Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution

$$X_i$$	0	1	2	3	4	5
$$f_i$$	$$k+2$$	$$2k$$	$$k^2-1$$	$$k^2-1$$	$$k^2+1$$	$$k-3$$

where $$\Sigma f_i = 62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$[\mu^2 + \sigma^2]$$ is equal to

$$\sum f_i = (k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 3k^2 + 4k - 2 = 62$$

$$3k^2 + 4k - 64 = 0$$. Using quadratic formula: $$k = \frac{-4 + \sqrt{16+768}}{6} = \frac{-4+28}{6} = 4$$.

Frequencies: 6, 8, 15, 15, 17, 1. Total = 62 ✓

$$\mu = \frac{0(6)+1(8)+2(15)+3(15)+4(17)+5(1)}{62} = \frac{0+8+30+45+68+5}{62} = \frac{156}{62} \approx 2.516$$

$$\sum f_ix_i^2 = 0+8+60+135+272+25 = 500$$. $$\sigma^2 = 500/62 - (156/62)^2 = 8.065 - 6.33 = 1.735$$.

$$\mu^2 + \sigma^2 = 6.33 + 1.735 = 8.065$$. $$[\mu^2 + \sigma^2] = 8$$.

The correct answer is Option 2: 8.