Question

In: Statistics and Probability

According to the Internal Revenue Service, the average length of time for an individual to complete...

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.41 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

A) In words, define the random variable X.

the length of time, in minutes, for an individual to complete IRS Form 1040the length of time, in hours, for an individual to complete IRS Form 1040    the number of individuals who complete IRS Form 1040the number of taxpayers sampled

B) In words, define the random variable

X.

the average length of time, in hours, for a sample of 36 taxpayers to complete IRS Form 1040the average income of a random sample of taxpayers    the average length of time, in minutes, for a sample of 36 taxpayers to complete IRS Form 1040the average length of time, in hours, for a sample of 100 taxpayers to complete IRS Form 1040

c)

Give the distribution of

X.

(Round your answers to two decimal places.)

X ~

D) Find the probability that the 36 taxpayers took an average of more than 12 hours to finish their Form 1040s. (Round your answer to four decimal places.)

E) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in a complete sentence.

No, because the probability is very close to 1.Yes, because the probability is very close to 0.   

Solutions

Expert Solution

a) option-B) the length of time, in hours, for an individual to complete IRS Form 1040

b) Option-A) the average length of time, in hours, for a sample of 36 taxpayers to complete IRS Form 1040

c) Mean of sampling distribution = µ = 10.41

Standard deviation of sampling distribution = sd / sqrt(n) = 2 / sqrt(36) = 0.33

X ~ N (10.41, 0.33)

d)

                            

                             = P(Z > 4.77)

                             = 0

e) Option-B) Yes, because the probability is very close to 0.  

orchestra answered 1 year ago
Next > < Previous

Related Solutions

According to the Internal Revenue Service, the average income tax refund for the 2011 tax year...
According to the Internal Revenue Service, the average income tax refund for the 2011 tax year was $2,913. Assume the refund per person follows the normal probability distribution with a standard deviation of $950. a. What is the probability that a randomly selected tax return refund from the 2011 tax year will be 1. more than $ 2000 2. between $1,600 and $2,500? 3. between $3,200 and $4,000? b. Confirm the answers to part a using Excel or PH Stat.
According to the Internal Revenue Service, 80% of all tax returns lead to a refund. A...
According to the Internal Revenue Service, 80% of all tax returns lead to a refund. A random sample of 100 tax returns is taken, and it was found that 83% of the tax returns in the sample require a refund. Using the sampling distribution of the proportion, calculate the probability that the sample proportion exceeds 85% in the sample of 100 tax returns? a. 0.1056 b. 0.2266 c. 0.2972 d. 0.0916 30. In a small town, there are 3,000 registered...
According to the Internal Revenue Service (IRS), the chances of your tax return being audited are...
According to the Internal Revenue Service (IRS), the chances of your tax return being audited are about 6 in 1,000 if your income is less than $50,000; 10 in 1,000 if your income is between $50,000 and $99,999; and 49 in 1,000 if your income is $100,000 or more (Statistical Abstract of the United States: 1995). If two taxpayers with incomes under $50,000 are randomly selected and two with incomes more than $100,000 are randomly selected, what is the probability...
Pay your taxes: According to the Internal Revenue Service, the proportion of federal tax returns for...
Pay your taxes: According to the Internal Revenue Service, the proportion of federal tax returns for which no tax was paid was =p0.326. As part of a tax audit, tax officials draw a simple sample of =n140 tax returns. Use Cumulative Normal Distribution Table as needed. Round your answers to at least four decimal places if necessary. Part 1 of 4 (a)What is the probability that the sample proportion of tax returns for which no tax was paid is less...
The Internal Revenue Service sampled 20 tax returns and found that the average tax refund was...
The Internal Revenue Service sampled 20 tax returns and found that the average tax refund was $425.39 with a standard deviation of $107.10. Construct a 99% confidence interval for the mean value of all tax refunds. Why do we have a use a t-confidence interval instead of a z-confidence interval for this problem?
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for...
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for taxpayers. One explanation of this figure is that taxpayers would rather have the government keep back too much money during the year than to owe it money at the end of the year. Suppose the average amount of tax at the end of a year is a refund of $1,332, with a standard deviation of $725. Assume that amounts owed or due on tax...
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for...
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for taxpayers. One explanation of this figure is that taxpayers would rather have the government keep back too much money during the year than to owe it money at the end of the year. Suppose the average amount of tax at the end of a year is a refund of $1,332, with a standard deviation of $725. Assume that amounts owed or due on tax...
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for...
According to the Internal Revenue Service, income tax returns one year averaged $1,332 in refunds for taxpayers. One explanation of this figure is that taxpayers would rather have the government keep back too much money during the year than to owe it money at the end of the year. Suppose the average amount of tax at the end of a year is a refund of $1,332, with a standard deviation of $725. Assume that amounts owed or due on tax...
The length of time for one individual to be served at a cafeteria is a random...
The length of time for one individual to be served at a cafeteria is a random variable having an exponential distribution with a mean of 5 minutes. What is the probability that a person is served in less than 2 minutes on at least 5 of the next 7 days?
The length of time for an individual to wait at a lunch counter is a random...
The length of time for an individual to wait at a lunch counter is a random variable whose density function is f(x) = (1/4)e^(-x/4) for x > 0 and = 0 otherwise a) Find the mean and the variance b) Find the probability that the random variable is within three standard deviations of the mean and compare Chebyshev's Theorem.
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT