Safari mathxl.com POX-1) EXAM Week 2: Quiz Do Take a Test SOLUTION: A EA FAMILY

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Safari mathxl.com POX-1) EXAM Week 2: Quiz Do Take a Test SOLUTION: A EA FAMILY Qianv Applied Managerial Statistics Quiz: Week 2 Quiz 7018(8 complete) This Question: 8 pts The chances of a tax return being audited are about 22 in 1,000 if an income is less than $100,000 and 32 in 1,000 f an income is $100,000 or more. Complete parts a through e a. What is the probability that a taxpayer with income less than $100,000 will be audited? With income of $100,000 or more? Ptaxpayer with income less than $100,000 is audited)- 0 022 Type an integer or a decimal.) What is the probability that a taxpayer with income of $100,000 or more will be audited? Ptaxpayer with income of $100,000 or higher is audited) 0.032 (Type an integer or a decimal.) b. If four taxpayers with incomes under $100,000 are randomly selected, what is the probablity that exactly one will be audited? That more than one will be audited? Ptx= 1)= (Round to four decinal places as needed.) What is the probability that more than one ill be audited? P(x 1)-(Round to four decirnal places as needed.) c. Repeat part b assuming that four taxpayers with incomes of $100,000 or more are randomly selected Click to select your answerls).

Explanation / Answer

Solution:-

a) The probability that a tax payer with income less than $100,000 is audited is 0.022.

The probability that a tax payer with income more than $100,000 is audited is 0.032.

b) The probability that exactly one will be audited is 0.0823

The probability that more than one will be audited is 0.00282.

p = 0.022, n = 4

x = 1

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 1) = 0.0823

P(x > 1) = 0.00282

c)

The probability that exactly one will be audited is 0.1161

The probability that more than one will be audited is 0.0059.

p = 0.032, n = 4

x = 1

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 1) = 0.1161

P(x > 1) = 0.0059

d) P(none of the taxpayer will be audted) = 0.8962

p = 0.022, n = 2

x = 0

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 0) = 0.9565

p = 0.032, n = 2

x = 0

By applying binomial distribution

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 0) = 0.9370

P(none of the taxpayer will be audted) = 0.9565 × 0.9370

P(none of the taxpayer will be audted) = 0.8962

e) (C)

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