Question

The mean and standard deviation of $50$ observations are $50$ and $10.$ Later on, it was decided to omit an observation that was incorrectly recorded as $99.$ If the variance of the remaining $49$ observations is $\frac{\lambda }{49},$ then the value of $\lambda$ is equal to

Answer 1

Answer: 2550

$\bar{x}=\frac{\sum x_1}{50}=50$
$\Rightarrow \sum x_i=2500$
Correct $\sum x_i=2500-99=2401$
Correct $\bar{x}=\frac{2401}{49}=49$
${\sigma }^2=\frac{\sum x_i^2}{n}-{\bar{x}}^2$
$\Rightarrow 100=\frac{\sum x_i^2}{50}-2500$
$\Rightarrow \sum x_i^2=130000$
Correct $\sum x_i^2=130000-99^2=120199$
Correct ${\sigma }^2=\frac{\sum x_i^2}{n}-{\bar{x}}^2$
$=\frac{120199}{49}-49^2$
$=\frac{120199-117649}{49}=\frac{2550}{49}=\frac{\lambda }{49}$
Hence, $\lambda =2550$