The Internal Revenue Service reports that the mean federal income tax paid in the year 2010 was $8040 . Assume that the standard deviation is $4800. The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law.

Question

(a) What is the probability that the sample mean tax is less than $8100? Round the answer to at least four decimal places.

(b) What is the probability that the sample mean tax is between $7400 and $8000? Round the answer to at least four decimal places.

(c) Find the 40th percentile of the sample mean. Round the answer to at least two decimal places.

Answer 1

To answer the questions, we will use the properties of the sampling distribution of the mean. When the sample size is large (n > 30), the sampling distribution of the sample mean can be approximated by a normal distribution.

Given that:

• Population Mean (\(\mu\)) = $8040
• Population Standard Deviation (\(\sigma\)) = $4800
• Sample Size (n) = 1000

The standard error of the mean (SEM) is given by: \[ SEM = \frac{\sigma}{\sqrt{n}} = \frac{4800}{\sqrt{1000}} \approx \frac{4800}{31.6228} \approx 151.37 \]

(a) Probability that the sample mean tax is less than $8100

We need to find: \[ P(\bar{X} < 8100) \]

Using the z-score formula: \[ z = \frac{\bar{X} - \mu}{SEM} = \frac{8100 - 8040}{151.37} \approx \frac{60}{151.37} \approx 0.395 \]

Now, we will look up the z-score of 0.395 in the standard normal distribution table or calculate using a calculator: \[ P(Z < 0.395) \approx 0.6532 \]

Thus, the probability that the sample mean tax is less than $8100: \[ \boxed{0.6532} \]

(b) Probability that the sample mean tax is between $7400 and $8000

We need to find: \[ P(7400 < \bar{X} < 8000) \]

We calculate the z-scores for 7400 and 8000.

For $7400: \[ z = \frac{7400 - 8040}{151.37} \approx \frac{-640}{151.37} \approx -4.23 \]

For $8000: \[ z = \frac{8000 - 8040}{151.37} \approx \frac{-40}{151.37} \approx -0.264 \]

Now we find: \[ P(Z < -4.23) \approx 0 \quad \text{(very small probability)} \] \[ P(Z < -0.264) \approx 0.3944 \]

Thus, we find: \[ P(7400 < \bar{X} < 8000) = P(Z < -0.264) - P(Z < -4.23) \approx 0.3944 - 0 \approx 0.3944 \]

So the probability that the sample mean tax is between $7400 and $8000: \[ \boxed{0.3944} \]

(c) Find the 40th percentile of the sample mean

To find the 40th percentile, we need the z-score corresponding to the 40th percentile. From the z-table or using statistical software, we find: \[ z \approx -0.2533 \quad \text{(40th percentile)} \]

Now we convert this z-score back to the sample mean: \[ \bar{X} = \mu + z \cdot SEM \] \[ \bar{X} = 8040 + (-0.2533) \cdot 151.37 \approx 8040 - 38.34 \approx 8001.66 \]

Thus, the 40th percentile of the sample mean: \[ \boxed{8001.66} \]

Answer 2

d) Would it be unusual if the sample mean were less than $7700? Round answer to at least four decimal places.
It [is or isnt?] unusual because the probability of the sample mean being less is [?]

E) Do you think it would be unusual for an individual to pay a tax of less than $7700? Explain. Assume the variable is normally distributed. Round the answer to at least four decimal places.

[yes or no?] because the probability to pay less than $7700 is [?]

Answer 3

To determine whether it would be unusual for the sample mean to be less than $7700 and for an individual to pay less than $7700, we will calculate the probabilities according to the normal distribution.

(d) Would it be unusual if the sample mean were less than $7700?

We need to find the probability that the sample mean is less than $7700: \[ P(\bar{X} < 7700) \]

Calculating the z-score: \[ z = \frac{\bar{X} - \mu}{SEM} = \frac{7700 - 8040}{151.37} \approx \frac{-340}{151.37} \approx -2.245 \]

Now we find the probability corresponding to this z-score: \[ P(Z < -2.245) \]

Using a z-table or calculator, we find: \[ P(Z < -2.245) \approx 0.0123 \]

Thus, it would be unusual if the sample mean were less than $7700 because the probability of the sample mean being less than $7700 is approximately \(0.0123\).

So, the final answer is:

• It is unusual because the probability of the sample mean being less is 0.0123.

(e) Do you think it would be unusual for an individual to pay a tax of less than $7700?

To find this probability, we need to calculate \( P(X < 7700) \) using the population distribution.

Calculating the z-score for an individual: \[ z = \frac{X - \mu}{\sigma} = \frac{7700 - 8040}{4800} \approx \frac{-340}{4800} \approx -0.0708 \]

Now we find the probability corresponding to this z-score: \[ P(Z < -0.0708) \]

Using a z-table or calculator, we find: \[ P(Z < -0.0708) \approx 0.4713 \]

Thus, it would not be unusual for an individual to pay a tax of less than $7700 because the probability to pay less than $7700 is approximately \(0.4713\).

So, the final answer is:

• No, because the probability to pay less than $7700 is 0.4713.