(i) To find the frequency density, use the formula:

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

For the class 3.20 ≤ x < 3.40, the class width is \(3.40 - 3.20 = 0.20\).

Given frequency is 8, so:

\(a = \frac{8}{0.20} = 40\)

(ii) To draw the histogram:

- Use uniform linear scales from at least 2.8 to 3.4 on the x-axis and 0 to 240 on the y-axis.

- Label both axes, with 'length (m)' on the x-axis and 'frequency density' on the y-axis.

- Draw bars with the following heights based on frequency density:

• 2.80 ≤ x < 3.00: height = 85
• 3.00 ≤ x < 3.10: height = 240
• 3.10 ≤ x < 3.20: height = 190
• 3.20 ≤ x < 3.40: height = 40

- Ensure correct widths with no gaps between bars.

To construct the histogram, we first determine the frequency of factory floor areas within each interval:

- Interval \(0 \leq x < 500\): 150, 350, 450 (3 factories)

- Interval \(500 \leq x < 1000\): 578, 595, 644, 722, 798, 802, 904 (7 factories)

- Interval \(1000 \leq x < 2000\): 1000, 1330, 1533, 1561, 1778, 1960 (6 factories)

- Interval \(2000 \leq x < 3000\): 2167, 2330, 2433 (3 factories)

- Interval \(3000 \leq x < 4000\): 3231 (1 factory)

The frequencies are therefore 3, 7, 6, 3, and 1 for each respective interval.

According to the mark scheme, the scaled frequencies are the same as the frequencies: 3, 7, 3, 1.5, 0.5.

The histogram should have bars with these heights, and the axes should be labeled appropriately with 'scaled f' and 'area, m\(^2\)'.

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

Mean = \(\frac{(2.5 \times 11 + 7.5 \times 20 + 15 \times 32 + 25 \times 18 + 35 \times 10 + 55 \times 6)}{97} = 18.4\) minutes.

To calculate the standard deviation, use:

\(\text{sd} = \sqrt{\frac{(2.5^2 \times 11 + 7.5^2 \times 20 + 15^2 \times 32 + 25^2 \times 18 + 35^2 \times 10 + 55^2 \times 6)}{97} - \text{mean}^2}\) = \(13.3\) minutes.

(ii) To draw the histogram, calculate the frequency densities:

• 0 ≤ t < 5: \(\frac{11}{5} = 2.2\)
• 5 ≤ t < 10: \(\frac{20}{5} = 4.0\)
• 10 ≤ t < 20: \(\frac{32}{10} = 3.2\)
• 20 ≤ t < 30: \(\frac{18}{10} = 1.8\)
• 30 ≤ t < 40: \(\frac{10}{10} = 1.0\)
• 40 ≤ t < 70: \(\frac{6}{30} = 0.2\)

Plot these frequency densities against the time intervals on the histogram.

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

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

Class widths: 20, 10, 10, 20, 30

Frequency densities: \(\frac{440}{20} = 22\), \(\frac{720}{10} = 72\), \(\frac{920}{10} = 92\), \(\frac{300}{20} = 15\), \(\frac{120}{30} = 4\)

Plot these on the histogram with time on the x-axis and frequency density on the y-axis.

(b) Calculate the standard deviation using the midpoints of each class interval:

Midpoints: 10, 25, 35, 50, 75

Variance formula:

\(\frac{440(10^2) + 720(25^2) + 920(35^2) + 300(50^2) + 120(75^2)}{2500} - 31.44^2\)

\(= \frac{3046000}{2500} - 31.44^2 = 229.9264\)

Standard deviation \(= \sqrt{229.9264} = 15.2\)

(c) The upper quartile lies in the class interval 30–40.

(d) The standard deviation stays the same because the data remains in the same intervals, so the overall distribution is unchanged.

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

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

Class Widths: 30, 15, 20, 10, 25

Frequency Densities: \(\frac{21}{30} = 0.7\), \(\frac{30}{15} = 2\), \(\frac{68}{20} = 3.4\), \(\frac{86}{10} = 8.6\), \(\frac{45}{25} = 1.8\)

Draw bars with these heights for each class interval.

(b) The median is the middle value of the ordered data set. Since there are 250 children, the median is the 125.5th value. Cumulative frequencies: 21, 51, 119, 205, 250. The median falls in the 66 – 75 interval.

(c) The mean is 61.05 minutes, which is different from the median interval. This indicates the distribution is not symmetrical, possibly skewed.