Question

Exam Grades

Exam grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of 11. Use the normal distribution to answer the following questions.

(a) What percent of students scored above a 90 ?

Round your answer to one decimal place.

(b) What percent of students scored below a 58 ?

Round your answer to one decimal place.

(c) If the lowest4%of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?

Round your answer to one decimal place.

Cutoff grade =

(d) If the highest11%of students will be given a grade of A, what is the cutoff to get an A?

Round your answer to one decimal place.

Cutoff grade =

Answer 1

Solution:

Given:

Examination grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of 11.

Thus and

Part a) What percent of students scored above a 90 ?

P( X > 90) =........?

Find z score for x = 90

Thus we get:

P ( X > 90) = P( Z > 1.64)

P ( X > 90) = 1 - P( Z < 1.64)

Look in z table for z = 1.6 and 0.04 and find corresponding area.

From z table we get:

P( Z < 1.64) = 0.9495

Thus

P ( X > 90) = 1 - P( Z < 1.64)

P ( X > 90) = 1 - 0.9495

P ( X > 90) = 0.0505

P ( X > 90) = 5.05%

P ( X > 90) = 5.1%

Part b) What percent of students scored below a 58 ?

P( X < 58) =..........?

Thus we get:

P( X < 58) = P( Z < -1.27)

Look in z table for z = -1.2 and 0.07 and find area.

From z table we get:

P( Z< -1.27) = 0.1020

Thus

P( X < 58) = P( Z < -1.27)

P( X < 58) = 0.1020

P( X < 58) = 10.20%

P( X < 58) = 10.2%

Part c) If the lowest 4%of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?

Find x value such that:

P( X < x ) =4%

P( X < x ) = 0.0400

Thus find z such that:

P( Z < z ) = 0.0400

Look in z table for Area = 0.0400 or its closest area and find z value.

Area 0.0401 is closest to 0.0400 and it corresponds to -1.7 and 0.05

Thus z = -1.75

Use following formula to find x value:

Cutoff grade = 52.8

Part d) If the highest 11% of students will be given a grade of A, what is the cutoff to get an A?

Find x value such that:

P( X > x ) =11%

P( X > x ) =0.1100

Thus find z such that:

P( Z > z ) = 0.1100

that is find z such that:

P( Z< z) = 1 - 0.1100

P( Z< z) = 0.8900

Look in z table for Area = 0.8900 or its closest area and find z value.

Area 0.8907 is closest to 0.8900

it corresponds to 1.2 and 0.03

Thus z = 1.23

Cutoff grade = 85.5