91Ó°ÊÓ

From the previous exercise, we have:

$$\begin{gathered} \mathrm{μ}=4.8 \\ \mathrm{σ}=1.0883 \\ x=2 \end{gathered}$$

The $z$ score is the value decreased by the mean and divided by the standard deviation. Then we have,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{2-4.8}{1.0883} \\ =-2.57 \end{gathered}$$

Now we will determine the value of the probability, that is:

$$\begin{gathered} P({\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_Wâ‰\yen 2)=P(Z>-2.57) \\ =1-P(Z<-2.57) \\ =1-0.0051 \\ =0.9949 \\ =00.49\% \end{gathered}$$

It is given that:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{2-4.8}{1.0883} \\ =-2.57 \end{gathered}$$

And the probability is $0.9949$.

Thus, a probability is considered to be small when the probability is less than $0.05$.

We then note that the probability of the mean difference${\stackrel{¯}{x}}_M-{\stackrel{¯}{x}}_W$ being at least two is large which means that it is unlikely that the mean difference is at least two and thus we would be not be surprised if the sample mean height for the young men is at least two inches greater than the sample mean height for the young women. Thus, we should not be surprised.