Let in a series of $2 \mathrm{n}$ observations, half of them are equal to a and remaining half are equal to -a. Also by adding a constant 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:
Correct Option: 1
Let observations are denoted by $x i$ 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 $\sigma_{\mathrm{x}}^{2}=\frac{\sum \mathrm{x}_{i}^{2}}{2 \mathrm{n}}-(\overline{\mathrm{x}})^{2}=\frac{\mathrm{a}^{2}+\mathrm{a}^{2}+\ldots+\mathrm{a}^{2}}{2 \mathrm{n}}-0=\mathrm{a}^{2}$
$\Rightarrow \sigma_{\mathrm{x}}=\mathrm{a}$
Now, adding a constant $b$ then $\bar{y}=\bar{x}+b=5$
$\Rightarrow \mathrm{b}=5$
and $\sigma_{y}=\sigma_{x}$ (No change in S.D.) $\Rightarrow a=20 \Rightarrow a^{2}+b^{2}=425$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.