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Date	November 2020	Marks available	3	Reference code	20N.1.SL.TZ0.S_8
Level	Standard Level	Paper	Paper 1 (without calculator)	Time zone	Time zone 0
Command term	Find	Question number	S_8	Adapted from	N/A

Question

Each athlete on a running team recorded the distance ( $M$ miles) they ran in $30$ minutes.

The median distance is $4$ miles and the interquartile range is $1.1$ miles.

This information is shown in the following box-and-whisker plot.

The distance in miles, $M$ , can be converted to the distance in kilometres, $K$ , using the formula $K = \frac{8}{5} M$ .

The variance of the distances run by the athletes is $\frac{16}{9} {km}^{2}$ .

The standard deviation of the distances is $b$ miles.

A total of $600$ athletes from different teams compete in a $5 km$ race. The times the $600$ athletes took to run the $5 km$ race are shown in the following cumulative frequency graph.

There were $400$ athletes who took between $22$ and $m$ minutes to complete the $5 km$ race.

Find the value of $a$ .

[2]

Write down the value of the median distance in kilometres (km).

[1]

Find the value of $b$ .

[4]

Find $m$ .

[3]

The first $150$ athletes that completed the race won a prize.

Given that an athlete took between $22$ and $m$ minutes to complete the $5 km$ race, calculate the probability that they won a prize.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

valid approach (M1)

eg $Q_{3} - Q_{1}, Q_{3} - 1.1, 4.5 - a = 1.1$

$a = 3.4$ A1 N2

[2 marks]

$\frac{32}{5} (= 6.4)$ (km) A1 N1

[1 mark]

METHOD 1 (standard deviation first)

valid approach (M1)

eg $standard deviation = \sqrt{variance}, \sqrt{\frac{16}{9}}$

standard deviation $= \frac{4}{3}$ (km) (A1)

valid approach to convert their standard deviation (M1)

eg $\frac{4}{3} \times \frac{5}{8}, \sqrt{\frac{16}{9}} = \frac{8}{5} M$

$\frac{20}{24}$ (miles) $(= \frac{5}{6})$ A1 N3

Note: If no working shown, award M1A1M0A0 for the value $\frac{4}{3}$ .
If working shown, and candidate’s final answer is $\frac{4}{3}$ , award M1A1M0A0.

METHOD 2 (variance first)

valid approach to convert variance (M1)

eg ${(\frac{5}{8})}^{2}, \frac{64}{25}, \frac{16}{9} \times {(\frac{5}{8})}^{2}$

variance $= \frac{25}{36}$ (A1)

valid approach (M1)

eg $standard deviation = \sqrt{variance}, \sqrt{\frac{25}{36}}, \sqrt{\frac{16}{9} \times {(\frac{5}{8})}^{2}}$

$\frac{20}{24}$ (miles) $(= \frac{5}{6})$ A1 N3

[4 marks]

correct frequency for $22$ minutes (A1)

eg $20$

adding their frequency (do not accept $22 + 400$ ) (M1)

eg $20 + 400, 420$ athletes

$m = 30$ (minutes) A1 N3

[3 marks]

$27$ (minutes) (A1)

correct working (A1)

eg $130$ athletes between $22$ and $27$ minutes, $P (22 < t < 27) = \frac{150 - 20}{600}, \frac{13}{60}$

evidence of conditional probability or reduced sample space (M1)

eg $P (A B), P (t < 27 22 < t < 30), \frac{P (22 < t < 27)}{P (22 < t < m)}, \frac{150}{400}$

correct working (A1)

eg $\frac{\frac{130}{600}}{\frac{400}{600}}, \frac{150 - 20}{400}$

$\frac{130}{400} (\frac{13}{40} = \frac{78000}{240000} = \frac{390}{1200} = 0.325)$ A1 N5

Note: If no other working is shown, award A0A0M1A0A0 for answer of $\frac{150}{400}$ .
Award N0 for answer of $\frac{3}{8}$ with no other working shown.

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.2—Histograms, CF graphs, box plots

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Hide related questions

Topic 4—Statistics and probability » SL 4.3—Mean, median, mode. Mean of grouped data, standard deviation. Quartiles, IQR

Topic 4—Statistics and probability » SL 4.5—Probability concepts, expected numbers

Topic 4—Statistics and probability

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Markscheme

Examiners report

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