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The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
$7$
$20$
$19$
$17$
Solution
We have
$\text { Variance }=\frac{\sum\limits_{ r =1}^{15} x _{ r }^{2}}{15}-\left(\frac{\sum\limits_{ r =1}^{15} x _{ r }}{15}\right)^{2}$
Now, as per information given in equation
$\frac{\sum x _{ r }^{2}}{15}-8^{2}=3^{2} \Rightarrow \sum x _{ T }^{2}=\log 5$
Now, the new $\sum x _{ r }^{2}=\log 5-5^{2}+20^{2}=1470$
And, new $\sum x _{ r }=(15 \times 8)-5+(20)=135$
Variance $=\frac{1470}{15}-\left(\frac{135}{15}\right)^{2}=98-81=17$
Similar Questions
What is the standard deviation of the following series
| class | $0-10$ | $10-20$ | $20-30$ | $30-40$ |
| Freq | $1$ | $3$ | $4$ | $2$ |
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 $…………$.
