Statistics — MATHS STD 11 Science — Question

1. s

Solution:

The given observations are a, b, c, d, e.

$\text{Mean}=\text{m}=\frac{\text{a+b+c+d+e}}{5}$

$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$

Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$

Now, consider the observations a + k, b + k, c + k, d + k, e + k.

New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$

$=\frac{\text{a+b+c+d+e+5k}}{5}$

$=\frac{5\text{m}+5\text{k}}{5}$

$=\text{m}+\text{k}$

$\therefore$ New standard deviation

$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$

$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}}$ [Using (1)]

$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$

$=\text{s}$

Hence, the correct answer is option (a).