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

diferira7c 2021-12-27 Answered
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.

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

usaho4w
Answered 2021-12-28 Author has 3084 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, \(\displaystyle{n}={20},{p}={0.02},{q}={0.98}{\left(={1}-{0.02}\right)}\).
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.
Not exactly what you’re looking for?
Ask My Question
0
 
braodagxj
Answered 2021-12-29 Author has 1288 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:
\(\displaystyle{S}={\left\lbrace{0},\ {1},\ {2},\cdots,\ {19},\ {20}\right\rbrace}\)
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 \(pdf(0.4,\ 0)=0.670\)

0
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) \(P(x>3)=1-p(x\le3)\)
\(=1-\sum_{x=0}^{3}((20),(x))(0.02)^{x}(1-0.02)^{20-x}\)
\(=1-0.9994\)
\(=0.0006\)
\(P(x>3)=0.0006\)

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-07-02
An investor plans to put $50,000 in one of four investments. The return on each investment depends on whether next year’s economy is strong or weak. The following table summarizes the possible payoffs, in dollars, for the four investments.
Certificate of deposit
Office complex
Land speculation
Technical school
amp; Strong amp;6,000 amp;15,000 amp;33,000 amp;5,500
amp; Weak amp;6,000 amp;5,000 amp;−17,000 amp;10,000
Let V, W, X, and Y denote the payoffs for the certificate of deposit, office complex, land speculation, and technical school, respectively. Then V, W, X, and Y are random variables. Assume that next year’s economy has a 40% chance of being strong and a 60% chance of being weak. a. Find the probability distribution of each random variable V, W, X, and Y. b. Determine the expected value of each random variable. c. Which investment has the best expected payoff? the worst? d. Which investment would you select? Explain.
asked 2021-09-27
If a seed is planted, it has a 90% chance of growing into a healthy plant.
If 7 seeds are planted, what is the probability that exactly 2 don't grow?
asked 2021-09-15
If a seed is planted, it has a 80% chance of growing into a healthy plant.
If 7 seeds are planted, what is the probability that exactly 3 don't grow?
asked 2021-08-21
a) The firm cost of capital before taxes.
Given: The borrowed money is $125000 at \(\displaystyle{8}\%\) rate.
The investment by the partners is $75000.
The required rate of return by the partner is \(\displaystyle{12}\%\).
b) The firm cost of capital after tax.
asked 2021-09-28
Whether it is surprising for more than 4 elk in the sample to survive to adulthood and calculate an appropriate probability.
Given information:
Number of trials, \(\displaystyle{n}={7}\)
Probability of success, \(\displaystyle{p}={44}\%={0.44}\)
asked 2021-09-15
a) Whether it is unusual to have more than five successes or not.
Given: The number of successes lying outside the range \(\displaystyle\mu-{2.5}\sigma\ \to\ \mu+{2.5}\sigma\) are considered as unusual. The success probability in a single trail is 0.2 and the number of trials is 10.
b) Whether one would be likely to get more than half of the questions correct or not.
Given: A multiple-choice exam consisting of 10 questions with 5 possible responses for each questions. Consider the explanation in part (a), it is unusual to get more than 5 successes when \(\displaystyle{n}={10}\ {\quad\text{and}\quad}\ {p}={0.2}\).
asked 2021-09-06
Rivers Casino did an audit of all of their blackjack tables and found that the amount of money collected per night by a certain table has a mean of $6,000, a standard deviation of $200, and is approximately normally distributed.
What is the probability that a table will collect between $5800 and $6200 tomorrow night?

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...