Let m be the mean and be the standard deviation of the distribution where f_i=62.

Question

Let m be the mean and $\sigma$ be the standard deviation of the distribution where $\sum f_i=62$. If $[x]$ denotes the greatest integer $\le x$, then $[\mu^2+\sigma^2]$ is equal to

Accepted Answer

An intermediate step solves $3k^2+16k-12k-64=0 \Rightarrow k=4$ (rejecting the negative root).

$$\mu = \frac{\sum f_ix_i}{\sum f_i} = \frac{8+2(15)+3(15)+4(17)+5(\ldots)}{62} = \frac{156}{62}$$

$$\sigma^2+\mu^2 = \frac{\sum f_ix_i^2}{\sum f_i} = \frac{500}{62}$$

$$[\sigma^2+\mu^2] = \left[\frac{500}{62}\right] = [8.06] = 8$$

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