#### Answer

a)0.8888
b)0.9265
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) $z=\frac{value-mean}{standard \ deviation}=\frac{195-205.5}{8.6}=-1.22$ Using the table, the probability of z being more than -1.22 is equal to 1 minus the probability of z being less than -1.22, which is: 1-0.1122=0.8888.
b) By using the Central Limit Theorem, the sample mean has a mean of $\mu$ and standard deviation of $\frac{\sigma}{\sqrt n}$. $z=\frac{value-mean}{standard \ deviation}=\frac{203-205.5}{\frac{8.6}{\sqrt{25}}}=-1.45.$ Using the table, the probability of z being more than -1.45 is equal to 1 minus the probability of z being less than -1.45, which is: 1-0.0735=0.9265.
c) The reason is that we know that the overhead reach distances are normally distributed and therefore the distribution of the sample mean is also normally distributed, according to the Central Limit Theorem.