Question

Descriptive analysis revealed that the mean Test 3 score of all 63 students in Dr. Bill’s statistics courses was an 80. Similarly, the standard deviation for all students’ Test 3 scores was found to be 16. Assume the Test 3 scores are approximately normally distributed.

1. Find the z-score that corresponds to a Test 3 score of 34. Round your solution to two decimal places.

2. What is the probability that a randomly selected student scored a 75 or above on the test? That is, calculate . Round your solution to four decimal places.

Answer 1

(1)

= 80

= 16

X = 34

Z = (34 - 80)/16

= - 2.88

So,

Answer is:

- 2.88

(2)

= 80

= 16

To find P(X 75):

Z = (75 - 80)/16

= - 0.3125

By Technology, Cumulative Area Under Standard Normal Curve = 0.3773

So,

P(X 75): = 1 - 0.3773 = 0.6227

So,

Answer is:

0.6227