A Level Mathematics 9709 Paper 51, May/June 2023

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9709/51 · May/June 2023

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A summary of 50 values of $x$ gives $Σ (x - q) = 700, Σ (x - q)^{2} = 14 235,$ where $q$ is a constant.

Find the standard deviation of these values of $x$ .

[2]

Given that $Σ x = 2865$ , find the value of $q$ .

[2]

Find the number of ways in which a committee of 6 people can be chosen from 6 men and 8 women if it must include 3 men and 3 women.

[2]

A different committee of 6 people is to be chosen from 6 men and 8 women. Three of the 6 men are brothers. Find the number of ways in which this committee can be chosen if there are no restrictions on the numbers of men and women, but it must include no more than two of the brothers.

[3]

Find the number of different arrangements of the 8 letters in the word COCOONED.

[1]

Find the number of different arrangements of the 8 letters in the word COCOONED in which the first letter is O and the last letter is N.

[2]

Find the probability that a randomly chosen arrangement of the 8 letters in the word COCOONED has all three Os together given that the two Cs are next to each other.

[3]

A mathematical puzzle is given to a large number of students. The times taken to complete the puzzle are normally distributed with mean 14.6 minutes and standard deviation 5.2 minutes.

In a random sample of 250 of the students, how many would you expect to have taken more than 20 minutes to complete the puzzle?

[4]

All the students are given a second puzzle to complete. Their times, in minutes, are normally distributed with mean $μ$ and standard deviation $σ$ . It is found that 20% of the students have times less than 14.5 minutes and 67% of the students have times greater than 18.5 minutes. Find the value of $μ$ and the value of $σ$ .

[5]

The populations of 150 villages in the UK, to the nearest hundred, are summarised in the table.

Population	$100 - 800$	$900 - 1200$	$1300 - 2000$	$2100 - 3200$	$3300 - 4800$
Number of villages	$8$	$12$	$50$	$48$	$32$

Draw a histogram to represent this information.

[4]

Write down the class interval which contains the median for this information.

[1]

Find the greatest possible value of the interquartile range for the populations of the 150 villages.

[2]

Eli has four fair 4-sided dice with sides labelled 1, 2, 3, 4. He throws all four dice at the same time. The random variable $X$ denotes the number of 2s obtained.

Show that $P (X = 3) = \frac{3}{64}$ .

[2]

Complete the following probability distribution table for $X$ .

$x$	$0$	$1$	$2$	$3$	$4$
$P (X = x)$	$\frac{81}{256}$			$\frac{3}{64}$	$\frac{1}{256}$

[2]

Find $E (X)$ .

[2]

Eli throws the four dice at the same time on 96 occasions. Use an approximation to find the probability that he obtains at least two 2s on fewer than 20 of these occasions.

[5]

A children’s wildlife magazine is published every Monday. For the next 12 weeks it will include a model animal as a free gift. There are five different models: tiger, leopard, rhinoceros, elephant and buffalo, each with the same probability of being included in the magazine. Sahim buys one copy of the magazine every Monday.

Find the probability that the first time that the free gift is an elephant is before the 6th Monday.

[2]

Find the probability that Sahim will get more than two leopards in the 12 magazines.

[3]

Find the probability that after 5 weeks Sahim has exactly one of each animal.

[3]

Mark scheme

9709/51 · May/June 2023

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1a · 2 marks

M1
$Var = [\frac{Σ (x - q)^{2}}{50} - {(\frac{Σ (x - q)}{50})}^{2}] = \frac{14235}{50} - {(\frac{700}{50})}^{2}$ $[= 284.7 - 196 = 88.7]$
A1
$[sd = \sqrt{88.7} =] 9.42$

1b · 2 marks

M1
$Σ x - 50 q = 700$ $[2865 - 50 q = 700]$
A1
$q = 43.3, 43 \frac{3}{10}$

2a · 2 marks

M1
$^{6} C_{3} \times^{8} C_{3}$
A1
1120

2b · 3 marks

B1
0 brothers $[^{3} C_{0}] \times^{11} C_{6} = 462$ , 1 brother $^{3} C_{1} \times^{11} C_{5} = 1386$ , 2 brothers $^{3} C_{2} \times^{11} C_{4} = 990$ (or $^{14} C_{6} -^{11} C_{3}$ )
M1
Add values of 3 correct scenarios (or $3003 - 165$ )
A1
2838

3a · 1 mark

B1
$[\frac{8!}{2! 3!} =] 3360$

3b · 2 marks

M1
$\frac{6!}{2! 2!}$
A1
180

3c · 3 marks

M1
$[P (O O O ∣ C C) = \frac{P (O O O \cap C C)}{P (C C)} =] 5!$
M1
$\frac{7!}{3!}$
A1
$= \frac{120}{840}, \frac{1}{7}, 0.143$

4a · 4 marks

M1
$P (Z > \frac{20 - 14.6}{5.2}) = P (Z > 1.03846)$
M1
$1 - 0.8504$
A1
0.150
B1
$[250 \times their 0.1496 =] 37, 38$

4b · 5 marks

B1
$z_{1} = \frac{14.5 - μ}{σ} = - 0.842$
B1
$z_{2} = \frac{18.5 - μ}{σ} = - 0.44$
M1
Use of the $\pm$ standardisation formula once with $μ, σ$ and a $z$ -value
M1
Solve, obtaining values for $μ$ and $σ$ .
A1
$μ = 22.9, σ = 9.95$

5a · 4 marks

M1
Calculation of frequency densities
A1
All heights correct on graph.
B1
Bar ends at 50, 850, 1250, 2050, 3250, 4850 read at the axis with a horizontal linear scale with at least 3 values indicated. $50 ⩽ horizontal scale ⩽ 4850$ .
B1
Axes labelled frequency density (fd) and population (pop) OE, or in a title. Linear vertical scale, with at least 3 values indicated. Vertical axis must cover at least the range $0 ⩽ vertical axis ⩽ 0.0625$ . Axes may be reversed.

5b · 1 mark

B1
2100 – 3200

5c · 2 marks

M1
3249 – 1250
A1
1999

6a · 2 marks

M1
$[P (X = 3) =] \frac{3}{4} \times {(\frac{1}{4})}^{3} \times 4$
A1
$= \frac{3}{64}$

6b · 2 marks

B1
Either $P (1) = \frac{27}{64}, 0.421875$ or $P (2) = \frac{27}{128}, 0.2109375$ correct to at least 3SF.
B1
Both values in table.

6c · 2 marks

M1
$[E (X) =] [0 \times \frac{81}{256}] + 1 \times their \frac{27}{64} + 2 \times their \frac{27}{128} + 3 \times \frac{12}{256} + 4 \times \frac{1}{256}$ $= 0 + \frac{27}{64} + \frac{54}{128} + \frac{36}{256} + \frac{4}{256}$
A1
$= 1$

6d · 5 marks

B1
$Mean = 96 \times \frac{67}{256} = 25.125$ $Var = 96 \times \frac{67}{256} \times \frac{189}{256} = 18.549$
M1
Substituting their $μ$ and $σ$ into $\pm$ standardisation formula (any number for 19.5).
M1
Using continuity correction 19.5 or 20.5 in their standardisation formula.
M1
$[= P (Z < - 1.306) = 1 - Φ (1.306) =] 1 - 0.9042 =$ Appropriate area $Φ$ , from final process.
A1
0.0958

7a · 2 marks

M1
$[P (X < 6) = P (X ⩽ 5) =] 1 - {0.8}^{5}$ (or geometric series sum for 5 terms)
A1
$= 0.672$

7b · 3 marks

M1
One term $^{12} C_{x} (p)^{x} (1 - p)^{12 - x}, 0 < p < 1, x \neq 0, 1, 2.$
A1
Correct expression, accept unsimplified, leading to final answer.
B1
$= 0.442$

7c · 3 marks

M1
$(0.2)^{5} \times s$ , $s$ a positive integer (or $(^{5} C_{1})^{5}$ or $5^{5}$ as denominator)
M1
$t \times 5!$ where $0 < t < 1$ (or $^{5} C_{1} \times^{4} C_{1} \times^{3} C_{1} \times^{2} C_{1} \times [^{1} C_{1}]$ or $5!$ as numerator)
A1
$= 0.0384, \frac{24}{625}$

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