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 77 beats per minute.

the probability is $square$.
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 77\), \(\mu=74.0\), and \(\sigma = 12.5\).

$$z=\frac{77 - 74.0}{12.5}=\frac{3}{12.5}=0.24$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt77)\), which is equivalent to \(P(Z\lt0.24)\) using the standard normal distribution (since \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\) and \(Z=\frac{X - \mu}{\sigma}\)).
Looking up the value of \(z = 0.24\) in the standard - normal distribution table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\lt0.24)=0.5948\)

Answer:

\(0.5948\)