Solve this following


Let in a series of $2 n$ observations, half of them are equal to a and remaining half are equal to $-\mathrm{a}$. Also by adding a constant $\mathrm{b}$ in each of these observations, the mean and standard deviation of new set become 5 and 20 , respectively. Then the value of $a^{2}+b^{2}$ is equal to :


  1. 425

  2. 650

  3. 250

  4. 925

Correct Option: 1


Let observations are denoted by $\mathrm{x}_{i}$ for $1 \leq i<$

$2 \mathrm{n}$

$\bar{x}=\frac{\sum x_{i}}{2 n}=\frac{(a+a+\ldots+a)-(a+a+\ldots+a)}{2 n}$

$\Rightarrow \bar{x}=0$

and $\sigma_{\mathrm{x}}^{2}=\frac{\sum \mathrm{x}_{i}^{2}}{2 \mathrm{n}}-(\overline{\mathrm{x}})^{2}=\frac{\mathrm{a}^{2}+\mathrm{a}^{2}+\ldots+\mathrm{a}^{2}}{2 \mathrm{n}}-0=\mathrm{a}^{2}$

$\Rightarrow \sigma_{x}=\mathrm{a}$

Now, adding a constant b then $\bar{y}=\bar{x}+b=5$

$\Rightarrow \mathrm{b}=5$

and $\sigma_{\mathrm{y}}=\sigma_{\mathrm{x}}$ (No change in S.D.) $\Rightarrow \mathrm{a}=20$

$\Rightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=425$


Leave a comment