Assume that womens heights are normally distributed with a m

Assume that women\'s heights are normally distributed with a mean given by Mu = 64.3 in, and a standard deviation given by

sigma = 2.1 in.

(a) If 1 woman is randomlyselected, find the probability that her height is less than 65in.

(b) If 32 women are randomlyselected, find the probability that they have a mean height less than

65in.

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    65      
u = mean =    64.3      
          
s = standard deviation =    2.1      
          
Thus,          
          
z = (x - u) / s =    0.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.333333333   ) =    0.63055866 [ANSWER]

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B)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    65      
u = mean =    64.3      
n = sample size =    32      
s = standard deviation =    2.1      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.885618083      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   1.885618083   ) =    0.970326781 [ANSWER]