(a) To find the standard deviation, we first calculate the variance using the formula:

\(\text{Var} = \frac{\Sigma (x - q)^2}{50} - \left(\frac{\Sigma (x - q)}{50}\right)^2\)

Substitute the given values:

\(\text{Var} = \frac{14235}{50} - \left(\frac{700}{50}\right)^2\)

\(\text{Var} = 284.7 - 196 = 88.7\)

The standard deviation is the square root of the variance:

\(\text{sd} = \sqrt{88.7} \approx 9.42\)

(b) Given \(\Sigma x = 2865\), we use the equation:

\(\Sigma x - 50q = 700\)

Substitute \(\Sigma x = 2865\):

\(2865 - 50q = 700\)

Solve for \(q\):

\(50q = 2865 - 700\)

\(50q = 2165\)

\(q = \frac{2165}{50} = 43.3\)

(i) Given the mean of the n values of x is 24.25, we have:

\(\frac{136}{n} + 20 = 24.25\)

Solving for \(n\):

\(24.25n - 20n = 136\)

\(4.25n = 136\)

\(n = \frac{136}{4.25} = 32\)

(ii) Using the coded information for variance:

\(\text{Variance} = \frac{2888}{32} - \left(\frac{136}{32}\right)^2\)

\(= 72.1875 \approx 72.19\)

Using the uncoded information for variance:

\(\text{Variance} = \frac{\Sigma x^2}{32} - 24.25^2\)

Equating the two expressions for variance:

\(\frac{\Sigma x^2}{32} - 24.25^2 = 72.1875\)

\(\frac{\Sigma x^2}{32} = 72.1875 + 24.25^2\)

\(\Sigma x^2 = 32 \times (72.1875 + 24.25^2)\)

\(\Sigma x^2 = 21128\)

First, calculate \(\Sigma(x - 45)\):

\(\Sigma(x - 45) = \Sigma x - 20 \times 45 = 1218 - 20 \times 45 = 318\)

Next, use the variance formula to find \(\Sigma(x - 45)^2\):

The variance formula is \(\frac{\Sigma(x - 45)^2}{20} - \left(\frac{\Sigma(x - 45)}{20}\right)^2 = 4.2^2\)

Substitute \(\Sigma(x - 45) = 318\):

\(\frac{\Sigma(x - 45)^2}{20} - \left(\frac{318}{20}\right)^2 = 4.2^2\)

Calculate \(4.2^2 = 17.64\) and \(\left(\frac{318}{20}\right)^2 = 252.81\):

\(\frac{\Sigma(x - 45)^2}{20} = 17.64 + 252.81 = 270.45\)

\(\Sigma(x - 45)^2 = 270.45 \times 20 = 5409\)

We know that \(\Sigma(x - 10) = 27\) and the mean number of emails is 11.5. The mean is given by \(\frac{\Sigma x}{n} = 11.5\).

First, express \(\Sigma x\) in terms of \(n\):

\(\Sigma x = 11.5n\).

We also have:

\(\Sigma(x - 10) = \Sigma x - 10n = 27\).

Substitute \(\Sigma x = 11.5n\) into the equation:

\(11.5n - 10n = 27\).

Simplify the equation:

\(1.5n = 27\).

Solving for \(n\):

\(n = \frac{27}{1.5} = 18\).

(i) We know \(\Sigma (x - k) = 315\) and there are 30 buns, so the mean of \(x - k\) is \(\frac{315}{30} = 10.5\). Since the mean weight of the buns is 50.5 grams, we have:

\(50.5 - k = 10.5\)

Solving for \(k\), we get:

\(k = 50.5 - 10.5 = 40\)

(ii) The variance is given by:

\(\text{var} = \frac{\Sigma (x - k)^2}{30} = \frac{4022}{30} = 134.067\)

We also know the variance formula:

\(\text{var} = \frac{\Sigma x^2}{30} - (\text{mean})^2\)

Substituting the mean \(50.5\), we have:

\(\text{var} = 134.067 - 50.5^2 = 23.817\)

The standard deviation is the square root of the variance:

\(\text{sd} = \sqrt{23.817} = 4.88\)

First, calculate \((x - 1760)\) for each value of \(x\):

1.6, -1.5, 2.3, 1.4, -0.6, -0.9, 2.5, 1.9, 2.4, 1.9, 2.8, 1.0

Calculate the mean of \((x - 1760)\):

\(\text{Mean} = \frac{1.6 + (-1.5) + 2.3 + 1.4 + (-0.6) + (-0.9) + 2.5 + 1.9 + 2.4 + 1.9 + 2.8 + 1.0}{12} = 1.23\)

Calculate the standard deviation of \((x - 1760)\):

\(\text{sd} = 1.39\)

Since \(x = (x - 1760) + 1760\), the mean of \(x\) is:

\(\text{Mean of } x = 1.23 + 1760 = 1761.23\)

The standard deviation remains the same, so:

\(\text{sd of } x = 1.39\)

(i) To find \(c\), use the equation for the mean: \(\frac{\Sigma x}{9} = 2205\). Therefore, \(\Sigma x = 2205 \times 9 = 19845\). Given \(\Sigma(x-c) = 1845\), we have \(\Sigma x - 9c = 1845\). Substituting \(\Sigma x = 19845\), we get:

\(19845 - 9c = 1845\)

\(9c = 19845 - 1845 = 18000\)

\(c = \frac{18000}{9} = 2000\)

(ii) The variance is given by \(\frac{\Sigma(x-c)^2}{9} - (\text{mean of } x)^2\). Substituting the given values:

Variance = \(\frac{477450}{9} - 2205^2\)

Variance = \(53050 - 2205^2\)

Variance = \(11025\)

(iii) With the new apartment, the mean becomes \(2120.50\) for 10 apartments. Therefore, the new total is:

\(2120.50 \times 10 = 21205\)

The rental price of the additional apartment is:

\(21205 - 19845 = 1360\)

Given that the mean of 10 values is 86.2, we have:

\(\frac{\Sigma x}{10} = 86.2\)

Therefore, \(\Sigma x = 86.2 \times 10 = 862\).

For part (ii), we know \(\Sigma(x-a) = 362\).

Expanding this, we have:

\(\Sigma x - 10a = 362\)

Substituting \(\Sigma x = 862\) from part (i):

\(862 - 10a = 362\)

Solving for \(a\):

\(10a = 862 - 362\)

\(10a = 500\)

\(a = 50\)

To find the standard deviation, we use the formula for the standard deviation of a sample:

\(\text{sd} = \sqrt{\frac{\Sigma(t - \bar{t})^2}{n}}\)

Given \(\Sigma(t - 2.5)^2 = 96.1\) and \(n = 250\), we first find the variance:

\(\text{variance} = \frac{96.1}{250} - (0.3)^2\)

where \(0.3\) is the coded mean difference \(2.8 - 2.5\).

\(\text{variance} = \frac{96.1}{250} - 0.09\)

\(\text{variance} = 0.3844 - 0.09 = 0.2944\)

Now, calculate the standard deviation:

\(\text{sd} = \sqrt{0.2944} = 0.543\)

We start with the equation \(\Sigma (x - 100) = 216\).

This can be expanded to \(\Sigma x - 100n = 216\).

We are given \(\Sigma x = 2416\), so substitute this into the equation:

\(2416 - 100n = 216\).

Rearrange to solve for \(n\):

\(2416 - 216 = 100n\)

\(2200 = 100n\)

\(n = \frac{2200}{100} = 22\).

To find the mean, we use the formula:

\(\bar{x} = 80 - \frac{\Sigma (x - 80)}{30}\)

\(\bar{x} = 80 - \frac{-147}{30} = 80 + 4.9 = 75.1\)

To find the standard deviation, we use the formula:

\(\text{sd} = \sqrt{\frac{\Sigma (x - 80)^2}{30} - \left(\frac{\Sigma (x - 80)}{30}\right)^2}\)

\(\text{sd} = \sqrt{\frac{952}{30} - \left(\frac{-147}{30}\right)^2}\)

\(\text{sd} = \sqrt{31.7333 - 4.9^2} = \sqrt{7.72} \approx 2.78\)

First, calculate \(\Sigma x\) using the given \(\Sigma(x - 20) = 35\):

\(\Sigma x - 50 \times 20 = 35\)

\(\Sigma x = 35 + 1000 = 1035\)

Now, find the mean \(\bar{x}\):

\(\bar{x} = \frac{\Sigma x}{50} = \frac{1035}{50} = 20.7\)

Use the formula for variance:

\(\text{Variance} = \frac{\Sigma x^2}{50} - \left(\frac{\Sigma x}{50}\right)^2\)

Substitute the values:

\(\text{Variance} = \frac{25036}{50} - \left(\frac{1035}{50}\right)^2\)

\(\text{Variance} = 500.72 - (20.7)^2\)

\(\text{Variance} = 500.72 - 428.49\)

\(\text{Variance} = 72.23\)

(i) Calculate \(\Sigma(x - 62)\):

\((62.7 - 62) + (59.6 - 62) + (64.2 - 62) + (61.5 - 62) + (68.3 - 62) + (66.9 - 62) + (62.0 - 62) + (62.3 - 62) = 0.7 - 2.4 + 2.2 - 0.5 + 6.3 + 4.9 + 0 + 0.3 = 11.5\)

(ii) Calculate \(\Sigma(x - 62)^2\):

\((0.7)^2 + (-2.4)^2 + (2.2)^2 + (-0.5)^2 + (6.3)^2 + (4.9)^2 + (0)^2 + (0.3)^2 = 0.49 + 5.76 + 4.84 + 0.25 + 39.69 + 24.01 + 0 + 0.09 = 75.13\)

(iii) Calculate the mean and variance:

Mean \(= \frac{62.7 + 59.6 + 64.2 + 61.5 + 68.3 + 66.9 + 62.0 + 62.3}{8} = \frac{507.5}{8} = 63.4375\)

Variance \(= \frac{75.13}{8} - \left(\frac{11.5}{8}\right)^2 = 9.39125 - 2.06125 = 7.32\)

First, calculate \(\Sigma x\) using the mean:

\(\text{Mean} = \frac{\Sigma x}{18} = \frac{58}{9}\)

\(\Sigma x = 18 \times \frac{58}{9} = 116\)

Now, find \(\Sigma (x - 5)\):

\(\Sigma (x - 5) = \Sigma x - 18 \times 5 = 116 - 90 = 26\)

Next, calculate \(\Sigma (x - 5)^2\) using the formula:

\(\Sigma (x - 5)^2 = \Sigma x^2 - 2 \times 5 \times \Sigma x + 18 \times 5^2\)

\(\Sigma (x - 5)^2 = 967 - 2 \times 5 \times 116 + 18 \times 25\)

\(\Sigma (x - 5)^2 = 967 - 1160 + 450 = 257\)

First, find the mean of the speeds:

\(\bar{x} = 50 + \frac{81.4}{22} = 53.7\)

Next, calculate the variance using the formula:

\(\text{Var} = \frac{\Sigma(x - 50)^2}{22} - (\bar{x} - 50)^2\)

\(\text{Var} = \frac{671}{22} - 3.7^2 = 16.81\)

To find \(\Sigma x^2\), use the expanded equation:

\(\Sigma x^2 = \Sigma(x - 50)^2 + 2 \times 50 \times \Sigma(x - 50) + 22 \times 50^2\)

\(\Sigma x^2 = 671 + 2 \times 50 \times 81.4 + 22 \times 2500\)

\(\Sigma x^2 = 671 + 8140 + 55000 = 63811\)

(i) To find the standard deviation, use the formula:

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

Substitute the given values:

\(\text{sd}^2 = \frac{1957.5}{30} - \left(\frac{234}{30}\right)^2\)

\(\text{sd}^2 = 65.25 - 6.084\)

\(\text{sd}^2 = 59.166\)

\(\text{sd} = \sqrt{59.166} \approx 2.1\)

(ii) Given the mean is 86, use the formula for the mean:

\(\bar{x} = \frac{\Sigma x}{n} = 86\)

\(86 = \frac{234}{30} + c\)

\(86 = 7.8 + c\)

\(c = 86 - 7.8\)

\(c = 78.2\)

We start with the equation \(\Sigma(x - 36) = -60\). Expanding this gives:

\(\Sigma x - 24 \times 36 = -60\)

\(\Sigma x = 24 \times 36 - 60 = 804\)

Next, we use \(\Sigma(x - 36)^2 = 227.76\). Expanding gives:

\(\Sigma x^2 - 2 \times 36 \times \Sigma x + 24 \times 36^2 = 227.76\)

Substitute \(\Sigma x = 804\):

\(\Sigma x^2 - 2 \times 36 \times 804 + 24 \times 1296 = 227.76\)

\(\Sigma x^2 - 57888 + 31104 = 227.76\)

\(\Sigma x^2 = 227.76 + 57888 - 31104 = 27011.76\)

(i) Let \(n\) be the number of children. The mean height is given by:

\(\frac{\Sigma x}{n} = 104.8\)

We know \(\Sigma(x - 100) = 72\), so:

\(\Sigma x - 100n = 72\)

Substitute \(\Sigma x = 104.8n\) into the equation:

\(104.8n - 100n = 72\)

\(4.8n = 72\)

\(n = \frac{72}{4.8} = 15\)

(ii) We need to find \(\Sigma(x - 104.8)^2\). We have:

\(\Sigma(x - 100)^2 = 499.2\)

\(\Sigma(x - 104.8)^2 = \Sigma(x - 100 + 100 - 104.8)^2\)

\(= \Sigma((x - 100) - 4.8)^2\)

Using the expansion \((a - b)^2 = a^2 - 2ab + b^2\), we have:

\(\Sigma(x - 104.8)^2 = \Sigma(x - 100)^2 - 2 \times 4.8 \times \Sigma(x - 100) + 15 \times 4.8^2\)

\(= 499.2 - 2 \times 4.8 \times 72 + 15 \times 4.8^2\)

\(= 499.2 - 691.2 + 345.6\)

\(= 153.6\)

First, calculate the mean \(\bar{x}\) using the formula:

\(\bar{x} = \frac{\Sigma x}{n} = \frac{645}{150} = 4.3\)

Next, use the formula for variance:

\(\text{Variance} = \frac{\Sigma x^2}{n} - \bar{x}^2 = \frac{8287.5}{150} - 4.3^2\)

Calculate \(\frac{8287.5}{150} = 55.25\) and \(4.3^2 = 18.49\).

Thus, the variance is:

\(55.25 - 18.49 = 36.76\)

The standard deviation is:

\(\sqrt{36.76} = 6.063\)

Finally, calculate \(\Sigma (x - \bar{x})^2\) using:

\(\Sigma (x - \bar{x})^2 = n \times \text{Variance} = 150 \times 36.76 = 5514\)

Rounding to the nearest whole number gives 5510.

(i) To find the standard deviation, first determine the number of data points \(n\) using the equation for the mean:

\(\frac{133}{n} + 25 = 28.325\)

Solving for \(n\):

\(\frac{133}{n} = 3.325\)

\(n = 40\)

Next, use the formula for variance with coded data:

\(\frac{3762}{40} - 3.325^2 = 82.99\)

The standard deviation is:

\(\sqrt{82.99} = 9.11\)

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

\(82.99 = \frac{\Sigma x^2}{40} - 28.325^2\)

\(\Sigma x^2 = (82.99 + 28.325^2) \times 40\)

\(\Sigma x^2 = 35412\)

Alternatively, using the expanded form:

\(\Sigma(x - 25)^2 = \Sigma x^2 - 50\Sigma x + 40 \times 25^2\)

\(\Sigma x^2 = 3762 + 50 \times 1133 + 25000\)

\(\Sigma x^2 = 35412\)

(i) To find the mean, use the formula:

\(\text{Mean} = 45 + \frac{-148}{36} = 40.9\)

To find the variance, use:

\(\text{Var} = \frac{3089}{36} - \left(\frac{-148}{36}\right)^2 = 68.9\)

The standard deviation is:

\(\text{sd} = \sqrt{68.9} = 8.30\)

(ii) With the new data value of 29 added:

New \(\Sigma(x - 45) = -148 + (29 - 45) = -164\)

New \(\Sigma(x - 45)^2 = 3089 + (29 - 45)^2 = 3345\)

The new standard deviation is:

\(\text{New sd} = \sqrt{\frac{3345}{37} - \left(\frac{-164}{37}\right)^2} = 8.41\)

(i) To find the mean speed, use the formula for the mean: \(\bar{x} = 60 + \frac{\Sigma(x - 60)}{70}\). Given \(\Sigma(x - 60) = 245\), we have:

\(\bar{x} = 60 + \frac{245}{70} = 60 + 3.5 = 63.5\)

(ii) To find \(\Sigma(x - 50)\), use the relationship:

\(\Sigma(x - 50) = \Sigma x - 70 \times 50\)

We know \(\Sigma(x - 60) = 245\), so:

\(\Sigma x = 245 + 70 \times 60\)

\(\Sigma x = 245 + 4200 = 4445\)

Then:

\(\Sigma(x - 50) = 4445 - 3500 = 945\)

(iii) To calculate \(\Sigma(x - 50)^2\), use the variance formula:

\(\text{Variance} = \frac{\Sigma(x - 50)^2}{70} - \left(\frac{\Sigma(x - 50)}{70}\right)^2\)

Given the standard deviation is 10.6, the variance is \(10.6^2 = 112.36\).

Let \(\frac{\Sigma(x - 50)}{70} = 13.5\) (from part (ii)), then:

\(112.36 = \frac{\Sigma(x - 50)^2}{70} - 13.5^2\)

\(112.36 = \frac{\Sigma(x - 50)^2}{70} - 182.25\)

\(\frac{\Sigma(x - 50)^2}{70} = 112.36 + 182.25 = 294.61\)

\(\Sigma(x - 50)^2 = 294.61 \times 70 = 20623\)

We start with the given equations:

\(\Sigma(x - 200) = 446\)

\(\Sigma x = 6846\)

We can express \(\Sigma(x - 200)\) as:

\(\Sigma x - \Sigma 200 = 446\)

Since \(\Sigma 200 = 200n\), we substitute:

\(\Sigma x - 200n = 446\)

Substitute \(\Sigma x = 6846\) into the equation:

\(6846 - 200n = 446\)

Solve for \(n\):

\(6846 - 446 = 200n\)

\(6400 = 200n\)

\(n = \frac{6400}{200}\)

\(n = 32\)

(i) To find \(\Sigma(x - 50)\), use the formula:

\(\Sigma(x - 50) = \Sigma x - 16 \times 50 = 824 - 800 = 24\)

To find \(\Sigma(x - 50)^2\), use the variance formula:

\(\frac{\Sigma(x - 50)^2}{16} - \left(\frac{\Sigma(x - 50)}{16}\right)^2 = 6.5^2\)

\(\frac{\Sigma(x - 50)^2}{16} - \left(\frac{24}{16}\right)^2 = 42.25\)

\(\Sigma(x - 50)^2 = 712\)

(ii) With the new mark, the total sum becomes \(824 + 72 = 896\) and the number of students is 17.

New mean:

\(\text{New mean} = \frac{896}{17} = 52.7\)

New variance:

\(\text{New variance} = \frac{712 + 22^2}{17} - \left(\frac{24 + (72 - 50)}{17}\right)^2\)

\(= \frac{712 + 484}{17} - \left(\frac{46}{17}\right)^2\)

\(= \frac{1196}{17} - \left(2.7059\right)^2\)

\(= 70.3529 - 7.3248 = 63.0281\)

New standard deviation:

\(\text{New standard deviation} = \sqrt{63.0281} \approx 7.94\)

To find the mean \(\bar{t}\), use the formula:

\(\bar{t} = \frac{\Sigma t}{n}\)

where \(n = 50\) and \(\Sigma t = 910\).

\(\bar{t} = \frac{910}{50} = 18.2\)

To find the standard deviation, use the formula:

\(\text{standard deviation} = \sqrt{\frac{\Sigma (t - \bar{t})^2}{n}}\)

where \(\Sigma (t - \bar{t})^2 = 876\).

\(\text{standard deviation} = \sqrt{\frac{876}{50}} = 4.19\)

(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:

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

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

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

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

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

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

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

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

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