The GPAS of all students enrolled at a college have an appro

The GPA\'S of all students enrolled at a college have an approx. normal distribution with a mean of 3.02 and a standard deviation of .29. Find the probablity that the mean of a selected sample of 20 selected students is between 2.95 and 3.11.

Solution

Mean ( u ) =3.02
Standard Deviation ( sd )= 0.29/ Sqrt ( 20 )
Number ( n ) = 20
Normal Distribution = Z= X- u / (sd/Sqrt(n) ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 2.95) = (2.95-3.02)/0.29/ Sqrt ( 20 )
= -0.07/0.0648
= -1.0795
= P ( Z <-1.0795) From Standard Normal Table
= 0.14019
P(X < 3.11) = (3.11-3.02)/0.29/ Sqrt ( 20 )
= 0.09/0.0648 = 1.3879
= P ( Z <1.3879) From Standard Normal Table
= 0.91742
P(2.95 < X < 3.11) = 0.91742-0.14019 = 0.7772