Question

The mean maximum aerobic power (VO2MAX) score for women ages 20 to 29 is 36 ml/min/kg with a standard deviation of 7 ml/min/kg. Find the probability of a woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.

Answer 1

0.099 is the probability of a woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.

Explanation:

We are given the following information in the question:

Mean, μ = 36 ml/min/kg

Standard Deviation, σ = 7 ml/min/kg

We assume that the distribution of aerobic power is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.)

P(x > 45)

`P( x > 45) = P( z > \displaystyle(45 - 36)/(7)) = P(z > 1.285)`

`= 1 - P(z \leq 1.285)`

Calculation the value from standard normal z table, we have,

`P(x > 45) = 1 - 0.901 =0.099 = 9.9\%`

0.099 is the probability of a woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.