Question

Suppose that the operations manager of a nose mask packaging delivery service is contemplating the purchase of a new fleet of trucks during this COVID-19 period. When packages are efficiently stored in the trucks in preparation for delivery, two major constraints have to be considered. The weight in pounds and volume in cubic feet for each item. Now suppose that in a sample of 200 packages the average weight is 26 pounds with a standard deviation of 3.9 pounds. In addition suppose that the average volume for each of these packages is 8.8 cubic feet with standard deviation of 2.2 cubic feet. How can we compare the variation of the weight and volume?

Answer 1

Answer:

Step-by-step explanation:

Coefficient of variation, CV = standard deviation/mean

$CV= \frac{\sigma}{\mu}$

Therefore, CV for weight = 3.9/26 = 0.15

CV for volume = 2.2/8.8 = 0.25

The coefficient of variation is higher for volume. So, volume is having higher variability