(a)The scores of 10 students in a test, in which the maximum marks were 50 as follows :
28,36,34,28,48,22,35,27,19,41
Find the variance .
(b)Latter on the maximum marks were increased to 100 , and accordingly each students score was doubled . Find the variance of the new scores.

Answer

To solve the problem, we will calculate the variance of the scores of the 10 students in two parts: (a) for the original scores and (b) for the new scores after doubling.

### Part (a): Finding the Variance of the Original Scores

1. **List the Scores**:
The scores of the 10 students are:
\[ 28, 36, 34, 28, 48, 22, 35, 27, 19, 41 \]

2. **Calculate the Mean (\( \bar{x} \))**:
\[ \bar{x} = \frac{\sum x_i}{n} \]
where \( n = 10 \) (the number of students).
\[ \sum x_i = 28 + 36 + 34 + 28 + 48 + 22 + 35 + 27 + 19 + 41 = 318 \]
\[ \bar{x} = \frac{318}{10} = 31.8 \]

3. **Calculate \( x_i - \bar{x} \)**:
\[ \begin{align*} 28 - 31.8 & = -3.8 \\ 36 - 31.8 & = 4.2 \\ 34 - 31.8 & = 2.2 \\ 28 - 31.8 & = -3.8 \\ 48 - 31.8 & = 16.2 \\ 22 - 31.8 & = -9.8 \\ 35 - 31.8 & = 3.2 \\ 27 - 31.8 & = -4.8 \\ 19 - 31.8 & = -12.8 \\ 41 - 31.8 & = 9.2 \\ \end{align*} \]

4. **Calculate \( (x_i - \bar{x})^2 \)**:
\[ \begin{align*} (-3.8)^2 & = 14.44 \\ (4.2)^2 & = 17.64 \\ (2.2)^2 & = 4.84 \\ (-3.8)^2 & = 14.44 \\ (16.2)^2 & = 262.44 \\ (-9.8)^2 & = 96.04 \\ (3.2)^2 & = 10.24 \\ (-4.8)^2 & = 23.04 \\ (-12.8)^2 & = 163.84 \\ (9.2)^2 & = 84.64 \\ \end{align*} \]

5. **Sum of Squares**:
\[ \sum (x_i - \bar{x})^2 = 14.44 + 17.64 + 4.84 + 14.44 + 262.44 + 96.04 + 10.24 + 23.04 + 163.84 + 84.64 = 691.6 \]

6. **Calculate Variance (\( \sigma^2 \))**:
\[ \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{691.6}{10} = 69.16 \]

### Part (b): Finding the Variance of the New Scores

1. **New Scores**: Each student's score is doubled.
\[ \text{New Scores} = 2 \times \{28, 36, 34, 28, 48, 22, 35, 27, 19, 41\} = \{56, 72, 68, 56, 96, 44, 70, 54, 38, 82\} \]

2. **New Mean**:
The new mean will be:
\[ \bar{x}' = 2 \times \bar{x} = 2 \times 31.8 = 63.6 \]

3. **Calculate New Variance**:
When all scores are doubled, the variance is multiplied by \( 4 \) (since variance is affected by the square of the scaling factor):
\[ \sigma'^2 = 4 \times \sigma^2 = 4 \times 69.16 = 276.64 \]

### Final Answers:
- (a) Variance of the original scores: **69.16**
- (b) Variance of the new scores: **276.64**