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Men have a normal distribution with mean $=188(mg/dl)$

Standard deviation $=41mg/dl$

$14$year-old boy has a mean $=170mg/dl$

Standard deviation$=30mg/dl$

Distribution M: Normal with ${渭}_M=188and{蟽}_M=41$

Distribution B: Normal with ${渭}_B=170and{蟽}_B=30$

If M and B are normally distributed, then their difference $M-B$is also normally distributed

Mean and standard deviation

${渭}_{aX+bY}=a{渭}_X+b{渭}_Y$

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

Therefore we get,

localid="1650362260757" $$\begin{gathered} {渭}_{M-B}={渭}_M-{渭}_B \\ =188-170 \\ =18 \end{gathered}$$

localid="1650362473508" $$\begin{gathered} {蟽}_{M-B}=\sqrt{{蟽}_M^2+{蟽}_B^2} \\ =\sqrt{{41}^2+{30}^2} \\ =\sqrt{2581} \\ 鈮�50.8035 \end{gathered}$$

So the distribution is normal with mean${渭}_{M-B}=18$and standard deviation${蟽}_{M-B}鈮�50.8035.$

Men have a normal distribution with mean $=188mg/dl$

Standard deviation$=41mg/dl$

$14$year-old boy has mean$=170mg/dl$

Standard deviation$=30mg/dl$

From part (a) answer,

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

$$\begin{gathered} z=\frac{x-渭}{蟽} \\ =\frac{0-18}{50.80} \\ =-0.35 \end{gathered}$$

Find the probability using table A

$$\begin{gathered} P(M-B<0)=P(Z<-0.35) \\ =0.3632 \end{gathered}$$.