Question

The professors who teach the Introduction to Psychology course at State University pride themselves on the normal distributions of exam scores. After the first exam, the current professor reports to the class that the mean for the exam was 73, with a standard deviation of 7.

a. What proportion of student would be expected to score above 80?

b What proportion of students would be expected to score between 55 and 75?

c. What proportion of students would be expected to score less than 65?

d. If the top 10% of the class receive an A for the exam, what score would be required for a student to receive an A?

e. If the bottom 10% of the class fail the exam, what score would earn a student a failing grade?

Answer 1

Solution :

Given that,

a.

P(x > 80) = 1 - P(x < 80)

= 1 - P[(x - ) / < (80 - 73) / 7)

= 1 - P(z < 1)

= 1 - 0.8413

= 0.1587

proportion = 0.1587

b.

P(55 < x < 75) = P[(55 - 73)/ 7) < (x - ) /  < (75 - 73) / 7) ]

= P(-2.57 < z < 0.29)

= P(z < 0.29) - P(z < -2.57)

= 0.6141 - 0.0051

= 0.609

proportion = 0.609

c.

P(x < 65) = P[(x - ) / < (65 - 73) / 7]

= P(z < -1.14)

= 0.1271

proportion = 0.1271

d.

Using standard normal table ,

P(Z > z) = 10%

1 - P(Z < z) = 0.1

P(Z < z) = 1 - 0.1

P(Z < 1.28) = 0.9

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 7 + 73 = 81.96

score would be required for a student to receive an A is 81.96

e.

Using standard normal table ,

P(Z < z) = 10%

P(Z < ) = 0.1

z = -1.28

Using z-score formula,

x = z * +

x =-1.28 * 7 + 73 = 64.04

score would earn a student a failing grade