Question

assume that females have pulse rates that are normally distributed with a mean of \\( \mu = 74.0 \\) beats per minute and a standard deviation of \\( \sigma = 12.5 \\) beats per minute. complete parts (a) through (c) below.
a. if 1 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute.
the probability is.6844.
(round to four decimal places as needed.)
b. if 4 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 beats per minute.
the probability is \\( \square \\).
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score for part b

The formula for the z - score of a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(\mu = 74.0\), \(\sigma=12.5\), \(n = 4\), and \(\bar{x}=80\).

$$z=\frac{80 - 74.0}{\frac{12.5}{\sqrt{4}}}=\frac{6}{\frac{12.5}{2}}=\frac{6}{6.25}=0.96$$

Step2: Find the probability using the standard normal distribution

We want to find \(P(\bar{X}<80)\). Looking up the z - score of \(z = 0.96\) in the standard normal distribution table (or using a calculator with a normal - distribution function \(P(Z<z)\)), we know that \(P(Z < 0.96)=0.8315\)

Answer:

\(0.8315\)