In: Statistics and Probability
In the current tax year, IRS, the internal revenue service of the United States, estimates that five persons of the many high network individual tax returns would be fraudulent. That is, they will contain errors that are purposely made to cheat the government. Although these errors are often well concealed, let us suppose that a thorough IRS audit will uncover them.
Given this information, if a random sample of 100 such tax returns are audited, what is the probability that exactly five fraudulent returns will be uncovered? Here, the number of trials is n=100. And p=0.05 is the probability of a tax return will be fraudulent. Answer the following questions.
Here we have
n=100 and p=0.05
Since np = 5 and n(1-p) = 95 both are greater than 5 so we can use normal approximation here.
Using normal approximation, X has approximately normal distribution with mean and SD as follows:
The z-score for X = 5-0.05 is
The z-score for X = 5+0.05 is
The probability that five fraudulent returns will be uncovered based on 100 IRS audits is
The exact probability using, excel function "=BINOMDIST(5,100,0.05,FALSE) is 0.1800.
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Here we have
n=250 and p=0.05
Since np = 12.5 and n(1-p) = 237.5 both are greater than 5 so we can use normal approximation here.
Using normal approximation, X has approximately normal distribution with mean and SD as follows:
The z-score for X = 15-0.05 is
The probability that the IRS will uncover at least 15 fraudulent errors is
The exact probability using, excel function "=1-BINOMDIST(14,100,0.05,TRUE)" is 0.2712.
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The z-score for X = 20+0.05 is
The probability that the IRS would uncover at least 15 fraudulent returns but at most 20 fraudulent returns is
The exact probability using, excel function "=BINOMDIST(20,250,0.05,TRUE)-BINOMDIST(14,250,0.05,TRUE)" is 0.2563.
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The z-score for X = 0-0.05 is
The z-score for X = 0+0.05 is
The probability that out of the 250 randomly selected high net worth tax returns no fraudulent return is uncovered is