Find the value of \( a \).

Given: Median = 4 miles, IQR = 1.1 miles, \( Q_3 = 4.5 \) miles (from box plot).

IQR = \( Q_3 – Q_1 = 1.1 \).

\[ 4.5 – a = 1.1 \]

\[ a = 4.5 – 1.1 = 3.4 \]

Answer: \( a = 3.4 \)

Write down the median distance in kilometres.

Median in miles = 4. Conversion: \( K = \frac{8}{5}M \).

\[ K = \frac{8}{5} \cdot 4 = \frac{32}{5} = 6.4 \text{ km} \]

Answer: \( \frac{32}{5} \) (6.4 km)

Find the standard deviation \( b \) in miles.

Given: Variance in km² = \( \frac{16}{9} \).

Method 1 (Standard Deviation First):

Standard deviation in km: \[ \sqrt{\frac{16}{9}} = \frac{4}{3} \text{ km} \]

Convert to miles: \( K = \frac{8}{5}M \), so \( M = \frac{5}{8}K \).

Standard deviation in miles: \[ \frac{4}{3} \cdot \frac{5}{8} = \frac{20}{24} = \frac{5}{6} \]

Method 2 (Variance First):

Convert variance: Variance in miles = Variance in km × \( \left( \frac{5}{8} \right)^2 \).

\[ \frac{16}{9} \cdot \frac{25}{64} = \frac{16 \cdot 25}{9 \cdot 64} = \frac{400}{576} = \frac{25}{36} \]

Standard deviation in miles: \[ \sqrt{\frac{25}{36}} = \frac{5}{6} \]

Answer: \( b = \frac{5}{6} \)

Find \( m \).

From the cumulative frequency graph, 20 athletes took 22 minutes or less.

400 athletes took between 22 and \( m \) minutes: cumulative frequency at \( m \) = \( 20 + 400 = 420 \).

From the graph, 420 corresponds to 30 minutes.

Answer: \( m = 30 \text{ minutes} \)

Calculate the probability that an athlete who took between 22 and \( m \) minutes won a prize.

First 150 athletes won a prize. From the graph, 150 corresponds to 27 minutes.

Between 22 and 27 minutes: \( 150 – 20 = 130 \) athletes.

Between 22 and \( m = 30 \) minutes: 400 athletes.

Probability: \[ P(\text{prize} \mid 22 < t < 30) = \frac{P(22 < t < 27)}{P(22 < t < 30)} = \frac{130}{400} = \frac{13}{40} \]

Answer: \( \frac{13}{40} \) (0.325)