(a) Calculate cumulative frequencies:

• 0 – 9: 15
• 10 – 14: 15 + 48 = 63
• 15 – 19: 63 + 66 = 129
• 20 – 30: 129 + 21 = 150

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

(b) To find the length d such that 40% of the fish have a length of d cm or more, calculate 60% of 150, which is 90. From the cumulative frequency graph, find the length corresponding to a cumulative frequency of 90. This is approximately 16.5 cm.

(c) Calculate the midpoints of each class:

• 0 – 9: 4.75
• 10 – 14: 12
• 15 – 19: 17
• 20 – 30: 25

\(Use the formula for variance:\)

\(\text{Var} = \frac{(4.75^2 \times 15) + (12^2 \times 48) + (17^2 \times 66) + (25^2 \times 21)}{150} - 15.295^2\)

Calculate:

\(\text{Var} = \frac{39449.4375}{150} - 15.295^2 = 29.1\)