Question

The distribution of heights of young women aged $18$to $24$is approximately $N(64.5,2.5)$

2. What percent of young women have heights greater than $67$inches? Show your work.

Short answer

The percentage of young women have heights greater than $67$inches are $5.6923\%.$

Step-by-step solution

Given information

Heights of young women aged between$=18to24.$

Percentage of young women heights greater than $67$inches$=?$

Explanation 

Normal distribution $=N(64.5,2.5)$

Here mean value $\mu =64.5$

Variance ${\sigma }^2=2.5$

Then$\sigma =1.58114$

Then $z-$score, $z=\frac{x-\mu }{\sigma }$

Here need to find the percentage of heights greater than $67$,

localid="1649920459043" $$\begin{gathered} P(x>67)=P\left(z>\frac{67-64.5}{\sqrt{2.5}}\right) \\ P(x>67)=p(z>1.58114) \end{gathered}$$

From the table of normal distribution,

$p(x>67)=0.056923$

Convert into percentage,

localid="1649920477292" $$\begin{gathered} =0.056923\times 100\% \\ p(x>67)=5.6923\% \end{gathered}$$

The normal distribution curve is,