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 \( 61 \):

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