(i) Plot the cumulative frequency points at the upper class boundaries: (20, 52), (30, 94), (40, 142), (50, 172), (70, 222), (100, 250). Draw a smooth curve or straight lines connecting these points.

(ii) To find \(k\), locate the cumulative frequency of 150 on the graph and read the corresponding rainfall value. From the graph, \(k \approx 42\) mm.

(iii) Calculate the frequencies for each class interval: \(52, 42, 48, 30, 50, 28\).

Find the midpoints of each class: \(10, 25, 35, 45, 60, 85\).

Calculate the mean: \(\bar{x} = \frac{(10 \times 52) + (25 \times 42) + (35 \times 48) + (45 \times 30) + (60 \times 50) + (85 \times 28)}{250} = \frac{9980}{250} = 39.9 \text{ mm}\)

Calculate the variance: \(s^2 = \frac{(10^2 \times 52) + (25^2 \times 42) + (35^2 \times 48) + (45^2 \times 30) + (60^2 \times 50) + (85^2 \times 28)}{250} - \bar{x}^2 = 539.59\)

Standard deviation: \(s = \sqrt{539.59} = 23.2 \text{ mm}\)

To draw the cumulative frequency graph, calculate the cumulative frequencies:

Plot these cumulative frequencies against the upper class boundaries on the graph.

To find the median, locate the point where the cumulative frequency is 450 (half of 900) on the graph. From the graph, the median time \(t\) is approximately 4.8 minutes.

To solve this problem, we first need to construct the cumulative frequency table from the given frequency distribution:

(i) Plot these cumulative frequencies against the upper class boundaries on the grid to draw the cumulative frequency graph.

(ii) To estimate the percentage of trees with circumference larger than 75 cm, we need to find the cumulative frequency at 75 cm. Using linear interpolation between 50 cm and 80 cm:

The total number of trees is 140. The number of trees with circumference larger than 75 cm is:

The percentage is then:

(i) To draw the cumulative frequency graph, calculate the cumulative frequencies:

• 3 – 5 seconds: Cumulative frequency = 10
• 6 – 8 seconds: Cumulative frequency = 10 + 15 = 25
• 9 – 11 seconds: Cumulative frequency = 25 + 17 = 42
• 12 – 16 seconds: Cumulative frequency = 42 + 4 = 46
• 17 – 25 seconds: Cumulative frequency = 46 + 2 = 48

Plot these cumulative frequencies against the upper class boundaries: (5.5, 10), (8.5, 25), (11.5, 42), (16.5, 46), (25.5, 48).

(ii) To find \(t\), note that 35 cars accelerated in more than \(t\) seconds, so 13 cars accelerated in \(t\) seconds or less. From the cumulative frequency graph, find the time corresponding to a cumulative frequency of 13. This gives \(t \approx 6.5\) seconds, so \(6 < t < 7\).

(i) The frequency for each class is calculated by multiplying the frequency density by the class width:

• For length 0-5 cm: Frequency density = 2, Class width = 5, Frequency = 2 × 5 = 10
• For length 5-10 cm: Frequency density = 8, Class width = 5, Frequency = 8 × 5 = 40
• For length 10-20 cm: Frequency density = 12, Class width = 10, Frequency = 12 × 10 = 120
• For length 20-25 cm: Frequency density = 6, Class width = 5, Frequency = 6 × 5 = 30

(ii) Cumulative frequencies are calculated as follows:

• Length < 5 cm: Cumulative frequency = 10
• Length < 10 cm: Cumulative frequency = 10 + 40 = 50
• Length < 20 cm: Cumulative frequency = 50 + 120 = 170
• Length < 25 cm: Cumulative frequency = 170 + 30 = 200

Plot these cumulative frequencies against the upper class boundaries to draw the cumulative frequency graph.

(iii) From the cumulative frequency graph:

• Median is the value at the 100th position (half of 200), which is approximately 14.2 cm.
• Lower quartile (LQ) is the value at the 50th position, approximately 10 cm.
• Upper quartile (UQ) is the value at the 150th position, approximately 18.5 cm.
• Interquartile range (IQR) = UQ - LQ = 18.5 - 10 = 8.5 cm.

(iv) Estimate the mean using midpoints and frequencies:

• Mean = \(\frac{(2.5 \times 10) + (7.5 \times 40) + (15 \times 120) + (22.5 \times 30)}{200}\)
• Mean = \(\frac{25 + 300 + 1800 + 675}{200} = \frac{2800}{200} = 14\)