Let X be a normally distributed random variable with 100 an

Let X be a normally distributed random variable with = 100 and No Image = 10. Find the probability that X is between 70 and 120. (Round your answer to the nearest whole number percent.)

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    70      
x2 = upper bound =    120      
u = mean =    100      
          
s = standard deviation =    10      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -3      
z2 = upper z score = (x2 - u) / s =    2      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.001349898      
P(z < z2) =    0.977249868      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.97589997 = 98% [ANSWER]