91Ó°ÊÓ

Calculate the sample mean's standard deviation.

Let $\mathrm{μ}$ be the population mean height of women in their $20s$ in the United States, which is $62.6$ inches, and $\left(\mathrm{σ}\right)$ be the population standard deviation, which is $2.88$ inches. The sample size $\left(n\right)$ is also $25$

Formula used:$\text{population mean and standard deviation:}{\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ}\text{and}{\mathrm{σ}}_{\stackrel{¯}{x}}=\mathrm{σ}/\sqrt{n}\text{.}$

Let $\mathrm{μ}$be the population mean height of women in their $20s$in the United States, which is $62.6$inches, and $\left(\mathrm{σ}\right)$be the population standard deviation, which is $2.88$inches. In addition, the sample size $n$is $25$.

The sample mean sampling distribution is a normal distribution with the sample mean as,

$$\begin{gathered} {\mathrm{μ}}_{\stackrel{¯}{x}}=\mathrm{μ} \\ =62.6 \end{gathered}$$

The sample standard deviation is,

$$\begin{gathered} {\mathrm{σ}}_{\stackrel{¯}{x}}=\frac{\mathrm{σ}}{\sqrt{n}} \\ =\frac{2.88}{\sqrt{25}} \\ =\frac{2.88}{5} \\ =0.576 \end{gathered}$$

Calculate the percentage of all samples of $25$women in their $20s$who have a mean height of at least $64.24$inches.

The $z-$score for $64.24$is,

$$\begin{gathered} z=\frac{\stackrel{¯}{x}-\mathrm{μ}}{{\mathrm{σ}}_{\stackrel{¯}{x}}} \\ =\frac{64.24-62.6}{0.576} \\ =\frac{1.64}{0.576} \\ =2.85 \end{gathered}$$

To find the area between the $z-$scores, use Table II: Areas under the standard normal curve.

$0.9978$is the area to the left of the $z-$score $2.85$

Area to right of $2.85=1$- Area to the left of $2.85$

$$\begin{gathered} =1-0.9978 \\ =0.0022 \end{gathered}$$

As a result, $0.0022$percent of all samples of $25$women in their $20s$have mean heights of at least $64.24$inches.