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

Midpoints: 3, 8, 15.5, 25.5, 40.5

Mean = \(\frac{(3 \times 59) + (8 \times 67) + (15.5 \times 38) + (25.5 \times 18) + (40.5 \times 11)}{193} = 11.4\)

To calculate the standard deviation, use:

\(\sigma^2 = \frac{(3^2 \times 59) + (8^2 \times 67) + (15.5^2 \times 38) + (25.5^2 \times 18) + (40.5^2 \times 11)}{193} - (11.4)^2\)

\(\sigma = 9.78 \text{ or } 9.79\)

(ii) For the histogram, calculate frequency densities:

Frequency density = \(\frac{\text{Frequency}}{\text{Class width}}\)

Class widths: 5, 5, 10, 10, 20

Frequency densities: \(\frac{59}{5} = 11.8, \frac{67}{5} = 13.4, \frac{38}{10} = 3.8, \frac{18}{10} = 1.8, \frac{11}{20} = 0.55\)

Draw bars with these heights at midpoints 5.5, 10.5, 20.5, 30.5, and 40.5.

(i) To find the median, calculate the cumulative frequency to find the middle value. The cumulative frequencies are: 19, 31, 59, 81, 99, 112. The median is the 56th value, which falls in the 15 < t ≤ 20 interval. The upper quartile (UQ) is the 84th value, which falls in the 25 < t ≤ 40 interval.

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

• 0 < t ≤ 10: fd = 19/10 = 1.9
• 10 < t ≤ 15: fd = 12/5 = 2.4
• 15 < t ≤ 20: fd = 28/5 = 5.6
• 20 < t ≤ 25: fd = 22/5 = 4.4
• 25 < t ≤ 40: fd = 18/15 = 1.2
• 40 < t ≤ 60: fd = 13/20 = 0.65

Draw bars with these heights and corresponding widths.

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

\(\frac{(5 \times 19) + (12.5 \times 12) + (17.5 \times 28) + (22.5 \times 22) + (32.5 \times 18) + (50 \times 13)}{112}\) =\( \frac{2465}{112} = 22.0 \text{ minutes}\)

Solution

(i) The median is the 110th value (since \( \dfrac{220}{2} = 110 \)). From the cumulative frequency table, the median lies in the interval \(45\text{–}50\ \text{g}\).

(ii) The lower quartile (LQ) is the 55th value, which lies in the interval \(40\text{–}45\ \text{g}\). The upper quartile (UQ) is the 165th value, which lies in the interval \(50\text{–}60\ \text{g}\). Therefore, the smallest interquartile range (IQR) could be \(5\ \text{g}\) (if \( \text{LQ} = 45 \) and \( \text{UQ} = 50 \)), and the largest could be \(20\ \text{g}\) (if \( \text{LQ} = 40 \) and \( \text{UQ} = 60 \)).

(iii) The number of sausages weighing between \(50\ \text{g}\) and \(60\ \text{g}\) is the difference in cumulative frequencies: \( 210 - 160 = 50 \).

(iv) To draw the histogram, first find class frequencies from cumulative frequencies, then compute frequency density \( \big( \text{fd} = \dfrac{\text{frequency}}{\text{class width}} \big) \):

\[ \begin{aligned} &<20:\ \text{fd} = \dfrac{0}{10} = 0,\\[4pt] &20\text{–}30:\ \text{fd} = \dfrac{20}{10} = 2,\\[4pt] &30\text{–}40:\ \text{fd} = \dfrac{30}{10} = 3,\\[4pt] &40\text{–}45:\ \text{fd} = \dfrac{50}{5} = 10,\\[4pt] &45\text{–}50:\ \text{fd} = \dfrac{60}{5} = 12,\\[4pt] &50\text{–}60:\ \text{fd} = \dfrac{50}{10} = 5,\\[4pt] &60\text{–}70:\ \text{fd} = \dfrac{10}{10} = 1. \end{aligned} \]

Plot these frequency densities against the corresponding weight intervals, with bar widths equal to class widths and no gaps.

(i) To find the frequencies, use the frequency density from the histogram. The frequency is calculated by multiplying the frequency density by the class width.

• For \( 2 < t \le 4 \): Frequency density \(= 10\), Width \(= 2\), Frequency \(= 10 \times 2 = 20\)
• For \( 4 < t \le 6 \): Frequency density \(= 22\), Width \(= 2\), Frequency \(= 22 \times 2 = 44\)
• For \( 6 < t \le 7 \): Frequency density \(= 34\), Width \(= 1\), Frequency \(= 34 \times 1 = 34\)
• For \( 7 < t \le 8 \): Frequency density \(= 30\), Width \(= 1\), Frequency \(= 30 \times 1 = 30\)
• For \( 8 < t \le 10 \): Frequency density \(= 15\), Width \(= 2\), Frequency \(= 15 \times 2 = 30\)
• For \( 10 < t \le 16 \): Frequency density \(= 6\), Width \(= 6\), Frequency \(= 6 \times 6 = 36\)

(ii) To estimate the mean time, calculate the midpoints of each class interval and use them to find the mean:

• Midpoints: \( 3,\ 5,\ 6.5,\ 7.5,\ 9,\ 13 \)
• \( \text{Mean} = \frac{ (3 \times 20) + (5 \times 44) + (6.5 \times 34) + {}\\ (7.5 \times 30) + (9 \times 30) + (13 \times 36) }{194}\) \(= \frac{1464}{194} = 7.55\ \text{minutes} \)

(i) The frequency density for the 1 – 10 interval is given by:

\(\text{Frequency Density} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{2x}{10}\)

We know the height is 3 cm, so:

\(3 = \frac{2x}{10}\)

Solving for \(x\):

\(3 \times 10 = 2x\)

\(30 = 2x\)

\(x = 15\)

For the 51 – 70 interval, the class width is 20, and the frequency is \(x = 15\).

\(\text{Frequency Density} = \frac{15}{20} = 0.75\)

Thus, the height of the 51 – 70 rectangle is 0.75 cm.

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

\(\text{Mean} = \frac{(5.5 \times 30) + (15.5 \times 60) + (23 \times 45) + (28 \times 75) + (40.5 \times 60) + (60.5 \times 15)}{285}\)

\(= \frac{165 + 930 + 1035 + 2100 + 2430 + 907.5}{285}\)

\(= \frac{7567.5}{285}\)

\(\approx 26.6 \text{ grams}\)