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  • 5 years ago

A watch company determines that each box of 500 watches has an average of 10 defective watches with standard deviation 3. Suppose that 1000 boxes are produced. Use Chebychev′s Theorem to estimate the number of boxes having between 0 and 20 defective watches. Round the answer to the nearest integer.

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  1. anonymous
    • 5 years ago
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    Chebychev′s Inequality states given standard deviation k no more than 1/k^2 of values will be more than k deviations from mean to find k we see 20 is 10 more than our mean of 10 watches 10 is 10/3 deviations k = 10/3 k^2 = 100/9 1/k^2 = 9/100 9/100 * 1000 = 90 Therefore no more than 90 boxes will have more than 20 defective watches 1000-90=910 910 boxes will have between 0 and 20 defective watches Hope this helps

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