Mathematics 9709/62 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The mass, in kilograms, of a block of cheese sold in a supermarket is denoted by the random variable M .
The masses of a random sample of 40 blocks are summarised as follows.
n = 40 Σ m = 20.50 Σ m2 = 10.7280
(a) Calculate unbiased estimates of the population mean and variance of M . [3]
(b) The price, $P, of a block of cheese of mass M kg is found using the formula P = 11M + 0.50.
Find estimates of the population mean and variance of P. [3]
Mark scheme
1(a) 20.5 B1 41 20.5
= 0.5125 Accept 0.513 or . Condone .
40 80 40
40 10.728 1 20.502 M1 887
2
39 40
− (
'0.5125' )
or
39
10.728 −
40
Biased variance (0.005544 or
160 000
) scores M0 A0.
887 A1 CAO
0.0056859 or 0.00569 (3 sf) or
156 000
3
1(b) 491 B1 FT FT their 0.5125
[11 × ‘0.5125’ + 0.5]) = 6.1375 or or 6.14 (3sf)
80
112 × ‘0.0056859’ M1 With nothing added.
Using their variance in (a) (no sd/var confusion)
0.688 (3sf) A1 CAO
3
Andy and Jessica are doing a survey about musical preferences. They plan to choose a representative sample of six students from the 256 students at their college.
(a) Andy suggests that they go to the music building during the lunch hour and choose six students at random from the students who are there.
Give a reason why this method is unsatisfactory. [1]
(b) Jessica decides to use another method. She numbers all the students in the college from 1 to 256.
Then she uses her calculator and generates the following random numbers.
204393 162007 204028 587119 207395
From these numbers, she obtains six student numbers. The first three of her student numbers are
204, 162 and 7.
Continue Jessica’s method to obtain the next three student numbers. [2]
Mark scheme
2(a) E.g. Bias towards students who play instruments or only music students B1 OE
Or any reason that some are excluded e.g. because it is
or e.g. the six will possibly be friends/have similar music preferences lunchtime or because the music building is chosen
or any suggestion that opinions may not be
independent.
Note: ‘not representative of all students’ needs
qualifying
1
2(b) 28, 119, 207 B1 B1 for 28, 119 (condone 028).
B1 B1 for 207 and only 3 values stated.
2
The probability that a certain spinner lands on red on any spin is p. The spinner is spun 140 times and it lands on red 35 times.
(a) Find an approximate 96% confidence interval for p. [3]
From three further experiments, Jack finds a 90% confidence interval, a 95% confidence interval and a 99% confidence interval for p.
(b) Find the probability that exactly two of these confidence intervals contain the true value of p.
[3]
Mark scheme
3(a) 0.25 ± z 0.25×0.75 M1 Expression of correct form (allow M1 for just one side
140
stated). Must be a z-value.
z = 2.054 or 2.055 B1
0.175 to 0.325 (3sf) A1 Must be an interval.
3
3(b) 0.90 × 0.95 × 0.01 M1 M1 M1 for one correct triple product.
+ 0.90 × 0.05 × 0.99 M1 for all correct and added.
+ 0.10 × 0.95 × 0.99
0.147 A1 SC If zero scored award B1 for a 2 or 3 term
expression of the form 0.90 × 0.95 [×c] OE.
(0 < c ⩽ 1)
3
A certain kind of firework is supposed to last for 30 seconds, on average, after it is lit. An inspector suspects that the fireworks actually last a shorter time than this, on average. He takes a random sample of 100 fireworks of this kind. Each firework in the sample is lit and the time it lasts is noted.
(a) Give a reason why it is necessary to take a sample rather than testing all the fireworks of this kind. [1]
It is given that the population standard deviation of the times that fireworks of this kind last is 5 seconds.
(b) The mean time lasted by the 100 fireworks in the sample is found to be 29 seconds.
Test the inspector’s suspicion at the 1% significance level. [5]
(c) State with a reason whether the Central Limit theorem was needed in the solution to part (b).
[1]
Mark scheme
4(a) Fireworks are destroyed when tested. B1
1
4(b) H0: Pop mean time lasted (or μ) = 30 B1 Not just ‘mean’.
H1: Pop mean time lasted (or μ) < 30
± 29 − 30 M1 For standardising. Must have 100 .
5
100
( 2900 − 3000 )
Use of totals N(3000,2500) giving
2500
scores M1. No mixed methods.
± –2 A1
–2 > –2.326 [Do not reject H0] M1 Accept –2.326 to –2.329.
Valid comparison or area comparison 0.0228>0.01 or
0.9772<0.99.
Accept CR method 28.837<29 or 30.163>30.
There is not enough evidence that mean time lasted is less than 30 seconds A1 FT In context (if used need mean or time / condone
OR Not enough evidence to support the inspector’s suspicion average instead of mean), not definite, e.g. not ‘mean
time lasted is not less than 30 seconds’.
No contradictions.
Note 2 tailed test can score B0 M1 A1 M1
(comparison with 2.574–2.579) A0 (no FT).
5
4(c) Yes. Because population distribution is unknown [condone not Normal]. B1 Both needed.
Condone X for parent population.
1
In a certain large document, typing errors occur at random and at a constant mean rate of 0.2 per page.
(a) Find the probability that there are fewer than 3 typing errors in 10 randomly chosen pages. [2]
(b) Use an approximating distribution to find the probability that there are more than 50 typing errors in 200 randomly chosen pages. [4]
In the same document, formatting errors occur at random and at a constant mean rate of 0.3 per page.
(c) Find the probability that the total number of typing and formatting errors in 20 randomly chosen pages is between 8 and 11 inclusive. [3]
Mark scheme
5(a) 2
e–2(1 + 2 + 22! ) M1 P(X < 3) any λ. Allow one end error.
0.677 (3sf) A1 Unsupported correct answer scores SC B1 only.
2
5(b) N(40, 40) M1 SOI
50.5 − 40 [= 1.660] M1 For standardising with their values.
40 Allow with wrong or no cc must have square root.
P(z > ‘1.660’) = 1 – Φ(‘1.660’) M1 Correct area consistent with their working.
0.0485 or 0.0484 (3sf) A1
4
5(c) λ = 10 B1 Condone mean = 10.
108 109 1010 1011 M1 Allow any λ (allow one end error).
e −10 + + +
8! 9! 10! 11!
0.477 (3sf) A1 Unsupported correct answer scores SC B2 only.
3
A machine is supposed to produce random digits. Bob thinks that the machine is not fair and that
1 . In order to test his suspicion he notes the the probability of it producing the digit 0 is less than 10 number of times the digit 0 occurs in 30 digits produced by the machine. He carries out a test at the
10% significance level.
(a) State suitable null and alternative hypotheses. [1]
(b) Find the rejection region for the test. [4]
(c) State the probability of a Type I error. [1]
It is now given that the machine actually produces a 0 once in every 40 digits, on average.
(d) Find the probability of a Type II error. [3]
(e) Explain the meaning of a Type II error in this context. [1]
Mark scheme
6(a) H0: P(0) = 101 B1 Accept p.
H1: P(0) < 101
1
6(b) For B(30,0.1) M1 Used not just stated.
P(X = 0) = 0.930 [= 0.0424] [<0.1] M1
P(X = 0 or 1) = 0.930 + 30×0.929×0.1=0.184 [>0.1] B1 Accept 0.184 or 0.183.
Rejection region is 0 zeros A1 Dependent on M1 M1 and at least one comparison, no
errors seen.
SC One unsupported correct answer 0.0424/0.184(or
0.183) and correct rejection region scores B1; with
comparison with 0.1 scores B2.
Two unsupported correct answers 0.0424 and 0.184(or
0.183) and correct rejection region scores B2 or if with
one comparison with 0.1 scores B3.
4
6(c) 0.0424 B1 FT their (b) must have a critical region (only follow
though Binomial), dependent on answer < 0.1.
1
6(d) Bin(30, 401 ) B1 SOI
1 – 0.97530 M1 FT their rr and with Bin(30, 1/40)).
0.532 (3dp) A1 SC Unsupported correct answer scores B2 only.
3
6(e) Not concluding that the probability is less than 101 , when in fact it is. B1 In context.
1
(a) The probability density function of the random variable X is given by
T kx 4 − x 0 ≤ x ≤ 2, f x =
0 otherwise, where k is a constant.
3.
(i) Show that k = 16 [3]
(ii) Find E X . [3]
(b) The random variable Y has the following properties.
• Y takes values between 0 and 5 only.
• The probability density function of Y is symmetrical.
Given that P Y < a = 0.2, find P 2.5 < Y < 5 − a illustrating your method with a sketch on the axes provided. [3]
y
0 5
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(a)(i) 2 M1 Attempt integral f(x) and = 1. Ignore limits
k (4 x − x 2 )dx = 1
0
(must see a power increase for attempted integration).
4 x 2 x3 2 A1 Correct integration and correct limits.
k − =1
2 3 0
16 3 A1 OE AG
k× = 1 k = Convincingly obtained. At least one interim step. No
3 16
errors seen.
3
7(a)(ii) 2 M1 Attempt integral xf(x). Ignore limits.
3
16 (4 x 2 − x 3 )dx (must see a power increase for attempted integration).
0 Condone missing k.
3 4 x 2 x3 2 A1 Correct integration and correct limits. Condone
− missing k.
16 2 3 0
5 A1 Unsupported correct answer scores SC B2 only.
4
3
7(b) Symmetrical frequency density graph, 0 to 5, showing area 0.2 to left of a B1 With a to the left of centre.
Either 0.2 between 5–a and 5 or 0.8 between 0 and 5–a B1 Shown on graph or stated (5–a seen or implied).
a must be non-numerical.
[P(2.5<Y<5–a)] = 0.3 B1 Must be clearly final answer.
a must be non-numerical.
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/62, October/November 2021 (9709_w21_ms_62), and the question paper (9709_w21_qp_62), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/62 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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