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

Accepted Solution

Answer:69 bpmStep-by-step 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 bpmz = ----------------- = -0.43073 = --------------------        std. dev.                                    15Thus,  x - 75 = -0.43073(15) = -6.461, so x = 75 - 6.6461, or approx. 68.54, or (to the nearest integer), approx 69 bpm