Question

QUESTION 1:

The proportion of observations from a standard Normal distribution that take values less than -0.98 is about?

QUESTION 2:

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(131, 16.6) distribution.

(a) What percent (±±0.1) of the runners have heart rates above 135 (use software)?

%.

(b) What percent (±±0.1) of the nonrunners have heart rates above 135 (use software)?

Answer 1

(1) using z table, find -0.9 in the left most column and 0.08 in the top row, then select the intersecting cell, we get

P(z<-0.98) = 0.1635

(2) (a) using normalcdf

setting lower = 135, upper = 9999, mean = 104 and sd = 12.5

we get

=normalcdf(lowe, upper, mean, sd)

= normalcdf(135,9999,104,12.5)

= 0.0066

=0.0066*100

=0.66%

(b) using normalcdf

setting lower = 135, upper = 9999, mean = 131 and sd = 16.6

we get

=normalcdf(lowe, upper, mean, sd)

= normalcdf(135,9999,131,16.6)

= 0.4048

= 0.4048*100

= 40.48%