From the data we collected the population mean height all my
Solution
1.
By central limit theorem:
a) It is approximately normally distributed.
b) It has the same mean, u(X) = 68.3 in.
c) It has a standard deviation of sigma/sqrt(n) = 3.7/sqrt(31) = 0.664539617 in.
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2.
We first get the z score for the critical value. As z = (x - u) / s, then as          
           
 x = critical value =    69      
 u = mean =    68.3      
           
 s = standard deviation =    3.7      
           
 Thus,          
           
 z = (x - u) / s =    0.189189189      
           
 Thus, using a table/technology, the right tailed area of this is          
           
 P(z >   0.189189189   ) =    0.42497227 [ANSWER]
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3.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
           
 x = critical value =    69      
 u = mean =    68.3      
 n = sample size =    31      
 s = standard deviation =    3.7      
           
 Thus,          
           
 z = (x - u) * sqrt(n) / s =    1.053360825      
           
 Thus, using a table/technology, the right tailed area of this is          
           
 P(z >   1.053360825   ) =    0.146087826 [ANSWER]
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4.
The probability that a sample of 31 students will have a mean of 69 in or more is 0.146088.

