Mathematics 9709/51 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
The score when two fair six-sided dice are thrown is the sum of the two numbers on the upper faces.
1.
(a) Show that the probability that the score is 4 is 12 [1]
The two dice are thrown repeatedly until a score of 4 is obtained. The number of throws taken is denoted by the random variable X .
(b) Find the mean of X . [1]
(c) Find the probability that a score of 4 is first obtained on the 6th throw. [1]
(d) Find P X < 8. [2]
Mark scheme
1(a) 3 1 B1
Prob of 4 (from 1,3, 3,1 or 2,2) = = AG
36 12
1
1(b) 1 B1
Mean = = 12
1
12
1
1(c) 5
11 1 161051 B1
× = 0.0539 or
12 12 2985984
1
1(d) 11
7 M1
1−
12
16344637 A1
0.456 or
35831808
2
(a) Find the number of different arrangements that can be made from the 9 letters of the word
JEWELLERY in which the three Es are together and the two Ls are together. [2]
(b) Find the number of different arrangements that can be made from the 9 letters of the word
JEWELLERY in which the two Ls are not next to each other. [4]
Mark scheme
2(a) 6! M1
720 A1
2
2(b) 9! M1
Total number: ( 30240 )
3!2!
8! M1
Number with Ls together = ( 6720 )
3!
9! 8! M1
Number with Ls not together = −
3!2! 3!
= 30 240 – 6720
23 520 A1
Alternative method for question 2(b)
7! 8 × 7
×
3! 2
7! × k in numerator, k integer ≥ 1 M1
8 × 7 × m in numerator or 8C2 × m, m integer ≥ 1 M1
3! in denominator M1
23 520 A1
4
A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses
3 sweets at random from a box.
(a) Draw up the probability distribution table for the number of jellies that Jemeel chooses. [4]
The company also produces large boxes of sweets. For any large box, the probability that it contains more jellies than chocolates is 0.64. 10 large boxes are chosen at random.
(b) Find the probability that no more than 7 of these boxes contain more jellies than chocolates. [3]
In a music competition, there are 8 pianists, 4 guitarists and 6 violinists. 7 of these musicians will be selected to go through to the final.
How many different selections of 7 finalists can be made if there must be at least 2 pianists, at least
1 guitarist and more violinists than guitarists? [4]
On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities
0.35, 0.44, 0.21 respectively. When she cooks a pizza, Rani has some fruit with probability 0.3.
When she cooks a burger, she has some fruit with probability 0.8. When she cooks a curry, she never has any fruit.
(a) Draw a fully labelled tree diagram to represent this information. [2]
(b) Find the probability that Rani has some fruit. [2]
(c) Find the probability that Rani does not have a burger given that she does not have any fruit. [4]
Mark scheme
5(a)
Fully correct labelled tree for method of transport with correct probabilities. B1
Fully correct labelled branches with correct probabilities for lateness with either 1 branch after W or 2 branches with the prob 0 B1
2
5(b) 0.35 × 0.3 + 0.44 × 0.8 (+ 0) M1
0.457 A1
2
5(c) P ( B'∩ F') M1
P(not B|not fruit) =
P ( F')
0.35 × 0.7 + 0.21 × 1 M1
1 − their ( b )
0.455 M1
0.543
(M1 for 1 – their (b) or summing three appropriate 2-factor probabilities, correct or consistent with their tree diagram as
denominator)
455 A1
0.838 or
543
4
The lengths of female snakes of a particular species are normally distributed with mean 54 cm and standard deviation 6.1 cm.
(a) Find the probability that a randomly chosen female snake of this species has length between
50 cm and 60 cm. [4]
The lengths of male snakes of this species also have a normal distribution. A scientist measures the lengths of a random sample of 200 male snakes of this species. He finds that 32 have lengths less than
45 cm and 17 have lengths more than 56 cm.
(b) Find estimates for the mean and standard deviation of the lengths of male snakes of this species.
[5]
Mark scheme
6(a) 50 − 54 60 − 54 M1
P <z< = P ( −0.6557 < Z < 0.9836 )
6.1 6.1
Both values correct A1
Φ (0.9836) – Φ (–0.6557) = Φ (0.9836) + Φ (0.6557) – 1 M1
= 0.8375 + 0.7441 – 1
(Correct area)
0.582 A1
4
6(b) 45 − μ B1
= −0.994
σ
56 − μ B1
= 1.372
σ
One appropriate standardisation equation with , , z-value (not probability) and 45 or 56. M1
11 = 2.366 σ M1
(M1 for correct algebraic elimination of µ or σ from their two simultaneous equations to form an equation in one variable)
σ = 4.65, μ = 49.6 A1
5
The numbers of chocolate bars sold per day in a cinema over a period of 100 days are summarised in the following table.
Number of chocolate bars sold 1 − 10 11 − 15 16 − 30 31 − 50 51 − 60
Number of days 18 24 30 20 8
(a) Draw a histogram to represent this information. [5]
(b) What is the greatest possible value of the interquartile range for the data? [2]
(c) Calculate estimates of the mean and standard deviation of the number of chocolate bars sold.
[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
7(a) Class widths: 10, 5, 15, 20, 10 M1
Frequency density = frequency/their class width: 1.8, 4.8, 2, 1, 0.8 M1
All heights correct on diagram (using a linear scale) A1
Correct bar ends B1
Bar ends: 10.5, 15.5, 30.5, 50.5, 60.5 B1
5
7(b) 11 – 15 and 31 – 50 B1
Greatest IQR = 50 – 11 = 39 B1
2
7(c) 18 × 5.5 + 24 × 13 + 30 × 23 + 20 × 40.5 + 8 × 55.5 2355 B1
Mean = = = 23.6
100 100
18 × 5.52 + 24 × 132 + 30 × 232 + 20 × 40.52 + 8 × 55.52 M1
Var = − mean 2
100
77917.5 A1
− mean 2 = 224.57
100
Standard deviation = 15.0 A1 FT
(FT their variance)
4
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Mathematics 9709/51 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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