Question

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.

1. What is the probability that five fraudulent returns will be uncovered based on 100 IRS audits ? (n=100, p=0.05)
2. If a random sample of 250 high net worth tax returns are audited, what is the probability that the IRS will uncover at least 15 fraudulent errors? (n=250 and P= 0.05)
3. If a random sample of 250 high net worth tax returns are audited, what is the probability that the IRS would uncover at least 15 fraudulent returns but at most 20 fraudulent returns? (n=250 and P= 0.05)
4. What is the probability that out of the 250 randomly selected high net worth tax returns no fraudulent return is uncovered? (n=250 and P= 0.05)
5. Aside from the ethics of tax fraud and based solely on your answers to questions 1-4, do you think it would be advisable to cheat on your tax return? Do you need more information to decide? What type of information?

Answer 1

Let X denotes Number of fraudulent returns will be uncovered.

n = number of tax return audited.

p = probability of tax return will be fraudulent.

The probability distribution of X is binomial with parameter n and p.

E(X) = np and SD(X) =sqrt(n*p* (1-p))

i) P ( Exactly five fraudulant return will be uncovered) = P ( X=5)

Since n is large and p is small.

Using Poisson Approximation to binomial distribution.

X ~ Poisson ( lambda = np )

Here n = 100 and p = 0.05

lambda = 100 *0.05 = 5.

X ~ Poisson ( 5)

Required Probability = P ( X = 5)

P ( Exactly five fraudulent return will be uncovered) = 0.1755

ii) n = 250 and p = 0.05

E(X) = np = 250 * 0.05 = 12.5

SD(X) = sqrt ( 250 * 0.05 *0.95) = 3.4460

Required Probability = P ( X > = 15) = P ( X > = 15.5) ( Using Continuity correction)

Since n is large and p is small

Using Normal approximation to binomial distribution

Z = ( X - E(X) )/ SD(X) ~ N(0,1).

Required Probability = P ( Z > 0.8706)

From normal prob. table

P ( Z > 0.8706) = 0.1920

iii) Required Probability = P ( 15 < = X < = 20)

= P ( X < =20) - P ( X < 15)

= P( X < = 20.5) - P ( X < 15.5)

= P ( Z < 2.3215) - P( Z< 0.8706)

= 0.9899 - 0.8080

= 0.1819

P ( The IRS would uncover at least 15 fraudulent returns but at most 20 fraudulent returns) = 0.1819.

iv) n = 250 and p = 0.05

lambda = n*p = 250 *0.05 = 12.5

Using Poisson approximation to binomial distribution.

X ~ Poisson ( 12.5)

Required Probability = P ( X =0)

= 3.7267 * 10 ^-6.

Since prob. of no fraudulent return is uncovered is very small. It would not be advisable to cheat on tax return.