Question

For a total population of a large southern city, where the mean family income is $34,000 with a standard deviation of $5,000, imagine taking a sample of 200 residents. What is the probability that the sample mean exceeds $37,000?

Answer 1

The probability that the sample mean exceeds $37,000, calculated using the Central Limit Theorem and standard error, is nearly zero because the z-score is extremely high, which corresponds to a negligible area under the normal distribution curve.

Step-by-step explanation:

To determine the probability that the sample mean exceeds $37,000, we can use the Central Limit Theorem (CLT). According to the CLT, the sampling distribution of the sample mean will be approximately normally distributed, because the sample size is large (n=200). The standard error (SE) of the sampling distribution is the standard deviation divided by the square root of the sample size.

First, calculate the standard error (SE):

SE = $5,000 / √200

SE = $5,000 / 14.14

SE = $353.55 (approximately)

Next, we need to find the z-score for a sample mean of $37,000:

Z = ($37,000 - $34,000) / SE

Z = $3,000 / $353.55

Z = 8.49 (approximately)

Using a z-table or a calculator, you'd find that the probability of a z-score of 8.49 or higher is extremely low—essentially, zero. So, in practical terms, the probability that the sample mean exceeds $37,000 is nearly zero.