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

Answer:

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.      

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

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

P(x > 45)

$P( x > 45) = P( z > \displaystyle\frac{45 - 36}{7}) = P(z > 1.285)$

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

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.