The full quiz is here. The answers appear below in bold. Comments, which are not part of the answers, are italicized.
Time: 20 minutes. This quiz is closed book, closed notes.
Below is a gallery of Q-Q plots. In each plot, the X-axis shows the "reference distribution" values and the Y-axis shows the values of some large batch of numbers. As usual, values along the X axis increase to the right and values along the Y axis increase to the top. The reference distribution is unimodal and symmetric (like the Normal distribution).
Each of problems 0 through 5 is a statement about the batch relative to the reference distribution. List all the Q-Q plots that match each description. There may be zero, one, or more than one answer to each question. Take off ten points for each incorrect answer and take off five points for each missing answer. As an example, the zeroth question is done for you.
0. The batch has (approximately) the same shape as the reference distribution: D, E.
1. The batch is long tailed (platykurtic) in both directions: A
2. The batch is short tailed (leptokurtic) in both directions: C
3. The batch is positively skewed: B, F
4. The batch has a large number of nondetect values with a common detection limit: B
5. The batch has two low outliers. None
| A | |
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B |
| C | |
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D |
| E | |
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F |
Mary is computing logarithms to an unusual base. To two decimal places, she knows that log(2) = -0.76, log(3) = -1.20, and log(10) = -2.51.
6. What is log(200)? Log(200) = log(2 * 10 * 10) = log(2) + log(10) + log(10) = -0.76 - 2.51 - 2.51 = -5.78
7. What is log(5)? Log(5) = log(10/2) = log(10) - log(2) = -2.51 - (-0.76) = -1.75.
8. What is log(1)? 0. Log(1) is always 0, regardless of the base.
9. What is log(-2)? Logarithms of negative numbers are undefined.
10. (Extra credit) What is the base of Mary's logarithms? The logarithm of the base is the unique value whose log is 1. Notice, from problem 7, that log(5) is almost exactly 1 less than log(2). Since log(2/5) = log(2) - log(5) = -0.76 - (-1.75) = 0.99 (which is essentially 1), the base of the logarithms is approximately 2/5 = 0.40. You can also solve directly. Let b be the unknown base. Then, by definition, -0.76 = log(2) = ln(2) / ln(b) = 0.69 / ln(b), approximately. Solving for ln(b) gives ln(b) = -0.69/0.76 = -0.9, approximately, so b = exp(-0.9) = exp(-1) * exp(0.1) = (approximately) 1/e * 1.1 = 1/(2.72/1.1) = (approximately) 1/2.5 = 0.4. As a check, note that log(1/2) = 0.76 < 1 and log(1/3) = 1.20 > 1, so the base b must lie between 1/3 = 0.33 and 1/2 = 0.50. Linear interpolation between these gives 0.41, which is very close.
Scoring: The passing score is 95. Incorrect "extra credit" answers will be deducted from your score; correct extra credit answers will add to your score.
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This page was created 10 February 2001.