Intuitive answer:
Since all the marks are multiplied by 3 and added by 7, the mean should be 4*3 + 7 = 19
The standard deviation is a measure of average squared difference from the mean and doesn't change when you add the same amount to each mark, it only changes when multiply all the marks by 3
Thus,
\sigma = 2.3 * 3 = 6.9
Variance = \sigma^2 = 6.9^2 = 47.61
Let n be the number of numbers where {n|n\in \mathbb{Z_+} }
in this case n= 5
Let \mu be the mean \text{var} be the variance and, let sigma be the standard deviation
Proof of mean: \mu_0 =\frac{ \sum _i^n x_i}{n} = 4
\sum _i^n x_i = 4n
\mu =\frac{ \sum _i^n (3x_i+7)}{n}
Applying the commutative property:
=\frac{3\sum _i^n x_i + \sum _i^n7}{n} = \frac{3 \sum _i^n x_i + 7n}{n}
= 3 \frac{\sum _i^n x_i}{n} + 7 = 3*4 + 7 = 19
Proof for standard deviation:
\text{var}_0 = \sigma^2 = 2.3^2 = 5.29
\text{var}_0 = \frac{\sum _i^n(x_i -\mu_0)^2}{n} = \frac{\sum _i^n(x_i -4)^2}{n} = 5.29
\text{var} = \frac{\sum _i^n(3x_i + 7 -19)^2}{n} = \frac{\sum _i^n(3x_i -12)^2}{n}
= \frac{\sum _i^n(3(x_i -4))^2}{n} = \frac{\sum _i^n9(x_i -4)^2}{n} = 9\frac{\sum _i^n(x_i -4)^2}{n}
\text{var} = 9*5.29 = 47.61
