(i) To find the mean weight of the 45 fruits, calculate the total weight of the oranges and lemons and divide by the total number of fruits:

\(\text{Mean} = \frac{(220 \times 20) + (118 \times 25)}{45} = \frac{4400 + 2950}{45} = \frac{7350}{45} = 163\)

(ii) First, find \(\Sigma x_o^2\) and \(\Sigma x_l^2\) using the variance formula:

For oranges:

\(\frac{\Sigma x_o^2}{20} - 220^2 = 32^2\)

\(\Sigma x_o^2 = 20 \times (32^2 + 220^2) = 988480\)

For lemons:

\(\frac{\Sigma x_l^2}{25} - 118^2 = 12^2\)

\(\Sigma x_l^2 = 25 \times (12^2 + 118^2) = 351700\)

\(Combine these to find the variance of all fruits:\)

\(\Sigma x_o^2 + \Sigma x_l^2 = 988480 + 351700 = 1340180\)

\(\text{Variance} = \frac{1340180}{45} - \left(\frac{7350}{45}\right)^2 = 3100 \text{ to } 3120\)

First, calculate the midpoints of each class interval:

• 101 – 120: midpoint = 110.5
• 121 – 130: midpoint = 125.5
• 131 – 135: midpoint = 133
• 136 – 145: midpoint = 140.5
• 146 – 160: midpoint = 153

Calculate the mean using the formula:

\(\bar{x} = \frac{\sum (f \times x)}{\sum f} = \frac{18 \times 110.5 + 48 \times 125.5 + 34 \times 133 + 32 \times 140.5 + 18 \times 153}{150}\)

\(= \frac{1989 + 6024 + 4522 + 4496 + 2754}{150} = 131.9\)

\(Calculate the variance using the formula:\)

\(s^2 = \frac{\sum (f \times x^2)}{\sum f} - (\bar{x})^2\)

\(= \frac{18 \times 110.5^2 + 48 \times 125.5^2 + 34 \times 133^2 + 32 \times 140.5^2 + 18 \times 153^2}{150} - (131.9)^2\)

\(= \frac{3200 + 41400 + 194400 + 157300 + 153600}{150} - 17397.61\)

\(= 137.54\)

\(Standard deviation \(s = \sqrt{137.54} = 11.7 \]\)

(i) For Ashfaq, a suitable statistical diagram is a stem and leaf diagram. This is because it effectively shows the shape, spread, and range of the data, and is particularly useful for a small number of prices, such as the 21 prices collected by Ashfaq. It also allows for easy calculation of the median.

(ii) For Kuljit, a suitable statistical diagram is a histogram. This is because it shows the shape, spread, and modal class of the data, and is suitable for a large number of prices, such as the 163 prices collected by Kuljit. A histogram is more appropriate than a stem and leaf diagram for such a large dataset.

To find the mean, use the formula:

\(\bar{x} = \frac{\sum x}{n}\)

\(\bar{x} = \frac{48.3 + 55.2 + 59.9 + 67.7 + 60.5 + 75.6 + 62.5 + 57.4 + 53.4 + 49.2 + 64.1}{11} = 59.4\)

To find the standard deviation, use the formula:

\(\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}\)

Calculate each squared deviation:

\((48.3 - 59.4)^2, (55.2 - 59.4)^2, (59.9 - 59.4)^2, \ldots, (64.1 - 59.4)^2\)

Sum these squared deviations:

\(\sum (x - \bar{x})^2 = 646.64\)

Then, \(\sigma = \sqrt{\frac{646.64}{11}} = 7.68\)

(i) To find the mean age of all 24 cars, we calculate the total sum of ages and divide by the total number of cars:

For Red Street Garage: \(\text{Total sum} = 3.6 \times 9 = 32.4\)

For Fairwheel Garage: \(\Sigma x = 64\)

Total sum of ages = \(32.4 + 64 = 96.4\)

\(Total number of cars = 9 + 15 = 24\)

Mean age = \(\frac{96.4}{24} = 4.02 \text{ years}\)

\((ii) To find the standard deviation of all 24 cars, we first find the variance:\)

Variance for Red Street Garage: \(\frac{\Sigma x^2}{9} - 3.6^2 = 1.925^2\)

\(\Sigma x^2 = 150\)

Total \(\Sigma x^2 = 150 + 352 = 502\)

\(Mean age = 4.02\)

Variance = \(\frac{502}{24} - 4.02^2 = 4.780\)

Standard deviation = \(\sqrt{4.780} = 2.19 \text{ years}\)