Almost all medical schools in the United States require stud

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT on your campus, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 5.5. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 25.0.

Solution

a) If you choose one student at random, what is the probability that the student\'s score is between 20 and 30?
Normal Distribution
Mean ( u ) =25
Standard Deviation ( sd )=5.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 20) = (20-25)/5.5
= -5/5.5 = -0.9091
= P ( Z <-0.9091) From Standard Normal Table
= 0.18165
P(X < 30) = (30-25)/5.5
= 5/5.5 = 0.9091
= P ( Z <0.9091) From Standard Normal Table
= 0.81835
P(20 < X < 30) = 0.81835-0.18165 = 0.6367                  

b) you sample 25 students. What is the sampling distribution of their average score?
Mean ( u ) =25
Standard Deviation ( sd )=5.5/ Sqrt ( 25 ) = 1.1  

c) what is the probability that the mean score of your sample is between 20 and 30? Show you work.
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 20) = (20-25)/5.5/ Sqrt ( 25 )
= -5/1.1
= -4.5455
= P ( Z <-4.5455) From Standard Normal Table
= 0
P(X < 30) = (30-25)/5.5/ Sqrt ( 25 )
= 5/1.1 = 4.5455
= P ( Z <4.5455) From Standard Normal Table
= 1
P(20 < X < 30) = 1-0 = 1