(i) The back-to-back stem-and-leaf diagram is constructed as follows:

Key: \( 5 \mid 8 \mid 4 \) means \( 0.85 \) m for flat screen.

(ii) To find the median of the conventional TVs, order the data: 0.65, 0.67, 0.69, 0.71, 0.74, 0.75, 0.77, 0.85, 0.86. The median is the middle value, 0.74.

The interquartile range (IQR) is calculated as the difference between the upper quartile (UQ) and the lower quartile (LQ). LQ = 0.68, UQ = 0.81, so IQR = 0.81 - 0.68 = 0.13.

(iii) For the flat screen TVs, calculate the mean:

Mean = \(\frac{0.85 + 0.94 + 0.91 + 0.96 + 1.04 + 0.89 + 1.07 + 0.92 + 0.76}{9} = 0.927\).

Calculate the standard deviation (SD):

SD = \(\sqrt{\frac{(0.85-0.927)^2 + (0.94-0.927)^2 + (0.91-0.927)^2 + (0.96-0.927)^2 + (1.04-0.927)^2 + (0.89-0.927)^2 + (1.07-0.927)^2 + (0.92-0.927)^2 + (0.76-0.927)^2}{9}}\)\( \approx 0.0882\).

(a) To find the median for the Cheetahs, list the times in order: 61, 63, 65, 78, 79, 82, 87, 88, 89, 90, 97, 98, 99, 100, 101, 102, 103, 104, 105. The median is the 10th value: 99 minutes.

For the interquartile range (IQR), find the lower quartile (LQ) and upper quartile (UQ). LQ is the 5th value: 83 minutes. UQ is the 15th value: 106 minutes. IQR = UQ - LQ = 106 - 83 = 23 minutes.

(b) The median time for the Cheetahs (99 minutes) is less than the median time for the Panthers (103 minutes), indicating that the Cheetahs are generally faster. The IQR for the Cheetahs (23 minutes) is greater than the IQR for the Panthers (14 minutes), indicating that the Cheetahs' times are more spread out.

(c) The total time for 20 Cheetahs is 99 minutes × 20 = 1980 minutes. The total time for the 19 Cheetahs is 1862 minutes. Kenny's time = 1980 - 1862 = 118 minutes.

(i) To create a back-to-back stem-and-leaf diagram, we use the integer part of the weights as the stem and the decimal part as the leaves.

Key: \( 1 \mid 196 \mid 2 \) means \( 1.961 \) kg for sugar and \( 1.962 \) kg for flour.

(ii) To find the median of the sugar weights, we arrange them in order: 1.961, 1.968, 1.977, 1.983, 1.984, 1.989, 1.994, 2.008, 2.011, 2.014, 2.017.

The median is the middle value: 1.989 kg.

To find the interquartile range, we find the lower quartile (LQ) and upper quartile (UQ).

LQ is the median of the first half: 1.977 kg.

UQ is the median of the second half: 2.011 kg.

\(Interquartile Range = UQ - LQ = 2.011 - 1.977 = 0.034 kg.\)

(i) To draw a stem-and-leaf diagram, separate each number into a 'stem' (the tens digit) and a 'leaf' (the units digit). Arrange the leaves in ascending order for each stem.

Key: 1 | 2 represents 12 people.

(ii) To find the median, order the data: 2, 5, 6, 8, 8, 12, 14, 16, 17, 17, 19, 21, 22, 23, 23, 23, 25, 26, 27, 31, 35. The median is the middle value, which is 19.

The lower quartile (LQ) is the median of the first half: 2, 5, 6, 8, 8, 12, 14, 16, 17, 17. LQ = 10.

The upper quartile (UQ) is the median of the second half: 21, 22, 23, 23, 23, 25, 26, 27, 31, 35. UQ = 24

The interquartile range (IQR) is UQ - LQ = 24 - 10 = 14.

(iii) The median is preferable because the mode could be any number which is duplicated more than twice, making it less reliable as a measure of central tendency.

To create a stem-and-leaf diagram, we first identify the stems and leaves from the data set. The stems are the tens digits, and the leaves are the units digits.

1. List the stems: 10, 11, 12, 13, 14, 15, 16.

2. Assign the leaves to each stem:

Key: \( 10 \mid 4 \) represents \( 104 \).