Q 7.74. - Elementary Statistics

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 $20 s$ was $62.6$ inches. Assume that the heights of today's women in their $20 s$ 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 $20 s$ have mean heights of at least $64.24$ inches?

Step-by-Step Solution

Answer

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

Step 1: Given information

Calculate the sample mean's standard deviation.

Let $\mu$ be the population mean height of women in their $20 s$ 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 $25$

Step 2: Concept

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

Step 3: 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,

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

Step 4: 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,

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

Step 5: 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.