Respuesta :
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. Ā Ā Ā
Step-by-step explanation:
We are given the following information in the question:
Mean, Ī¼ = 36 ml/min/kg
Standard Deviation, Ļ = 7 ml/min/kg
We assume that the distribution of aerobic power is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(woman between the ages of 20 to 29 having a VO2MAX score of greater than 45 ml/min/kg.)
P(x > 45)
[tex]P( x > 45) = P( z > \displaystyle\frac{45 - 36}{7}) = P(z > 1.285)[/tex]
[tex]= 1 - P(z \leq 1.285)[/tex]
Calculation the value from standard normal z table, we have, Ā
[tex]P(x > 45) = 1 - 0.901 =0.099 = 9.9\%[/tex]
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.
