JEE Math Practice Question

Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} $${x_i}$$ 0 1 2 3 4 5 $${f_i}$$ $$k + 2$$ $$2k$$ $${k^2} - 1$$ $${k^2} - 1$$ $${k^2} + 1$$ $$k - 3$$ where $$\sum f_{i}=62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\left[\mu^{2}+\sigma^{2}\right]$$ is equal to :

1. A.9
2. B.8
3. C.6
4. D.7