(i) To find the mean height \(\bar{x}\), use the formula:

\(\bar{x} = 130 - \frac{\Sigma(x - 130)}{82}\)

Substitute the given values:

\(\bar{x} = 130 - \frac{-287}{82} = 130 + \frac{287}{82}\)

\(\bar{x} = 130 + 3.5 = 126.5 \text{ cm}\)

(ii) To find \(\Sigma(x - 130)^2\), use the formula for variance:

\(\frac{\Sigma(x - 130)^2}{82} - (\bar{x} - 130)^2 = 6.9^2\)

Substitute \(\bar{x} - 130 = -3.5\):

\(\frac{\Sigma(x - 130)^2}{82} - (-3.5)^2 = 6.9^2\)

\(\frac{\Sigma(x - 130)^2}{82} = 6.9^2 + 3.5^2\)

\(\frac{\Sigma(x - 130)^2}{82} = 47.61 + 12.25 = 59.86\)

\(\Sigma(x - 130)^2 = 82 \times 59.86 = 4908.52\)

Thus, \(\Sigma(x - 130)^2 \approx 4908.5 \text{ cm}\).

(i) We know that \(\Sigma(x-a) = -73.2\). Since there are 24 observations, the mean of \(x-a\) is \(\frac{-73.2}{24} = -3.05\). Given that the mean of \(x\) is 8.95, we have:

\(a = 8.95 + 3.05 = 12\)

(ii) The standard deviation is calculated using:

\(\text{Standard deviation} = \sqrt{\frac{\Sigma(x-a)^2}{24} - (\text{mean of } x-a)^2}\)

\(= \sqrt{\frac{2115}{24} - (-3.05)^2}\)

\(= \sqrt{88.125 - 9.3025}\)

\(= \sqrt{78.8225}\)

\(= 8.88\)

To find the mean, we use the formula:

\(\text{Mean} = 35 - \frac{\Sigma(t - 35)}{12}\)

Substitute the given value:

\(\text{Mean} = 35 - \frac{-15}{12} = 35 + \frac{15}{12} = 33.75\)

Rounding to one decimal place, the mean is 33.8 minutes.

To find the standard deviation, we use the formula:

\(sd = \sqrt{\frac{\Sigma(t - 35)^2}{12} - \left(\frac{\Sigma(t - 35)}{12}\right)^2}\)

Substitute the given values:

\(sd = \sqrt{\frac{82.23}{12} - \left(\frac{-15}{12}\right)^2}\)

Calculate each part:

\(\frac{82.23}{12} = 6.8525\)

\(\left(\frac{-15}{12}\right)^2 = \left(-1.25\right)^2 = 1.5625\)

\(sd = \sqrt{6.8525 - 1.5625} = \sqrt{5.29}\)

\(sd \approx 2.3 \text{ minutes}\)

First, calculate the coded mean:

\(\frac{-47.2}{30} = -1.573\)

Then, calculate the actual mean:

\(\bar{x} = 110 - 1.573 = 108.4\)

Rounding gives the mean as 108.

Next, calculate the standard deviation:

\(\text{Standard deviation} = \sqrt{\frac{5460}{30} - (-1.573)^2}\)

\(= \sqrt{182 - 2.475}\)

\(= \sqrt{179.525}\)

\(\approx 13.4\)

(a) The mean of the 40 values of \(x\) is given by:

\(\frac{\Sigma x}{40} = 34\)

We know \(\Sigma(x-k) = 520\), so:

\(\frac{\Sigma x - 40k}{40} = \frac{520}{40}\)

\(34 - k = 13\)

\(k = 34 - 13 = 21\)

(b) The variance is calculated using:

\(\text{Var} = \frac{\Sigma(x-k)^2}{40} - \left(\frac{\Sigma(x-k)}{40}\right)^2\)

Substitute the given values:

\(\text{Var} = \frac{9640}{40} - \left(\frac{520}{40}\right)^2\)

\(\text{Var} = 241 - 13^2\)

\(\text{Var} = 241 - 169 = 72\)