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Household incomes in a certain region have mean $58,300 and standard deviation $16,500. The probability that the mean of a random sample of 3600 household incomes would be within $200 of the population mean is about: (A)0.2263

(B)0.5329
(C)0.2553
(D) 0.2481
(D) 0.5599

User Yvoyer
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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.

User Leonardo Cardoso
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