Answer
a)0.0668
b)0.6985.
c)The overhead reach distances are normally distributed and therefore the distribution of the sample mean is also normally distributed.
Work Step by Step
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
