Given that the population mean is 132 and the population sta

Given that the population mean is 132 and the population standard deviation is 26 on a continuous normally distributed scale. Find P(98 sx s 150) Given probabilities accurate to four decimals, that is, 0.12345 or 12.345% would be given as 0.1235.

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    98      
x2 = upper bound =    150      
u = mean =    132      
          
s = standard deviation =    26      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.307692308      
z2 = upper z score = (x2 - u) / s =    0.692307692      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.095488847      
P(z < z2) =    0.75562794      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.660139093 = 0.6601 [ANSWER]