The article \"Expectation Analysis of the Probability of Failure for Water Suppl

Question

The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"t proposed using the Poisson distribution to model the number of failures in pipelines of various types. Suppose that for cast-iron pipe of a particular length, the expected number of failures is 1 (very close to one of the cases considered in the article). Then X, the number of failures, has a Poisson distribution with -1. (Round your answers to three decimal places.) (a) Obtain P(X 4) by using the Cumulative Poisson Probabilities table in the Appendix of Tables. (b) Determine P(X -1) from the pmf formula Px-1) Determine P(X - 1) from the Cumulative Poisson Probabilities table in the Appendix of Tables. P(X = 1) = (c) Determine P(1 s X S 3). (d) What is the probability that X exceeds its mean value by more than one standard deviation? You may need to use the appropriate table in the Appendix of Tables to answer this question.

Explanation / Answer

POSSION DISTRIBUTION
pmf of P.D is = f ( k ) = e- x / x!
where
= parameter of the distribution.   
x = is the number of independent trials
I.
mean =
= 1
a.
LESS THAN EQUAL
P( X < = 4) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-1 * 1 ^ 4 / 4! + e^-1 * 0 ^ 3 / 3! + e^-1 * ^ 2 / 2! + e^-1 * ^ 1 / 1! + e^-1 * ^ 0 / 0!
= 0.99634
b.
P( X = 1 ) = e ^-1 * 1^1 / 1! = 0.36788
c.
P( X = 2 ) = e ^-1 * 1^2 / 2! = 0.18394
P( X = 3 ) = e ^-1 * 1^3 / 3! = 0.06131

P(1<= X <=3) = 0.36788+0.18394+0.06131 = 0.613113

d.
the standard deviatio value for the given possion distribution
will be sqrt(variance) = sqrt(1) = 1
and the diffrence it exceed mean by one standard deviation value will be P(X>2)

P( X < = 2) = P(X=2) + P(X=1) + P(X=0) +   
= e^-1 * 1 ^ 2 / 2! + e^-1 * 3 ^ 1 / 1! + e^-1 * ^ 0 / 0! +
= 0.9197,
P( X > 2) = 1 -P ( X <= 2) = 1 - 0.9197 = 0.0803

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