Mathematics 9709/61 October/November 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
It is known that, on average, 1 in 300 flowers of a certain kind are white. A random sample of
200 flowers of this kind is selected.
(a) Use an appropriate approximating distribution to find the probability that more than 1 flower in the sample is white. [3]
(b) Justify the approximating distribution used in part (a). [1]
The probability that a randomly chosen flower of another kind is white is 0.02. A random sample of
150 of these flowers is selected.
(c) Use an appropriate approximating distribution to find the probability that the total number of white flowers in the two samples is less than 4. [3]
Mark scheme
1(a) 2 B1 Poisson with correct mean stated (to at least 3 sf) or implied in
Po working.
3
2
− 2 M1 1 – P(X = 0 or 1); allow incorrect λ; allow one end error
1 − e 3 1 +
3
= 0.144 (3 sf) A1 SC B1 for use of binomial or no working shown leading to correct
final answer.
3
1(b) 2 B1 Accept p or np clearly stated in part (a).
n > 50 and np = < 5 or n > 50 and p = 1 < 0.1 Do not accept n is large and p is small.
3 300
3
1(c) 11 B1 Poisson with correct mean stated (to at least 3sf) or implied in
Po working.
3
M1 P(X = 0, 1, 2, 3); allow incorrect λ; allow one end error. Must not
11
2 3
11
be multiplied by any additional values.
11 3
11
3
+
−
e 3 1+ +
3 2! 3!
= 0.501 (3 sf) A1 As final answer.
3
In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel.
(a) It was found that 102 adults in the sample travel by bus. Find an approximate 90% confidence interval for the proportion of all the adults in Fromleigh who travel by bus. [3]
(b) The survey included a question about the amount, x dollars, spent on travel per year. The results are summarised as follows. n = 250 Σ x = 50 460 Σ x2 = 19 854 200
Find unbiased estimates of the population mean and variance of the amount spent per year on travel. [3]
A councillor wanted to select a random sample of houses in Fromleigh. He planned to select the first house on each of the 143 streets in Fromleigh.
(c) Explain why this would not provide a random sample. [1]
Mark scheme
2(a) 102 250 − 102 M1 Any z but must be a z value. One side of the interval scores M1.
×
250 250 (= 0.000966144)
250
102
± z '0.00096614'
250
z = 1.645 B1
Confident Interval is 0.357 to 0.459 (3 sf) A1 Must be an interval.
3
2(b) 50 460 B1 Allow without units. Allow 3s.f. $202.
Estimate of mean = $201.84
250
250 19854 200 50 460
2
1 504602 M1
− or 19854200 −
249 250 250 249 250
Estimate of variance = 38 832.75 dollars2 or 38 800 (3 sf) A1 Allow with missing units. (Calculation of biased gives 38 700
scores M0A0)
3
2(c) e.g. Every house doesn’t have an equal chance of being selected B1 Or other similar e.g. Houses in streets with few houses are more
or most houses have no chance of being selected. likely to be selected. Not just ‘biased’, OE, without explanation
1
The masses, in kilograms, of female and male animals of a certain species have the distributions
N 102, 272 and N 170, 552 respectively.
Find the probability that a randomly chosen female has a mass that is less than half the mass of a randomly chosen male. [6]
Mark scheme
3 F – 0.5M M1 SOI
~ N(17, 272 + 0.25 × 552) B1 for 102 – 0.5(170) (= 17) or 34
B1 for 272 + 0.25 × 552 (= 1485.25) or 22 × 27 2 + 552 (= 5941)
0 − '17 ' M1 Must have an attempt at combining F and M. No standard
(= –0.4411) deviation/variance errors.
'1485.25'
P(F – 0.5M < 0) = ɸ('-0.4411') = 1 – ɸ('0.4411') M1 Correct area consistent with their figures.
= 0.330 (3 sf) A1 Allow 0.33 if no greater accuracy given
Alternative method for question 3
2F – M
~ N(34, 22 × 27 2 + 552 ) B1 for 102 – 0.5(170) (= 17) or 34
B1 for 272 + 0.25 × 552 (= 1485.25) or 22 × 27 2 + 552 (= 5941)
0 − '34' M1 Must have an attempt at combining F and M. No standard
(= –0.4411) deviation/variance errors.
'5941'
P(2F – M < 0) = ɸ('-0.4411') = 1 – ɸ('0.4411') M1 Correct area consistent with their figures.
= 0.330 (3 sf) A1 Allow 0.33 if no greater accuracy given
6
f x
k
x
0 a
The diagram shows the probability density function, f x, of a random variable X . For 0 ≤ x ≤ a, f x = k; elsewhere f x = 0.
(a) Express k in terms of a. [1]
(b) Given that Var X = 3, find a. [4]
Mark scheme
4(a) 1 B1
(k =)
a
1
4(b) a2 a B1 FT OE seen. FT their k
(Mean =) their k × =
2 2
' ' M1 Attempt at correct integral and use of limits. Accept in terms of k
1 2 a2
a
a0x dx =
3
or incorrect k.
2
' a ' a2 M1 For subtracting mean2, allow if integration not complete.
− = FT incorrect values of k.
2 12
a2 A1 Can be in terms of k.
= 3 a = 6
12
4
The number of absences per week by workers at a factory has the distribution Po 2.1.
(a) Find the standard deviation of the number of absences per week. [1]
(b) Find the probability that the number of absences in a 2-week period is at least 2. [3]
(c) Find the probability that the number of absences in a 3-week period is more than 4 and less than 8. [2]
Following a change in working conditions, the management wished to test whether the mean number of absences has decreased. They found that, in a randomly chosen 3-week period, there were exactly
2 absences.
(d) Carry out the test at the 10% significance level. [5]
(e) State, with a reason, which of the errors, Type I or Type II, might have been made in carrying out the test in part (d). [2]
Mark scheme
5(a) √2.1 or 1.45 (3 sf) B1
1
5(b) λ = 4.2 B1
1 – e–4.2(1 + 4.2) M1 1 – P(X ⩽ 1) any λ, allow one end error.
= 0.922 (3 sf) A1
3
5(c) λ = 6.3 M1 P(X = 5, 6, 7) any λ, allow one end error.
6.35 6.36 6.37
e−6.3 + +
5! 6! 7!
= 0.455 (3 sf) A1
2
5(d) H0: λ = 6.3 B1 Accept µ, accept 2.1 (per week)
H1: λ < 6.3
M1
−6.3 6.32
P(X ⩽ 2) = e 1 + 6.3 +
2!
= 0.0498 or 0.0499 A1 Accept 0.0499
‘0.0498’ < 0.1 M1 For valid comparison.
For CV method the comparison can be ‘2 lies in CR of X ⩽ 2’
There is evidence that mean number of absences has decreased. A1 FT In context, not definite, e.g. not ‘Mean number of absences has
decreased.’ No contradictions.
5
5(e) H0 rejected *B1 FT OE
Hence Type I error possible DB1 FT
2
The time, in minutes, for Anjan’s journey to work on Mondays has mean 38.4 and standard deviation
6.9.
(a) Find the probability that Anjan’s mean journey time for a random sample of 30 Mondays is between 38 and 40 minutes. [5]
Anjan wishes to test whether his mean journey time is different on Tuesdays. He chooses a random sample of 30 Tuesdays and finds that his mean journey time for these 30 Tuesdays is 40.2 minutes.
Assume that the standard deviation for his journey time on Tuesdays is 6.9 minutes.
(b) (i) State, with a reason, whether Anjan should use a one-tail or a two-tail test. [1]
(ii) Carry out the test at the 10% significance level. [5]
(iii) Explain whether it was necessary to use the Central Limit theorem in part (b)(ii). [1]
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) 40 − 38.4 38 − 38.4 M1 M1 for either correct expression must have √30 (condone
= 1.270 = –0.3175 continuity correction)
6.9 6.9
30 30
A1 A1 for ±1.270 or for 1.27 or AWRT
A1 A1 for ± (–0.3175) must be opposite sign or for 0.317 or 0.318 or
AWRT
Φ(‘1.270’) – (1 – ɸ('0.3175') M1 For correct method consistent with their values
= 0.523 (3 sf) or 0.522 A1
5
6(b)(i) 2-tail because looking for ‘change’, not decrease or increase B1 OE
1
6(b)(ii) H0: Population mean journey time (or μ) = 38.4 B1 Not just ‘mean journey time’
H1: Population mean journey time (or μ) ≠ 38.4
40.2 − 38.4 M1 For standardising (must have √30 )
6.9
30
= 1.429 A1
‘1.429’ < 1.645 M1 For valid comparison (area comparison 0.0765 > 0.05 )
There is no evidence that mean journey time has changed. A1 FT In context. Not definite (e.g. not ‘mean journey time has not
changed’). No contradictions. FT their ‘1.429’
(Note use of 1-tail test scores B0 M1A1M1(comparison with
1.282) A0 max)
Alternative method for question 6(b)(ii) – critical values method
H0: Population mean journey time (or μ) = 38.4 B1 Not just ‘mean journey time’
H1: Population mean journey time (or μ) ≠ 38.4
6.9 M1
38 + 1.645
30
= 40.47 A1
40.2 < 40.47 M1 For valid comparison
There is no evidence that mean journey time has changed. A1 FT In context. Not definite (e.g. not ‘mean journey time has not
changed’). No contradictions.
5
6(b)(iii) Yes, because population distribution unknown. B1 Allow: Yes, because population distribution not normal.
1
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Mathematics 9709/61 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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