Question

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 greater than 70 beats per minute.

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute.

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

Short answer

a. The probability that the pulse rate of the selected female is greater than 70 beats per minute is equal to 0.6255.

b.The probability that for a sample of 25 females, their mean pulse rate is greater than 70 beats per minute is equal to 0.9452.

c. The sample mean female pulse rate follows the normal distribution because the population of female pulse rates follows the normal distribution. As a result, in part (b), the normal distribution can be used to compute the probability.

Step-by-step solution

Given information

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.

Conversion of a sample value to a z-score

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:

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

Probability values

a.

The sample value given has a value equal to x=70 beats per minute.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{70-74.0}{12.5} \\ =-0.32 \end{gathered}$$

Thus, the required probability value is computed using the standard normal table as follows:

$$\begin{gathered} P\left(z>-0.32\right)=P\left(z<0.32\right) \\ =0.6255 \end{gathered}$$

Therefore, the probability that the pulse rate of the selected female is greater than 70 beats per minute is equal to 0.6255.

b.

Let the sample size be equal to n = 25.

The sample mean is equal to $\overline{x}=70\text{beats}\text{per}\text{minute}$.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} \\ =\frac{70-74.0}{\frac{12.5}{\sqrt{25}}} \\ =-1.6 \end{gathered}$$

Thus, the required probability is computed using the standard normal table as follows:

$$\begin{gathered} P\left(z>-1.6\right)=P\left(z<1.6\right) \\ =0.9452 \end{gathered}$$

Therefore, the probability that for a sample of 25 females, their mean pulse rate is greater than 70 beats per minute is equal to 0.9452.

Sampling distribution of the sample mean

c.

Although the sample size (25) is less than 30, it is given that the population of female pulse rates is normally distributed.

Hence, the sample mean female pulse rate can be assumed to follow the normal distribution.