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 \( \$ 4700 \). The IRS plans to draw a sample of 1000 tax returns to study the effect of a new tax law. Part: \( 0 / 5 \) Part 1 of 5 (a) What is the probability that the sample mean tax is less than \( \$ 8000 \) ? Round the answer to at least four decimal places. The probability that the sample mean tax is less than \( \$ 8000 \) is

To find the probability that the sample mean tax is less than \( \$8000 \), we can use the Central Limit Theorem. Since the sample size is 1000, the sampling distribution of the sample mean can be approximated as normal. The mean of the sampling distribution is \( \$8040 \) and the standard deviation (also called the standard error) is calculated by dividing the population standard deviation by the square root of the sample size: \[ \text{Standard Error} = \frac{\sigma}{\sqrt{n}} = \frac{4700}{\sqrt{1000}} \approx 148.33 \] Next, we need to calculate the z-score for the sample mean of \( \$8000 \): \[ z = \frac{\bar{x} - \mu}{\text{Standard Error}} = \frac{8000 - 8040}{148.33} \approx -0.27 \] Now we can look up this z-score in the standard normal distribution table or use a calculator to find the probability. A z-score of \( -0.27 \) corresponds to an area of approximately \( 0.3936 \). Thus, the probability that the sample mean tax is less than \( \$8000 \) is \( 0.3936 \) when rounded to four decimal places: **The probability that the sample mean tax is less than \( \$8000 \) is 0.3936.**