A machine fills 750 ml bottles of whiskey with a measured amount of whiskey. The

Question

A machine fills 750 ml bottles of whiskey with a measured amount of whiskey. The volume of whiskey in each bottle has a normal distribution with a known mean and standard deviation. By law, the contents of any product must be contained in 90% of the products sold. In this case, 90% of the 750 ml bottles of whiskey must contain at least 750 ml of whiskey. Assuming that the set point of the filling machine can be varied without changing the standard deviation of 6.0 ml, and that the cost of the whiskey at bottling time is $34 per liter, what is the annual cost of the whiskey to satisfy the 90% rule? Assume the bottling machine operates 24 hours per day for 180 days per year and that filling a bottle requires 2 seconds

Explanation / Answer

Annual bottling capacity of the bottling operation = 24 hours/ day x 180 days x 3600 seconds per hour / 2 seconds = 7776000

The bottled quantity follows a normal distribution

It is given that 90% of the bottles must at least contain 750 ml . Therefore, probability that bottled quantity is 750 will be 0.90

It is also given that standard deviation ( Sd ) of quantity filled in each bottle = 6 ml

Z value of probability 0.90 = NORMSINV ( 0.90) = 1.2815

Let the mean value of the quantity filled in the bottle = M     ml

Hence,

M + Z x Sd = 750

Or M + 1.2815 X 6 = 750

Or, M + 7.689 = 750

Or, M = 750 – 7.689 = 742.31

Thus, mean bottled quantity = 742.31 ml

Hence total bottled quantity = Annual bottling capacity x mean bottled quantity = 7776000 x 742.31 ml = 7776 x 742.31 L

Therefore , annual cost of whiskey , $

= Total bottled quantity ( Litre) x Cost of whiskey ( L )

= 7776 x 742.31 x 34

= $196254887.04

ANNUAL COST OF WHISKEY = $196254887.04

ANNUAL COST OF WHISKEY = $196254887.04

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