Mathematics 9709/62 February/March 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability and Statistics 2, February/March 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A construction company notes the time, t days, that it takes to build each house of a certain design.
The results for a random sample of 60 such houses are summarised as follows.
Σ t = 4820 Σ t2 = 392 050
(a) Calculate a 98% confidence interval for the population mean time. [6]
(b) Explain why it was necessary to use the Central Limit theorem in part (a). [1]
Mark scheme
1(a) B1
Est(μ) = 4820 or 241 or 80.3 (3 sf)
60 3
M1 Use of biased (80.72) score M0 A0.
Est(σ2) = 60 ( 392 050 − ( 4820 ) 2 )
59 60 60
A1
82.0904
14530
to 82.635 or SD = 9.0604 to 9.0904 (3sf)
177
z = 2.326 B1
4820 ± z× '82.0904' M1 Expression of the correct form – must be z value.
60 60
77.6 to 83.1 (3 sf) A1 CWO
Use of biased 77.6 to 83.0(3) can score B1M1A1 (max
4/6).
6
1(b) Population distribution of times unknown B1 Accept ‘not normal’.
1
f x
1
2 k
x
O k
The diagram shows the graph of the probability density function, f, of a random variable X .
(a) Find the value of the constant k. [2]
(b) Using this value of k, find f x for 0 ≤ x ≤ k and hence find E X . [3]
(c) Find the value of p such that P p < X < 1 = 0.25. [4]
Mark scheme
2(a) 1 × 1 k×k = 1 M1 k
1 1
2 2 Or use of − x + k dx = 1 and attempt at integral.
0
2 2
k=2 A1 Unsupported answers M0 A0. Do not accept ±2 .
2
2(b) B1 FT 1 1
f(x) = – 1 x + 1 FT their k from y = − x + k .
2 2 2
2 x3 x 2 2 M1 Attempt integration of xf(x) limits 0 to k.
1 2
( − x + x ) dx = − +
2 6 2 0 FT their f(x).
0 Could be in terms of k.
2 or 0.667 (3 sf) A1
3
3
2(c) 1 M1 FT their equation of line; correct integral and limits
1
(− x + 1)dx [= 0.25] (could be reversed) stated or 12 (1 − p)(1 − 12 p + 12 ) [=
2
p
0.25].
x2 1 M1 Attempt substitution of correct limits (not reversed) into
− 4 + x p = 0.25 their integral or attempt expand must equal 0.25.
OE
p2
–1 +1+ – p = 0.25
4 4
p2 – 4p + 2 = 0 M1 Obtain 3-term quadratic set equal to 0, obtain at least 1
solution.
p = 2 – √2 or 0.586 A1 CAO
4
An architect wishes to investigate whether the buildings in a certain city are higher, on average, than buildings in other cities. He takes a large random sample of buildings from the city and finds the mean height of the buildings in the sample. He calculates the value of the test statistic, z, and finds that z = 2.41.
(a) Explain briefly whether he should use a one-tail test or a two-tail test. [1]
(b) Carry out the test at the 1% significance level. [3]
Mark scheme
3(a) One-tail because investigating whether "higher" B1 OE. Must have both parts.
1
3(b) H0: Population mean (or μ) in city same as for others B1 FT If (a) two-tail:
H1: Population mean (or μ) in city greater than for others H0: Pop mean (or μ) in city same as for others.
H1: Pop mean (or μ) in region different from others.
2.41 > 2.326 or 0.008 < 0.01 or 0.992 > 0.99 M1 If (a) two-tail: 2.41 < 2.576 or 0.992 < 0.995.
There is evidence that buildings are higher [on average]. A1 FT In context, not definite. No contradictions.
If (a) two-tail: There is no evidence that the [average]
height of buildings is different.
3
On average, 1 in 400 microchips made at a certain factory are faulty. The number of faulty microchips in a random sample of 1000 is denoted by X .
(a) State the distribution of X , giving the values of any parameters. [1]
(b) State an approximating distribution for X , giving the values of any parameters. [2]
(c) Use this approximating distribution to find each of the following.
(i) P X = 4. [2]
(ii) P 2 ≤ X ≤ 4. [2]
(d) Use a suitable approximating distribution to find the probability that, in a random sample of 700 microchips, there will be at least 1 faulty one. [3]
Mark scheme
4(a) B1 1
B(1000, 1 ) Accept Bin and n = 1000, p = .
400 400
1
4(b) Po(2.5) B2 B1 for Po. B1 for λ = 2.5 .
2
4(c)(i) 4 M1 FT their (b) for Normal must have a continuity
e-2.5× 2.5 correction.
4!
Allow any λ
0.134 (3 sf) A1 CWO
2
4(c)(ii) 2 3 4 M1 FT their (b) for Normal must have a continuity
e-2.5( 2.5 + 2.5 + 2.5 ) correction.
2! 3! 4!
Allow with one term extra or omitted or wrong. Allow
any λ.
0.604 (3 sf) A1 CWO
2
4(d) 1
λ = 2.5 × 0.7 or λ = 700× 400 [= 1.75] M1 Must see λ or use of Poisson.
1 – e–1.75 M1 Allow any λ. Allow 1 – P(0,1).
0.826 A1 SC B1 Use of B(700,0.0025) leading to 0.826.
3
The volumes, in litres, of juice in large and small bottles have the distributions N 5.10, 0.0102 and
N 2.51, 0.0036 respectively.
(a) Find the probability that the total volume of juice in 3 randomly chosen large bottles and
4 randomly chosen small bottles is less than 25.5 litres. [5]
(b) Find the probability that the volume of juice in a randomly chosen large bottle is at least twice the volume of juice in a randomly chosen small bottle. [5]
Mark scheme
5(a) E(L1+L2+L3+S1+S2+S3+S4) = 3 × 5.10 + 4 × 2.51 [= 25.34] B1 OE (E(3L + 4S – 25.5) = -0.16)
Var(L1+L2+L3+S1+S2+S3+S4) = 3 × 0.0102 + 4 × 0.0036 [= 0.045] B1 3 2
or SD = = 0.2121 .
20
25.5−'25.34' [= 0.754] M1 No SD/variance mix.
'0.045' Standardising with their values (must be from a
combination attempt).
Φ(‘0.754’) M1 For the correct area consistent with their working.
0.775 (3 sf) A1
5
5(b) E(L – 2S) = 5.10 − 2 × 2.51 [= 0.08] B1 OE
Var(L – 2S) = 0.0102 + 22 × 0.0036 [= 0.0246] B1 Or SD = 0.1568 .
0 − ' 0.08 ' [= –0.510] M1 No SD/variance mix.
' 0.0246 ' Standardising with their values (must be from a
combination attempt).
P(Z > '–0.510') = ɸ(‘0.510’) M1 For the correct area consistent with their working.
0.695 (3 sf) A1
5
It is known that 8% of adults in a certain town own a Chantor car. After an advertising campaign, a car dealer wishes to investigate whether this proportion has increased. He chooses a random sample of 25 adults from the town and notes how many of them own a Chantor car.
(a) He finds that 4 of the 25 adults own a Chantor car.
Carry out a hypothesis test at the 5% significance level. [5]
(b) Explain which of the errors, Type I or Type II, might have been made in carrying out the test in part (a). [2]
Later, the car dealer takes another random sample of 25 adults from the town and carries out a similar hypothesis test at the 5% significance level.
(c) Find the probability of a Type I error. [3]
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
6(a) H0: population proportion = 0.08 OE B1 Allow ‘p = 0.08’ etc.
H1: population proportion > 0.08 OE
P(X > 4) = 1 – P(X < 3) = M1 Allow 1 – (one term omitted or extra or wrong).
1 – (0.9225 + 25×0.9224×0.08 +25C2×0.9223×0.082 + 25C3×0.9222×0.083)
0.135 (3 sf) A1
0.135 > 0.05 M1 Valid comparison.
Note: ‘0.865'<0.95 can score M1 A1 and can recover
previous M1 A1 for 0.865.
There is no evidence that proportion owning Chantor has increased A1 FT In context. Not definite, e.g. not ‘Proportion not
increased’. No contradictions.
5
6(b) H0 was not rejected. *B1 FT H0 was rejected (consistent with (a)).
Hence Type II might have been made. DB1 FT Type I error.
2
6(c) P(X > 5) = 1 – P(X < 4) *M1 Attempted.
( )
= 1 − (1 − 0.1351) + 25 C 4 × 0.9221 × 0.084 [= 0.0451] Note: If critical region method used in (a) marks can be
awarded here.
0.0451 < 0.05 A1 Comparison of 0.045[1] with 0.05.
Note: If critical region method used in (a) marks can be
awarded here.
P(Type I error) = 0.0451 or 0.0452 A1 Dependent on M1* only.
SC Unsupported answers score: B1 for 0.0451<0.05
and B1 for final answer 0.0451 only.
3
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Mathematics 9709/62 for February/March 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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