The times of the finishers of the New York City 10km run are

The times of the finishers of the New York City 10km run are normally distributed with a mean of \\mu =61=61 minutes and a standard deviation of \\sigma = 9 =9 minutes. Find the probability that it took a randomly selected finisher between 50 minutes and 70 minutes to finish.

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    50      
x2 = upper bound =    70      
u = mean =    61      
          
s = standard deviation =    9      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.222222222      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.110811801      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.730532945   [ANSWER]