The attendance at football games averages 500 with standard

The attendance at football games averages 500 with standard deviation of 100. What is the probability that attendance will be between 550 and 600 at this coming championship game?

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    550      
x2 = upper bound =    600      
u = mean =    500      
          
s = standard deviation =    100      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    0.5      
z2 = upper z score = (x2 - u) / s =    1      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.691462461      
P(z < z2) =    0.841344746      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.149882285   [answer]