91影视

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 ${渭}_M=69.3and{蟽}_M=2.8$

Distribution W: Normal with ${渭}_W=64.5and{蟽}_W=2.5$

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

Properties mean and standard deviation

${渭}_{aX+bY}=a{渭}_X+b{渭}_Y$

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

Then we get,

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

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

Therefore the distribution is normal with${渭}_{M-W}=4.8and{蟽}_{M-W}鈮�3.7537.$

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:

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

Find the probability using table A

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

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