A machine fills a 3 pound coffee can with a measured amount of coffee. the weigh

Question

A machine fills a 3 pound coffee can with a measured amount of coffee. the weight of coffee in each can has a known normal distribution and a standard deviation. By law, the contents of any product must be contained in 90% of the products sold.In this case, 90% of the 3 pound coffee cans must contain at least 3 pounds of coffee. assuming that the set point of the filling machine can be varied without changing the standard deviationof 2.9 ounces, and that the coffee costs $0.75 per pound at time of filling. What is the annual cost of the coffee needed to satisfy the 90% rule? assume the filling machine operates 24 hours per day for 250 days per year and that filling a can requires 2 seconds.

Explanation / Answer

1 pound is 16 ounces

So, 3 pounds is equal to 48 ounces.

Given that = 2.9 ounces

For 90% confidence that 3 pounds of cans contain 3 pounds of coffee with a standard deviation of 2.9 ounces

Z = (x )/

1.645 = (48 )/2.9

= 43.23 ounces which is the mean weight of coffee in each can.

Total cans filled in 250 days operating 24 hrs a day = (250 x 24 x 60 x 60)/ 2 = 10,800,000

Total weight of coffee filled = 10,800,000 x 43.23 = 466,884,000 pounds

Total cost = 466,884,000 x 0.75 = 350,163,000 dollars, which is the annual cost of coffee needed to satisfy 90% rule.

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