Question

Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketball player is 79 inches tall. Approximately what percent of the adult male population is taller than the average basketball player?

Answer 1

`\mathbb P(X>79)=\mathbb P\left(\frac{X-70}3>\frac{79-70}3\right)=\mathbb P(Z>3)`

Recall that the empirical rule for normal distributions says that approximately 99.7% of the distribution falls within three standard deviations of the mean. Therefore 0.3% must lie outside that range, with 0.15% on either side.

So,
`\mathbb P(Z>3)\approx0.15\%`.