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

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

Class widths are: 20, 10, 5, 10, 15.

Calculate frequency densities:

\(\frac{18}{20} = 0.9\)

\(\frac{48}{10} = 4.8\)

\(\frac{34}{5} = 6.8\)

\(\frac{32}{10} = 3.2\)

\(\frac{18}{15} = 1.2\)

Draw bars with these heights on the histogram, ensuring the axes are labeled with 'Frequency Density' and 'Time in minutes'.

(i) 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 10–29: \(\frac{10}{20} = 0.5\)

For 30–39: \(\frac{24}{10} = 2.4\)

For 40–49: \(\frac{30}{10} = 3\)

For 50–59: \(\frac{14}{10} = 1.4\)

For 60–89: \(\frac{12}{30} = 0.4\)

Draw bars with these heights on the graph, with bar ends at 9.5, 29.5, 39.5, 59.5, 89.5. Label axes with 'Frequency density (fd)' and 'Speed/km h^-1'.

(ii) To calculate the mean speed, use the midpoints of each class interval:

Midpoints: 19.5, 34.5, 44.5, 54.5, 74.5

Calculate the weighted mean:

\(\frac{19.5 \times 10 + 34.5 \times 24 + 44.5 \times 30 + 54.5 \times 14 + 74.5 \times 12}{90}\)

\(= \frac{195 + 828 + 1335 + 763 + 894}{90}\)

\(= \frac{4015}{90} \approx 44.6 \text{ km h}^{-1}\)

(i) To find the lower quartile, we need to determine the position of the 25th percentile in the cumulative frequency distribution. The cumulative frequencies are:

- 10–14: 6

\(- 15–19: 6 + 12 = 18\)

\(- 20–24: 18 + 14 = 32\)

\(- 25–34: 32 + 10 = 42\)

\(- 35–59: 42 + 8 = 50\)

The 25th percentile corresponds to the 12.5th value. Since the cumulative frequency reaches 18 in the 15–19 interval, the lower quartile is in the 15–19 kg class interval.

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

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

- 10–14: \(\frac{6}{5} = 1.2\)

- 15–19: \(\frac{12}{5} = 2.4\)

- 20–24: \(\frac{14}{5} = 2.8\)

- 25–34: \(\frac{10}{10} = 1.0\)

- 35–59: \(\frac{8}{25} = 0.32\)

Draw bars with these heights on the histogram, ensuring the horizontal axis ranges from at least 9.5 to 59.5 and the vertical axis is labeled with frequency density.

Solution:

(1) The total frequency is 242, so:

\(14 + 46 + 102 + a + 40 = 242\)

\(a = 40\)

(2) To estimate the mean, take midpoints of each class: 0.5, 1.5, 3.5, 7.5, 20.

Total \(f \times x = 7 + 69 + 357 + 300 + 800 = 1533\).

Mean \(= \dfrac{1533}{242} \approx 6.3\) minutes.

(3) The histogram can be drawn with the following class widths and frequencies per unit:

(i) 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 \(0 \leq t < 20\): \(\text{FD} = \frac{320}{20} = 16\)

For \(20 \leq t < 40\): \(\text{FD} = \frac{280}{20} = 14\)

For \(40 \leq t < 60\): \(\text{FD} = \frac{220}{20} = 11\)

For \(60 \leq t < 100\): \(\text{FD} = \frac{220}{40} = 5.5\)

For \(100 \leq t < 140\): \(\text{FD} = \frac{100}{40} = 2.5\)

Draw bars with these heights on the histogram.

(ii) To estimate the mean, use the midpoints of each class interval:

\(\text{Mean} = \frac{\sum (\text{Midpoint} \times \text{Frequency})}{\text{Total Frequency}}\)

\(= \frac{(10 \times 320) + (30 \times 280) + (50 \times 220) + (80 \times 220) + (120 \times 100)}{1140}\)

\(= \frac{3200 + 8400 + 11000 + 17600 + 12000}{1140}\)

\(= \frac{52200}{1140}\)

\(= 45.8\)