The full quiz is here. The answers appear below. Comments, which are not part of the answers, are italicized.
1. Estimate the probability that a Binomial(225, 0.20) variable will not exceed 38.
B(225, 0.20) has a mean of 225 * 0.20 = 45 and a variance of 45 * 0.80 = 36, so its standard deviation is sqrt(36) = 6.
The Normal approximation to the desired probability is the CDF of N(45, 6) at the value 38.5. The Z-score for 38.5 is (38.5 - 45)/6 = -6.5/6 = -1.083. The value of the CDF here, based on the values we memorized, should be about 0.15. (A more accurate value is 13.93%.)
2. A box has six hundred tickets. Three hundred of them have a zero, one hundred have a one, one hundred have a two, and one hundred have a three. a) Compute the fifth moment of the distribution defined by this box.
An equivalent box has three tickets with a 0, and one each with a 1, 2, and 3. Its mean is (0 + 0 + 0 + 1 + 2 + 3)/6 = 1. The residuals are therefore (-1, -1, -1, 0, 1, 2). Their fifth powers are (-1, -1, -1, 0, 1, 32). Each probability is 1/6. Therefore the fifth moment is (-1 + -1 + -1 + 0 + 1 + 32)/6 = 5.
b) One of the tickets with a zero is removed and replaced by a ticket with an 11. To approximate the change in fifth moment, note that the mean increases only by (11 - 0)/600 = 0.0183..., which is quite small. So assume the mean is essentially unchanged and, based on that assumption, estimate the new fifth moment.
We have to work in the box with 600 tickets to solve this one, because it cannot be simplified (you cannot divide the single ticket having an 11). The only change in the fifth moment calculation occurs because a single ticket with a 0 (residual -1) changes to an 11 (residual 10). The moment calculation, recall, is a sum of fifth powers of residuals, each weighted by the ticket's probability of being drawn: 1/600. Therefore, to estimate the change in fifth moment (which started at 5, as computed in problem 2), we must remove a value of -15 = -1 and replace it by a value of 105 = 100,000. This is multiplied by the probability of the ticket, which is 1/600. Therefore the fifth moment increases by 100,001/600 = 166.7, approximately. Our estimated answer is 5 + 166.7 = 171.7.
The correct value for the new fifth moment is 169.844, so our approximation was close. This problem illustrates how sensitive higher moments (powers 4, 5, 6, ...) are to single small outlying values. The value of 11 is not that different from 0, 1, 2, 3, and it occurs in only 1/600 of the tickets, but it dominates the calculation of the fifth moment.
Scoring: The passing score is 90.
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This page was created 24 February 2001 and modified 3 March 2001 to clarify the answer to problem 2b.