Question

Taller Young Women. In the document Anthropometric Reference Data for Children and Adults, C. Fryer et al. present data from the National Health and Nutrition Examination Survey on a variety of human body measurements. A half-century ago, the mean height of (U.S.) women in their $20s$ was $62.6$ inches. Assume that the heights of today's women in their $20s$ are approximately normally distributed with a standard deviation of $2.88$ inches. If the mean height today is the same as that of a half-century ago, what percentage of all samples of $25$ of today"s women in their $20s$ have mean heights of at least $64.24$ inches?

Short answer

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

Step-by-step solution

Given information

Calculate the sample mean's standard deviation.

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

Concept

Formula used:$\text{population mean and standard deviation:}{\mu }_{\overline{x}}=\mu \text{and}{\sigma }_{\overline{x}}=\sigma /\sqrt{n}\text{.}$

Calculation

Let $\mu$be the population mean height of women in their $20s$in the United States, which is $62.6$inches, and $\left(\sigma \right)$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,

$$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =62.6 \end{gathered}$$

The sample standard deviation is,

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{2.88}{\sqrt{25}} \\ =\frac{2.88}{5} \\ =0.576 \end{gathered}$$

Calculation

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,

$$\begin{gathered} z=\frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}} \\ =\frac{64.24-62.6}{0.576} \\ =\frac{1.64}{0.576} \\ =2.85 \end{gathered}$$

Calculation

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.