SOLUTION: Leaks occur in a pipeline at a mean rate of 1 leak per 1,000 meters. In a 2,500-meter section of pipe, (a) Using the Poisson approximation to the binomial, what is the probabil

Question 1191375: Leaks occur in a pipeline at a mean rate of 1 leak per 1,000 meters. In a 2,500-meter section of pipe,

(a) Using the Poisson approximation to the binomial, what is the probability of no leaks? (Round your answer to 4 decimal places.)
Probability: ???

(b) Using the Poisson approximation to the binomial, what is the probability of three or more leaks? (Round your answer to 4 decimal places.)
Probability: ???

(c) What is the expected number of leaks? (Round your answer to 1 decimal place.)
Expected number of leaks: ???

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
Here's how to solve this problem using the Poisson approximation to the binomial distribution:
**1. Calculate the Mean (λ):**
The mean number of leaks for the 2,500-meter section is proportional to the given rate.
λ = (1 leak / 1000 meters) * 2500 meters = 2.5 leaks
**2. (a) Probability of No Leaks (P(X=0)):**
The Poisson probability formula is:
P(X=k) = (e^(-λ) * λ^k) / k!
For no leaks (k=0):
P(X=0) = (e^(-2.5) * 2.5^0) / 0!
P(X=0) = (0.0821) * 1 / 1
P(X=0) ≈ 0.0821
**3. (b) Probability of Three or More Leaks (P(X≥3)):**
It's easier to calculate the complement and subtract from 1:
P(X≥3) = 1 - [P(X=0) + P(X=1) + P(X=2)]
* P(X=1) = (e^(-2.5) * 2.5^1) / 1! ≈ 0.2052
* P(X=2) = (e^(-2.5) * 2.5^2) / 2! ≈ 0.2565
P(X≥3) = 1 - (0.0821 + 0.2052 + 0.2565)
P(X≥3) = 1 - 0.5438
P(X≥3) ≈ 0.4562
**4. (c) Expected Number of Leaks:**
The expected number of leaks in a Poisson distribution is equal to the mean (λ).
Expected number of leaks = 2.5
**Answers:**
* (a) Probability of no leaks: 0.0821
* (b) Probability of three or more leaks: 0.4562
* (c) Expected number of leaks: 2.5

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