Mathematics 9709/61 May/June 2023: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
In a certain country, 20 540 adults out of a population of 6 012 300 have a degree in medicine.
(a) Use an approximating distribution to calculate the probability that, in a random sample of 1000 adults in this country, there will be fewer than 4 adults who have a degree in medicine. [4]
(b) Justify the approximating distribution used in part (a). [2]
Mark scheme
1(a) 20540/6012300 = 0.0034163 B1
[1000 × 0.0034163 = 3.4163]
Po(3.4163) B1 Could be implied by expression seen.
2 3 M1 Allow any λ. Allow with one end error. Must see
e–their '3.4163'(1 + 3.4163 + 3.4163 + 3.4163 ) OR
2! 3! expression.
e–their '3.4163'(1 + 3.4163 + 5.8356+6.6453) or 0.03283 + 0.1122 +0.1916 +
0.21819)
= 0.555 (3sf) A1 CAO
SC No working: B1 B1 (Po must be stated) B1 correct
answer (max 3/4).
SC Binomial: B1 B0 B1 correct answer (max 2/4).
4
1(b) n = 1000 > 50 B1 Must show comparison with 50.
np = 3.4163 < 5 B1 Must show comparison with 5.
2 SC B1: n > 50 (or n large), np < 5.
SC B1: n large, p small.
(a) f x
1
x
−1 0 1 2 3
The graph of the function f is a straight line segment from 0, 0 to 2, 1.
Show that f could be a probability density function. [2]
(b) g x
x
0
The graph of the function g is a semicircle, centre 0, 0, entirely above the x-axis.
Given that g is a probability density function, find the radius of the semicircle. [2]
(c) h x
x
0 10 20
The time, X minutes, taken by a large number of students to complete a test has probability density function h, as shown in the diagram.
(i) Without calculation, use the diagram to explain how you can tell that the median time is less than 15 minutes. [1]
It is now given that d 40 1
2
− 10 ≤ x ≤ 20, h x = x 10
0 otherwise.
(ii) Find the mean time. [3]
Mark scheme
2(a) 2 B1 Calculation and result.
1
2
× 2 × 1 or
0
1
2
xdx = 1, which is the correct area under a pdf.
f(x) ⩾ 0 B1 Condone f(x) > 0 or ‘Line is above x-axis’ OE.
2
2(b) 1
πr2 = 1 M1 Area of semi-circle equated to 1 OE.
2
Missing factor of ½ gets M1A0.
r= 2
or 0.798 (3sf) A1
π
2
2(c)(i) Area to the left of 15 is greater than 0.5 B1 OE, e.g. ‘The distribution of X is skewed to the right /
positively skewed, suggesting the median will be less
than the mid-point of the interval.’ or ‘The distribution
of X is skewed to the right / positively skewed’ or ‘It is
a decreasing function suggesting the median will be less
than the mid-point of the interval’.
1
2(c)(ii) 20 M1 Integration of xh(x) attempted. Ignore limits.
10
( 40x − 10x )dx
20 A1 Correct integration and limits (can be implied by final
40ln x − 20
x2
answer).
10
= 40ln 2 – 15 or 12.7 (3sf) A1
3
In the past, the annual amount of wheat produced per farm by a large number of similar sized farms in a certain region had mean 24.0 tonnes and standard deviation 5.2 tonnes. Last summer a new fertiliser was used by all the farms, and it was expected that the mean amount of wheat produced per farm would be greater than 24.0 tonnes. In order to test whether this was true, a scientist recorded the amounts of wheat produced by a random sample of 50 farms last summer. He found that the value of the sample mean was 25.8 tonnes.
Stating a necessary assumption, carry out the test at the 1% significance level. [6]
Mark scheme
3 Assume SD still = 5.2 B1 OE i.e. ‘Assume the SD remains unchanged’.
H0: μ = 24.0 H1: μ > 24.0 B1 Or population mean; not just mean.
25.8− 24.0
5.2
M1 For standardising (could be implied). Must have √50.
50
= 2.448 A1 Or P( X > 25.8) = 0.0071 .
‘2.448’ > 2.326 M1 Or 0.0071 < 0.01 .
For valid comparison.
[Reject H0] There is evidence that (mean) amount of wheat is greater. A1FT OE. FT their zcalc.
In context, not definite, eg not ‘Mean amount of wheat
is greater’ No contradictions
CV method: CV= 25.71 M1A1 25.71<25.8 M1 A1FT
or CV=24.09 M1 A1 24.09>24 M1 A1FT.
6
A certain train journey takes place every day throughout the year. The time taken, in minutes, for the journey is normally distributed with variance 11.2.
(a) The mean time for a random sample of n of these journeys was found. A 94% confidence interval for the population mean time was calculated and was found to have a width of 1.4076 minutes, correct to 4 decimal places.
Find the value of n. [3]
(b) A passenger noted the times for 50 randomly chosen journeys in January, February and March.
Give a reason why this sample is unsuitable for use in finding a confidence interval for the population mean time. [1]
(c) A researcher took 4 random samples and a 94% confidence interval for the population mean was found from each sample.
Find the probability that exactly 3 of these confidence intervals contain the true value of the population mean. [2]
Mark scheme
4(a) z× 11.2
= 1.4076 ÷ 2 M1 Any z, but must be a z.
n
z = 1.881 or 1.882 B1
[n = ( 0.7038 ) × 11.2 ] A1 Must be a whole number.
1.881 2
n = 80
3
4(b) Jan, Feb and March not typical of whole year. B1 Or, e.g., weather is different at different times of year.
1
4(c) 0.943 × 0.06 × 4 M1
= 0.199 (3 sf) A1
2
Large packets of rice are packed in cartons, each containing 20 randomly chosen packets. The masses of these packets are normally distributed with mean 1010 g and standard deviation 3.4 g. The masses of the cartons, when empty, are independently normally distributed with mean 50 g and standard deviation 2.0 g.
(a) Find the variance of the masses of full cartons. [2]
Small packets of rice are packed in boxes. The total masses of full boxes are normally distributed with mean 6730 g and standard deviation 15.0 g. The masses of the boxes and cartons are distributed independently of each other.
(b) Find the probability that the mass of a randomly chosen full carton is more than three times the mass of a randomly chosen full box. [5]
Mark scheme
5(a) 2.02 + 20 × 3.42 M1
= 235.2 A1
2
5(b) E(C – 3B) = 50 + 20×1010 – 3×6730 or 60 B1
Var(C – 3B) = ‘235.2’ + 9×152 or 2260.2 M1 FT their values from (a).
[C – 3B ~ N(‘60’, ‘2260.2’)] M1 Standardising with their values (could be implied).
= 0 − 60 [= –1.262]
2260.2
1 – Φ(‘–1.262’) = Φ(‘1.262’) M1 Probability area consistent with their values.
= 0.897 (3 sf) A1
5
A sample of 5 randomly selected values of a variable X is as follows:
1 2 6 1 a where a > 0.
Given that an unbiased estimate of the variance of X calculated from this sample is 11
2
, find the value of a. [3]
Mark scheme
6 5 1 + 22 + 62 + 1 + a 2 1 + 2 + 6 + 1 + a 11
−
2
( 5 ) 5
M1* OE attempted or e.g., 42+ a 2 − 10+ a 2 = 22 . Allow use of
=
5
4 5 5 2 biased i.e., without 54 .
1 (10 + a 2 ) 11
or (42 + a 2
) − =
4 5 2
4a2 – 20a + 0 = 0 or a2 – 5a + 0 = 0 DM1 Two- or three-term quadratic equation in a, with at least
two terms correct.
a=5 A1 Ignore a = 0, if seen.
3
The number of accidents per week at a certain factory has a Poisson distribution. In the past the mean has been 1.9 accidents per week. Last year, the manager gave all his employees a new booklet on safety. He decides to test, at the 5% significance level, whether the mean number of accidents has been reduced. He notes the number of accidents during 4 randomly chosen weeks this year.
(a) State suitable null and alternative hypotheses for the test. [1]
(b) Find the critical region for the test and state the probability of a Type I error. [6]
(c) State what is meant by a Type I error in this context. [1]
(d) During the 4 randomly chosen weeks there are a total of 3 accidents.
State the conclusion that the manager should reach. Give a reason for your answer. [2]
(e) Assuming that the mean remains 1.9 accidents per week, use a suitable approximation to calculate the probability that there will be more than 100 accidents during a 52-week period. [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) H0: λ = 7.6 [or 1.9] B1 Or Population mean = 7.6 or µ (not just ‘mean’).
H1: λ < 7.6 [or 1.9] Or Population mean < 7.6 or µ.
1
7(b) Mean = 7.6 B1 Seen.
2 M1 OE.
P(X ⩽ 2) = e-7.6 (1 + 7.6 + 7.6 ) [= 0.0188 or 0.0187]
2
2 3 M1 OE. Expression must be seen in at least one probability
P(X ⩽ 3) = e-7.6(1 + 7.6 + 7.6 + 7.6 ) [= 0.0554 or 0.0553]
2 3! calculation.
0.0188 or 0.0187 and 0.0554 or 0.0553 A1 A1 for both values.
Critical region is X ⩽ 2 A1 Dep on both M marks.
SC No Poisson expression seen in either prob scores B1
for 0.0188 or 0.0187 and B1 for 0.0554 or 0.0553 and
B1 for CR.
P(Type I error) = P(X ⩽ 2) = 0.0188 or 0.0187 (3 sf) B1FT FT their P(X ⩽ 2) or their CR.
6
7(c) Concluding that the (mean) no. of accidents has reduced when it has not. B1 OE. Must be in context.
Accept: ‘It is believed that the booklet has helped to
improve safety when actually it has not’.
1
7(d) 3 not in critical region. M1 FT their CR
or P(X < 3) = 0.0554 > 0.05 .
No evidence mean number of accidents has decreased. A1FT In context.
Cannot be a definite statement, e.g., ‘mean number
accidents has not decreased’.
2
7(e) N(98.8, 98.8) B1 May be implied.
100.5 − 98.8 M1 For standardising (could be implied by correct answer).
[= 0.171] Allow with wrong or no continuity correction.
98.8
1 – Φ(‘0.171’) M1 For probability area consistent with their working.
= 0.432 (3 sf) A1
4
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Mathematics 9709/61 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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