Question

Taxes: The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040. Assume that the standard deviation is $4500. The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law. Part 1 of 5 (a) What is the probability that the sample mean tax is less than $7800? Round the answer to at least four decimal places. The probability that the sample mean tax is less than $7800 is . Part 2 of 5 (b) What is the probability that the sample mean tax is between $7400 and $7900? Round the answer to at least four decimal places. The probability that the sample mean tax is between $7400 and $7900 is Part 3 of 5 (c) Find the 10th percentile of the sample mean. Round the answer to at least two decimal places. The 10th percentile of the sample mean is $0. Part 4 of 5 (d) Would it be unusual if the sample mean were less than $7600? Round answer to at least four decimal places. unusual because the probability of the sample mean being less than It (Choose one) $7600 is Part 5 of 5 (e) Do you think it would be unusual for an individual to pay a tax of less than $ 7600? Explain. Assume the variable is normally distributed. Round the answer to at least four decimal places. (Choose one) 7, because the probability that an individual to pays a tax less than $ 7600 is

Answer 1

To find probabilities related to the sample mean tax, we need to standardize the values using the Z-score formula. The probability of the sample mean tax being less than $7800 is approximately 0.0038. The probability of the sample mean tax being between $7400 and $7900 is approximately 0.7995. The 10th percentile of the sample mean tax is $7449.47. It would not be unusual for the sample mean to be less than $7600. However, it would be unusual for an individual to pay a tax of less than $7600.

Step-by-step explanation:

To solve the given problems, we need to use the concept of the sampling distribution of the sample mean. The sampling distribution is the distribution of the sample means for all possible samples of a certain size taken from a population.

(a) Probability that the sample mean tax is less than $7800:

To find this probability, we need to standardize the sample mean using the formula:

Z = (X - µ) / (σ / √n)

Where:

X is the sample mean ($7800)

µ is the population mean ($8040)

σ is the population standard deviation ($4500)

n is the sample size (1000)

Substituting the given values into the formula:

Z = (7800 - 8040) / (4500 / √1000) = -2.67

Using a Z-table or calculator, we can find that the probability of a Z-score less than -2.67 is approximately 0.0038.

The probability that the sample mean tax is less than $7800 is approximately 0.0038.

(b) Probability that the sample mean tax is between $7400 and $7900:

To find this probability, we need to find the Z-scores for $7400 and $7900:

Z1 = ($7400 - 8040) / (4500 / √1000) = -4.09

Z2 = ($7900 - 8040) / (4500 / √1000) = -0.79

Using a Z-table or calculator, we can find the probability of a Z-score between -4.09 and -0.79 is approximately 0.7995.

The probability that the sample mean tax is between $7400 and $7900 is approximately 0.7995.

(c) 10th percentile of the sample mean:

To find the 10th percentile, we need to find the Z-score corresponding to the cumulative probability of 0.10:

Z = invNorm(0.10) = -1.28

Substituting the Z-score into the formula:

X = µ + (Z * (σ / √n))

X = 8040 + (-1.28 * (4500 / √1000)) = $7449.47

The 10th percentile of the sample mean is $7449.47.

(d) Unusualness of the sample mean less than $7600:

To determine if it would be unusual for the sample mean to be less than $7600, we need to calculate the Z-score for $7600 using the same formula as in part (a):

Z = (7600 - 8040) / (4500 / √1000) = -1.44

The probability of a Z-score less than -1.44 is approximately 0.0749.

Therefore, it would not be unusual for the sample mean to be less than $7600.

(e) Unusualness of an individual paying a tax of less than $7600:

To determine if it would be unusual for an individual to pay a tax of less than $7600, we need to compare the individual tax payment to the population mean and standard deviation.

Since we do not have information about the population distribution, we cannot determine the exact probability.

However, based on the information provided, the probability would be less than 0.0749, the probability calculated in part (d) for the sample mean.

Therefore, it would be unusual for an individual to pay a tax of less than $7600.