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The mean and standard deviation of $15$ observations were found to be $12$ and $3$ respectively. On rechecking it was found that an observation was read as $10$ in place of $12$ . If $\mu$ and $\sigma^2$ denote the mean and variance of the correct observations respectively, then $15\left(\mu+\mu^2+\sigma^2\right)$ is equal to$...................$
$2521$
$3562$
$1245$
$2356$
Solution
Let the incorrect mean be $\mu^{\prime}$ and standard deviation be $\sigma^{\prime}$
We have
$\mu^{\prime}=\frac{\Sigma x_i}{15}=12 \Rightarrow \Sigma x_i=180$
As per given information correct $\Sigma x_i=180-10+12$
$\Rightarrow \mu(\text { correct mean })=\frac{182}{15}$
Also
$ \sigma^{\prime}=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295 $
$\text { Correct } \Sigma \mathrm{x}_{\mathrm{i}}^2=2295-100+144=2339 $
$ \sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}$
Required value
$ =15\left(\mu+\mu^2+\sigma^2\right) $
$ =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) $
$ =15\left(\frac{182}{15}+\frac{2339}{15}\right) $
$ =2521$
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 $……….$.