Question

A town has 500 real estate agents. The mean value of the properties sold in a year by these agents is $600,000, and the standard deviation is $200,000. A random sample of 100 agents is selected, and the value of the properties they sold in a year is recorded. a. What is the standard error of the sample mean? b. What is the probability that the sample mean exceeds $616,000? c. What is the probability that the sample mean exceeds $592,000? d. What is the probability that the sample mean is between $581,000 and $612,000? Click on the icon to view the standard normal table. ... a. The standard error of the sample mean is (Round to the nearest integer as needed.)

Answer 1

a. The standard error of the sample mean is $20,000. b. The probability that the sample mean exceeds $616,000 is approximately 21.19%. c. The probability that the sample mean exceeds $592,000 is approximately 34.46%. d. The probability that the sample mean is between $581,000 and $612,000 is approximately 72.57%.

Step-by-step explanation:

a. The standard error of the sample mean can be calculated using the formula:

Standard Error = standard deviation / square root of sample size

Given that the standard deviation is $200,000 and the sample size is 100, the standard error is:

Standard Error = $200,000 / √100 = $200,000 / 10 = $20,000

So, the standard error of the sample mean is $20,000 (rounded to the nearest integer).

b. To calculate the probability that the sample mean exceeds $616,000, we need to convert the sample mean into a z-score. The formula for calculating the z-score is:

z = (sample mean - population mean) / standard error

Given that the population mean is $600,000, the sample mean is $616,000, and the standard error is $20,000, the z-score can be calculated as:

z = ($616,000 - $600,000) / $20,000 = $16,000 / $20,000 = 0.8

Using the standard normal table, we can find the probability associated with a z-score of 0.8. The probability that the sample mean exceeds $616,000 is approximately 0.2119 or 21.19%.

c. To calculate the probability that the sample mean exceeds $592,000, we follow the same steps as in part b. The z-score is calculated as:

z = ($592,000 - $600,000) / $20,000 = -$8,000 / $20,000 = -0.4

Looking up the probability associated with a z-score of -0.4 in the standard normal table, we find that the probability is approximately 0.3446 or 34.46%.

d. To calculate the probability that the sample mean is between $581,000 and $612,000, we need to calculate the z-scores for both values and find the difference between the two probabilities. The z-score for $581,000 is:

z = ($581,000 - $600,000) / $20,000 = -$19,000 / $20,000 = -0.95

The z-score for $612,000 is:

z = ($612,000 - $600,000) / $20,000 = $12,000 / $20,000 = 0.6

Using the standard normal table, we can find the probabilities associated with these z-scores. The probability that the sample mean is between $581,000 and $612,000 is the difference between these two probabilities, which is approximately 0.7257 or 72.57%.