(a) Draw a cumulative frequency graph with the given cumulative frequencies at the upper class boundaries: (200, 16), (300, 46), (500, 88), (900, 122), (1200, 134), (1600, 140).

(b) To find the interquartile range (IQR), use the cumulative frequency graph:

• Lower quartile (LQ) at 25% of 140: 35
• Upper quartile (UQ) at 75% of 140: 105
• IQR = UQ - LQ = 105 - 35 = 70

(c) Calculate the mean and standard deviation:

• Class midpoints: 100, 250, 400, 700, 1050, 1400
• Frequencies: 16, 30, 42, 34, 12, 6
• Mean: \(\frac{16 \times 100 + 30 \times 250 + 42 \times 400 + 34 \times 700 + 12 \times 1050 + 6 \times 1400}{140} = 505\)
• Variance: \(\frac{16 \times 100^2 + 30 \times 250^2 + 42 \times 400^2 + 34 \times 700^2 + 12 \times 1050^2 + 6 \times 1400^2}{140} - 505^2 = 324^2\)
• Standard deviation: \(\sqrt{324^2} = 324\)

(a) To estimate the number of plants with heights less than 100 cm, locate 100 cm on the x-axis and find the corresponding cumulative frequency on the y-axis. The graph shows approximately 60 plants.

(b) The 65th percentile corresponds to 65% of 160 plants, which is 104 plants. Locate 104 on the y-axis and find the corresponding height on the x-axis. The graph shows this height to be approximately 136 cm.

(c) The interquartile range (IQR) is the difference between the upper quartile (Q3) and the lower quartile (Q1). For 160 plants, Q1 is at the 40th plant (25% of 160) and Q3 is at the 120th plant (75% of 160). From the graph, Q1 is approximately 76 cm and Q3 is approximately 150 cm. Therefore, the IQR is 150 - 76 = 74 cm.

(a) To draw the cumulative frequency graph:

• Calculate the cumulative frequencies: 12, 28, 60, 126, 146, 150.
• Plot these cumulative frequencies against the upper boundaries of the intervals: 4.5, 10.5, 20.5, 30.5, 40.5, 60.5.
• Draw a smooth curve through the points.

(b) To find \(d\) for 30% of journeys:

• Calculate 70% of 150: \(0.7 \times 150 = 105\).
• Find the distance corresponding to a cumulative frequency of 105 on the graph. \(d \approx 27 \text{ km}\).

(c) To calculate the mean distance:

• Find midpoints of each interval: 2.25, 7.5, 15.5, 25.5, 35.5, 50.5.
• Calculate the mean: \(\bar{x} = \frac{(2.25 \times 12) + (7.5 \times 16) + (15.5 \times 32) + (25.5 \times 66) + (35.5 \times 20) + (50.5 \times 4)}{150}\)
• \(\bar{x} = \frac{3238}{150} \approx 21.6 \text{ km}\).