Mathematics 9709/72 October/November 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 7, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
On average, 1 in 150 components made by a certain machine are faulty. The random variable X denotes the number of faulty components in a random sample of 500 components.
(i) Describe fully the distribution of X . [2]
(ii) State a suitable approximating distribution for X , giving a justification for your choice. [2]
(iii) Use your approximating distribution to find the probability that the sample will include at least
3 faulty components. [3]
Mark scheme
1(i) Binomial B1
1 B1 1
n = 500 and p = or 0.00667 Or B 500, for B1B1
150 150
2
1(ii) Poisson B1
10 B1 Accept n > 50
n large and mean = or 3.3 or better, which is < 5
3
2
1(iii) M1 1-P(X = 0, 1, 2)
10
2
10 3
10
1 − e 3 × 1 + +
−
3 2
= 1 – 0.353 A1 Correct expression with λ=3.3 or better
= 0.647 (3 sf) A1 SC Use of Binomial scores B1 for 0.648. Use of Normal
scores B1 for 0.67(0) to 0.677
3
2(i)(a) Assume standard deviation for the region is 7.1 B1 Or standard deviation is same as for whole population OE
63.2 − 65.2 M1 Attempt to find correct equation (accept +2.182)
7.1
= −2.182
n
n = {–2.182 × 7.1 ÷ (–2)}2 A1 Any correct expression for n or n . SOI
n = 60 A1 CWO. Must be an integer
4
2(i)(b) H0: population mean (or µ) = 65.2 B1 Not just ‘mean’
H1: population mean (or µ) < 65.2
2.182 > 1.751 M1 Or valid area comparison.
There is evidence that animals are shorter in this region A1 CWO. No contradictions
3
The heights of a certain species of animal have been found to have mean 65.2 cm and standard deviation 7.1 cm. A researcher suspects that animals of this species in a certain region are shorter on average than elsewhere. She takes a large random sample of n animals of this species from this region and finds that their mean height is 63.2 cm. She then carries out an appropriate hypothesis test.
(i) She finds that the value of the test statistic z is −2.182, correct to 3 decimal places.
(a) Stating a necessary assumption, calculate the value of n. [4]
(b) Carry out the hypothesis test at the 4% significance level. [3]
(ii) Explain why it was necessary to use the Central Limit theorem in carrying out the test. [1]
Mark scheme
2(ii) Population unknown or population not given as normal B1 Allow population not normal. Accept distribution of X
unknown.
1
The masses, in grams, of bags of flour are normally distributed with mean -. The masses, m grams, of a random sample of 50 bags are summarised by Σ m = 25 110 and Σ m2 = 12 610 300.
(i) Calculate a 96% confidence interval for -, giving the end-points correct to 1 decimal place. [6]
Another random sample of 50 bags of flour is taken and a 99% confidence interval for - is calculated.
(ii) Without calculation, state whether this confidence interval will be wider or narrower than the confidence interval found in part (i). Give a reason for your answer. [1]
Mark scheme
3(i) 25110 B1
est(µ) = (= 502.2)
50
12610300 25110 50 58
2
M1 OE
( ) 50
est σ 2 =
49 50
− = ×
50 49 50
= 1.1836
58 A1 Accept SD = 1.0879
1.18 (3 sf) or
49
z = 2.054 or 2.055 B1
'1.1836' M1 Must be of correct form.
502.2 ± z ×
50
501.9 to 502.5 (1dp) A1 CWO. Must be in interval.
SC accept use of biased variance (1.16) for M1 A1
6
3(ii) More confident or z would be greater, B1 OE
Hence wider. Reason needed
1
A random variable X has probability density function given by
T1 x 0 ≤ x ≤ a, f x = 2
0 otherwise, where a is a constant.
(i) Find a. [2]
(ii) Show that E X = 43 . [3]
The median of X is denoted by m.
(iii) Find P E X < X < m . [4]
Mark scheme
4(i) 1 a 1
a M1 Attempt at triangle area or integral f(x) and = 1,
2
× a × = 1 or
2 20∫xdx = 1
a2
= 1 OE
4
a=2 A1
2
4(ii) 2
1 2 M1 Attempt integral xf(x)
20∫x dx
x3 2 M1 Correct integral and limits 0 to their ‘a’
=
6 0
8 4 A1 AG
= = CWO
6 3
3
4(iii) 4 M1 Attempt integral f(x) between correct limits
3
4 1
P X < =
3 2 0 ∫
xdx
4 A1 5
= or
9 9
1 ' 4' M1 5 1
P ( E ( X ) < X < m) = − or −
9 2
2 9
1 A1
18
Alternative method for question 4(iii)
Attempt to find m M1
m= 2 A1
4 M1
Integrate f(x) between and ' 2 '
3
1 A1
18
4
The masses, in grams, of large boxes of chocolates and small boxes of chocolates have the distributions
N 325, 6.1 and N 167, 5.6 respectively.
(i) Find the probability that the total mass of 10 randomly chosen large boxes of chocolates is less than 3240 g. [4]
(ii) Find the probability that the mass of a randomly chosen large box of chocolates is more than twice the mass of a randomly chosen small box of chocolates. [5]
Mark scheme
5(i) mean = 3250 var. = 61 B1 6.1
Or mean = 325 var. =
10
3240 − 3250 M1 Standardise with their values (no mixed methods)
( = −1.280 )
61
φ('–1.280') = 1 – φ('1.280) M1 Area consistent with their figures
0.100 A1 Allow 0.1
4
5(ii) E(D) = 325 – 2 × 167 = –9 B1 Accept ±9
Var(D) = 6.1 + 22 × 5.6 (= 28.5) B1
0 − ( −9 ) M1 Standardising with their values. Must have a combination
( = 1.686 ) attempt on denominator and
28.5
1 − φ('1.686') M1 Area consistent with their figures
0.0459 A1
5
The number of accidents per month, X , at a factory has a Poisson distribution. In the past the mean has been 1.1 accidents per month. Some new machinery is introduced and the management wish to test whether the mean has increased. They note the number of accidents in a randomly chosen month and carry out a hypothesis test at the 1% significance level.
(i) Show that the critical region for the test is X ≥ 5. Given that the number of accidents is 6, carry out the test. [6]
Later they carry out a similar test, also at the 1% significance level.
(ii) Explain the meaning of a Type I error in this context and state the probability of a Type I error.
[2]
(iii) Given that the mean is now 7.0, find the probability of a Type II error. [2]
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(i) H0: Pop mean (or λ or µ) is 1.1 B1
H1: Pop mean (or λ or µ) is more than 1.1
1.12 1.13 M1 Correct expression for either P(X ⩾ 4) or P(X ⩾ 5)
P(X ⩾ 4) = 1 −e–1.1 1 + 1.1 + 2 + 3!
0.0257 A1 Correct value of either P(X ⩾ 4) or P(X ⩾ 5)
4
1.1 B1 B1 for the other value
P ( X . 5 ) = 0.0257 − e−1.1 × = 0.00544
4! (Note use of P(X < 4) = 0.9743 and P(X < 5) = 0.99456 can
score only if comparison with 0.99 seen)
0.00544 < 0.01 < 0.0257 M1 OE stated (valid comparison)
There is evidence mean has increased B1 SC P(X ⩾ 6) = 0.000968 M1A1
Conclusion B1
6
6(ii) Concluding mean has increased when it has not B1 In context
‘0.00544’ B1FT FT their P(X ⩾ 5), dep < 0.01
2
6(iii) 7 2 73 7 4 M1 Correct expression for P(X ⩽ 4 | λ = 7.0)
e−7.0 1 + 7 + + +
2 3! 4!
0.173 (3 sf) A1
2
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Mathematics 9709/72 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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