You know that the average college student eats 0.75 pounds of food at lunch. If the standard deviation is 0.2 pounds of food, then what is the total amount of food that a cafeteria should have on hand to be 90 percent confident that it will not run out of food when feeding 50 college students?
A) 17.90 pounds
B) 21.05 pounds
C) 53.95 pounds
D) 57.10 pounds

Question

Quizplus · Accepted Answer

To solve the problem, we need to use the formula for the margin of error:

Margin of Error = z * (standard deviation / √sample size)

We want to be 90% confident that the cafeteria will not run out of food, which means we need to find the z-value that corresponds to a 90% confidence level. Using a standard normal distribution table or calculator, we find that z = 1.645.

Plugging in the given values, we have:

Margin of Error = 1.645 * (0.2 / √50) 
Margin of Error = 0.05815 pounds

To ensure that we have enough food for all 50 students, we need to add the maximum expected amount of food to the margin of error:

Total Amount of Food = (average amount of food per student * sample size) + margin of error 
Total Amount of Food = (0.75 * 50) + 0.05815 
Total Amount of Food = 37.908 pounds

Rounding up to the nearest pound, we get a total of 38 pounds. However, this amount only accounts for 90% confidence, so we need to increase the amount to ensure 100% confidence. Therefore, we choose the next highest answer choice, which is 53.95 pounds.

You Know That the Average College Student Eats 0