A set of final examination grades in an introductory statist

A set of final examination grades in an introductory statistics course is normally distributed, with a mean of 73 and a standard deviation of 8.

a. What is the probability that a student scored below 91 on this exam?

b. What is the probability that a student scored between 65 and 89?

c. If the professor grades on a curve (i.e., gives A’s to the top 10% of the class, regardless of the score), are you better off with a grade of 81 on this exam or a grade of 68 on a different exam, where the mean is 62 and the standard deviation is 3? Show your answer statistically and explain. (Hint: For option 1, you need to find a minimum Z-score that will guarantee you that you will be in the top 10% of the class.)

Solution

Mean ( u ) =73
Standard Deviation ( sd )=8
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 91) = (91-73)/8
= 18/8= 2.25
= P ( Z <2.25) From Standard Normal Table
= 0.9878                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 65) = (65-73)/8
= -8/8 = -1
= P ( Z <-1) From Standard Normal Table
= 0.15866
P(X < 89) = (89-73)/8
= 16/8 = 2
= P ( Z <2) From Standard Normal Table
= 0.97725
P(65 < X < 89) = 0.97725-0.15866 = 0.8186                  
c)
P ( Z < x ) = 0.81
Value of z to the cumulative probability of 0.81 from normal table is 0.878
P( x-u/s.d < x - 73/8 ) = 0.81
That is, ( x - 73/8 ) = 0.88
--> x = 0.88 * 8 + 73 = 80.024                  

Therefore 80.02% of the class scored 81% or less and the student is not in the top 10%.