Question

Suppose that the height, in inches, of a $25$-year-old man is a normal random variable with parameters role="math" localid="1646741074533" $\mu =71\text{and}{\sigma }^2=6.25$. What percentage of $25$-year-old men are taller than $6$ feet, $2$ inches? What percentage of men in the $6$-footer club are taller than $6$ feet, $5$ inches?

Short answer

The percentage of $25$-year-old men are taller than $6$ feet, $2$inches is $11.51$and the percentage of men in the $6$-footer club are taller than $6$ feet, $5$inches is $2.38$.

Step-by-step solution

Step 1. Given information.

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" $$\begin{gathered} \mu =71 \\ {\sigma }^2=6.25 \\ \therefore \sigma =\sqrt{6.25}=2.5 \end{gathered}$$

Step 2. Find the percentage of 25-year-old men are taller than 6 feet, 2 inches.

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

So, the required percentage is,

$$\begin{gathered} P(X>74)=P\left(\frac{X-\mu }{\sigma }>\frac{74-\mu }{\sigma }\right) \\ =P\left(z>\frac{74-71}{2.5}\right) \\ =1-P\left(z\le 1.2\right) \\ =1-0.8849 \\ =0.1151 \\ =11.51\% \end{gathered}$$

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

Step 3. Calculate the percentage of men in the 6-footer club are taller than 6 feet, 5 inches.

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

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

$$\begin{gathered} P(X>77)=P\left(\frac{X-\mu }{\sigma }>\frac{77-\mu }{\sigma }\right) \\ =P\left(z>\frac{77-71}{2.5}\right) \\ =1-P\left(z\le 2,4\right) \\ =1-0.9918 \\ =0.0082 \end{gathered}$$

$$\begin{gathered} P(X>72)=P\left(\frac{X-\mu }{\sigma }>\frac{72-\mu }{\sigma }\right) \\ =P\left(z>\frac{72-71}{2.5}\right) \\ =1-P\left(z\le 0.4\right) \\ =1-0.6554 \\ =0.3466 \end{gathered}$$

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

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