At a large university the mean age of the students is mu 22

At a large university, the mean age of the students is mu = 22.3 years and the standard deviation is sigma =4 years. If we select a random sample of n=36 students, and consider the distribution of the sample mean x , find: the mean of x distribution, mux = the standard deviation of x distribution, sigma x = the probability that sample mean x is 23 years or more.

Solution

a)

It has the same mean,

u(X) = 22.3

******************

b)

It has a standard deviation of

sigma(X) = sigma/sqrt(n) = 4/sqrt(36) = 0.666666667 [ANSWER]

***************

c)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    23      
u = mean =    22.3      
          
s = standard deviation =    0.666666667      
          
Thus,          
          
z = (x - u) / s =    1.05      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.05   ) =    0.146859056 [ANSWER]