There are 8000 students at the University of Tennessee at Ch

There are 8,000 students at the University of Tennessee at Chattanooga. The average age of all the students is 24 years with a standard deviation of 9 years. A random sample of 36 students is selected. Determine the standard error of the mean What is the probability that the sample mean will be larger than 19.5? What is the probability\' that the sample mean will be between 25.5 and 27 years?

Solution

a)
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size

Standard deviation( sd )=9
Sample Size(n)=36
Standard Error = ( 9/ Sqrt ( 36) )
= 1.5

b)
P(X > 19.5) = (19.5-24)/9/ Sqrt ( 36 )
= -4.5/1.5= -3
= P ( Z >-3) From Standard Normal Table
= 0.9987                  
c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 25.5) = (25.5-24)/9/ Sqrt ( 36 )
= 1.5/1.5
= 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(X < 27) = (27-24)/9/ Sqrt ( 36 )
= 3/1.5 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(25.5 < X < 27) = 0.97725-0.84134 = 0.1359