The heights of young men follow a Normal distribution with mean of $69.3$inches and standard deviation $2.8$inches. The heights of young women follow a Normal distribution with mean $64.5$inches and standard deviation $2.5$inches.

(a) Let M $=$the height of a randomly selected young man and W $=$the height of a randomly selected young woman. Describe the shape, center, and spread of the distribution of $M-W$

(b) Find the probability that a randomly selected young man is at least $2$inches taller than a randomly selected young woman. Show your work.

Heights of women have a mean$=69.3in$

Standard deviation$=2.8in$

Heights of men have a mean$=64.5in$

Standard deviation$=2.5in$

Distribution M: Normal with ${\mu }_M=69.3and{\sigma }_M=2.8$

Distribution W: Normal with ${\mu }_W=64.5and{\sigma }_W=2.5$

If M and W are normally distributed, then there difference $M-W$is also normally distributed

Properties mean and standard deviation

${\mu }_{aX+bY}=a{\mu }_X+b{\mu }_Y$

${\sigma }_{aX+bY}=\sqrt{a^2{\sigma }_X^2+b^2{\sigma }_Y^2}$

Then we get,

localid="1650362575481" $$\begin{gathered} {\mu }_{M-W}={\mu }_M-{\mu }_W \\ =69.3-64.5 \\ =4.8 \end{gathered}$$

localid="1650362603775" $$\begin{gathered} {\sigma }_{M-W}=\sqrt{{\sigma }_M^2+{\sigma }_W^2} \\ =\sqrt{2.8^2+2.5^2} \\ \approx 3.7537 \end{gathered}$$

Therefore the distribution is normal with${\mu }_{M-W}=4.8and{\sigma }_{M-W}\approx 3.7537.$

Heights of women have a mean$=69.3in$

Standard deviation$=2.8in$

Heights of men have a mean$=64.5in$

Standard deviation$=2.5in$

From the result of part (a)

The z-value is the difference between the population mean and the standard deviation, divided by the population mean:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{2-4.8}{3.7537} \\ =-0.75 \end{gathered}$$

Find the probability using table A

$$\begin{gathered} P(M-W>2)=P(Z>-0.75) \\ =P(Z<0.75) \\ =0.7734 \end{gathered}$$

$P(M-W>2)=0.7734.$