## Elementary Statistics (12th Edition)

a) $z=\frac{value-mean}{standard \ deviation}=\frac{218.4-205.5}{8.6}=1.5$ Using the table, P(z>1.5)=1-P(z<1.5)=1-0.9332=0.0668. b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z=\frac{value-mean}{standard \ deviation}=\frac{204-205.5}{\frac{8.6}{\sqrt{9}}}=-0.52.$ Using the table, P(z>-0.52)=1-P(z)=1-0.3015=0.6985