Mathematics 9709/62 February/March 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics 2, February/March 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The lengths, X cm, of a sample of 100 insects of a certain type were summarised as follows.
n = 100 / x = 36.8 / x 2 = 17.34
(a) Calculate unbiased estimates for the population mean and variance of X. [3]
(b) State a necessary condition for the estimates found in part (a) to be reliable. [1]
Mark scheme
1(a) 46 B1 Oe.
est(μ) = 0.368 =
125
100 17.34 1 36.82
M1 For use of a correct formula (ft their µ).
2
est(σ2 ) = − their '0.368' or 17.34 –
99 100 99 100
= 0.0384 (3 sf) A1
3
1(b) Must be a random sample B1 E.g.
• Values must have been randomly selected.
• Sample should be representative of the
population.
• All values should have equal chance of being
selected.
• It should be an unbiased sample.
• Independent sample/insect lengths are
independent of one another.
ISW
1
A random sample of 250 people living in Barapet was chosen. It was found that 78 of these people owned a BETEC phone.
(a) Calculate an approximate 98% confidence interval for the proportion of people living in Barapet who own a BETEC phone. [3]
(b) Manjit claims that more than 40% of the people living in Barapet own a BETEC phone.
Use your answer to part (a) to comment on this claim. [1]
Mark scheme
2(a) 78 ×(1− 78 ) M1 Use of a correct formula (any z).
78 ± z × 250 250
250 250
z = 2.326 B1
= 0.244 to 0.38[0] (3 sf) A1 Must be an interval.
3
2(b) Unlikely to be true because confidence interval does not contain 0.4. B1 ft FT their confidence interval.
Must include this reason and ‘unlikely’, oe.
Allow “not true because 0.4 is not in the confidence
interval.”
But “Confidence interval only goes up to 0.38 so
not true” and “‘it’ lies outside the confidence
interval” both score B0.
1
In a certain lottery, on average 1 in every 10 000 tickets is a prize-winning ticket. An agent sells 6000 tickets.
(a) Use a suitable approximating distribution to find the probability that at least 3 of the tickets sold by the agent are prize-winning tickets. [3]
(b) Justify the use of your approximating distribution in this context. [1]
Mark scheme
3(a) [λ =] 0.6 B1 Mean = 0.6 seen.
0.62 M1 Any λ Allow one end error.
1 – e─0.6 1 + 0.6 + Must see expression.
2
Accept correct Σ notation.
or 1 – e-0.6 (1 + 0.6 + 0.18)
or 1 – (0.5488 + 0.3293 + 0.09879)
= 0.0231 A1 SC 0.0231 and no working scores B1 (could be
implied).
SC use of binomial scores M1A1 for 0.0231.
3
3(b) 6000 > 50 and either np = 0.6 < 5 or 1 < 0.1 B1 Must state values of n and either np or p.
10000 Note: ‘n large, p small’ is insufficient.
1
Each year a transport firm uses X litres of gasoline and Y litres of diesel fuel, where X and Y have the independent distributions X + N (10 700, 950 2) and Y + N (13 400, 1210 2) .
(a) Find the probability that in a randomly chosen year the firm uses more gasoline than diesel fuel.
[5]
The costs per litre of gasoline and diesel fuel are $0.80 and $0.85 respectively.
(b) Find the probability that the total cost of gasoline and diesel fuel in a randomly chosen year is between $20 000 and $22 000. [5]
Mark scheme
4(a) E(X − Y) = 10700 – 13400 [=−2700] B1 Oe, e.g. (Y − X).
Var(X − Y) = 9502 + 12102 [= 2366600] M1
0 − (their '− 2700') M1 For standardising with their E and Var.
[= 1.755]
their '2366600'
1 − Φ(their ‘1.755’) M1 For area consistent with their values.
= 0.0396 or 0.0397 (3 sf) A1
5
4(b) E(Total) = 10700 × 0.8 + 13400 × 0.85 [=19950] B1
Var(Total) = 9502 × 0.82 + 12102 × 0.852 [= 1635412.25] M1
22000 −their '19950' [= 1.603] or 20000 −their '19950' [= 0.0391] M1 For one standardisation with their E and Var.
their '1635412.25' their '1635412.25'
Φ(their ‘1.603’) − Φ(their ‘0.0391’) = 0.9455 – 0.5156 M1 For area consistent with their values.
= 0.43[0] (3 sf) A1
5
A teacher models the numbers of girls and boys who arrive late for her class on any day by the independent random variables G + Po(0.10) and B + Po(0.15) respectively.
(a) Find the probability that during a randomly chosen 2-day period no girls arrive late. [1]
(b) Find the probability that during a randomly chosen 5-day period the total number of students who arrive late is less than 3. [3]
(c) It is given that the values of P(G = r) and P(B = r) for r H 3 are very small and can be ignored.
Find the probability that on a randomly chosen day more girls arrive late than boys. [3]
Following a timetable change the teacher claims that on average more students arrive late than before the change. During a randomly chosen 5-day period a total of 4 students are late.
(d) Test the teacher’s claim at the 5% significance level. [5]
Mark scheme
5(a) [e−0.2 ] = 0.819 (3 sf) B1 Accept e–0.2 as final answer.
1
5(b) λ = 1.25 B1
1.252 M1 Any λ Allow one end error.
e−1.25 1 + 1.25 + Must see expression (in any form).
2
Accept correct Σ notation.
or e−1.25 (1 + 1.25 + 0.78125)
or 0.2865 + 0.3581 + 0.2238
= 0.868 (3 sf) A1 SC Answer with no working seen scores B1 (could
be implied).
3
5(c) e−0.15 × e−0.1 (0.1) = 0.077879 M1 P(B = 0) × P(G = 1) 0.8607 × 0.09048
0.12 P(B = 0) × P(G = 2) 0.8607 × 0.004524
e−0.15 × e−0.1
2
= 0.003894 P(B = 1) × P(G = 2) 0.1291 × 0.004524
0.12 Note: P(B = 0) × P(G = 2) and P(B = 1) × P(G = 2)
e−0.15 × 0.15 × e−0.1 × = 0.0005841
2 may be seen within P(G = 2) × P(B < 2).
For one expression seen.
0.12 −0.15 0.12 M1 P(B = 0) × P(G = 1) + P(B = 0) × P(G = 2)
e−0.15 × e−0.1 (0.1) + e−0.15 × e−0.1 +e × 0.15 × e−0.1 × + P(B = 1) × P(G = 2).
2 2
For the three Poisson terms added (must be from a
= 0.077879 + 0.00389036 + 0.0005841 complete attempt at all 3 terms).
= 0.0824 (3 sf) A1
Alternative method for Question 5(c)
P(B = 0) × P(G > 0) M1 For one expression seen.
e−0.15 × (1 − e−0.1 )
P(B = 1) × P(G > 1)
e−0.15 × 0.15 × (1 − e−0.1 (1 + 0.1))
e−0.15 × (1 − e−0.1 ) + e−0.15 × 0.15 × (1 − e−0.1 (1 + 0.1)) M1 For adding their expressions.
= 0.0824 (3 sf) A1
3
5(d) H0 : λ = 1.25 or 0.25[per day] B1 Or µ or ‘population mean’.
H1 : λ > 1.25 or 0.25[per day]
2 3 M1 Any λ. No end errors. Expression must be seen (in
P(> 4 late) = 1 − e−1.25 1 + 1.25 + 1.25 + 1.25 any form). Accept correct Σ notation.
2 3!
or 1 – e−1.25 (1 + 1.25 + 0.7813 + 0.3255)
or 1 – (0.2865 + 0.3581 + 0.2238 + 0.09326)
= 0.0383 A1 SC 0.0383 with no working scores B1.
0.0383 < 0.05 M1 For a valid comparison.
[Reject H0 ] A1 FT No contradictions. In context and not definite,
‘Hence there is sufficient evidence to suggest that the teacher’s claim is true’ e.g. not ‘More students are late’ or ‘Claim is
or ‘There is sufficient evidence to suggest that more students are late on correct’.
average’. Ft their 0.0383.
5
The graph of the probability density function f of a random variable X is symmetrical about the line x = 2 . It is given that P (2 1 X 1 5) = 117
256 .
(a) Using only this information show that P (X 2- 1) = 245
256 . [2]
It is now given that, for x in a suitable domain,
f (x) = k (12 + 4x - x 2) , where k is a constant.
(b) Find the value of k. [3]
(c) A different random variable X has probability density function g (x) = 29 `2 + x - x 2j . The domain of
X is all values of x for which g (x) H 0 .
Find Var(X ). [5]
Mark scheme
6(a) 1 − 117 11 M1 For use of symmetry about x = 2, oe.
= 256
2 256 E.g. (1 – 2 × 117 ) ÷ 2 or 1 + 117 .
256 2 256
11 117 11 A1 Any correct numerical expression seen leading to
1− or 2 × + AG.
256 256 256
245
= AG
256
2
6(b) 5 M1 Attempt to integrate f(x) with any limits.
2
k (12 + 4 x − x 2 )dx
3 5
x
= k 12 x + 2 x −
2
3
2
117 M1 Use of limits 2 and 5 and equating their integration
39k =
256 117
attempt to .
256
Or limits –2 and 6 equated to 1.
245
Or limits –1 and 6 equated to .
256
234
Or limits –1 to 5 equated to .
256
Oe.
No mixed methods.
3 A1
k= or 0.0117
256
3
6(c) [2 + x − x2 = 0] B1
x = −1 and x = 2 seen or implied
[Domain is −1 ⩽ x ⩽ 2]
Mean = 0.5 B1
2 *M1 Attempt to integrate x2 g(x) with any limits.
2
(2 x 2 + x3 − x 4 )dx
9
−1
4 5 2
= 2 2 x3 + x − x
9 3 4 5
−1
[= 0.7]
their ‘0.7’ – their ‘0.5’2 DM1 Subtract their mean2 from their ∫ x2 g(x)dx
(both must be numerical).
9 A1
= 0.45 or
20
5
The heights, in centimetres, of adult females in Litania have mean n and standard deviation v. It is known that in 2004 the values of n and v were 163.21 and 6.95 respectively. The government claims that the value of n this year is greater than it was in 2004. In order to test this claim a researcher plans to carry out a hypothesis test at the 1% significance level. He records the heights of a random sample of
300 adult females in Litania this year and finds the value of the sample mean.
(a) State the probability of a Type I error. [1]
You should assume that the value of v after 2004 remains at 6.95 .
(b) Given that the value of n this year is actually 164.91, find the probability of a Type II error. [5]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7(a) 0.01 or 1% B1 Note: x ⩽ 0.01 scores B0.
1
7(b) h − 163.21 M1 Accept any z (±).
2.326 =
6.95 ÷ 300
h = 164.14 A1 Accept 3 sf accuracy here.
[Rejection region is h > 164.14]
[P(Type II) = P( h < 164.14 | µ = 164.91)]
their '164.14'− 164.91 M1 For standardising 164.91 with their 164.14 (could
[= −1.919]
\
6.95 ÷ 300 be 163.21).
Φ(their ‘−1.919’) = 1 − Φ(their ‘1.919’) M1 For area attempt consistent with their values.
= 0.0275 or 0.0276 or 0.028[0] (3.s.f) A1 Accept anything in range 0.0275 to 0.028[0].
5
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Mathematics 9709/62 for February/March 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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