The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
Time: 20 minutes. This quiz is open book, open notes. You must provide reasons to get credit for your answers.
Estimated problem-solution times are provided as a hint.
1. Estimate the probability that a Binomial(400, 0.05) variable will not exceed 30. Justify your approximation method. (Three minutes.)
The Normal approximation will work because 400 * 0.05 = 20 >> 5 and 400 * (1-0.05) = 380 >> 5. It is important to check this. Other checks may be necessary, too. For example, the Normal approximation becomes poorer as we look further into the tails of the distribution. See #4 for an example of how bad the Normal approximation can be at the extreme end of the distribution.
The mean is 400 * 0.05 = 20 and the variance is 400 * 0.05 * 0.95 = 20 * 0.95 = 19.
We need to estimate the probability that the variable is less than or equal to 30.5, which is (30.5 - 20)/sqrt(19) = 2.4 standard deviations above the mean. Using tables we get 99.2% as the answer. Interpolating linearly from memorized values (2 sds corresponds to 100 - 2.3%, 3 sds corresponds to 100-0.15%) gives about 98.5%, which we know is rough but is a good check of the answer.
The commonest mistake was to forget what question you were answering and to provide 0.8% as the answer. This is the approximate probability that the outcome will be 31 or more.
The figure shows PDFs for four distributions. Number 1 is symmetric, number 2 has the lowest peak, number 3 is discontinuous, and number 4 has many modes. Distributions #1 and #2 have infinite support, but almost all their probability is concentrated within the region shown here and their tails continue to decrease rapidly outside the region shown. #3 and #4 have finite support contained within the region shown here.
#1 is Normal, #2 is Lognormal, #3 is Uniform, and #4 is a Beta distribution that has been jittered randomly.
2. Which distribution has the smallest third central moment? Which one the largest? (One minute.)
Distributions 1 and 3 are symmetric and so have zero odd central moments. Distribution 2 is strongly skewed positively and distribution 4 has a slight, but definite, negative skewness. Therefore #4 has the smallest third central moment and #2 has the largest.
The third central moment is not the skewness, but it is related to the skewness. (You have to divide the third central moment by the cube of the standard deviation to get the skewness.) Therefore what you learned about skewness early in the course can help you reason about third moments.
3. Of these four distributions, which one(s) will have the largest 12th central moment? (15 seconds.)
Distribution #2, because of its obviously heavy right tail compared to the other distributions.
4. (Extra credit) Estimate the probability that a Binomial(400, 0.05) variable will equal 0. Justify your estimate. (Three minutes.)
The Normal approximation will be unreliable because 0 is too extreme. The probability we seek is (1-0.05)400. The logarithm of this number is 400*ln(1-0.05) ~ 400*(-0.05 - 0.052/2 - ...) = -20 - 0.5 - ..., so this number is about exp(-20.5), which is very small. To estimate it, remember exp(2.3) ~ 10 and note that 20.7 = 9 * 2.3 is close to 20.5, so exp(-20.5) = exp(0.2 - 20.7) = exp(0.2) * exp(2.3 * -20.7/2.3) = exp(0.2) * exp(2.3 * -9) ~ (1 + 0.2 + 0.22/2! + ...) * 10-9 = 1.22 * 10-9 = 0.000000122%.
If we try to use the Normal approximation, then as in problem 1 we compute that -0.5 is 20.5/sqrt(19) = 4.70 sds below the mean and +0.5 is 19.5/sqrt(19) = 4.47 sds below the mean. The Normal approximation gives the probability as 3.85 * 10-6 - 1.28 * 10-6 = 2.57 * 10-6, which is over 2000 times the correct value.
Scoring: The passing score is 90.
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This page was created 5 March 2001 and updated 7 March 2001 to add more comments.