Answer to Question #204094 in Statistics and Probability for Lauren Veitz

Question

The heights of WNBA basketball players are normally distributed. What percentage of these heights is within 2 standard deviations of the mean? 

Answer 1

Use the 68–95–99.7 rule

Let $X$ be a height $X\sim N(\mu, \sigma^2),$ where $\ \mu$ is the mean of the distribution, and $\sigma$ is its standard deviation:

$$P(\mu-2\sigma\leq X\leq \mu+2\sigma)\approx0.9545$$

$$P(\mu-2\sigma\leq X\leq \mu+2\sigma)$$

$$=P( X\leq \mu-2\sigma)-P( X< \mu+2\sigma)$$

$$=P( Z\leq \dfrac{\mu+2\sigma-\mu}{\sigma})-P( Z< \dfrac{\mu-2\sigma-\mu}{\sigma})$$

$$=P( Z\leq 2)-P( Z< -2)$$

$$\approx0.977249868-0.022750132\approx0.9545$$