An industrial psychologist observed 8 drill-press operators for one working day. She recorded the number of times each operator pressed the "faster" button instead of the "stop" button to determine whether the design of the control panel was contributing to the high rate of accidents in the plant. The resulting scores were as follows: 5, 2, 8, 2, 3, 2, 4, 12 The variance for this distribution is _________. Assume sample data.
A) 33.75
B) 11.19
C) 12.79
D) 15.75

Question

Quizplus · Accepted Answer

To calculate the variance, we need to first find the mean:
Mean = (5 + 2 + 8 + 2 + 3 + 2 + 4 + 12)/8 = 4.25

Next, we find the deviation from the mean for each score:
(5-4.25), (2-4.25), (8-4.25), (2-4.25), (3-4.25), (2-4.25), (4-4.25), (12-4.25)
= 0.75, -2.25, 3.75, -2.25, -1.25, -2.25, -0.25, 7.75

Then, we square each deviation:
0.56, 5.06, 14.06, 5.06, 1.56, 5.06, 0.06, 60.06

To find the variance, we sum up these squared deviations and divide by n-1:
(0.56 + 5.06 + 14.06 + 5.06 + 1.56 + 5.06 + 0.06 + 60.06)/7
= 12.79

Therefore, the variance for this distribution is 12.79.

An Industrial Psychologist Observed 8 Drill-Press Operators for One Working