Originally Posted by
Prove It You are told $\displaystyle \displaystyle \begin{align*} \frac{t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49}}{50} = 38 \implies t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} = 1900 \end{align*}$. Two of these terms are 45 and 55, so removing them gives
$\displaystyle \displaystyle \begin{align*} t_k + t_{k + 1} + t_{k + 2} + \dots + t_{k + 49} - 45 - 55 &= 1900 - 45 - 55 \\ &= 1800 \end{align*}$
There are now 48 terms, so the mean of these 48 terms will be $\displaystyle \displaystyle \begin{align*} \frac{1800}{48} = 37.5 \end{align*}$.