The mean and standard deviation of 15 observa | vedclass

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Question

We have

$	ext { Variance }=\frac{\sum\limits_{ r =1}^{15} x _{ r }^{2}}{15}-\left(\frac{\sum\limits_{ r =1}^{15} x _{ r }}{15}ight)^{2}$

Now,

Vedclass · Accepted Answer

$17$

Vedclass · Answer

We have

$	ext { Variance }=\frac{\sum\limits_{ r =1}^{15} x _{ r }^{2}}{15}-\left(\frac{\sum\limits_{ r =1}^{15} x _{ r }}{15}ight)^{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 $\sum x _{ r }=(15 	imes 8)-5+(20)=135$

Variance $=\frac{1470}{15}-\left(\frac{135}{15}ight)^{2}=98-81=17$

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......

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