Question

A professor at a local university noted that the grades of her students were normally distributed with a mean of 73 and a standard deviation of 11. All probabilities should be to four decimal places. The professor has informed us that 7.93 percent of her students received grades of A. What is the minimum score needed to receive a grade of A? (Round to two decimal places) Answer Students who made 57.93 or lower on the exam failed the course. What percent of students failed the course? (Round to two decimal places) Answer

Answer 1

Let , X be the scores of the student's

X follows Normal Distribution with mean = = 73 and standard deviation = = 11

1)

The professor has informed us that 7.93 percent of her students received grades of A.

That is P( X > a ) = 7.93% = 0.0793 .

We have to find a .

P( X > a ) = 7.93% = 0.0793

So, P( X <= a ) = 1 - 0.0793 = 0.9207

Using Excel function ,   =NORMINV( probability , , )

a = NORMINV( 0.9207, 73 , 11 ) = 88.51

The minimum score needed to receive a grade of A is 88.51

2)

We have to find P( X <= 57.93 )

Using Excel function , =NORMDIST( x, , , 1 )

P( X <= 57.93 ) =NORMDIST( 57.93 , 73 , 11 , 1 ) = 0.0853 = 8.53%

8.53 percent of students failed the course