Use the normal distribution for this question The mean maxim

Use the normal distribution for this question. The mean maximum aerobic power (VO 2MAX 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 between 30 ml/min/kg and 42 ml/min/kg

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    30      
x2 = upper bound =    42      
u = mean =    36      
          
s = standard deviation =    7      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.857142857      
z2 = upper z score = (x2 - u) / s =    0.857142857      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.195682969      
P(z < z2) =    0.804317031      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.608634062   [ANSWER]  

Use the normal distribution for this question. The mean maximum aerobic power (VO 2MAX score for women ages 20 to 29 is 36 ml/min/kg with a standard deviation o

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