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 #
