Assume that the random variabble x is normally distributed w

Assume that the random variabble x is normally distributed with mean =50 and standard deviation = 10. Compute the probability P(38 < x<_ 55)

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    38      
x2 = upper bound =    55      
u = mean =    50      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.2      
z2 = upper z score = (x2 - u) / s =    0.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.11506967      
P(z < z2) =    0.691462461      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.576392791   [ANSWER]