Question

The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 20-year period. Assume each year is independent.

a. In words, define the random variable X.

b. List the values that X may take on.

c. Give the distribution of X. X ~ _____(_____,_____)

d. How many audits are expected in a 20-year period?

e. Find the probability that a person is not audited at all.

f. Find the probability that a person is audited more than twice

Short answer

a. The random variable X is the number of audit in a $20$year period.

b. The values that X may take on are $0,1,2,........,20.$

c. The distribution of X is $X~B(20,0.02)$

d. $0.4$audits are expected in a $20$-year period

e. The probability that a person is not audited at all is $0.6676$

f. The probability that a person is audited more than twice is$0.0071$

Step-by-step solution

Content Introduction

The binomial distribution determines the probability of looking at a specific quantity of a success results in a specific quantity of trials.

Part (a) Step 1: Explanation

We are given,

The chance of an IRS audit for a tax return with over $\$25,000$in income is about $2\%$per year with a $20$-year period.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case the random variable X is the number of audits in$20$year period.

Part (b) Step 1: Explanation

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand $20$then X is

$X=0,1,2,.......,20.$

Part (c) Step 1: Explanation

The random variable is distributed by the data provided X will keep the track of online offerings. The probability distribution of binomial distribution has two parameters $n=numberoftrialsandp=probabilityofsuccess$.

The binomial distribution is of the form: $X~B(n,p)$

Therefore, The distribution of X is

$X~B(20,0.02)$

Part (d) Step 1: Explanation

The expected binomial distribution is calculated as:

$\mu =np$

$\mu$is the number of audits expected in $20$year period,

$n=20$

$p=0.02$

Therefore, the number of audits expected in$20$ year period is:

$$\begin{gathered} \mu =np \\ \mu =20\times 0.02 \\ \mu =0.4 \end{gathered}$$

Part (e) Step 1: Explanation

The probability that a person is audited is $0.02$

Therefore, the probability that a person is not audited at all will be

$$\begin{gathered} =1-0.02 \\ =0.98 \end{gathered}$$

using Binompdf on TI calculator we can calculate the binomial distribution.

role="math" localid="1649089492311" $Binompdf(n,p,c)$where, role="math" localid="1649089497672" $n$is number of trials, $p$is probability of success, $c$is the probability of c success, for some number c.

The values are: $n=20,p=0.02,c=0$

$Binompdf(20,0.02,0)=0.6676$

Part (f) Step 1: Explanation

The probability that a person is audited more than twice is$0.0071$