Final answer:
To find the probability that the mean of a random sample of 3600 household incomes would be within $200 of the population mean, we can use the Central Limit Theorem. The probability is approximately 0.5336.
Step-by-step explanation:
To find the probability that the mean of a random sample of 3600 household incomes would be within $200 of the population mean, we can use the Central Limit Theorem. According to the theorem, the distribution of sample means follows a normal distribution with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
Let's calculate the standard deviation of the sample mean:
Standard deviation of the sample mean = Standard deviation of the population / Square root of the sample size
= $16,500 / √3600
= $16,500 / 60
= $275
Since we want the mean of the random sample to be within $200 of the population mean, we have a margin of error of $200. To find the z-score corresponding to this margin of error, we can use the formula:
Z-score = (Margin of error) / (Standard deviation of the sample mean)
= $200 / $275
= 0.7273
To find the probability corresponding to this z-score, we can use a standard normal distribution table or a calculator. It gives us a probability of approximately 0.2668. However, since we want the probability that the mean is within $200 from both sides, we need to double this probability:
Probability = 2 * 0.2668
= 0.5336
Therefore, the probability that the mean of a random sample of 3600 household incomes would be within $200 of the population mean is approximately 0.5336, which is closest to option (B) 0.5329.