In a study of exercise, a large group of male runners walk on a treadmill for six 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:a) 1.88% percent of the runners have heart rates above 130.b) 50% percent of the nomrunners have heart rates above 130.Step-by-step explanation:Problems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.(a) What percent of the runners have heart rates above 130? In a study of exercise, a large group of male runners walk on a treadmill for six minutes. Their heart rates in beats per minute at the end vary from runner to runner according to the N(104,12.5) distribution. This means that [tex]\mu = 104, \sigma = 12.5[/tex].This percentage is 1 subtracted by the pvalue of Z when [tex]X = 130[/tex].[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{130 - 104}{12.5}[/tex][tex]Z = 2.08[/tex][tex]Z = 2.08[/tex] has a pvalue of 0.9812.This means that 1-0.9812 = 0.0188 = 1.88% percent of the runners have heart rates above 130.(b) What percent of the nonrunners have heart rates above 130?The heart rates for male nonrunners after the same exercise have the N(130, 17) distribution. This means that [tex]\mu = 130, \sigma = 17[/tex].This percentage is 1 subtracted by the pvalue of Z when [tex]X = 130[/tex].[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{130 - 130}{17}[/tex][tex]Z = 0.00[/tex][tex]Z = 0.00[/tex] has a pvalue of 0.5000.This means that 1-0.50 = 0.50 = 50% percent of the nomrunners have heart rates above 130.