Reply Sun 24 Jan, 2016 01:13 pm
A company’s manufacturing process uses 500 gallons of water at a time, a scrubbing machine then removes most of the chemical pollutant before pumping the water into a lake. Legally the treated water should contain no more than 80 parts per million of the chemical, but the machine isn’t perfect and it is costly to operate. Since there are fines if the discharged water exceeds the legal maximum, the company sets the machine to attain an average of 75 ppm. They believe the machine’s output can be described by a normal model with standard deviation 4.2 ppm.

(1) what percent of the batches of water discharged exceed the 80ppm standard?
(2) Company's lawyer insists that they have not more than 2% of the water over the limit. to what mean value should the company set the scrubbing machine? Assume the standard deviation does not change.
(3)achieving a mean that low would raise the cost too much, they decided to leave the mean set at 75 ppm and try to reduce the standard deviation to achieve the "only 2% over" goal. Fine the new standard deviation.
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Reply Sun 24 Jan, 2016 05:18 pm
@chitownbulls,
1) The Z score for a value of 80 when the mean is 75 and the standard deviation is 4.2 is:

z = (80-75)/4.2 = 1.19

Look up the cumulative distribution value for 1.19 (here is an online calculator, leave the mean as zero and the standard deviation as one). You get 88.3%.

2) Find the z score for 98% (2% failing). You can use the same calculator from above by clearing out the z score, and putting .98 in for the cumulative probability. You get 2.054. Using the formula above:

2.054 = (80 - x)/4.2
x = 73.5

3) In order to get 2%, you need a z score of 2.054, so using the same formula

2.054 = (80-75)/x where x is the new standard deviation.
x = 2.43
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