A university placement office conducted a survey of 100 engineers who had graduated from a local university. For these engineers, the mean salary was computed to be $72,000 with a standard deviation of $8,000. Which of the following best characterizes the percentage of these engineers who earn either more than $96,000 or less than $48,000?
A) approximately 2.3%
B) at least 5.6% (1/18 of the engineers)
C) at most 5.6% (1/18 of the engineers)
D) at most 11.1% (1/9 of the engineers)

Question

Quizplus · Accepted Answer

To find the percentage of engineers who earn either more than $96,000 or less than $48,000, we need to find the percentage of engineers who earn more than $96,000 and the percentage of engineers who earn less than $48,000 separately and then add them up.

To find the percentage of engineers who earn more than $96,000, we need to standardize this value by finding the z-score:

z = (x - μ) / σ
z = (96000 - 72000) / 8000
z = 3

Using a standard normal table or calculator, we find that the percentage of engineers who earn more than $96,000 is approximately 0.27%.

To find the percentage of engineers who earn less than $48,000, we again need to standardize this value:

z = (x - μ) / σ
z = (48000 - 72000) / 8000
z = -3

Using a standard normal table or calculator, we find that the percentage of engineers who earn less than $48,000 is approximately 0.27%.

Adding the two percentages together, we get a total of:

0.27% + 0.27% = 0.54%

To convert this decimal to a percentage, we multiply by 100:

0.54% = 0.54 x 100 = 5.4%

Therefore, the percentage of engineers who earn either more than $96,000 or less than $48,000 is at most 11.1%, which is answer choice D (since all other answer choices are less than 5.4%).

A University Placement Office Conducted a Survey of 100 Engineers