Demand for gallons of milk (a perishable item) is normally distributed with a weekly mean of 50 and standard deviation of 15. How many additional gallons of milk must be stocked to increase the service level from 50% to 80%?

What will be an ideal response?

Z for 50% is 0, Z for 80% is .84. Thus, the increase in milk stock would be 15(.84 ) - 15(0 ) = 12.6 = 13 gallons extra.

Business

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