The lifetime of a sewage treatment plant is 30 years. It is determined that, on

Question

The lifetime of a sewage treatment plant is 30 years. It is determined that, on the average, a devastating storm will occur every 50 years. When this storm arrives, the treatment plant will be overloaded and dump raw sewage into a nearby river. For questions (a) - (e), solve two ways. For the first, use the Binomial distribution or some simplification of the Binomial Distribution. For the second, use the Poisson distribution or the Exponential distribution (whichever is most appropriate). What is the probability that the plant will not be overloaded during the first 10 years of operation? What is the probability of the first overload occurring during the 10th year of operation? What is the probability of the second overload occurring during the 10th year of operation? On the average, when will the first and second overloads occur? What are your standard deviations of the answers in (d)? You will notice that your results using the Binomial (or some variation) and the Poisson (or some variation) are always extremely close, but never identical. That is because they have two different interpretations. What are those interpretations? Which do you feel is (more) correct and why?

Explanation / Answer

a. Probability of storm occuring in one year = 1/50 = 0.02 so so P( not overload) = 0.98

so (i) by bionomial distribution P( 0 <X<10; 0.98) = 10C10 * (0.98)10 = 0.8170

(ii) by poission distribution : average number of storms can come in first 10 years = 10 * 0.02 = 0.2

so by poission distribution P(storm= 0; 0.2) = by poission distribution = (e-) (x) / x! = e-0.2 * 0.20/0! = 0.8187

b. Probability of first overload occuring in 10th year P(stormin a given year) = 0.02

so (i) by bionomial distribtuion P( x= 10; 0.02) = 10C10 * (0.98)9 * ( 0.02) = 0.01667

(ii) By poisson ditribution P(storm in 1oth year; 0.02) = (e-) (x) / x!, where = 0.02 and x = 1

so P(poisson distribution) = 0.0196

(c) P( that second overload will occur in 10 years and one in the 10th year) = ?

(i) By bionomial distribution P(storm come 2 times in 10 ) = 9C1 * ( 0.98)8 * (0.02)2 = 0.00306

(ii) By poisson distribution P( storm come 2 times in 10 years and 2nd time it come in 10 th year)

= P( 1 storm success in 9 year) * P( storm in 10th year) = 0.1503 * 0.0196 = 0.00295

(d) On average, when will the first and second overload occur E(first overload) = 25 years

and P( second ovrload) = 25 + 50 = 75 years.

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