Chapter 6 - The Normal Distribution - 6-3 The Central Limit Theorem - Exercises 6-3 - Page 353: 25

$n\approx60$

Given $\mu=508,\sigma=72,P(X>520)=0.0985$ we have $P(X>520)=1-P(X=520)=0.0985$ so $P(X=520)=0.9015$ Use Table E or a calculator, the corresponding z-value is $z=1.2901$. Assume the sample size is $n$, we have $z=\frac{520-508}{72/\sqrt n}=1.2901$. We can find $n\approx60$