Question

In a study of exercise, a large group of male runners walk on a treadmill for 6 minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the N(104, 12.5) distribution. The heart rates for male nonrunners after the same exercise have the N(130, 17) distribution.

(a) What percent of the runners have heart rates above 130?
(b) What percent of the nonrunners have heart rates above 130?

Answer 1

For nonrunners the answer is simply 50% or 0.50 since a Normal distribution has 50% of the probability above and below the mean. For nonrunners the mean is 130.

For runners: P(X>130) = P[z > (130 - 104)/12.5] = P(z > 2.08), now simply look this up in a Standard Normal table

P(z > -2.08) = 0.0188 or 1.88%

Hope that helped

Answer 2

Thank you so much!