Question

A tax auditor is selecting a sample of 5 tax returns for an audit. If 2 or more of these returns are​"improper," the entire population of 50 tax returns will be audited. Complete parts​ (a) through​ (e) below.

Q. What is the probability that the entire population will be audited if the true number of improper returns in the population is:

a) 15

b) 20

c) 5

d) 10

Answer 1

Part a

Q. What is the probability that the entire population will be audited if the true number of improper returns in the population is 15.

We are given N = 50, X = 15, so p = X/N = 15/50 = 0.3

Here, we have to use binomial distribution for finding required probability.

So, we have n = 5, p = 0.3, and we have to find P(X≥2)

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≤1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.3^0*(1 – 0.3)^(5 – 0)

P(X=0) = 0.16807

P(X=1) = 5C1*0.3^1*(1 – 0.3)^(5 – 1)

P(X=1) = 0.36015

P(X≤1) = P(X=0) + P(X=1)

P(X≤1) = 0.16807 + 0.36015

P(X≤1) = 0.52822

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≥2) = 1 – 0.52822

P(X≥2) = 0.47178

Required probability = 0.47178

Part b

Q. What is the probability that the entire population will be audited if the true number of improper returns in the population is 20.

We are given N = 50, X = 20, so p = X/N = 20/50 = 0.4

Here, we have to use binomial distribution for finding required probability.

So, we have n = 5, p = 0.4, and we have to find P(X≥2)

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≤1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.4^0*(1 – 0.4)^(5 – 0)

P(X=0) = 0.07776

P(X=1) = 5C1*0.4^1*(1 – 0.4)^(5 – 1)

P(X=1) = 0.2592

P(X≤1) = P(X=0) + P(X=1)

P(X≤1) = 0.07776 + 0.2592

P(X≤1) = 0.33696

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≥2) = 1 – 0.33696

P(X≥2) = 0.66304

Required probability = 0.66304

Part c

Q. What is the probability that the entire population will be audited if the true number of improper returns in the population is 5.

We are given N = 50, X = 5, so p = X/N = 5/50 = 0.1

Here, we have to use binomial distribution for finding required probability.

So, we have n = 5, p = 0.1, and we have to find P(X≥2)

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≤1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.1^0*(1 – 0.1)^(5 – 0)

P(X=0) = 0.59049

P(X=1) = 5C1*0.1^1*(1 – 0.1)^(5 – 1)

P(X=1) = 0.32805

P(X≤1) = P(X=0) + P(X=1)

P(X≤1) = 0.59049 + 0.32805

P(X≤1) = 0.91854

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≥2) = 1 – 0.91854

P(X≥2) = 0.08146

Required probability = 0.08146

Part d

Q. What is the probability that the entire population will be audited if the true number of improper returns in the population is 10.

We are given N = 50, X = 10, so p = X/N = 10/50 = 0.2

Here, we have to use binomial distribution for finding required probability.

So, we have n = 5, p = 0.2, and we have to find P(X≥2)

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≤1) = P(X=0) + P(X=1)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.2^0*(1 – 0.2)^(5 – 0)

P(X=0) = 0.32768

P(X=1) = 5C1*0.2^1*(1 – 0.2)^(5 – 1)

P(X=1) = 0.4096

P(X≤1) = P(X=0) + P(X=1)

P(X≤1) = 0.32768 + 0.4096

P(X≤1) = 0.73728

P(X≥2) = 1 – P(X<2) = 1 – P(X≤1)

P(X≥2) = 1 – 0.73728

P(X≥2) = 0.26272

Required probability = 0.26272