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:

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

(i) Let the cost of a sweater be \(x\).

The total cost of the items is given by:

\(4 \times 5.5 + 3x + 90 = 8 \times 29\)

\(22 + 3x + 90 = 232\)

\(112 + 3x = 232\)

\(3x = 120\)

\(x = 40\)

Thus, the cost of a sweater is $40.

(ii) Since the standard deviation is $0, all items cost the same. Therefore, the cost of each item is $26. Diksha buys 4 shirts, so the total cost for the shirts is:

\(4 \times 26 = 104\)

Thus, Diksha spends $104 on shirts.

To estimate the mean, we first find the midpoints of each class interval:

• 1 – 25: midpoint = 13
• 26 – 35: midpoint = 30.5
• 36 – 45: midpoint = 40.5
• 46 – 55: midpoint = 50.5
• 56 – 90: midpoint = 73

Next, we calculate the mean using the formula:

\(\bar{x} = \frac{\sum f x}{\sum f} = \frac{4 \times 13 + 24 \times 30.5 + 38 \times 40.5 + 34 \times 50.5 + 20 \times 73}{120} = \frac{5500}{120} = 45.8\)

\(To find the variance, we use:\)

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

\(\sum f x^2 = 4 \times 13^2 + 24 \times 30.5^2 + 38 \times 40.5^2 + 34 \times 50.5^2 + 20 \times 73^2 = 278620\)

\(s^2 = \frac{278620}{120} - 45.8^2 = 2321.8333 - 2097.64 = 224.1933\)

The standard deviation is:

\(s = \sqrt{224.1933} = 14.9\)

(i) To find the mean weight of the 27 birds, use the formula for the combined mean:

\(\text{new mean} = \frac{9 \times 7.1 + 18 \times 5.2}{27} = 5.83 \text{ kg}\)

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

For turkeys: \(1.45^2 = \frac{\Sigma x_t^2}{9} \Rightarrow \Sigma x_t^2 = 472.6125\)

For geese: \(0.96^2 = \frac{\Sigma x_g^2}{18} - 5.2^2 \Rightarrow \Sigma x_g^2 = 503.3088\)

Now, find the combined standard deviation:

\(\text{New } s^2 = \frac{472.6125 + 503.3088}{27} - 5.83^2 = 2.117\)

\(\text{New } s = \sqrt{2.117} = 1.46 \text{ kg}\)

To find the mean, sum all the data points and divide by the number of data points:

\(\text{Mean} = \frac{5 + (-2) + 12 + 7 + (-3) + 2 + (-6) + 4 + 0 + 8}{10} = \frac{27}{10} = 2.7\)

\(To find the variance, use the formula:\)

\(\text{Variance} = \frac{(5^2 + (-2)^2 + 12^2 + 7^2 + (-3)^2 + 2^2 + (-6)^2 + 4^2 + 0^2 + 8^2)}{10} - (\text{Mean})^2\)

\(= \frac{35.1}{10} - 2.7^2 = 3.51 - 7.29 = 27.8\)

(i) To find the mean height of the remaining group of 27 people, we use the formula for the mean:

\(\text{New mean} = \frac{172.6 \times 28 - 161.8}{27} = 173 \text{ cm}\)

(ii) First, find \(\Sigma x^2\) for the original group of 28 people using the formula:

\(\Sigma x^2 = (4.58^2 + 172.6^2) \times 28 = 834728.6\)

Then, for the remaining group of 27 people:

\(\text{Remaining } \Sigma x^2 = 834728.6 - 161.8^2 = 808549.36\)

Finally, calculate the standard deviation for the remaining group:

\(\text{sd of remaining} = \sqrt{\frac{808549.36}{27} - 173^2} = 4.16 \text{ cm}\)

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

(i) To find the mean \(\bar{x}\), sum the times and divide by the number of students:

\(\bar{x} = \frac{15 + 10 + 48 + 10 + 19 + 14 + 16}{7} = \frac{132}{7} = 18.9\)

To find the standard deviation \(sd\), use the formula:

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

Calculate each squared deviation:

• \((15 - 18.9)^2 = 15.21\)
• \((10 - 18.9)^2 = 79.21\)
• \((48 - 18.9)^2 = 841.21\)
• \((10 - 18.9)^2 = 79.21\)
• \((19 - 18.9)^2 = 0.01\)
• \((14 - 18.9)^2 = 24.01\)
• \((16 - 18.9)^2 = 8.41\)

Sum of squared deviations: \(15.21 + 79.21 + 841.21 + 79.21 + 0.01 + 24.01 + 8.41 = 1057.27\)

\(sd = \sqrt{\frac{1057.27}{7}} = \sqrt{151.04} = 12.3\)

(ii) The median is the most appropriate measure of central tendency because it is not affected by the outlier (48).

(iii) The mode is inappropriate because it is 10, which is the lowest value and not representative of the data. The mean is inappropriate because it is affected by the outlier (48), which skews the average.

(i) Since the standard deviation is $0, all rides must cost the same, which is the mean of $2.50. Therefore, the roller coaster and the water slide each cost $2.50 per ride.

(ii) Let the cost of the revolving drum be $x. The equation for the mean cost is:

\(1 \times 2.5 + 3 \times 2.5 + 6 \times x = 3.76 \times 10\)

\(6x = 37.6 - 10\)

\(x = 4.6\)

\(The cost of the revolving drum is $4.60. Now, calculate the variance:\)

\(\sigma^2 = \frac{(2.5^2 \times 1 + 2.5^2 \times 3 + 4.6^2 \times 6)}{10} - 3.76^2\)

\(\sigma = 1.03\)

To find the mean, sum all the lengths and divide by the number of lengths:

\(\bar{x} = \frac{32 + 35 + 45 + 37 + 38 + 44 + 33 + 39 + 36 + 45}{10} = \frac{384}{10} = 38.4 \text{ mm}\)

\(To find the standard deviation, first calculate the variance:\)

\(s^2 = \frac{1}{10} \left( (32 - 38.4)^2 + (35 - 38.4)^2 + (45 - 38.4)^2 + (37 - 38.4)^2 + (38 - 38.4)^2 + (44 - 38.4)^2 + (33 - 38.4)^2 + (39 - 38.4)^2 + (36 - 38.4)^2 + (45 - 38.4)^2 \right)\)

\(s^2 = \frac{1}{10} (40.96 + 11.56 + 43.56 + 1.96 + 0.16 + 31.36 + 29.16 + 0.36 + 5.76 + 43.56)\)

\(s^2 = \frac{1}{10} (208.4) = 20.84\)

Then, the standard deviation is:

\(s = \sqrt{20.84} \approx 4.57 \text{ mm}\)

(i) In the first round, 32 teams play, resulting in 16 matches. Therefore, 16 teams lose and play only 1 match.

(ii) In the second round, 16 teams play, resulting in 8 matches. Therefore, 8 teams lose and play exactly 2 matches.

(iii) The frequency table is constructed as follows:

(iv) To calculate the mean and variance:
Total matches played = 16(1) + 8(2) + 4(3) + 2(4) + 2(5) = 62
Mean = \(\frac{62}{32} = 1.9375 \approx 1.94\)
For variance, calculate \(\sum f x^2 = 16(1^2) + 8(2^2) + 4(3^2) + 2(4^2) + 2(5^2) = 166\)
Variance = \(\frac{166}{32} - \left(\frac{62}{32}\right)^2 = 1.43\)

To find the median, first arrange the data in ascending order: 40, 40, 41, 42, 45, 47, 48, 49, 50, 51, 352.

The median is the middle value, which is the 6th value in this ordered list: 47.

The median is $47,000.

The mean is not suitable because the data contains an outlier (352), which skews the mean. The mode is not suitable because it does not represent the central tendency effectively in this context.

(i) To find the mean age of the whole group, use the formula for the combined mean:

\(\bar{x} = \frac{(\bar{x}_w \times 10) + (\bar{x}_m \times 13)}{23} = \frac{(41.2 \times 10) + (46.3 \times 13)}{23} = 44.1\)

(ii) First, find \(\Sigma x_w^2\) and \(\Sigma x_m^2\) using the given standard deviations:

\(15.1^2 = \frac{\Sigma x_w^2}{10} - 41.2^2\)

\(\Sigma x_w^2 = 19254.5\)

\(12.7^2 = \frac{\Sigma x_m^2}{13} - 46.3^2\)

\(\Sigma x_m^2 = 29964.74\)

Calculate the total sum of squares:

\(\Sigma = 19254.5 + 29964.74 = 49219.24\)

Finally, find the standard deviation for the whole group:

\(\text{sd} = \sqrt{\frac{49219.24}{23} - 44.1^2} = 14.0\)

To illustrate the data diagrammatically, follow these steps:

1. Choose either pie charts or bar charts for the representation.
2. If using pie charts, create two separate pie charts: one for males and one for females. Divide each pie chart into three sectors representing 'Young', 'Middle-aged', and 'Elderly' with angles proportional to the percentages given (e.g., for males: Young = 40%, Middle-aged = 35%, Elderly = 25%).
3. If using bar charts, create two sets of bars: one set for males and one set for females. Each set should have three bars representing 'Young', 'Middle-aged', and 'Elderly'. Ensure the height of each bar is proportional to the percentage given.
4. Label each chart or set of bars with the appropriate categories and percentages. Include labels for 'Males' and 'Females', as well as 'Young', 'Middle-aged', and 'Elderly'.

To estimate the mean number of hours of sunshine, we first find the midpoints of each class interval:

• For \(30 \leq x < 60\), midpoint = 45
• For \(60 \leq x < 90\), midpoint = 75
• For \(90 \leq x < 110\), midpoint = 100
• For \(110 \leq x < 140\), midpoint = 125
• For \(140 \leq x < 180\), midpoint = 160
• For \(180 \leq x < 240\), midpoint = 210

Next, we calculate the weighted sum of the midpoints:

\(4 \times 45 + 8 \times 75 + 14 \times 100 + 25 \times 125 + 7 \times 160 + 2 \times 210 = 6845\)

The total number of years is 60. Therefore, the estimated mean is:

\(\text{Mean} = \frac{6845}{60} = 114.08\overline{3}\)

To find \(f\) and the total number of children:

The midpoints of the intervals are: 5, 15, 30, and 60.

The equation for the mean is:

\(\frac{5 \times 2 + 15f + 30 \times 11 + 60 \times 4}{17 + f} = 27.5\)

\(10 + 15f + 330 + 240 = 27.5(17 + f)\)

\(15f + 580 = 467.5 + 27.5f\)

\(580 - 467.5 = 27.5f - 15f\)

\(112.5 = 12.5f\)

\(f = 9\)

Total number of children = \(17 + f = 26\).

To find the standard deviation:

Calculate \(\sigma\) using the formula:

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

\(\sigma = 16.1\)

(i) To find the mean:

\(\bar{x} = \frac{\Sigma x}{n} = \frac{745}{18} = 41.4\)

To find the standard deviation:

\(s = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2} = \sqrt{\frac{33951}{18} - \left(\frac{745}{18}\right)^2} = 13.2\)

(ii) Let the age of the person who left be \(x\).

The sum of the ages of the remaining 17 people is:

\(17 \times 41 = 697\)

Therefore, \(745 - x = 697\), giving \(x = 48\).

To find the standard deviation of the remaining 17 people:

\(s = \sqrt{\frac{\Sigma x^2 - x^2}{17} - (41)^2} = \sqrt{\frac{33951 - 48^2}{17} - 41^2} = 13.4\)

(i) To find the mean for Team A:

Mean, \(\bar{x}_A = \frac{150 + 220 + 77 + 30 + 298 + 118 + 160 + 57}{8} = 139\).

To find the standard deviation for Team A:

\(First, calculate the variance:\)

Variance, \(s_A^2 = \frac{(150-139)^2 + (220-139)^2 + (77-139)^2 + (30-139)^2 + (298-139)^2 + (118-139)^2 + (160-139)^2 + (57-139)^2}{8}\).

\(s_A^2 = \frac{121 + 6561 + 3844 + 11881 + 25281 + 441 + 441 + 6724}{8} = 6901.25\).

Standard deviation, \(\sigma_A = \sqrt{6901.25} = 83.1\).

(ii) Team B has a smaller standard deviation (29.63) compared to Team A (83.1), indicating that Team B's scores are more consistent.

Let the random variable \(X\) take values 0 and 2 with frequencies 23 and 17 respectively.

The total number of observations is 40.

The mean \(\bar{x}\) is calculated as:

\(\bar{x} = \frac{(0 \times 23) + (2 \times 17)}{40} = \frac{34}{40} = 0.85\)

The variance \(s^2\) is calculated using:

\(s^2 = \frac{(0^2 \times 23) + (2^2 \times 17)}{40} - (\bar{x})^2\)

\(s^2 = \frac{(0 \times 23) + (4 \times 17)}{40} - (0.85)^2\)

\(s^2 = \frac{68}{40} - 0.7225\)

\(s^2 = 1.7 - 0.7225 = 0.9775\)

\(Thus, the variance is approximately 0.978.\)

(a) To find the mean, sum all the estimates and divide by the number of estimates:

\(\text{Mean} = \frac{4.1 + 4.2 + 4.4 + 4.5 + 4.6 + 4.8 + 5.0 + 5.2 + 5.3 + 5.4 + 5.5 + 5.8 + 6.0 + 6.2 + 6.3 + 6.4 + 6.6 + 6.8 + 6.9 + 19.4}{20} = \frac{123.4}{20} = 6.17\)

(b) To find the median, arrange the data in ascending order and find the middle value. Since there are 20 values, the median is the average of the 10th and 11th values:

\(\text{Median} = \frac{5.4 + 5.5}{2} = 5.45\)

(c) The median is more suitable because the mean is unduly influenced by the extreme value of 19.4, which skews the average higher than the central tendency of the majority of the data.

(a) To find the mean height of all 17 members, we use the formula for the mean:

\(\text{Mean height} = \frac{\Sigma x + \Sigma y}{6 + 11} = \frac{1050 + 1991}{17} = \frac{3041}{17} = 178.9 \text{ cm}\)

\((b) To find the standard deviation, we first calculate the variance:\)

\(\Sigma x^2 + \Sigma y^2 = 193700 + 366400 = 560100\)

\(\text{Variance} = \frac{560100}{17} - \left(\frac{3041}{17}\right)^2 = \frac{560100}{17} - 178.88^2\)

\(\text{Variance} = 32947.0588 - 32008.289 = 938.7698\)

\(\text{Standard deviation} = \sqrt{938.7698} = 30.8 \text{ cm}\)

To find the median and interquartile range, first arrange the data in ascending order: 30, 38, 40, 40, 45, 50, 52, 55, 55, 60, 62, 110.

Median: The median is the average of the 6th and 7th values: \(\frac{50 + 52}{2} = 51\).

Interquartile Range (IQR): The lower quartile \(Q_1\) is the 3rd value: 40. The upper quartile \(Q_3\) is the 9th value: 57.5. Thus, \(\text{IQR} = Q_3 - Q_1 = 57.5 - 40 = 17.5\).

Disadvantage of using the mean: The mean is disproportionately affected by the extreme value 110, which is an outlier.

Given: Mean of 20 values \(\bar{x} = 60\) and standard deviation \(s = 4\).

(i) To find \(\Sigma x\) and \(\Sigma x^2\):

\(\Sigma x = 60 \times 20 = 1200\)

\(Using the formula for variance:\)

\(s^2 = \frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2\)

\(4^2 = \frac{\Sigma x^2}{20} - 60^2\)

\(16 = \frac{\Sigma x^2}{20} - 3600\)

\(\frac{\Sigma x^2}{20} = 3616\)

\(\Sigma x^2 = 3616 \times 20 = 72320\)

(ii) For the combined 30 values:

\(\Sigma x = 1200 + 550 = 1750\)

\(\Sigma x^2 = 72320 + 40500 = 112800\)

Mean of 30 values:

\(\bar{x} = \frac{1750}{30} = 58.3\)

Variance of 30 values:

\(s^2 = \frac{112800}{30} - \left(\frac{1750}{30}\right)^2\)

\(s^2 = 3760 - 3412.89 = 357.11\)

Standard deviation:

\(s = \sqrt{357.11} \approx 18.9\)

To determine the most suitable measure of central tendency, we consider the mean, mode, and median:

• Mean: The mean is affected by extreme values. In this dataset, the value 768 is much larger than the others, skewing the mean upwards. Therefore, the mean is not suitable.
• Mode: The mode is the most frequently occurring value. In this dataset, all values are different, so there is no mode. Thus, the mode is not suitable.
• Median: The median is the middle value when the data is ordered. It is not affected by extreme values. Ordering the data: 180, 194, 227, 228, 229, 235, 238, 242, 246, 275, 311, 332, 768. The median is 238, which is a representative measure of central tendency for this dataset.

Therefore, the median is the most suitable measure of central tendency for these times.

First, calculate the new sums:

\(\Sigma x = 1923 + 166 + 172 + 182 = 2443\)

\(\Sigma x^2 = 337221 + 166^2 + 172^2 + 182^2 = 427485\)

Calculate the mean:

\(\text{Mean} = \frac{\Sigma x}{14} = \frac{2443}{14} = 174.5\)

Calculate the variance:

\(\text{Variance} = \frac{\Sigma x^2}{14} - \left(\frac{\Sigma x}{14}\right)^2 = \frac{427485}{14} - \left(\frac{2443}{14}\right)^2\)

\(= 30534.64 - 30480.25 = 54.39\)

Calculate the standard deviation:

\(\text{Standard deviation} = \sqrt{54.39} = 9.19\)