Mathematics 9709/51 May/June 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Find the number of different arrangements of the 8 letters in the word DECEIVED in which all three Es are together and the two Ds are together. [2]
(b) Find the number of different arrangements of the 8 letters in the word DECEIVED in which the three Es are not all together. [4]
Mark scheme
1(a) 5! M1 k! where k = 5, 6 or 7
Condone × 1 OE
120 A1
2
1(b) 8! M1 8!
[Total no of ways =] [= 3360] , a = 1, 2 b = 1,3 a ≠ b
2!3! a !b!
6! M1 6!
[With 3Es together =] [= 360] , c = 1, 2 seen in an addition/subtraction
2! c!
[With 3Es not together] = 3360 – 360 M1 8! 6!
− where d , f = 1, 2 & e = 1,3
d !e! f !
3000 A1
4
There are 6 men and 8 women in a Book Club. The committee of the club consists of five of its members. Mr Lan and Mrs Lan are members of the club.
(a) In how many different ways can the committee be selected if exactly one of Mr Lan and Mrs Lan must be on the committee? [2]
(b) In how many different ways can the committee be selected if Mrs Lan must be on the committee and there must be more women than men on the committee? [4]
Mark scheme
2(a) 12
C4 × 2 M1 g
C4 × h g = 12, 13, h = 1,2
990 A1
Alternative method for question 2(a)
[total – both on – neither on] 14C5 – ( 12C3 + 12C5) = [2002 – 220 – 792] M1 k
C5 – (aC3 + aC5)
a = 12, 13 and k = 13, 14
990 A1
2
2(b) [Mrs Lan plus] M1 7
Cr × 6C4-r for r = 2, 3 or 4
2W 2M 7C2 × 6C2 = 315
3W 1M 7C3 × 6C1 = 210 B1 Outcome for one identifiable scenario correct,
4W 7
C4 = 35 accept unevaluated
M1 Add outcomes for 3 identifiable correct scenarios
Note: if scenarios not labelled, they may be
identified by seeing7Cr × 6Cs r + s = 4 to imply
r women and s men for both B & M marks only
[Total =] 560 A1
4
The times taken to travel to college by 2500 students are summarised in the table.
Time taken (t minutes) 0 ≤ t < 20 20 ≤ t < 30 30 ≤ t < 40 40 ≤ t < 60 60 ≤ t < 90
Frequency 440 720 920 300 120
(a) Draw a histogram to represent this information. [4]
From the data, the estimate of the mean value of t is 31.44.
(b) Calculate an estimate of the standard deviation of the times taken to travel to college. [3]
(c) In which class interval does the upper quartile lie? [1]
It was later discovered that the times taken to travel to college by two students were incorrectly recorded. One student’s time was recorded as 15 instead of 5 and the other’s time was recorded as 65 instead of 75.
(d) Without doing any further calculations, state with a reason whether the estimate of the standard deviation in part (b) would be increased, decreased or stay the same. [1]
Mark scheme
3(a) M1 At least 4 frequency densities calculated
Class 20 10 10 20 30
(Frequency ÷ class width, e.g.
width
440 440 440
condone ,
Frequency 22 72 92 15 4 20 19.5 20.5
density Accept unsimplified, may be read from graph using
their scale
A1 All heights correct on graph NOT FT
B1 Bar ends at [0,] 20, 30, 40, 60, 90 at axis with a
horizontal linear scale with at least 3 values
indicated. 0 ⩽ horizontal scale ⩽ 90
B1 Axes labelled frequency density (fd), time (t) and
minutes (mins) or in a title.
Linear vertical scale, with at least 3 values
indicated 0 ⩽ vertical axes ⩽ 92 (condone
90 used).
4
3(b) B1 At least 4 correct midpoints seen
Midpoints 10 25 35 50 75
[Mean = 31.44 given] M1 Correct formula for variance or standard deviation
440 × 102 + 720 × 252 + 920 × 352 + 300 × 502 + 120 × 752 (− mean2 included with their midpoints (not upper
[Variance = − 31.442 ] bound, lower bound, class width, frequency
2500
density, frequency or cumulative frequency) and
44000 + 450000 + 1127000 + 750000 + 675000
= − 31.442 their ∑f if calculated.
2500 Condone 1 data error.
3046000
[= − 31.442 = 229.9264 ]
2500
Or
Variance =
440(10 − 31.44) 2 + 720(25 − 31.44) 2 + 920(35 − 31.44) 2 + 300(50 − 31.44) 2 + 120(75 − 31.44) 2
2500
202 256 + 29860 + 11659 + 103342 + 227 697 574814
= = = 229.9264
2500 2500
Standard deviation = 15.2 A1 WWW, allow 15.16[3…]
3
3(c) 30‒40 B1
1
3(d) Stays the same, data still in same intervals B1 Frequencies unchanged
1
Jacob has four coins. One of the coins is biased such that when it is thrown the probability of obtaining
7 . The other three coins are fair. Jacob throws all four coins once. The number of heads a head is 10 that he obtains is denoted by the random variable X . The probability distribution table for X is as follows.
x 0 1 2 3 4
P X = x 3 a b c 7
80 80
(a) Show that a = 15 and find the values of b and c. [4]
(b) Find E X . [1]
Jacob throws all four coins together 10 times.
(c) Find the probability that he obtains exactly one head on fewer than 3 occasions. [3]
(d) Find the probability that Jacob obtains exactly one head for the first time on the 7th or 8th time that he throws the 4 coins. [2]
Mark scheme
4(a) 1 B1 Clear statement of unevaluated correct
a = P(1 head) = 0.7 × ( 0.5 ) + 0.3 × ( 0.5 ) × 3 =
3 3
5 1
calculation = . AG
5
3 M1 Clear statement of unevaluated calculation for
b = 0.7 × 0.53 × 3 + 0.3 × 0.53 × 3 = either b or c
8
3
c = 0.7 × 0.53 × 3 + 0.3 × 0.53 = A1 For either b or c correct
10
27 B1 FT 27
or c = 40 − b their b + their c =
40
4
4(b) 3 × 0 + 16 × 1 + 30 × 2 + 24 × 3 + 7 × 4 176 B1 FT Correct or accept unsimplified calculation using
E ( X ) = 80
= or 2.2 their values for b and c seen
80
(sum of probabilities = 1)
1
4(c) [P(0, 1, 2) = ]10C0 0.20 0.810 + 10C1 0.21 0.89 + 10C2 0.22 0.88 M1 One term 10Cx p x (1 − p )10− x , for
0 < x < 10, 0 < p < 1
0.107374 + 0.268435 + 0.301989 A1 Correct expression, accept unsimplified leading to
final answer
0.678 B1 0.677 < p ⩽ 0.678
Alternative method for question 4(c)
1 – [10C10 0∙2100.80 + 10C9 0∙290.81 + 10C8 0∙280.82 + 10C7 0∙270.83 +10C6 0∙260.84 + 10C5 0∙250.85 M1 One term 10Cx p x (1 − p )10− x , for
+ 10C4 0∙240.86 + 10C3 0∙230.87]
0 < x < 10, 0 < p < 1
A1 Correct expression, accept unsimplified
0.678 B1 0.677 < p ≤ 0.678
4
4(d) 0.86 × 0.2 + 0.87 × 0.2 = 0.0524288 + 0.041943 M1 p l × (1 − p ) + p m × (1 − p ) , l = 6, 7
m = l + 1, 0 < p < 1
0.0944 A1 0.09437 ⩽ p ⩽ 0.0944
2
The lengths, in cm, of the leaves of a particular type are modelled by the distribution N 5.2, 1.52 .
(a) Find the probability that a randomly chosen leaf of this type has length less than 6 cm. [2]
The lengths of the leaves of another type are also modelled by a normal distribution. A scientist measures the lengths of a random sample of 500 leaves of this type and finds that 46 are less than 3 cm long and 95 are more than 8 cm long.
(b) Find estimates for the mean and standard deviation of the lengths of leaves of this type. [5]
(c) In a random sample of 2000 leaves of this second type, how many would the scientist expect to find with lengths more than 1 standard deviation from the mean? [4]
Mark scheme
5(a) 6 − 5.2 M1 6, 5.2, 1.5 substituted into ± standardisation
P(X < 6) = P(Z < ) = P( Z < 0.5333) formula, condone 1.52, continuity correction ±0.5
1.5
0.703 A1
2
5(b) 3− μ B1 1.328 < z1 ⩽ 1.329 or
z1 = = −1.329 −1.329 ⩽ z1 < −1.328
σ
8−μ
z2 = = 0.878 B1 0.877 < z2 ⩽ 0.878 or
σ −0.878 ⩽ z2 < −0.877
Solve to find at least one unknown: M1 Use of the ± standardisation formula once with μ,
3− μ σ, a z-value (not 0.8179, 0.7910, 0.5367, 0.5753,
= −1.329 0.19, 0.092 etc.) and 3 or 8, condone continuity
σ
8− μ correction but not σ2 or √σ
= 0.878
σ M1 Use either the elimination method or the
substitution method to solve their two equations in
μ and σ
σ = 2.27, μ = 6.01 A1 2.26 ⩽ σ ⩽ 2.27, 6.01 ⩽ μ ⩽ 6.02
5
5(c) [P(Z < −1) + P(Z >1)] Φ(1) – Φ(−1) = M1 Identify 1 and –1 as the appropriate z-values.
= 2 – 2 Φ (1)
M1 Calculating the appropriate area from stated phis of
= 2 – 2 × 0.8413 z-values which must be ± the same number
0.3174 A1 Accept AWRT 0.317
Number of leaves: 2000 × 0.3174 = 634.8 so 634 or 635 B1 FT FT their 4 s.f. (or better) probability, final answer
must be positive integer no approximation or
rounding stated
4
Janice is playing a computer game. She has to complete level 1 and level 2 to finish the game. She is allowed at most two attempts at any level.
• For level 1, the probability that Janice completes it at the first attempt is 0.6. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.3.
• If Janice completes level 1, she immediately moves on to level 2.
• For level 2, the probability that Janice completes it at the first attempt is 0.4. If she fails at her first attempt, the probability that she completes it at the second attempt is 0.2.
(a) Show that the probability that Janice moves on to level 2 is 0.72. [1]
(b) Find the probability that Janice finishes the game. [3]
(c) Find the probability that Janice fails exactly one attempt, given that she finishes the game. [4]
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) 0.6 + 0.4 × 0.3 = 0.72 or 1 – 0.4 x 0.7 = 0.72 B1 Clear identified calculation AG
1
6(b) 0.72 × ( 0.4 + 0.6 × 0.2 ) M1 0.72 × u, 0 < u < 1
M1 v × (0.4 + 0.6 × 0.2), or
v × (1 – 0.6 × 0.8) 0 < v ⩽ 1 no additional terms
SC B1 for 0.72 × (0.4 + 0.12) or 0.72 × (1 – 0.48)
0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only
3
Alternative method for question 6(b)
[p(P1P2) + p(F1P1P2) + p(P1F2P2) + p(F1P1F2P2)] = M1 Any two terms unsimplified and correct
0.6 × 0.4 + 0.4 × 0.3 × 0.4 + 0.6 × 0.6 × 0.2 + 0.4 × 0.3 × 0.6 × 0.2
M1 Summing 4 appropriate scenarios by listing or on a
tree diagram
SC B1 for 0.24 + 0.048 + 0.072 + 0.0144
0.3744 A1 WWW. Condone 0.374. SC B1 for 0.3744 only
3
6(c) P (fails first or second level finishes game) M1 Either 0.6 × 0.6 × 0.2 or 0.4 × 0.3 × 0.4 seen
P (fails first or second level | finishes game) =
their (b) Condone 0.072 or 0.048 if seen in (b)
A1 Both correct accept unsimplified expression.
Numerator = P(S SF) + P(FS S) = 0.6 × 0.6 × 0.2 + 0.4 × 0.3 × 0.4 = 0.072 + 0.048 = 0.12 No additional terms
0.12 M1 Their sum of two 3-term probabilities as numerator
Required probability =
their(b) their (b) or correct
25 A1 0.3205 < p ⩽ 0.321
0.321 or
78
4
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Mathematics 9709/51 for May/June 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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