CS 6553 Performance Evaluation, Spring 2004 Assignment 1 Comments

The first problem was worth 50 points and each of the others was 10 points.

Here are some comments:

1. We will discuss this problem in class on February 12.
2. You should have found that the success rate is multiplicative, not the failure rate.
3. Most people found 13 errors. If you found less than this, make sure you did it correctly.
4. The expected number should be 1/p. It uses a sum like the one in class. The variance needs a sum like Σk²q^k = q(1+q)/(1-q)³.
5. As most people realized, the answer in the book for part f) should have been negative.
6. All of the answers in the book are incorrect. The correct answers are:
   a) 0.0013, b) 0.8413, c) 81.85%, d) 6.645.
7. There is not one correct answer for this. Either the mean or the median would be appropriate. I don't believe in using the ratio of max to min as a test.
8. Here are some of the measures of dispersion you should have calculated:
   • range: 9 to 45. Note that the range consists of two numbers and is not the difference of the max and min.
   • variance: 90.49
   • standard deviation: 9.49
   • C.O.V = 0.35
   • 10th and 90th percentiles: 14 and 38 (use the 4th and 30th values)
   • SIQR: 6.5 (use the 9th and 25th values)
   • mean square deviation: 8.03
9. Everyone seems to have eventually gotten the correct graph. The result is not quite normal but probably close enough.