We are given that a sample of $N=7$ has mean $\bar{X} = 5$, and a sample variance of $s^2 = 9$. One person, with a score of $11$ is removed from the data. What is the mean and variance for the new set of scores?
We can quite easily compute $\mu_{new} = 4$, but $s_{new}^2$ is a slightly more tedious to calculate.
Here's my approach, where I make the assumption that the $7$th score is $X_7 = 11$. $$ 63 = \sum_{i=1}^7 (X-5)^2 = \sum_{i=1}^6(X-5)^2 + 36 $$ So that tells us that $\sum_{i=1}^6(X-5)^2 = 27$.
$$ 27 = \sum_{i=1}^6(X^2 - 10x + 25) = \sum_{i=1}^6 X^2 -10 \sum_{i=1}^6 X + 150 \\ -123 + 2\sum_{i=1}^6 X + 96 = \sum_{i=1}^6 X^2 -8 \sum_{i=1}^6 X + 96 \\ -27 + 2\sum_{i=1}^6 X = 21 = \sum_{i=1}^6(X-4)^2 $$
Now, we divide each side by 5 and the RHS becomes $s_{new}^2$, and the left becomes $4.2$.
That is apparently not the correct answer. The answer should be $2.4$.
Hint: Sample Variance is bias corrected.
$$\begin{align}s^2 =~& \frac 1{\color{purple}{n-1}}\sum_{k=1}^n \left(X_k-\frac 1 n\sum_{k=1}^n X_k\right)^2 \\ =~& \frac 1 {n-1}\sum_{k=1}^n X_k^2-\frac n {(n-1)}\bar X^2 \end{align}$$
So substituting then rearranging: $$\begin{align}9 =~& \tfrac 1 6(11^2+\sum_{k=1}^6X_k^2)-\tfrac 76 (5^2) \\ 6(9) =~& 11^2+\sum_{k=1}^6 X_k^2-7(5^2) \\ \tfrac 1 5\sum_{k=1}^6 X_k^2 -\tfrac 6 5 (4^2) =~& \tfrac 65(9)-\tfrac 1 511^2+\tfrac 75(5^2)-\tfrac 6 5 (4^2) \\ s_{\rm new}^2 =~& \tfrac {12}5 \\ =~& 2.4 \end{align}$$