Using the Central Limit Theorem. In Exercises 5–8, assume that females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 “Body Data” in Appendix B).

a. If 1 adult female is randomly selected, find the probability that her pulse rate is between 78 beats per minute and 90 beats per minute.

b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean between 78 beats per minute and 90 beats per minute.

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

The population of female pulse rates is normally distributed with mean equal to 74.0 beats per minute and standard deviation equal to 12.5 beats per minute.

Let the population mean pulse rate be $\mu =74.0\text{beats}\text{per}\text{minute}$.

Let the population standard deviation of beats per minute $\sigma =12.5\text{beats}\text{per}\text{minute}$.

The z-score for a given sample observation has the following expression:

$z=\frac{x-\mu }{\sigma }$

The z-score for the sample mean has the following expression:

$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$

a.

The sample value should lie between 78 beats per minute and 90 beats per minute.

The probability that the sample value will lie between 78 beats per minute and 90 beats per minute is computed below:

$$\begin{gathered} P\left(78<x<90\right)=P\left(\frac{78-\mu }{\sigma }<\frac{x-\mu }{\sigma }<\frac{90-\mu }{\sigma }\right) \\ =P\left(\frac{78-74}{12.5}<z<\frac{90-74}{12.5}\right) \\ =P\left(0.32<z<1.28\right) \\ =P\left(z<1.28\right)-P\left(Z<0.32\right) \end{gathered}$$

Using the standard normal table, the required probability can be computed as:

$$\begin{gathered} P\left(78<x<90\right)=P\left(z<1.28\right)-P\left(Z<0.32\right) \\ =0.8997-0.6255 \\ =0.2742 \end{gathered}$$

Therefore, the probability that the pulse rate of the selected female is between 78 beats per minute and 79 beats per minute is equal to 0.2742.

b.

Let the sample size be equal to n = 16.

The sample mean should lie between 78 beats per minute and 90 beats per minute.

The probability that the sample mean will lie between78 beats per minute and 90 beats per minute is equal to:

$$\begin{gathered} P\left(78<\overline{x}<90\right)=P\left(\frac{78-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}<\frac{90-\mu }{\frac{\sigma }{\sqrt{n}}}\right) \\ =P\left(\frac{78-74}{\frac{12.5}{\sqrt{16}}}<z<\frac{90-74}{\frac{12.5}{\sqrt{16}}}\right) \\ =P\left(1.28<z<5.12\right) \\ =P\left(z<5.12\right)-P\left(Z<1.28\right) \end{gathered}$$

Using the standard normal table, the required probability can be computed as:

$$\begin{gathered} P\left(78<\overline{x}<90\right)=P\left(z<5.12\right)-P\left(Z<1.28\right) \\ =1-0.8997 \\ =0.1003 \end{gathered}$$

Therefore, the probability that for a sample of 16 females, their mean pulse rate is between 78 beats per minute and 90 beats per minute is equal to 0.1003.

c.

Despite the fact that the sample size of 4 is less than 30, it is also mentioned that the population of female pulse rates is normally distributed.

As a result, the sample mean female pulse rate can be expected to follow a normal distribution.