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 less than 80 beats per minute.

b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 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 less than 80 beats per minute is equal to 0.6844.

b.The probability that for a sample of 16 females, their mean pulse rate is less than 80 beats per minute is equal to 0.9726.

c. Since the population of female pulse rates is given to follow the normal distribution, the sample mean female pulse rate also follows the normal distribution. Thus, the normal distribution can be used to compute the probability in part (b).

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 value 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=80 beats per minute.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{80-74.0}{12.5} \\ =0.48 \end{gathered}$$

Thus, the following probability needs to be determined:

$P\left(z<0.48\right)$

The corresponding left tailed probability value for the z-score equal to 0.48 can be observed from the table and has a value equal to 0.6844.

Therefore, the probability that the pulse rate of the selected female is less than 80 beats per minute is equal to 0.6844.

b.

Let the sample size be equal to n = 16.

The sample mean is equal to $\overline{x}=80\text{beats}\text{per}\text{minute}$.

The corresponding z-score is equal to:

$$\begin{gathered} z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}} \\ =\frac{80-74.0}{\frac{12.5}{\sqrt{16}}} \\ =1.92 \end{gathered}$$

Thus, the following probability needs to be determined:

$P\left(z<1.92\right)$

The corresponding left tailed probability value for the z-score equal to 1.92 can be observed from the table and has a value equal to 0.9726.

Therefore, the probability that for a sample of 16 females, their mean pulse rate is less than 80 beats per minute is equal to 0.9726.

Sampling distribution of the sample mean

c.

Here, the sample size (16) is less than 30 but 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.