Mathematics 9709/72 February/March 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 7 Probability and Statistics, February/March 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The masses of a certain variety of plums are known to have standard deviation 13.2 g. A random sample of 200 of these plums is taken and the mean mass of the plums in the sample is found to be
62.3 g.
(i) Calculate a 98% confidence interval for the population mean mass. [3]
(ii) State with a reason whether it was necessary to use the Central Limit theorem in the calculation in part (i). [1]
Mark scheme
1(i) z = 2.326 B1
62.3 ± z 13.2 M1 Any z. Expression of correct form. Must be a ‘z’
200
60.1 to 64.5 (3 sfs) A1 Must be an interval
3
1(ii) Yes, because pop not (given to be) normal, or pop distribution unknown B1 No contradictions
1
At factory A the mean number of accidents per year is 32. At factory B the records of numbers of accidents before 2018 have been lost, but the number of accidents during 2018 was 21. It is known that the number of accidents per year can be well modelled by a Poisson distribution. Use an approximating distribution to test at the 2% significance level whether the mean number of accidents at factory B is less than at factory A. [6]
Mark scheme
3 H0: λ = 32 B1
H1: λ < 32 Accept ‘population mean’ (µ)
X ~ N(32, 32) B1 seen or implied
21.5 − 32 M1 Standardise with their values. Allow with no or wrong cc
32
= –1.856 A1
cv of z = –2.054 (or –2.055 or –2.053)
‘1.856’ < 2.054 M1 Valid comparison
or comp ɸ(“1.856”) with 0.98 i.e. 0.9682 < 0.98 oe
No evidence that fewer accidents at B than at A A1f No contradictions
Note Use of CV method
x = 20.38 M1 A1 comparison 21.5 > 20.38 M1 conc A1
6
The lifetimes, X hours, of a random sample of 50 batteries of a certain kind were found. The results are summarised by Σ x = 420 and Σ x2 = 27 530.
(i) Calculate an unbiased estimate of the population mean of X and show that an unbiased estimate of the population variance is 490, correct to 3 significant figures. [3]
(ii) The lifetimes of a further large sample of n batteries of this kind were noted, and the sample mean,
X , was found. Use your estimates from part (i) to find the value of n such that P X > 5 = 0.9377.
[4]
Mark scheme
4(i) x = 420 = 8.4 B1
50
s2 = 50 ( 27530 − ( 420 )2 ) M1 Or 1/49(27530 – (420)2/50)
49 50 50
= 489.8(36....) A1 Must see ⩾ 4 sf
3
4(ii) Φ –1(0.9377) = 1.536 B1
5− '8.4'
490
= –1.536 M1 Attempting to standardise – must have correct form
n
n = ( 1.536 )2 × 490 (= 100.0048) M1 Attempting numerical expression for n or √n (must have used
3.4
a ‘z’ value) may be implied by answer
n = 100 A1 No errors seen. Must be whole number
4
The number of eagles seen per hour in a certain location has the distribution Po 1.8. The number of vultures seen per hour in the same location has the independent distribution Po 2.6.
(i) Find the probability that, in a randomly chosen hour, at least 2 eagles are seen. [2]
(ii) Find the probability that, in a randomly chosen half-hour period, the total number of eagles and vultures seen is less than 5. [3]
Alex wants to be at least 99% certain of seeing at least 1 eagle.
(iii) Find the minimum time for which she should watch for eagles. [3]
Mark scheme
5(i) 1– e–1.8(1 + 1.8) M1 Accept any λ. Accept 1–P(0,1,2)
= 0.537 (3 sf) A1
2
5(ii) λ = 2.2 B1
2 3 4
e–2.2(1 + 2.2 + 2.22! + 2.23! + 2.24! ) M1 Attempt expr'n for
P(X ⩽ 4),allow one end error, allow any λ
= 0.928 (3 sf) or 0.927 A1
3
5(iii) 1 – e–1.8t ⩾ 0.99 or 1 – e–λ ⩾ 0.99 M1 Condone = signs/incorrect inequality signs
e–1.8t ⩽ 0.01 or e–λ ⩽ 0.01 M1 Valid attempt take logs (must have single term on each side)
–1.8t ⩽ ln0.01
t ⩾ 2.56 A1 or 2 hours, 34 mins or better. No errors seen
She must watch for at least 2.56 (hours)
3
The time taken by volunteers to complete a certain task is normally distributed. In the past the time, in minutes, has had mean 91.4 and standard deviation 6.4. A new, similar task is introduced and the times, t minutes, taken by a random sample of 6 volunteers to complete the new task are summarised by Σ t = 568.5. Andrea plans to carry out a test, at the 5% significance level, of whether the mean time for the new task is different from the mean time for the old task.
(i) Give a reason why Andrea should use a two-tail test. [1]
(ii) State the probability that a Type I error is made, and explain the meaning of a Type I error in this context. [2]
You may assume that the times taken for the new task are normally distributed.
(iii) Stating another necessary assumption, carry out the test. [7]
Mark scheme
6(i) Test is for “difference” oe B1 Test is not for ‘increase’ or ‘decrease’ oe No contradictions
1
6(ii) 0.05 B1
Conclude mean time is different when it is not B1 oe, in context
2
6(iii) Assume σ = 6.4 B1
H0: pop mean = 91.4 B1 Allow µ, but not ‘mean’
H1: pop mean ≠ 91.4
x = 568.5 (= 94.75) B1
6
'94.75' −91.4 M1 Must have √6
6.4
6
= 1.282 A1
cv of z = 1.96
‘1.282’ < 1.96 M1 Valid comparison
or comp ɸ(“1.282”) with 0.975 0.9(001) < 0.975 or 0.0999
(or 0.1) > 0.025
consistent use of one tail test can score M1 for comparison
with 1.645oe but not A1ft
oe. No contradictions. ft their z.
No evidence mean time different A1 ft CV method x = 96.52 M1 A1 94.75 < 96.52 M1 Conc A1
7
f x
x
1
0 40
A random variable X has probability density function given by
T ï2 cos x 0 ≤ x ≤ 14 0, f x =
0 otherwise, as shown in the diagram.
(i) Find P X > 16 0. [2]
(ii) Find the median of X . [4]
(iii) Find E X . [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(i) π
M1 Attempt integ f(x) with correct limits
∫
4
2 π cos xdx
6
π
= 2 [sinx] π4
6
= 2−2 2 oe or 0.293 (3 sf) A1 SC Final answer of 0.707 scores B1sc
2
7(ii) m M1 Attempt to integ f(x) & = 0.5. Ignore limits. Condone missing
2 ∫ cosxdx = 0.5
0 √2
m A1 Correct integral and limits 0 to unknown & = 0.5 Condone
2 [sinx] = 0.5 missing √2
0
2 sinm = 0.5
sinm = 2 1 2 oe M1 For rearranging their expression to the form sinm=… (sinm =
0.35355.... or 0.354)
seen or implied
m = 0.361 (3 sfs) A1 No errors seen
(Note 20.705 can score M1 A1 M1 A0)
4
7(iii) π
M1 Attempt to integ xf(x). Ignore limits. Condone missing √2
∫ x cos xdx
4
2
0
π π M1 Attempt to integ by parts leading to expression of form
∫ sin xdx }
4
= 2 {[x(sinx)] 4 – ±xsinx±cosx with correct limits
0 0
π A1 For √2(xsinx –(–cosx)) with correct limits
= 2 { 4π 2 – 0 – [–cosx] 4 }
0
= 2 { 4π 2 + cos π4 – 1} A1
= π4 + 1 – 2 oe or 0.371 (3 sf)
4
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Mathematics 9709/72 for February/March 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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