## Elementary Statistics (12th Edition)

a) $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{231.3-205.5}{8.6}=3$ $z_{2}=\frac{value-mean}{standard \ deviation}=\frac{179.7-205.5}{8.6}=-3$ Using the table, the probability of z being between -3 and 3 is equal to the probability of z being less than 3 minus the probability of z being less than -3, which is: 0.9987-0.0013=0.9974. b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{206-205.5}{\frac{8.6}{\sqrt{40}}}=0.37.$ $z_{1}=\frac{value-mean}{standard \ deviation}=\frac{204-205.5}{\frac{8.6}{\sqrt{40}}}=-1.1.$ Using the table, the probability of z being between 0.37 and -1.1 is equal to the probability of z being less than 0.37 minus the probability of z being less than -1.1, which is: 0.6443-0.1357=0.5086.