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anonymous
 one year ago
Problem: IRS will audit 2 million taxpayers. Plus 1 million taxpayers will be audited by mail. Assume 100 million total taxpayers. Find the probability that in a random sample of 50 taxpayers there will be one audit and one mailaudit.
My approach:
For the audit: Used Poisson distribution with mu = 1 and x = 1.
P(1) = 1^1 * e^(1) / 1! = 0.368.
For the mailaudit, I assumed the audit and mailaudit are independent, so I can use Poisson again with mu = 0.5 and x = 1. I get 0.303.
For both events, multiply, yielding 0.112.
What if I assume that the IRS will exclude auditees from maila
anonymous
 one year ago
Problem: IRS will audit 2 million taxpayers. Plus 1 million taxpayers will be audited by mail. Assume 100 million total taxpayers. Find the probability that in a random sample of 50 taxpayers there will be one audit and one mailaudit. My approach: For the audit: Used Poisson distribution with mu = 1 and x = 1. P(1) = 1^1 * e^(1) / 1! = 0.368. For the mailaudit, I assumed the audit and mailaudit are independent, so I can use Poisson again with mu = 0.5 and x = 1. I get 0.303. For both events, multiply, yielding 0.112. What if I assume that the IRS will exclude auditees from maila

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perl
 one year ago
Best ResponseYou've already chosen the best response.0hmm , does the problem say to use Poisson

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not explicitly, but based on the section where the problem was presented and the rules given in the text, it seems that Poisson was an acceptable approximation to the Binomial. n = 50. Although p = 0.01 is a little close to zero.
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