We start with the equation \(\Sigma (x - 50) = 144\).

This can be expanded to \(\Sigma x - 50n = 144\).

We are also given \(\Sigma x = 944\).

Substitute \(\Sigma x = 944\) into the expanded equation:

\(944 - 50n = 144\).

Rearrange to solve for \(n\):

\(944 - 144 = 50n\)

\(800 = 50n\)

\(n = \frac{800}{50}\)

\(n = 16\)

(iii) To find \(\Sigma(x - 60)^2\), calculate each \((x - 60)^2\) for the given times:

\((-15)^2, (-13)^2, (-7)^2, (-4)^2, (-4)^2, 1^2, 4^2, 6^2, 9^2, 13^2, 23^2, 15^2, 18^2\).

Summing these gives:

\(225 + 169 + 49 + 16 + 16 + 1 + 16 + 36 + 81 + 169 + 529 + 225 + 324 = 1856\).

Thus, \(\Sigma(x - 60)^2 = 1856\).

(iv) The variance is calculated using the formula:

\(\text{Var} = \frac{\Sigma(x - 60)^2}{n} - \left(\frac{\Sigma(x - 60)}{n}\right)^2\).

Given \(\Sigma(x - 60) = 46\) and \(n = 13\), substitute the values:

\(\text{Var} = \frac{1856}{13} - \left(\frac{46}{13}\right)^2\).

Calculate each part:

\(\frac{1856}{13} = 142.7692 \ldots\)

\(\left(\frac{46}{13}\right)^2 = 12.5385 \ldots\)

Thus, \(\text{Var} = 142.7692 - 12.5385 = 130.2307 \ldots \approx 130\).

(i) Calculate \(\Sigma(t - 120)\):

\(\Sigma(t - 120) = (95 - 120) + (126 - 120) + (117 - 120) + (135 - 120) + (120 - 120) + (125 - 120) + (114 - 120) + (119 - 120) + (136 - 120)\)

\(= -25 + 6 - 3 + 15 + 0 + 5 - 6 - 1 + 16 = 7\)

Calculate \(\Sigma(t - 120)^2\):

\(\Sigma(t - 120)^2 = (-25)^2 + 6^2 + (-3)^2 + 15^2 + 0^2 + 5^2 + (-6)^2 + (-1)^2 + 16^2\)

\(= 625 + 36 + 9 + 225 + 0 + 25 + 36 + 1 + 256 = 1213\)

(ii) Calculate the variance of \(t\):

Variance \(= \frac{\Sigma(t - 120)^2}{9} - \left(\frac{\Sigma(t - 120)}{9}\right)^2\)

\(= \frac{1213}{9} - \left(\frac{7}{9}\right)^2\)

\(= 134.2\)

(i) The formula for the standard deviation \(\sigma\) is:

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

Given \(\sigma = 3.2\), \(n = 40\), and \(\Sigma (x-c)^2 = 3099.2\), we have:

\(3.2^2 = \frac{3099.2}{40} - \left(\frac{\Sigma (x-c)}{40}\right)^2\)

\(10.24 = 77.48 - \left(\frac{\Sigma (x-c)}{40}\right)^2\)

\(\left(\frac{\Sigma (x-c)}{40}\right)^2 = 67.24\)

\(\Sigma (x-c) = 40 \times \sqrt{67.24} = 328\)

(ii) Given \(c = 50\), the mean \(\bar{x}\) is:

\(\Sigma x - 40c = 328\)

\(\Sigma x = 328 + 40 \times 50 = 2328\)

\(\text{Mean} = \frac{2328}{40} = 58.2\)

First, calculate \(\Sigma(x - 10)\):

\(\Sigma(x - 10) = \Sigma x - 12 \times 10 = 186 - 120 = 66\)

Next, use the formula for variance to find \(\Sigma(x - 10)^2\):

Given the standard deviation \(\sigma = 4.5\), the variance is \(\sigma^2 = 4.5^2 = 20.25\).

The formula for variance is:

\(\frac{\Sigma(x - 10)^2}{12} - \left(\frac{\Sigma(x - 10)}{12}\right)^2 = 20.25\)

Substitute \(\Sigma(x - 10) = 66\):

\(\frac{\Sigma(x - 10)^2}{12} - \left(\frac{66}{12}\right)^2 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} - 5.5^2 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} - 30.25 = 20.25\)

\(\frac{\Sigma(x - 10)^2}{12} = 50.5\)

\(\Sigma(x - 10)^2 = 606\)