OK, problem has sample n = 37, $\displaystyle \sum x = 277.875$, $\displaystyle \sum x^2 = 2093.109375$. Estimate with 99% confidence the mean.

I get $\displaystyle \mu = \frac{\sum x}{n} \Rightarrow \frac{277.875}{37} = 7.510$

and $\displaystyle \sigma = \sqrt{\sum x^2P(x) - \mu^2} = \sqrt {\frac{2093.109375}{37} - 7.51^2} \approx .413$

For a 99% confidence rating, $\displaystyle z_{.005} = 2.576$. Therefore, the confidence interval should be

$\displaystyle \mu \pm 2.576\frac {\sigma}{\sqrt n} \Rightarrow 7.51 \pm 2.576 \frac {.413}{\sqrt 37} \Rightarrow 7.51 \pm .175$.

The book gives answer $\displaystyle 7.51 \pm .176$. Not much of a difference, I know; I'm just wondering whether the difference might be due to rounding assumptions or whether I've made a more fundamental mistake.