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Let in a series of $2 n$ 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 ....... .
$425$
$650$
$250$
$925$
Solution
Let observations are denoted by $x _{i}$ for $1 \leq i< 2 n$
$\bar{x}=\frac{\sum x_{i}}{2 n}=\frac{(a+a+\ldots+a)-(a+a+\ldots+a)}{2 n}$
$\Rightarrow \overline{ x }=0$
and $\sigma_{ x }^{2}=\frac{\sum x _{i}^{2}}{2 n }-(\overline{ x })^{2}=\frac{ a ^{2}+ a ^{2}+\ldots+ a ^{2}}{2 n }-0= a ^{2}$
$\Rightarrow \sigma_{x}=a$
Now, adding a constant $b$ then $\overline{ y }=\overline{ x }+ b =5$
$\Rightarrow b =5$
and $\sigma_{y}=\sigma_{x}$ (No change in S.D.) $\Rightarrow a=20$
$\Rightarrow a^{2}+b^{2}=425$
Similar Questions
Let the mean of the data
| $X$ | $1$ | $3$ | $5$ | $7$ | $9$ |
| $(f)$ | $4$ | $24$ | $28$ | $\alpha$ | $8$ |
be $5.$ If $m$ and $\sigma^2$ are respectively the mean deviation about the mean and the variance of the data, then $\frac{3 \alpha}{m+\sigma^2}$ is equal to $……….$.
