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13.Statistics
medium
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......
A
$7$
B
$20$
C
$19$
D
$17$
(JEE MAIN-2022)
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 $=\frac{1470}{15}-\left(\frac{135}{15}\right)^{2}=98-81=17$
Standard 11
Mathematics
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