed with a different colour. Susma and Sanjay both wish to test whether the spinner is biased so that it lands on red on fewer spins than it would if it were fair. Susma spins the spinner 40 times. She finds that it lands on red exactly 4 times.
(a) Use a binomial distribution to carry out the test at the 5% significance level. [5]
Sanjay also spins the spinner 40 times. He finds that it lands on red r times.
(b) Use a binomial distribution to find the largest value of r that lies in the rejection region for the test at the 5% significance level. [3]
Mark scheme
2(a) H0: P(red) = 0.2 H1: P(red) < 0.2 B1 Allow H0: p = 0.2 H1: p < 0.2 .
P(X < 4) = 0.840 + 40×0.839×0.2 + 40C2×0.838×0.22 M1 For full expression seen.
+ 40C3×0.837×0.23 + 40C4×0.836×0.24 Allow one term omitted, incorrect or extra.
0.0759 A1 SC 0.0759 without working B1.
their ‘0.0759’ > 0.05 M1 Valid comparison (from binomial probs) of their P(X ⩽ 4)
with 0.05.
[Do not reject H0 ]. Not enough evidence that it lands on red fewer times A1 FT FT their 0.0759.
than if it were fair or not enough evidence to suggest that the spinner is In context, not definite, no contradictions.
biased
5
2(b) P(X ⩽ 3) = ` 0.0759 ` – 40C4×0.836×0.24 M1 OE Attempted. Must be using B(40, 0.2).
Method could be implied by correct answer here.
= 0.0285 or 0.0284 *A1
Largest value of r is 3 DA1
3
Drops of water fall randomly from a leaking tap at a constant average rate of 5.2 per minute.
(a) Find the probability that at least 3 drops fall during a randomly chosen 30-second period. [3]
(b) Use a suitable approximating distribution to find the probability that at least 650 drops fall during a randomly chosen 2-hour period. [4]
Mark scheme
3(a) λ = 5.2 ÷ 2 [= 2.6] B1
1 – e–2.6(1 + 2.6 + 2.62 ) or 1 – e-2.6( 1 + 2.6 + 3.38 )
2
M1 Allow any λ.
Allow one end error.
or 1- ( 0.07427 + 0.1931 + 0.2510 )
Must see expression.
= 0.482 (3 sf) B1
3
3(b) N(120×5.2, 120×5.2) B1 Stated or implied. Give at early stage.
649.5 − their '624' M1 Allow with no or wrong continuity correction.
[= 1.021]
their '624'
1– Φ(their ‘1.021’) M1 For area consistent with their working.
= 0.154 (3 sf) A1
4
Each month a company sells X kg of brown sugar and Y kg of white sugar, where X and Y have the independent distributions N 2500, 1202 and N 3700, 1302 respectively.
(a) Find the mean and standard deviation of the total amount of sugar that the company sells in
3 randomly chosen months. [3]
The company makes a profit of $1.50 per kilogram of brown sugar sold and makes a loss of $0.20 per kilogram of white sugar sold.
(b) Find the probability that, in a randomly chosen month, the total profit is less than $3000. [5]
Mark scheme
4(a) Mean = [3 (2500 + 3700)] = 18600 (kg) B1
Var(Total profit) = 3(1202 + 1302) or 93900 M1 or √ of this stated.
sd = 306 (kg) (3 sf) A1
3
4(b) E(1.5X – 0.2Y) = 1.5x2500 – 0.20x3700 = [3010] B1 Give at early stage.
Var(1.5X – 0.2Y) = 1.52×1202 + 0.22×1302 [= 33076] B1 Correct expression or result or sd = 182 (3 sf) seen.
3000 − 3010 M1 Ignore continuity correction attempts.
[= –0.055] E(X) and Var must come from a combination attempt.
their '33076'
Can be implied.
Φ(their ‘–0.055’) = 1 – Φ(their ‘0.055’) M1 For area consistent with their values.
Can be implied.
= 0.478 (3 sf) A1
5
A builders’ merchant sells stones of different sizes.
(a) The masses of size A stones have standard deviation 6 grams. The mean mass of a random sample of 200 size A stones is 45 grams.
Find a 95% confidence interval for the population mean mass of size A stones. [3]
(b) The masses of size B stones have standard deviation 11 grams. Using a random sample of size 200, an !% confidence interval for the population mean mass is found to have width 4 grams.
Find !. [4]
Mark scheme
5(a) 6 M1 For expression of correct form, any z.
45 ± z × Accept one side of interval for M1.
200
z = 1.96 B1 Must be seen.
44.2 to 45.8 (3 sf) A1 Must be an interval.
3
5(b) 11 M1 Or … = 4 for M1
z× =2
200
z = 2.571 A1 Accept 3sf if nothing better seen.
ɸ(their '2.571') = 0.9949 M1 OE For area consistent with their values.
and their '0.9949' – (1 – their '0.9949') [= 0.9898] Must be seen.
α = 99.0 (3 sf) A1 Allow 99.
cwo
Final answer of 0.99 scores A0.
4
The diagram shows the graph of the probability density function of a random variable X that takes values between −1 and 3 only. It is given that the graph is symmetrical about the line x = 1. Between x = −1 and x = 3 the graph is a quadratic curve.
The random variable S is such that E S = 2 × E X and Var S = Var X .
(a) On the grid below, sketch a quadratic graph for the probability density function of S. [1]
The random variable T is such that E T = E X and Var T = 14 Var X .
(b) On the grid below, sketch a quadratic graph for the probability density function of T . [2]
It is now given that
T
3 3 + 2 x − x2 −1 ≤ x ≤ 3, f x = 32
0 otherwise.
(c) Given that P 1 − a < X < 1 + a = 0.5, show that a3 − 12a + 8 = 0. [3]
(d) Hence verify that 0.69 < a < 0.70. [1]
Mark scheme
6(a) Curve of similar shape, x = 0 to x = 4, with highest point (2, 0.375) B1 Not straight lines, not bell shaped. Must be correct at x = 0
and x = 4, highest point must be at x = 2, y value ± ¼
square.
Must not go below the x-axis.
1
6(b) Curve of similar shape, from x = 0 to x = 2, highest point at x = 1 B1 Not straight lines, not bell shaped. Must be correct at x = 0
and x = 2. Highest point must be at x = 1.
Highest point (1, 0.75) B1
2
6(c) 3 1+ a M1 OE
(3 + 2 x − x )dx = 14 (3 + 2 x − x 2 )dx = 12
3 2 3
32
or 32 Attempt to integrate f(x) and correct limits with correct
1+ a 1− a RHS.
3 3 1+ a A1 Correct integration.
3
3x + x 2 − x3 = 14 or 323 3x + x 2 − x3 =1
3
32
1+ a 1− a 2
a3 – 12a + 8 = 0 A1 AG Substitute limits and correctly obtain equation.
May see 3/32(6a+4a–6a/3–2a3/3) = 0.5
No errors seen..
3
6(d) 0.693 – 12×0.69 + 8 = 0.049 (2 sf) > 0 B1 AG
0.703 – 12×0.70 + 8 = –0.057 (2 sf) < 0 Must state either the correct expression and > 0 and < 0
Hence 0.69 < a < 0.70 or both answers to 2 sf.
Both answers correct and conclusion.
Accept equivalent expressions.
OR: a = 0.695 (3 sf) which is between 0.69 & 0.70.
1
In the past Laxmi’s time, in minutes, for her journey to college had mean 32.5 and standard deviation 3.1. After a change in her route, Laxmi wishes to test whether the mean time has decreased.
She notes her journey times for a random sample of 50 journeys and she finds that the sample mean is 31.8 minutes. You should assume that the standard deviation is unchanged.
(a) Carry out a hypothesis test, at the 8% significance level, of whether Laxmi’s mean journey time has decreased. [5]
Later Laxmi carries out a similar test with the same hypotheses, at the 8% significance level, using another random sample of size 50.
(b) Given that the population mean is now 31.5, find the probability of a Type II error. [5]
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: Population mean time (or μ) = 32.5 B1 Not just “mean”.
H1: Population mean time (or μ) < 32.5
31.8 − 32.5 M1 Must have 50 .
±
3.1 ÷ 50 Could be implied.
= ± –1.597 A1
‘–1.597’ < –1.406 [or ‘1.597’ > 1.406] M1 Valid comparison of their zcalc with ±1.406.
or 0.0551 < 0.08 (or 0.0552 < 0.08).
[ reject H0 ] A1 FT In context, not definite, no contradictions.
There is evidence that [population] [mean ] time has decreased
Note: Accept critical value method 31.88 (31.9) M1 A1
and 31.8 < 31.88 M1 conclusion A1.
5
7(b) a − 32.5 M1 Standardise with 32.5 and 50 and z value on RHS.
= −1.406
3.1 ÷ 50
a = 31.88 or 31.9 A1 May be seen in part (a). Can score M1A1 here as well
using a similar approach to (a).
their '31.88 − 31.5 M1 Standardise with their cv and mean = 31.5.
[= 0.8668 to 0.8760]
3.1 ÷ 50 Must have 50 .
1 – Φ(‘0.8668') M1 For area consistent with their working.
= 0.190 to 0.193 (3 sf) A1
5
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Mathematics 9709/63 for October/November 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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