A Level Mathematics 9709 Paper 52, Oct/Nov 2024

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9709/52 · Oct/Nov 2024

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At a college, the students choose exactly one of tennis, hockey or netball to play. The table shows the numbers of students in Year 1 and Year 2 at the college playing each of these sports.

	Tennis	Hockey	Netball
Year 1	16	22	12
Year 2	24	18	28

One student is chosen at random from the 120 students. Events $X$ and $N$ are defined as follows: $X$ : the student is in Year 1 $N$ : the student plays netball.

Find $P (X ∣ N)$ .

[1]

Find $P (N ∣ X)$ .

[1]

Determine whether or not $X$ and $N$ are independent events.

[1]

One of the students who plays netball takes 8 shots at goal. On each shot, the probability that she will succeed is 0.15, independently of all other shots. Find the probability that she succeeds on fewer than 3 of these shots.

[3]

Find the number of different arrangements of the 9 letters in the word ALGEBRAIC.

[1]

Find the number of different arrangements of the 9 letters in the word ALGEBRAIC in which there are no more than two letters between the two As.

[3]

A fair coin and an ordinary fair six-sided dice are thrown at the same time. The random variable $X$ is defined as follows. • If the coin shows a tail, $X$ is twice the score on the dice. • If the coin shows a head, $X$ is the score on the dice if the score is even and $X$ is 0 otherwise.

Draw up the probability distribution table for $X$ .

[3]

Find $Var (X)$ .

[3]

The heights, in metres, of white pine trees are normally distributed with mean 19.8 and standard deviation 2.4 . In a certain forest there are 450 white pine trees.

How many of these trees would you expect to have height less than 18.2 metres?

[4]

The heights, in metres, of red pine trees are normally distributed with mean 23.4 and standard deviation $σ$ . It is known that 26% of red pine trees have height greater than 25.5 metres. Find the value of $σ$ .

[3]

In a class of 21 students, there are 10 violinists, 6 guitarists and 5 pianists. A group of 7 is to be chosen from these 21 students. The group will consist of 4 violinists, 2 guitarists and 1 pianist.

In how many ways can the group of 7 be chosen?

[2]

On another occasion a group of 5 will be chosen from the 21 students. The group must contain at least 2 violinists, at least 1 guitarist and at most 1 pianist. In how many ways can the group of 5 be chosen?

[4]

Teams of 15 runners took part in a charity run last Saturday. The times taken, in minutes, to complete the course by the runners from the Falcons and the runners from the Kites are shown in the table.

Falcons	38	39	42	44	46	48	50	51	52	56	58	59	64	69	76
Kites	32	40	40	45	47	48	52	54	58	59	59	60	61	63	65

Draw a back-to-back stem-and-leaf diagram to represent this information, with the Falcons on the left-hand side.

[4]

Find the median and the interquartile range of the times for the Falcons.

[3]

Let $x$ and $y$ denote the times, in minutes, of a runner from the Falcons and a runner from the Kites respectively. It is given that $\sum x = 792, \sum x^{2} = 43 504, \sum y = 783, \sum y^{2} = 42 223.$ Find the mean and the standard deviation of the times taken by all 30 runners from the two teams.

[3]

In a game, players attempt to score a goal by kicking a ball into a net. The probability that Leno scores a goal is 0.4 on any attempt, independently of all other attempts. The random variable $X$ denotes the number of attempts that it takes Leno to score a goal.

Find $P (X = 5)$ .

[1]

Find $P (3 ⩽ X ⩽ 7)$ .

[2]

Find the probability that Leno scores his second goal on or before his 5th attempt.

[3]

Leno has 75 attempts to score a goal. Use a suitable approximation to find the probability that Leno scores more than 28 goals but fewer than 35 goals.

[5]

Mark scheme

9709/52 · Oct/Nov 2024

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1a · 1 mark

B1
[P(X|N) =] 12/40

1b · 1 mark

B1
[P(N|X) =] 12/50

1c · 1 mark

B1
P(N ∩ X) = 12/120, P(N) = 40/120, P(X) = 50/120 40/120 × 50/120 = 5/36, 0.138[8...] ≠ 12/120, 0.1 Not independent

1d · 3 marks

M1
One term of form ^8C_x (p)^x (1 – p)^(8–x) with 0 < p < 1, x ≠ 0 or 8
A1
Correct unsimplified expression, no terms omitted leading to final answer
B1
= 0.895

2a · 1 mark

B1
[9! / 2! =] 181440

2b · 3 marks

B1
Correct outcome/value for 1 identified scenario, accept unsimplified
M1
Add values of 3 correct scenarios, no incorrect/repeated scenarios (or their 2(a), or correct, subtract values of 5 correct scenarios, no incorrect/repeated scenarios)
A1
= 105 840

3a · 3 marks

B1
Table with correct x values and at least 2 correct probabilities
B1
Four more probabilities correctly linked to the correct x value, need not be in table, accept unsimplified
B1
7 correct probabilities linked with correct outcomes, may not be in table

3b · 3 marks

M1
[E(X) =] [0 +] 2 × 2/12 + 4 × 2/12 + 6 × 2/12 + 8 × 1/12 + 10 × 1/12 + 12 × 1/12
M1
Appropriate variance formula using their (E(X))^2 value
A1
= 14.75, 14 3/4

4a · 4 marks

M1
Use of ± standardisation formula with 18.2, 19.8 and 2.4 substituted appropriately, no continuity correction
M1
Calculating the appropriate probability areas (leading to their final answer, expect < 0.5)
A1
= 0.252(3)
1 mark
[Expected number = 0.2523 × 450 = 113.5, ] = 113 or 114

4b · 3 marks

B1
0.643 ⩽ z ⩽ 0.6435 or −0.6435 ⩽ z ⩽ −0.643 seen
M1
± standardisation formula with 25.5, 23.4, σ equating to a z-value, (not 1 – their z-value…)
A1
σ = 3.27

5a · 2 marks

M1
^10C_a × ^6C_b × ^5C_c , a + b + c = 7, a,b,c integers
A1
[= 210 × 15 × 5] = 15750

5b · 4 marks

M1
One product using 2 or 3 combinations with upper numbers correct and lower numbers summing to 5 and linked to a correct identified scenario
B1
2 identified outcomes evaluated accurately, accept unsimplified
M1
Add values of 5 correct scenarios, no incorrect/repeated scenarios
A1
Total = 10935

6a · 4 marks

B1
Correct stem, ignore extra values (not in reverse, not split)
B1
Correct Falcons labelled on left, leaves in order from right to left and lined up vertically, no commas or other punctuation
B1
Correct Kites labelled on same diagram, leaves in order and lined up vertically, no commas or other punctuation
B1
Correct key, for their diagram, need both teams names and ‘mins’ at least once here, or in leaf headings or title

6b · 3 marks

B1
Median = 51 [minutes]
M1
[IQR =] 59 – 44
A1
= 15 [minutes]

6c · 3 marks

B1
[mean = (792 + 783)/30 = 1575/30 =] 52.5, 52 1/2, 105/2
M1
Accept unsimplified variance formula. FT their mean. Ignore any square root leading to sd for this mark
A1
σ = (√101.3167 =) 10.1

7a · 1 mark

B1
[(0.6)^4 × 0.4 =] 0.0518[4], 162/3125

7b · 2 marks

M1
(1 – p^7) – (1 – p^2) or p^2 – p^7 seen, 0 < p < 1 (or (1 – p) × p^2 + (1 – p) × p^3 + (1 – p) × p^4 + (1 – p) × p^5 + (1 – p) × p^6 seen, 0 < p < 1 or 0.144(1 – p^5)/(1 – p) seen 0 < p < 1)
A1
= 0.332[0... ], 25938/78125

7c · 3 marks

M1
2 correct unsimplified outcomes (or at least 2 correct unsimplified terms)
M1
Add values for 4 identified correct scenarios (or add values for 4 terms of the form ^5C_a(0.4)^a(0.6)^(5–a) or 1 – sum of 2 terms of the form ^5C_a(0.4)^a(0.6)^(5–a))
A1
= 0.663, 2072/3125

7d · 5 marks

B1
[Mean = 75 × 0.4 =] 30 [Variance = 75 × 0.4 × 0.6 =] 18
M1
Substituting their µ and positive σ into one ±standardising formula (any number for 28.5 or 34.5), not σ^2, not √σ
M1
Using continuity corrections 27.5 or 28.5 and 34.5 or 35.5 in their 2 separate standardisation formula
M1
Appropriate area Φ, from final process. Must be a probability
A1
= 0.494

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