Answer:
384.67
Explanation:
The margin of error for the 99% confidence interval for the true mean monthly salary of all former students can be calculated using the formula:
Margin of error = z * (standard deviation / sqrt(sample size))
where z is the z-score associated with the desired confidence level (in this case, 99%), standard deviation is the population standard deviation (given as $550), and sqrt(sample size) is the square root of the sample size (15).
The z-score for a 99% confidence level can be found using a standard normal distribution table or calculator, and it is approximately 2.576.
Plugging in the values, we get:
Margin of error = 2.576 * ($550 / sqrt(15))
≈ $384.67
Therefore, the margin of error of the 99% confidence interval for the true mean monthly salary of all former students is approximately $384.67.