Continious Distribution Sample Assignment

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Continious Distribution Sample Assignment

Questions: - A manufacturing firm has been involved in statistical quality control for several years. As part of the production process, parts are randomly selected and tested. From the records of these tests, it has been established that a defective part occurs in a pattern that is Poisson distributed on the average of 1.38 defects every 20 minutes during production runs. Use this information to determine the probability that less than 15min will elapse between any two defects.

Solutions: - The value of λ is 1.38 defects per 20-minutes interval. The value of μ can be determined by

            μ = 1/ λ = 1/1.38 = .7246

On the average, it is .7246 of the interval, or (.7246)(20 minutes) = 14.49 minutes, between defects. The value of x0 represents the desired number of intervals between arrivals of occurrences for the probability questions. In this problem, the probability question involves 15minutes and the interval is 20minutes. Thus x0 is 15/20, or .75 of an interval. The question here is to determine the probability of there being less than 15 minutes between defects. The probability formula always yields the right tail of the distribution – in this case, the probability of there being 15 minutes of more between arrivals. By using the value of x0 and the value of λ, the probability of there being 15 minutes or more between defects can be determined.

    P(x > x0 ) = P(x > .75) = e-λx0 = e(-1.38)(.75) = e-1.035 = .3552

The probability of .3552 is the probability that at least 15 minutes will elapse between defects. To determine the probability of there being less than 15 minutes between defects, compute 1- P(x). In this case, 1 - .3552 = .6448. There is a probability of .6448 that less than 15minutes will elapse between two defects when there is an average of 1.38 defects per 20-minute interval or an average of 14.49 minutes between defects.

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