The chance of an IRS audit for a tax return reporting more t

diferira7c

diferira7c

Answered question

2021-12-27

The chance of an IRS audit for a tax return reporting more than 25,000 in income is about 2% per year
a. give the distribution of X. XB(__,__)
b. how many audits are expected in a 20 year period?
c. find the probability that a person is not audited at all.
d. find the probability that a person is all the dead more than twice.

Answer & Explanation

usaho4w

usaho4w

Beginner2021-12-28Added 39 answers

Step 1
Since you have posted a question with multiple sub-parts, we will solve first three sub- parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved
(a) Determine the distribution of X.
The distribution of X is determined below as follows:
Let X denotes the number of audits a person with the income more than 25,000 which follows binomial distribution with the probability of success 0.02 and the number of years selected is 20.
That is, n=20,p=0.02,q=0.98(=10.02).
Therefore,
XB(20,0.02)
Step 2
(b) Obtain the expected number of audits in a 20 year period.
The expected number of audits in a 20 year period is obtained below as follows:
The required value is,
E(x)=np
=20×0.02
=0.40
The expected number of audits in a 20 year period is 0.40.
Step 3
(c) Obtain the probability that a person is not audited at all.
The probability that a person is not audited at all is obtained below as follows:
The required probability
Use Excel to obtain the probability value for x equals 0.
Follow the instruction to obtain the P-value:
1. Open EXCEL
2.Go to Formula bar.
3. In formula bar enter the function as“=BINOMDIST”
4. Enter the number of success as 0.
5. Enter the Trails as 20
6. Enter the probability as 0.02
7. Enter the cumulative as False.
8. Click enter
EXCEL output:
From the Excel output, the P-value is 0.6676
The probability that a person is not audited at all is 0.6676.
braodagxj

braodagxj

Beginner2021-12-29Added 38 answers

Step 1
a) In words, dene the Random Variable X.
Solution:
X = number of audits a person with income over $25,000 per annum has in a 20 year period.
b) List the values that X may take on.
Solution:
S={0, 1, 2,, 19, 20}
c) Give the distribution of X.
Solution:
XB(20, 0.02)
or we could use the mean value calculated below and approximate with the Poisson distribution
XP(0.4)
d) How many audits are expected in a 20 year period?
Solution: (0.98)20 or binom pdf(20, 0.02, 0)=0.668 or poisson pdf(0.4, 0)=0.670

karton

karton

Expert2022-01-04Added 613 answers

Step 1
Given:
n=20
p=0.02
X=the number of audits a person with income over $25,000 per annum has in a 20 year period and the values of X are 0, 1, 2, , 19, 20
Here X follows the Binomial distribution with the parameters n=20 and p=0.02
XBinomial(n=20, p=0.02)
a) The number of audits are expected in a 20-year period is
Mean, μ=np=20×0.02=0.4
b) Standard deviation,
SD=np(1p)=20×0.02(10.02)=0.6261
The standard deviation is 0.6261
c) The probability function of Binomial Distribution is given by
P(x=x)=((n),(x))px(1p)nx
So probability of being audited troice is
P(x=2)=((20),(2))(0.02)2(10.02)202
=0.0528
P(x=2)=0.0528
d) P(x>3)=1p(x3)
=1x=03((20),(x))(0.02)x(10.02)20x
=10.9994
=0.0006
P(x>3)=0.0006

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