The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
1. A distribution is an equal mixture of a Binomial(1, 0.5) distribution and an N(0, 1) distribution. Plot its CDF. Clearly indicate, to the nearest percent, values attained by the CDF at 0 and 1.
We know the CDF for N(0, 1) has approximately the following values, because we memorized them. It is continuous (contains no leaps):
| X | CDF (X) - CDF(-X) | CDF(-X) | CDF(X) |
| 0 | 0% | 50% | 50% |
| 1 | 68% | 16% | 84% |
| 2 | 95.4% | 2.3% | 97.7% |
| 3 | 99.7% | 0.1% | 99.9% |
We know the CDF for B(1, 0.5) has exactly the following values, by definition. It contains leaps of 50% at 0 and 1:
| X | CDF(X) |
| < 0 | 0 |
| 0 | 50% |
| between 0 and 1 | 50% |
| 1 | 100% |
| > 1 | 100% |
The CDF of the equal mixture will be the average of these. It therefore contains leaps of 25% at 0 and 1, and otherwise varies like one-half the CDF for N(0, 1). To pin it down more precisely, we can combine values from the two tables by averaging them:
| X | N(0, 1) CDF | B(0, 0.5) CDF | Average |
| -3 | 0.1% | 0% | almost 0 |
| -2 | 2.3% | 0% | about 1% |
| -1 | 16% | 0% | 8% |
| 0- | 50% | 0% | 25% |
| 0 | 50% | 50% | 50% |
| 1- | 84% | 50% | 67% |
| 1 | 84% | 100% | 92% |
| 2 | 97.7% | 100% | about 99% |
| 3 | 99.9% | 100% | almost 100% |
Here is a sketch. Compare it to the solution of a similar problem illustrated at the notes for class 5.
2. Find, to two decimal places, the number whose natural logarithm is -0.08. Show your work.
Solution 1: Remembering that ln(1 + x) = x - x2/2 + ..., a first-order approximation will be x = -0.08. Let's check this: ln(1 - 0.08) = -0.08 - (-0.08)2/2 + ... = -0.08 - 0.0032 + ... = -0.08 to two decimal places. Furthermore, ln(1 - 0.07) = -0.07 - 0.00245 - 0.000114 - ... = -0.0726, approximately, showing that 1-0.08 = 0.92 is much closer to the correct answer than is 1-0.07. Therefore the correct answer is 0.92.
Solution 2: This problem is asking for exp(-0.08). From a listserver e-mail (23 January 2001) recall that exp(x) = 1 + x + x2/2! + x3/3! + ..., so that exp(-0.08) = 1 - 0.08 + 0.0032 - ... = 0.92, to two decimal places. (Because the series is alternating, we know the error we make is no greater than the first omitted term, 0.0032).
Scoring: The passing score is 90.
Return to the Environmental Statistics home page
This page is copyright (c) 2001 Quantitative Decisions. Please cite it as
This page was created 25 February 2001.