Boxes are filled with sugar by a machine that is set todeliver an average of 20.

Question

Boxes are filled with sugar by a machine that is set todeliver an average of 20.3 ounces. The weights are normallydistributed with a standard deviation of 0.3 ounces. If a consumeradvocate buys a package of the product and then proceeds to weighthe contents, what is the probability that the package she haspurchased will have a content weight between 20 and 21 ounces? ifthe consumer advocate plans to purchase 100 boxes of the product,then file suit against the company if more than 5 of theboxes have content weights below the 20-ounce stated weight on thepackage, what is the probability that she will end up filing suitagainst the company? What should be the average quantity of sugar in a box be inorder to make the probability of a law suit be only 5%? Boxes are filled with sugar by a machine that is set todeliver an average of 20.3 ounces. The weights are normallydistributed with a standard deviation of 0.3 ounces. If a consumeradvocate buys a package of the product and then proceeds to weighthe contents, what is the probability that the package she haspurchased will have a content weight between 20 and 21 ounces? ifthe consumer advocate plans to purchase 100 boxes of the product,then file suit against the company if more than 5 of theboxes have content weights below the 20-ounce stated weight on thepackage, what is the probability that she will end up filing suitagainst the company? What should be the average quantity of sugar in a box be inorder to make the probability of a law suit be only 5%?

Explanation / Answer

1. Figure out this probability by finding the z-scores of 20 and21. They are -1 and 2.33333, respectively. Go to a z-score table tofind the respective probabilities. Z-score tables give thepercentage of the area that is below that z-score so for the2.3333, find the area that is below -2.3333 and that will be whatis above 2.3333 because normal distributions are symmetric. Take 1- (1st probability+ 2nd probability). 2. This sounds like a 95% confidence interval. To find this, takethe mean +/- the margin of error = mean +/- 1.96(standarddeviation/sqrt(n))

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