First, calculate the frequencies for each class interval:

• 40 < x ≤ 60: 14
• 60 < x ≤ 65: 24
• 65 < x ≤ 70: 22
• 70 < x ≤ 85: 46
• 85 < x ≤ 100: 14

Next, determine the midpoints for each class interval:

• 40 < x ≤ 60: 50
• 60 < x ≤ 65: 62.5
• 65 < x ≤ 70: 67.5
• 70 < x ≤ 85: 77.5
• 85 < x ≤ 100: 92.5

Calculate the mean:

\(\text{Mean} = \frac{0 \times 20 + 14 \times 50 + 24 \times 62.5 + 22 \times 67.5 + 46 \times 77.5 + 14 \times 92.5}{120} = \frac{8545}{120} = 71.2 \text{ kg}\)

\(Calculate the variance:\)

\(\text{Variance} = \frac{0 \times 20^2 + 14 \times 50^2 + 24 \times 62.5^2 + 22 \times 67.5^2 + 46 \times 77.5^2 + 14 \times 92.5^2}{120} - 71.2^2\)

\(= \frac{625062.5}{120} - 71.2^2 = 138.23\)

Calculate the standard deviation:

\(\text{Standard deviation} = \sqrt{138.23} = 11.8 \text{ kg}\)

(i) To find the mean age of all 32 members, calculate the total age of all members and divide by the total number of members:

\(\text{Total age of Juniors} = 15.5 \times 12 = 186\)

\(\text{Total age of Seniors} = 910\)

\(\text{Total age of all members} = 186 + 910 = 1096\)

\(\text{Mean age} = \frac{1096}{32} = 34.25 \text{ years}\)

\((ii) To find the standard deviation of all 32 members, first find the variance:\)

For Juniors, \(\text{variance} = \frac{\Sigma x^2}{12} - 15.5^2 = 1.2^2\)

\(\Sigma x^2 = 2900.28\)

For the whole group:

\(\Sigma z^2 = \Sigma x^2 + \Sigma y^2 = 2900.28 + 42850 = 45750\)

\(\text{Variance} = \frac{45750}{32} - (34.25)^2 = 256.63\)

\(\text{Standard deviation} = \sqrt{256.63} = 16.0 \text{ years}\)

(i) To find the mean price of all 51 holidays, use the formula for the combined mean:

\(\text{Mean} = \frac{30 \times 1500 + 21 \times 2400}{51} = \frac{45000 + 50400}{51} = 1870.59\)

Rounding to 3 significant figures, the mean is 1870.

(ii) First, find \(\sum x_F^2\) and \(\sum x_L^2\) using the given standard deviations and means:

For Farfield Travel:

\(230^2 = \frac{\sum x_F^2}{30} - 1500^2\)

\(\sum x_F^2 = 30 \times (230^2 + 1500^2) = 69\,087\,000\)

For Lacket Travel:

\(160^2 = \frac{\sum x_L^2}{21} - 2400^2\)

\(\sum x_L^2 = 21 \times (160^2 + 2400^2) = 121\,497\,600\)

\(Now, calculate the combined variance:\)

\(\text{New variance} = \frac{69\,087\,000 + 121\,497\,600}{51} - 1870.59^2 = 237\,853\)

\(The standard deviation is the square root of the variance:\)

\(\text{New standard deviation} = \sqrt{237\,853} \approx 488\)

To solve the problem, we first organize the data in ascending order: 36, 37, 38, 38, 38, 39, 40, 40, 42, 45.

1. Mode: The mode is the number that appears most frequently. Here, 38 appears three times, which is more frequent than any other number. Thus, the mode is 38.
2. Median: The median is the middle value of the ordered data. Since there are 10 data points, the median is the average of the 5th and 6th values: \(\frac{38 + 39}{2} = 38.5\).
3. Interquartile Range (IQR): The IQR is the difference between the upper quartile \(Q_3\) and the lower quartile \(Q_1\).
   • \(Q_1\) is the median of the first half of the data: 36, 37, 38, 38, 38. The median is 38.
   • \(Q_3\) is the median of the second half of the data: 39, 40, 40, 42, 45. The median is 40.
   • Thus, \(\text{IQR} = Q_3 - Q_1 = 40 - 38 = 2\).

(i) To find the mean age of all 19 people, use the formula for the combined mean:

\(\bar{x} = \frac{(48.7 \times 12) + (38.1 \times 7)}{19} = \frac{584.4 + 266.7}{19} = \frac{851.1}{19} = 44.8 \text{ years}\)

(ii) First, find \(\Sigma x^2\) and \(\Sigma y^2\) using the given standard deviations and means:

For the first group:

\(7.65^2 = \frac{\Sigma x^2}{12} - 48.7^2\)

\(58.5225 = \frac{\Sigma x^2}{12} - 2371.69\)

\(\Sigma x^2 = 12 \times (58.5225 + 2371.69) = 29162.55\)

For the second group:

\(4.2^2 = \frac{\Sigma y^2}{7} - 38.1^2\)

\(17.64 = \frac{\Sigma y^2}{7} - 1451.61\)

\(\Sigma y^2 = 7 \times (17.64 + 1451.61) = 10284.75\)

\(Now, find the combined variance:\)

\(\text{Combined variance} = \frac{29162.55 + 10284.75}{19} - 44.79^2\)

\(= \frac{39447.3}{19} - 2006.0641\)

\(= 2076.1737 - 2006.0641 = 70.1096\)

Finally, the standard deviation is:

\(\sigma = \sqrt{70.1096} \approx 8.37 \text{ or } 8.36 \text{ years}\)