91Ó°ÊÓ

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 ${\mathrm{μ}}_M=188and{\mathrm{σ}}_M=41$

Distribution B: Normal with ${\mathrm{μ}}_B=170and{\mathrm{σ}}_B=30$

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

Mean and standard deviation

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

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

Therefore we get,

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

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

So the distribution is normal with mean${\mathrm{μ}}_{M-B}=18$and standard deviation${\mathrm{σ}}_{M-B}≈50.8035.$

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:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{0-18}{50.80} \\ =-0.35 \end{gathered}$$

Find the probability using table A

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