91Ó°ÊÓ

Here, it is given that the height (in inches) of a $25$-year-old man is a normal random variable with parameters:

role="math" localid="1646741641325" $$\begin{gathered} \mathrm{μ}=71 \\ {\mathrm{σ}}^2=6.25 \\ ∴\mathrm{σ}=\sqrt{6.25}=2.5 \end{gathered}$$

$$\begin{gathered} 1\text{foot}=12\text{inches} \\ ∴6\text{feet}2\text{inches}=6\left(12\right)+2=74\text{inches} \end{gathered}$$

So, the required percentage is,

$$\begin{gathered} P(X>74)=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{74-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(z>\frac{74-71}{2.5}\right) \\ =1-P\left(z≤1.2\right) \\ =1-0.8849 \\ =0.1151 \\ =11.51\% \end{gathered}$$

Therefore, the percentage of $25$-year-old men are taller than $6$feet, $2$inches is$11.51$.

$$\begin{gathered} 1\text{foot}=12\text{inches} \\ ∴6\text{feet}5\text{inches}=6\left(12\right)+5=77\text{inches} \\ 6\text{feet}=6\left(12\right)=72\text{inches} \end{gathered}$$

Let the required percentage be $P\left(X>77/X>72\right)$.

$$\begin{gathered} P(X>77)=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{77-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(z>\frac{77-71}{2.5}\right) \\ =1-P\left(z≤2,4\right) \\ =1-0.9918 \\ =0.0082 \end{gathered}$$

$$\begin{gathered} P(X>72)=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{72-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(z>\frac{72-71}{2.5}\right) \\ =1-P\left(z≤0.4\right) \\ =1-0.6554 \\ =0.3466 \end{gathered}$$

role="math" localid="1646744610424" $$\begin{gathered} ∴P\left(X>77/X>72\right)=\frac{0.0082}{0.3466}=0.0238 \\ =2.38\% \end{gathered}$$

Therefore, the percentage of men in the $6$-footer club are taller than $6$ feet, $5$ inches is $2.38$.