Question

A sample of midterm test results is distributed normally where the mean = 77 and the variance = 12.

1. What percentage of these scores is between 70 and 80?

2. What raw score is the cutoff for the top 10% of scores?

3. What is the probability of obtaining a score lower than 70?

4. What is the probability of obtaining a score greater than 90?

Answer 1

Solution :

Given that ,

mean = = 77

variance = = 12

standard deviation = = 3.46

1

P(70 < x < 80) = P((70 - 77)/ 3.46) < (x - ) /  < (80 -77) / 3.46) )

= P(-2.02 < z <0.87)

= P(z < 0.87) - P(z < -2.02)

= 0.8078 - 0.0217

= 0.7861

Probability = 0.7861

2

P(Z > z ) = 10%

1 - P(Z < z ) = 0.10

P(Z < z) =1- 0.10

P(Z < 1.28) = 0.90

z = 1.28

Using z-score formula,

X = z* +

= 1.28*3.46 + 77

= 48.43

3

P(x < 70) = P((x - ) / < (70 - 77) / 3.46)

= P(z < -2.02)

Probability = 0.0217

4

P(x >90) = 1 - P(x < 90)

= 1 - P((x - ) / < (90 - 77) / 3.46)

= 1 - P(z < 3.76)

= 1 - 0.9999   

= 0.0001

Probability = 0.0001