(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.