Question

Suppose that the average number of equipment breakdowns at a factory per week is 5. Later we’ll show that the average is our best guess at the expected value of ? = ?ℎ? ?????? ?? ????????? ?????????? ?? ? ????? ???? (assuming that this distribution stays the same across multiple weeks.) It’s also known that, through some other sample statistic, that our best guess at the standard deviation of X given the data is 0.8. a. With the help of our boy Chebyshev, find an interval such that we can be at least 90% that the number of breakdowns next week will fall within that range. b. Suppose that the supervisor promises the board of directors that the number of breakdowns will rarely exceed 8 in a one- week period. Is the supervisor safe in making this claim? Why or why not? c. Google Chebyshev and just stare at this guy’s magnificent beard for a few seconds. You don’t need to write anything down. You’re welcome.

Answer 1

Chebyshev's inequality:

= 5

= 0.8

A.

1-1/k² = 0.9

k = 3.162

So, as per the formula given above:

P(5-3.162*0.8 < X < 5+3.162*0.8) 0.9

P(2.47 < X < 7.53) 0.9

So, the corresponding interval is [2.47, 7.53] [3,8]

B.

As per the previous part, the supervisor can be assured with   90% probability that no. of breakdowns will remain between [3,8], thus supervisor is almost safe in making this claim. Of course, if we want to increase the confidence by calculating the interval at a higher probability range then the interval will widen but 90% is a good enough confidence to be assured of a claim.

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