(i) To draw the histogram, calculate the frequency density for each class interval. The class widths are 20, 10, 10, 10, 10, 15. The frequency densities are calculated as follows:

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

For 1–20: \(\frac{40}{20} = 2.0\)

For 21–30: \(\frac{34}{10} = 3.4\)

For 31–40: \(\frac{56}{10} = 5.6\)

For 41–50: \(\frac{54}{10} = 5.4\)

For 51–60: \(\frac{29}{10} = 2.9\)

For 61–75: \(\frac{21}{15} = 1.4\)

Draw the histogram with these frequency densities.

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

Midpoints: 10.5, 25.5, 35.5, 45.5, 55.5, 68

Mean = \(\frac{\sum xf}{234} = \frac{8769.5}{234} = 37.5\)

To calculate the standard deviation, first find the variance:

Variance = \(\frac{\sum x^2f}{234} - \text{mean}^2\)

Standard Deviation = \(\sqrt{\text{Variance}} = 16.9\)

(i) To draw the histogram, calculate the frequency density for each class interval. Frequency density is given by \(\text{Frequency} / \text{Class Width}\).

For \(0.1 \leq t \leq 0.5\), Frequency = 11, Class Width = 0.4, Frequency Density = \(11 / 0.4 = 27.5\).

For \(0.6 \leq t \leq 1.0\), Frequency = 15, Class Width = 0.4, Frequency Density = \(15 / 0.4 = 37.5\).

For \(1.1 \leq t \leq 2.0\), Frequency = 18, Class Width = 0.9, Frequency Density = \(18 / 0.9 = 20\).

For \(2.1 \leq t \leq 3.0\), Frequency = 30, Class Width = 0.9, Frequency Density = \(30 / 0.9 = 33.33\).

For \(3.1 \leq t \leq 4.5\), Frequency = 21, Class Width = 1.4, Frequency Density = \(21 / 1.4 = 15\).

Draw bars with these heights on the histogram.

(ii) Calculate the mean time spent:

Find the midpoints of each interval: \(0.3, 0.8, 1.55, 2.55, 3.8\).

Multiply each midpoint by its frequency and sum these products: \((0.3 \times 11) + (0.8 \times 15) + (1.55 \times 18) + (2.55 \times 30) + (3.8 \times 21)\) \(= 199.5\).

Divide by the total frequency: \(199.5 / 95 = 2.1\).

Thus, the estimated mean time spent is 2.1 hours.

To calculate the standard deviation, we first find the midpoints of each class interval:

Midpoints: 10, 30, 45, 55, 80

Using the formula for variance:

\(\text{Var} = \frac{32 \times 10^2 + 46 \times 30^2 + 96 \times 45^2 + 52 \times 55^2 + 24 \times 80^2}{250} - 43.2^2\)

\(= \frac{549900}{250} - 43.2^2\)

\(= 2199.6 - 1866.24\)

\(= 333.36\)

Standard deviation \(\text{Sd} = \sqrt{333.36} = 18.3\)

To construct the grouped frequency table, we need to divide the range from \( 41.5 \) kg to \( 61.5 \) kg into five equal intervals.

\( \text{Total range} = 61.5 - 41.5 = 20 \) kg

\( \text{Class width} = \dfrac{20}{5} = 4 \) kg

The intervals are:

• 41.5–45.5
• 45.5–49.5
• 49.5–53.5
• 53.5–57.5
• 57.5–61.5

Next, count the number of weights in each interval:

• 41.5–45.5: 45, 44, 45 (4 weights)
• 45.5–49.5: 47, 49, 46, 48, 49 (7 weights)
• 49.5–53.5: 50, 51, 52, 50, 51, 52, 51, 53, 52, 53 (10 weights)
• 53.5–57.5: 55, 55, 54, 56, 57 (5 weights)
• 57.5–61.5: 58, 59, 60, 61 (4 weights)

Thus, the frequencies for each interval are \( 4,\ 7,\ 10,\ 5,\ 4 \) respectively.

(i) The modal age group is the one with the highest frequency density. From the histogram, the age group 30-35 years has the highest frequency density.

(ii) To find the number of fathers between 25 and 30 years old, multiply the frequency density by the class width. The frequency density for 25-30 years is 4.8, and the class width is 5 years. Therefore, the number of fathers is:

\(4.8 \times 5 = 24\)

(iii) To find the total number of fathers in the sample, sum the frequencies for all age groups. The frequencies are calculated by multiplying the frequency density by the class width for each group:

\(4 \times 1 + 3.6 \times 5 + 4.8 \times 5 + 5.6 \times 5 + 5.2 \times 5 + 2 \times 5 = 4 + 18 + 24 + 28 + 26 + 10 = 110\)