diferira7c

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. $X\sim B$(__,__)
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.

usaho4w

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, $n=20,p=0.02,q=0.98\left(=1-0.02\right)$.
Therefore,
$X\sim B\left(20,0.02\right)$
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,
$E\left(x\right)=np$
$=20×0.02$
$=0.40$
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

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: c) Give the distribution of X. Solution: or we could use the mean value calculated below and approximate with the Poisson distribution $X\sim P\left(0.4\right)$ d) How many audits are expected in a 20 year period? Solution: ${\left(0.98\right)}^{20}$ or binom or poisson 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
Here X follows the Binomial distribution with the parameters n=20 and p=0.02

a) The number of audits are expected in a 20-year period is
Mean, $\mu =np=20×0.02=0.4$
b) Standard deviation,
$SD=\sqrt{np\left(1-p\right)}=\sqrt{20×0.02\left(1-0.02\right)}=0.6261$
The standard deviation is 0.6261
c) The probability function of Binomial Distribution is given by
$P\left(x=x\right)=\left(\left(n\right),\left(x\right)\right){p}^{x}\left(1-p{\right)}^{n-x}$
So probability of being audited troice is
$P\left(x=2\right)=\left(\left(20\right),\left(2\right)\right)\left(0.02{\right)}^{2}\left(1-0.02{\right)}^{20-2}$
$=0.0528$
$P\left(x=2\right)=0.0528$
d) $P\left(x>3\right)=1-p\left(x\le 3\right)$
$=1-\sum _{x=0}^{3}\left(\left(20\right),\left(x\right)\right)\left(0.02{\right)}^{x}\left(1-0.02{\right)}^{20-x}$
$=1-0.9994$
$=0.0006$
$P\left(x>3\right)=0.0006$

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