Question 1176232: I need help please.
Q:
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 the 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 deviation of 2.2 ounces, and that the coffee costs $0.75 per pound at filling time, 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.
A:____________
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i understand this problem as follows:
since the distribution of the pounds of coffee is normal, then you can use the z-score formula to help solve this problem.
the z-score formula is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation.
the critical x-score for 10% of thde normal distribution to be less than it, is found by looking up an area of .10 in the z-score table, or using a normal distribution calculator to find the same z-score.
i used the z-score calculator from the ti-84 plus scientific calculator.
i looked for the inverse z-score for 10% of the area under the normal distribution curve to be equal to 10%.
the calculator told me that the z-score that has 10% of the area under the normal distribution curve to the left of it was equal to -1.281551567.
that was my critical z-score.
the raw score associated with that z-score was 3 pounds.
the standard deviation was given as 2.2 ounces which is equal to 2.2 / 16 = .175 pounds.
i used the z-score formula to find the mean.
the formula of z = (x - m) / s became:
-1.281551567 = (3 - m) / .1375.
i solved for m to get:
m = 3 - .1375 * -1.281551567 = 3 + .1762133404 = 3.1762133404.
that is the average pounds of coffee required per can so that 90% of the cans will have at least 3 pounds in them.
the mean is the set point.
the annual cost of the coffee is calculated as follows.
the filling machine operates continuously for 24 hours a day for 250 days each year.
this means that it is operating continuously for 24 * 250 = 6,000 hours each year.
since there are 60 * 60 = 3600 seconds each hour, this means that it is operating continuously for 6000 * 3600 = 21,600,000 seconds each year.
since it takes 2 seconds to fill a can, then it is filling 21,600,000 / 2 = 10,800,000 cans each year.
since each can requires an average of 3.17621334 pounds of coffee, then it is requiring 10,800,000 * 3.17621334 = 34,303,104.08 pounds of coffee each year.
since the coffee costs .75 per pound, then the total cost of the coffee each year is equal to .75 * 34,303,104.08 = 25,727,328.06 each year.
an annual cost of 25,727,318.06 each year is your solution.
graphically, the probability looks loke this.
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