Answer: Option D. -> 1
:
D
We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of "$k^2$".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4.$\frac{1}{4}$ = 1

Answer: Option D. -> 14
:
D
$Mean=\frac{1\times 5+2\times 4+3\times 6+4\times f}{15+f}$
$i.e.,3=\frac{5+8+18+4f}{15+f}\Rightarrow 45+3f=31+4f$
$\Rightarrow 45-31=f\Rightarrow f=14$

Answer: Option A. -> √2
:
A
$Here\overline{x}=\frac{1+2+3+4+5}{5}=3$
$\begin{array}{ccc}x& x-\overline{x}& (x-{\overline{x}}^2)\\ 1& -2& 4\\ 2& -1& 1\\ 3& 0& 0\\ 4& 1& 1\\ 5& 2& 4\end{array}\therefore \sum (x-\overline{x}{)}^2=10$
$\therefore S.D=\sigma =\sqrt{\frac{(x-\overline{x})}{N}}=\sqrt{\frac{10}{5}}=\sqrt{2}$

Answer: Option D. -> 52.25
:
D
Total number of students $=40+35+45+42=162$
Total marks obtained
$=(40\times 50)+(35\times 60)+(45\times 55)+(42\times 45)$
$=8465$
Overall average of marks per students $=\frac{8465}{162}=52.25$

Answer: Option A. -> 36.6,22.6             
:
A
$\text{We know that}\text{C.V =}\frac{\sigma \times 100}{\overline{x}}or\overline{x}=\frac{\sigma }{c.v}\times 100\therefore \text{Mean of first series =}\frac{21.2\times 100}{58}=\text{36.6}\text{Mean of second series =}\frac{15.6\times 100}{69}=\text{22.6}$