To solve these problems, we will use the concept of the sampling distribution of the sample mean. Given the population mean (\(\mu\)) is $8040, and the standard deviation (\(\sigma\)) is $4500, we can calculate the standard deviation of the sample mean (standard error) using the formula:
\[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{4500}{\sqrt{1000}} \approx 142.94 \]
Let’s proceed with each part:
(a) What is the probability that the sample mean tax is less than $8000?
We need to find the z-score first:
\[ z = \frac{\bar{x} - \mu}{\text{SE}} = \frac{8000 - 8040}{142.94} \approx \frac{-40}{142.94} \approx -0.279 \]
Now we will use the z-table or standard normal distribution to find the probability:
\[ P(Z < -0.279) \approx 0.3897 \]
Answer: The probability that the sample mean tax is less than $8000 is approximately 0.3897.
(b) What is the probability that the sample mean tax is between 7400 and 8000?
We will find the z-scores for both amounts:
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For $7400: \[ z = \frac{7400 - 8040}{142.94} \approx \frac{-640}{142.94} \approx -4.48 \]
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For $8000 (already calculated): \[ z \approx -0.279 \]
Now we find the corresponding probabilities:
- \(P(Z < -4.48) \approx 0\) (very close to zero)
- \(P(Z < -0.279) \approx 0.3897\)
Now subtract those probabilities to get the area between:
\[ P(7400 < \bar{x} < 8000) = P(Z < -0.279) - P(Z < -4.48) \approx 0.3897 - 0 \approx 0.3897 \]
Answer: The probability that the sample mean tax is between $7400 and $8000 is approximately 0.3897.
(c) Find the 60th percentile of the sample mean.
The z-score for the 60th percentile is approximately 0.253 (using z-tables).
Now, we calculate the corresponding sample mean:
\[ \bar{x} = \mu + z \cdot \text{SE} = 8040 + 0.253 \cdot 142.94 \approx 8040 + 36.179 \approx 8076.18 \]
Answer: The 60th percentile of the sample mean is approximately 8076.18.
(d) Would it be unusual if the sample mean were less than 7600?
Calculate the z-score for $7600:
\[ z = \frac{7600 - 8040}{142.94} \approx \frac{-440}{142.94} \approx -3.08 \]
Now, find the corresponding probability:
\[ P(Z < -3.08) \approx 0.001 \]
Since this probability is very small (less than 0.05), it can be considered unusual.
So, the answer is:
**It would be unusual because the probability is approximately 0.0010.
(e) Do you think it would be unusual for an individual to pay a tax of less than 7600?
We use the standard deviation of the population for individuals, which is $4500. We calculate the z-score:
\[ z = \frac{7600 - 8040}{4500} \approx \frac{-440}{4500} \approx -0.0978 \]
Now, find the corresponding probability:
\[ P(Z < -0.0978) \approx 0.4615 \]
This probability is not considered unusual (close to 0.5).
The answer is:
No, because the probability is approximately 0.4615.