Question

The resting heart rate of US females 20 years old or older is normally distributed with a mean of 75 bpm (beats per minute). Suppose the standard deviation was 15 bpm. If one randomly selected female's resting heart rate falls in the bottom 33% of all women, at most what could her rate have been?

A. 70.05
B. 68.4
C. 79.95
D. 84.439
E. .6
F. 65.561

Answer 1

69 bpm

Explanation:

Here we start out finding the z-score corresponding to the bottom 33% of the area under the standard normal curve. Using the invNorm( function on a basic TI-83 Plus calculator, I found that the z-score associated with the upper end of the bottom 33% is -0.43073.

Next we use the formula for z score to determine the x value representing this woman's heart rate:

x - mean x - 75 bpm

z = ----------------- = -0.43073 = --------------------

std. dev. 15

Thus, x - 75 = -0.43073(15) = -6.461, so x = 75 - 6.6461, or approx. 68.54, or (to the nearest integer), approx 69 bpm