(a) To draw the cumulative frequency graph, plot the points (40, 0), (60, 14), (65, 38), (70, 60), (85, 106), and (100, 120) on a graph with the x-axis representing weight (kg) and the y-axis representing cumulative frequency. Connect these points with a smooth curve.

(b) To find the weight W such that 35% of the students weigh more than W kg, calculate 65% of the total cumulative frequency (since 100% - 35% = 65%).

Calculate 65% of 120: \(120 \times 0.65 = 78\).

Using the cumulative frequency graph, find the weight corresponding to a cumulative frequency of 78. This weight is approximately 76 kg.

(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:

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

(i) From the cumulative frequency graph, the cumulative frequency at 14 cm is approximately 55, and at 24 cm is approximately 156. Therefore, the number of leaves between 14 and 24 cm is \(156 - 55 = 101\).

(ii) 10% of 160 leaves is 16 leaves. Therefore, 144 leaves have a length less than \(L\). From the graph, \(L\) corresponds to a cumulative frequency of 144, which is approximately 22 cm.

(iii) The median is at the 80th leaf (\(\frac{160}{2}\)), which corresponds to approximately 15.6 cm on the graph. The lower quartile \(Q_1\) is at the 40th leaf, approximately 12.7 cm, and the upper quartile \(Q_3\) is at the 120th leaf, approximately 18.8 cm. The interquartile range is \(Q_3 - Q_1 = 18.8 - 12.7 = 6.1\).

(iv) The median for Ransha's data is higher than Sharim's data. The interquartile range for Ransha's data is lower than Sharim's data, indicating less spread.

To find the median, locate the 700th pupil on the cumulative frequency graph (since there are 1400 pupils in total). For Mathematics, the median is at 40 marks, and for English, it is at 55 marks.

To find the interquartile range (IQR), locate the 350th and 1050th pupils on the cumulative frequency graph. For Mathematics, the IQR is from 12 to 80, giving an IQR of 68. For English, the IQR is from 42 to 62, giving an IQR of 20.

Thus, the median of English is larger than the median of Mathematics, indicating a higher central tendency for English marks. The IQR for Mathematics is larger than for English, indicating a greater spread in Mathematics marks.

(i) Plot the cumulative frequency points: (0, 0), (10, 16), (20, 50), (30, 106), (50, 146), (70, 176), (90, 200). Join these points with a smooth curve or straight lines.

(ii) The median is the value at the \(\frac{200}{2} = 100\)th position. From the graph, this corresponds to approximately 29 minutes.

(iii) For 80 drivers, the time is at least \(T\) minutes, meaning \(200 - 80 = 120\) drivers took less than \(T\) minutes. From the graph, \(T\) corresponds to approximately 37 minutes.

(iv) Calculate the estimated mean using midpoints: Frequencies are 16, 34, 56, 40, 30, 24. Midpoints are 5, 15, 25, 40, 60, 80. Estimated mean \(= \frac{5 \times 16 + 15 \times 34 + 25 \times 56 + 40 \times 40 + 60 \times 30 + 80 \times 24}{200} = \frac{7310}{200} = 36.55\) minutes.