Mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking i

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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......

$7$

$20$

$19$

$17$

(JEE MAIN-2022)

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

medium

hard

(AIEEE-2012)

hard

hard

hard

(JEE MAIN-2018)

Classes	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequencies	$5$	$8$	$15$	$16$	$6$