Chapter 8.8, Problem 63E

Normal distribution of heights The heights of U.S. men are normally distributed with a mean of 69 inches and a standard deviation of 3 inches. This means that the fraction of men with a height between a and b (with a < b) inches is given by the integral

$\frac{1}{3\sqrt{2\pi }}{\displaystyle {\int }_a^b{e}^{-{((x-69)/3)}^2/2}dx}.$

What percentage of American men are between 66 and 72 inches tall? Use the method of your choice and experiment with the number of subintervals until you obtain successive approximations that differ by less than 10⁻³.