Question

(3.32) In a study of exercise, a large group of male runners walks 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(103, 12.6 ) distribution. The heart rates for male nonrunners after the same exercise have the N(129, 16.5 ) distribution.

(a) What percent (±±0.01) of the runners have heart rates above 135 (use software)?

%.______
(b) What percent (±±0.01) of the nonrunners have heart rates above 135 (use software)?

%._____

Answer 1

Solution :

Given that ,

a) mean = = 103

standard deviation = = 12.6

P(x > 135) = 1 - p( x< 135)

=1- p P[(x - ) / < (135 - 103) / 12.6 ]

=1- P(z < 2.54)

Using z table,

= 1 - 0.9945

= 0.0055

The percentage is = 0.55%

b) mean = = 129

standard deviation = = 16.5

P(x > 135) = 1 - p( x< 135)

=1- p P[(x - ) / < (135 - 129) / 16.5 ]

=1- P(z < 0.36)

Using z table,

= 1 - 0.6406

= 0.3594

The percentage is = 35.94%