Assume that womens heights are normally distributed with a m

Assume that women\'s heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 90 women are randomly selected, find the probability that they have a mean height between 62.9 inches and 64.0 inches. (A) 0.7248 (B) 0.1739 (C) 0.9318 (D) 0.0424

Solution

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    62.9      
x2 = upper bound =    64      
u = mean =    63.6      
n = sample size =    90      
s = standard deviation =    2.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -2.656313235      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.517893277      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.00395001      
P(z < z2) =    0.935479347      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.9318 [answer, C]