In a small country town, the ages of the population are normally distributed. The mean age is 38 years and the standard deviation is 12 years. What percentage of the population lies between the ages of 38-62 is closest?
$$z=\frac{38-38}{12}$$
= 0
$$z=\frac{62-38}{12}$$
= 2
since 95% is between -2 and 2, so 0 to 2 is half of that
47.5%