IT 223 Winter 2006

Assignment 2 Solutions

Part I

Note: many final calculations are rounded values.

1. The z-score for Tonya’s SAT score:

(1318 - 1026) / 209 = 1.40

The z-score for Jermaine’s ACT score:

            (27 - 20.8) / 4.8 = 1.29

Tonya has the higher score on the standard scale.

2. The cumulative proportion for the z-score of Tonya’s SAT is .9192. This can be looked up in the table at the front of the text or by using the NORMSDIST function in Excel. The percentile is 91.92.

3. The z-score for the SAT score of 1287:

(1287 - 1026) / 209 = 1.25

Solve to find the ACT score that has a z-score of 1.25:

(score - 20.8) / 4.8 = 1.25

score = 1.25 * 4.8 + 20.8 = 26.8

4. The top 10% has the same z-score as a cumulative proportion of .9. The z-score for a CP of .9 is 1.28. This can be looked up in the table or by using the NORMSINV function. Solve to find the ACT score:

(score - 20.8) / 4.8 = 1.28

score = 1.28 * 4.8 + 20.8 = 26.9

5. The z-score for 15:

(15 - 20.8) / 4.8 = -1.21

and the cumulative proportion is .1131

The z-score for 25:

(25 - 20.8) / 4.8 = .86

and the cumulative proportion is .8051

The proportion between 15 and 25:

.8051 - .1131 = .692

Part II

Some answers will vary for part II because the timings differ from assignment to assignment.

But here are some general observations:

· Timing data is often skewed to the right.

· Assuming a normal distribution with 30 data points, approximately 20 data points (68% of 30) will be within one standard deviation of the mean. This is often the case even if the data do not fit a normal distribution very well.

· Often a log transformation (natural log or base-10) will transform timing data so that it fits a normal distribution better.