Mathematics 9709/61 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 6 Probability & Statistics 2, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Accidents at two factories occur randomly and independently. On average, the numbers of accidents per month are 3.1 at factory A and 1.7 at factory B.
Find the probability that the total number of accidents in the two factories during a 2-month period is more than 3. [4]
Mark scheme
1 λ = (3.1 + 1.7) × 2 M1 Attempt combined mean. Allow 3.1 + 1.7 for M1
= 9.6 A1 Correct mean
2 3
1 – e–9.6 (1 + 9.6 + 9.62 + 9.63! ) M1 Allow incorrect mean. Allow one end error.
= 0.986 (3 sf) A1 SC If 9.6 seen and unsupported 0.986 M1A1B1.
SC Unsupported correct answer of 0.986 only if 9.6
also not seen scores B2 only.
4
The time, in minutes, taken by students to complete a test has the distribution N 125, 36.
(a) Find the probability that the mean time taken to complete the test by a random sample of
40 students is less than 123 minutes. [3]
(b) Explain whether it was necessary to use the Central Limit theorem in the solution to part (a).
[1]
Mark scheme
2(a) ± 123 − 125 [= –2.108...] M1 Must have √40
6 No standard deviation/variance mix. Ignore any
40 continuity correction attempts for this mark.
P(z < ‘–2.108’) = 1 – Φ(‘2.108’) M1 For correct probability area consistent with their
working.
= 0.0175 or 0.0176 (3 sf) A1
3
2(b) No, population is normal B1 Need both.
1
The graph of the probability density function of a random variable X is symmetrical about the line x = 4.
Given that P X < 5 = 20
27
, find P 3 < X < 5. [2]
Mark scheme
3 20 20 1 M1 For either expression seen.
1– or –
27 27 2
20 20 20 1
– 1 − or −
27 27 27 2
13 A1 OE. Accept 0.481 or 0.482.
27
2
100 randomly chosen adults each throw a ball once. The length, l metres, of each throw is recorded.
The results are summarised below. n = 100 Σ l = 3820 Σ l2 = 182 200
Calculate a 94% confidence interval for the population mean length of throws by adults. [6]
Mark scheme
4 3820 B1
[= 38.2]
100
100 182200 1 38202 M1 Use of biased (362.76) scores M0
2
− '38.2 ' or 182200 −
99 100 99 100
12092 A1 Accept SD=19.1422 or 19.1(3sf)
= or 366.424 or 366 (3 sf)
33
'366.424' M1 Expression of the correct form must be a z-value.
‘38.2’ ± z×
100
z = 1.881 or 1.882 B1 Seen.
34.6 to 41.8 (3 sf) A1 Allow use of biased giving (34.6,41.8)
Must be an interval.
6
On average, 1 in 75 000 adults has a certain genetic disorder.
(a) Use a suitable approximating distribution to find the probability that, in a random sample of
10 000 people, at least 1 has the genetic disorder. [3]
(b) In a random sample of n people, where n is large, the probability that no-one has the genetic disorder is more than 0.9.
Find the largest possible value of n. [4]
Mark scheme
5(a) 2 M1 SOI. Allow Po(0.133).
Po
15
−
2 M1 Allow incorrect λ allow one end error
P(X ⩾ 1) = 1 – e 15
= 0.125 (3 sf) A1 SC Partially unsupported final answer:
2
Po stated B1 then unsupported 0.125 B1
15
SC Use of Binomial (0.1248) B1 only
Use of Normal scores M0
3
5(b) n B1
λ=
75000
−
n M1 Allow ‘=’
e 75000 > 0.9 Allow incorrect λ
n M1 Attempt ln both sides
– > ln 0.9 [n < 7902.04]
75000
Largest value of n is 7902 A1 CWO. Must be an integer.
Alternative method for Question 5(b)
e- μ > 0.9 M1 Allow ‘=’
–μ > ln 0.9 [μ < 0.10536] M1 Attempt ln both sides
n = μ × 75000 B1
Largest value of n is 7902 A1 CWO. Must be an integer.
Alternative method for Question 5(b)
74999 B1
75000
74999
n M1
> 0.9
75000
74999 M1 Attempt ln or log both sides
nln > ln 0.9
75000
Largest value of n is 7901 A1 CWO Must be an integer
4
The probability density function, f, of a random variable X is given by
T k 6x − x2 0 ≤ x ≤ 6, f x =
0 otherwise, where k is a constant.
State the value of E X and show that Var X = 95 . [6]
Mark scheme
6 E(X) = 3 B1 N.B. E(X)=108k is B0 until correct k substituted in.
6 M1 Attempt integration of f(x) and =1.
k (6 x − x 2 )dx = 1
0
Ignore limits at this stage.
x3 6
k 3x 2 − [= 1]
3 0
216 A1
k 108 − =1
3
3 1
k= or
108 36
3
6 *M1 Attempt integration of their k × x2f(x).
'
' (6 x3 − x 4 )dx
108 0
Ignore limits at this stage. Accept in terms of k.
3 3 x 4 x5 6
= − = 10.8
108 2 5 0
'10.8' – '3'2 DM1 Their 10.8 (from use of limits 0 and 6) minus their
(E(X))2.
Accept in terms of k: 388.8k–(108k)2
9 A1 CWO. Must be convincingly obtained as AG.
or 1.8
5
6
The masses, in kilograms, of large and small sacks of flour have the distributions N 55, 32 and
N 27, 2.52 respectively.
(a) Some sacks are loaded onto a boat. The maximum load of flour that the boat can carry safely is
340 kg.
Find the probability that the boat can carry safely 3 randomly chosen large sacks of flour and
6 randomly chosen small sacks of flour. [5]
(b) Find the probability that the mass of a randomly chosen large sack of flour is greater than the total mass of two randomly chosen small sacks of flour. [5]
Mark scheme
7(a) E(T)= 3 × 55 + 6 × 27 [ = 327 ] B1 OE. Accept unsimplified.
Var(T) = 3×32 + 6×2.52 [= 64.5] B1 Accept unsimplified.
340 − '327' M1 Must have √
[= 1.619]
'64.5'
P(z < '1.619') = Φ('1.619') M1 Correct probability area consistent with their
working.
0.947 (3 sf) A1
5
7(b) E(L–S1–S2) = 55 – 2 × 27 [=1] B1 OE e.g. E(S1+S2 – L)= –1. Accept unsimplified.
Var(L– S1–S2) = 32 + 2 × 2.52 [= 21.5] B1 Accept unsimplified.
0 − '1' M1 Standardising with their values. Must come from a
[= –0.216] combination attempt.
'21.5'
P(L–S1–S2 > 0) = Φ(‘0.216’) M1 Correct probability area consistent with their
working.
0.586 or 0.585 (3 sf) A1
5
At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15. The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced.
(a) It is suggested that a random sample of the students in Grade 12 should be used for the test.
Give a reason why this would not be an appropriate sample. [1]
A suitable sample of 50 students is selected and the number not wearing correct uniform is noted.
This figure is used to carry out a test at the 5% significance level.
(b) State suitable null and alternative hypotheses. [1]
(c) Use a binomial distribution to find the probability of a Type I error. You must justify your answer fully. [5]
(d) In fact 4 students out of the 50 are not wearing correct uniform.
State the conclusion of the test, explaining your answer. [2]
(e) State, with a reason, which of the errors, Type I or Type II, may have been made. [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
8(a) Not representative (of all students in the school) B1 OE idea of ‘not being representative’ e.g. different
grades in the school have different
characteristics/proportions....
Don’t accept ‘not random’ or ‘biased’ without
further explanation.
1
8(b) H0: P(not correct uniform) = 0.15 B1 Allow "p"
H1: P(not correct uniform) < 0.15
1
8(c) Any two probs attempted using B(50,0.15) M1
P(X ⩽ 3) = 0.8550 + 50 × 0.8549 × 0.15 + 50C2 × 0.8548 × 0.152 + 50C3 × 0.8547 × M1 Attempt the tail probability P(0,1,2,3) with
0.153 B(50,0.15) must be added.
P(X ⩽ 4) = 0.04605 + 50C4×0.8546×0.154 M1 OE. Their P(X ⩽ 3) + P(X = 4) or P(0,1,2,3,4) with
B(50,0.15) must be added.
P(X ⩽ 3) = 0.0460 or 0.0461 [<0.05] A1 Both correct.
P(X ⩽ 4) = 0.112 or [>0.05] OR if P(X ⩽ 4) not seen; P(4)=0.06606 and
0.06606>0.05 and P(X ⩽ 3)=0.0460 scores M1 A1
P(Type I) = 0.0460 or 0.0461 (3 sf) A1 Dependent on second M1.
SC If M1M1M1A0 scored allow A1FT for
incorrect P(X ⩽ 3) as long as <0.05
5
8(d) 4 is outside critical region (⩽3) OE or P(X ⩽ 4) = 0.112 which is > 0.05 M1 FT working from (c).
No evidence that proportion not wearing the correct uniform has decreased A1 In context not definite, e.g. not ‘Proportion has not
(Accept Ho) decreased’. No contradiction.
2
8(e) Not rejected H0 *B1 FT FT If Reject H0 in (d)
Type II DB1 FT FT Type I
2
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/61 May/June 2021 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/61, May/June 2021 (9709_s21_ms_61), and the question paper (9709_s21_qp_61), to read or download, with each question's marking points written out underneath.
How many marks is 9709/61 out of?+
Mathematics 9709/61 for May/June 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/61 answers marked?+
Yes. Upload your answers to 9709/61/M/J/21 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.