Computing the proportion of a normally distributed population between specified values of the standard deviation

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Ok I am stuck on the second part of this question.

(a) what percentage of the population is within ± $0.5$ standard deviations of the mean.

(b) what percentage of the population is more than 1 standard deviation above the mean?

For part (a), I got $38.3$% using $P_r(-0.5<z<0.5)$

However, I am stuck on part (b). Is it similar to part (a)?

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The empirical rule tells use that $68.2\%$ of the population is within $1$ standard deviation of the mean. We know that the normal distribution is symmetrical about the mean; thus, we can deduce that $34.1\%$ of the population is between $0$ and $1$ standard deviations above the mean. Thus, by symmetry, the percentage of the population that is over $1$ standard deviation above the mean is

$50\% - 34.1\% = 15.9\%$

The empirical rule is very useful for reasoning about normal distributions and is something you should commit to memory. If you need an overview check out this page.