(i) The number of athletes who ran between 10.5 and 11.0 seconds is the difference in cumulative frequencies:

\(10 - 4 = 6.\)

(ii) To draw the histogram, calculate the frequency for each interval:

• 10.0 to 10.5: 4 athletes
• 10.5 to 11.0: 6 athletes
• 11.0 to 12.0: 30 athletes
• 12.0 to 12.5: 9 athletes
• 12.5 to 13.5: 8 athletes

Use these frequencies to draw the histogram with appropriate widths and heights.

(iii) Calculate the mean using midpoints:

\(E(X) = \frac{(10.25 \times 4) + (10.75 \times 6) + (11.5 \times 30) + (12.25 \times 9) + (13 \times 8)}{57} = 11.7\)

Calculate the variance:

\(\text{Var}(X) = \frac{(10.25^2 \times 4) + (10.75^2 \times 6) + (11.5^2 \times 30) + (12.25^2 \times 9) + (13^2 \times 8)}{57} - (11.7)^2 = 0.547\)

(a) To draw the histogram, calculate the frequency density for each class interval using the formula:

\(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\)

For each interval:

• 100 – 800: \(\frac{8}{800} = 0.01\)
• 900 – 1200: \(\frac{12}{400} = 0.03\)
• 1300 – 2000: \(\frac{50}{800} = 0.0625\)
• 2100 – 3200: \(\frac{48}{1200} = 0.04\)
• 3300 – 4800: \(\frac{32}{1600} = 0.02\)

Draw bars with these heights on the histogram.

(b) The median is the 75th value. Cumulative frequencies are:

• 100 – 800: 8
• 900 – 1200: 20
• 1300 – 2000: 70
• 2100 – 3200: 118
• 3300 – 4800: 150

The median falls in the 2100 – 3200 interval.

(c) To find the interquartile range, calculate the lower quartile (LQ) and upper quartile (UQ):

LQ is the 37.5th value, which is in the 1300 – 2000 interval.

UQ is the 112.5th value, which is in the 2100 – 3200 interval.

Greatest possible value of IQR is \(3200 - 1250 = 1999\).

Solution

(i) First, find the class widths from the table:

• \( 1–5 \): \( 5 - 1 = 4 \), but since both ends are included, width \( = 5 \)
• \( 6–20 \): \( 20 - 6 + 1 = 15 \)
• \( 21–35 \): \( 35 - 21 + 1 = 15 \)
• \( 36–60 \): \( 60 - 36 + 1 = 25 \)
• \( 61–80 \): \( 80 - 61 + 1 = 20 \)

Then calculate frequency density using

\( \displaystyle \text{fd} = \frac{\text{Frequency}}{\text{Class Width}} \)

\( \displaystyle \text{fd} = \frac{24}{5},\ \frac{9}{15},\ \frac{21}{15},\ \frac{15}{25},\ \frac{42}{20} = 4.8,\ 0.6,\ 1.4,\ 0.6,\ 2.1 \).

Draw bars with these heights on the histogram; align class boundaries accurately and ensure correct widths with no gaps.

(ii) To estimate the mean, use class midpoints: \( 3,\ 13,\ 28,\ 48,\ 70.5 \).

\( \displaystyle \text{mean} = \frac{ (3 \times 24) + (13 \times 9) + (28 \times 21) + (48 \times 15) + (70.5 \times 42) }{111} = \frac{4458}{111} \approx 40.2\ \text{errors}. \)

(iii) Cumulative frequencies are \( 24,\ 33,\ 54,\ 69,\ 111 \).

Lower quartile position: \( \displaystyle 0.25 \times 111 = 27.75 \) → in class \( 6–20 \).
Upper quartile position: \( \displaystyle 0.75 \times 111 = 83.25 \) → in class \( 61–80 \).

For least possible IQR, take maximum possible LQ at \( 20 \) and minimum possible UQ at \( 61 \):

\( \displaystyle \text{IQR}_{\min} = 61 - 20 = 41 \).

1. In a correct histogram, the bars should be adjacent to each other without gaps, as the data is continuous. The diagram shows gaps between the bars, which is incorrect.

2. In a histogram, the area of each bar should represent the frequency of the data. This requires using frequency density, calculated as \(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}}\). The diagram does not use frequency density, leading to incorrect bar heights.

(i) The frequency density for the interval 200 to 250 is 1.5. The width of the interval is 50. Therefore, the number of people is given by:

\(1.5 \times 50 = 75\)

(ii) The frequencies for each interval are 10, 25, 50, 75, 30. The mid-points of the intervals are 125, 162.5, 187.5, 225, 300.

Mean calculation:

\(\text{Mean} = \frac{(10 \times 125) + (25 \times 162.5) + (50 \times 187.5) + (75 \times 225) + (30 \times 300)}{190}\)

\(= \frac{40562.5}{190} = 213\)

Standard deviation calculation:

\(sd^2 = \frac{10 \times 125^2 + 25 \times 162.5^2 + 50 \times 187.5^2 + 75 \times 225^2 + 30 \times 300^2}{190} - (213)^2\)

\(sd = 46.5 \text{ or } 46.6\)

(iii) The answers are estimates because the mid-points of each interval were used instead of the raw data.