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

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. $$\displaystyle{X}\sim{B}$$(__,__)
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.

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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, $$\displaystyle{n}={20},{p}={0.02},{q}={0.98}{\left(={1}-{0.02}\right)}$$.
Therefore,
$$\displaystyle{X}\sim{B}{\left({20},{0.02}\right)}$$
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,
$$\displaystyle{E}{\left({x}\right)}={n}{p}$$
$$\displaystyle={20}\times{0.02}$$
$$\displaystyle={0.40}$$
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.
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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: $$\displaystyle{S}={\left\lbrace{0},\ {1},\ {2},\cdots,\ {19},\ {20}\right\rbrace}$$ c) Give the distribution of X. Solution: $$\displaystyle{X}\sim{B}{\left({20},\ {0.02}\right)}$$ or we could use the mean value calculated below and approximate with the Poisson distribution $$\displaystyle{X}\sim{P}{\left({0.4}\right)}$$ d) How many audits are expected in a 20 year period? Solution: $$\displaystyle{\left({0.98}\right)}^{{{20}}}$$ or binom $$pdf(20,\ 0.02,\ 0)=0.668$$ or poisson $$pdf(0.4,\ 0)=0.670$$ karton Answered 2022-01-04 Author has 9103 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 $$0,\ 1,\ 2,\ \cdots,\ 19,\ 20$$
Here X follows the Binomial distribution with the parameters n=20 and p=0.02
$$X\sim\text{Binomial}(n=20,\ p=0.02)$$
a) The number of audits are expected in a 20-year period is
Mean, $$\mu=np=20\times0.02=0.4$$
b) Standard deviation,
$$SD=\sqrt{np(1-p)}=\sqrt{20\times0.02(1-0.02)}=0.6261$$
The standard deviation is 0.6261
c) The probability function of Binomial Distribution is given by
$$P(x=x)=((n),(x))p^{x}(1-p)^{n-x}$$
So probability of being audited troice is
$$P(x=2)=((20),(2))(0.02)^{2}(1-0.02)^{20-2}$$
$$=0.0528$$
$$P(x=2)=0.0528$$
d) $$P(x>3)=1-p(x\le3)$$
$$=1-\sum_{x=0}^{3}((20),(x))(0.02)^{x}(1-0.02)^{20-x}$$
$$=1-0.9994$$
$$=0.0006$$
$$P(x>3)=0.0006$$