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Let the mean and variance of $12$ observations be $\frac{9}{2}$ and $4$ respectively. Later on, it was observed that two observations were considered as $9$ and $10$ instead of $7$ and $14$ respectively. If the correct variance is $\frac{m}{n}$, where $m$ and $n$ are co-prime, then $m + n$ is equal to
$316$
$314$
$317$
$315$
Solution
$\frac{\sum x }{12}=\frac{9}{2}$
$\sum x =54$
$\frac{\Sigma x ^2}{12}-\left(\frac{9}{2}\right)^2=4$
$\sum x ^2=291$
$\sum x _{\text {new }}=54-(9+10)+7+14=56$
$\sum x _{\text {new }}^2=291-(81+100)+49+196=355$
$\sigma_{\text {new }}^2=\frac{355}{12}-\left(\frac{56}{12}\right)^2$
$\sigma_{\text {new }}^2=\frac{281}{36}=\frac{ m }{ n }$
$m + n =317$
Similar Questions
If the mean and variance of the frequency distribution
| $x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
| $f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $…………$.
