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

$\bar{x}=\frac{\sum x_{1}}{50}=50$
$\Rightarrow \sum x_{i}=2500$
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, $\lambda=2550$