Question

1. Suppose that the average grades of an introduction to marketing course final is 70 with a standard deviation of 5. Chris got 60 in this exam, please calculate Chris’ z-score. How do we interpret that number?

2. Cassie got 85 in this exam, please calculate Cassie’s z-score. How do we interpret that number?

Now calculate what percentile Cassie’s score lies.

3. Find the interval of scores within which 95% of the students fall (z-score = ±1.96) in this exam.

Answer 1

(1) Given that mean = 70 and standard deviation = 5

x value for Chris is 60

z score = (x-mean)/standard deviation = (60-70)/5 = -10/5 = -2

Chris score is 2 standard deviations below the mean exam score.

(2) Given that mean = 70 and standard deviation = 5

x value for Cassie is 85

z score = (x-mean)/standard deviation = (85-70)/5 = 15/5 = 3

Cassie score is 3 standard deviations above the mean exam score.

(3) Confidence interval is given as