The number of traffic accidents per day on a certain section of highway is thought to be Poisson distributed with a mean equal 2.19. Then the probability of no accidents on this section of highway during one day period is approximately:

A) 0.457
B) 0.112
C) 0.318
D) 0.296
E) 0.211

Question

natalie braley · Accepted Answer

Final Answer : B, Explanation :   The Poisson distribution with mean $\lambda$ is given by: P ( X = k ) = e − λ λ k k ! P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!} P ( X = k ) = k ! e − λ λ k ​ In this case, we have $\lambda=2.19$ and we want to find $P(X=0)$: P ( X = 0 ) = e − 2.19 ( 2.19 ) 0 0 ! ≈ 0.112 P(X=0)=\frac{e^{-2.19}(2.19)^0}{0!}\approx 0.112 P ( X = 0 ) = 0 ! e − 2.19 ( 2.19 ) 0 ​ ≈ 0.112 Therefore, the best choice is (B) with probability approximately 0.112.

The number of traffic accidents per day on a certain section of highway is thought to be Poisson distributed with a mean equal 2.19. Then the probability of no accidents on this section of highway during one day period is approximately:

A) 0.457
B) 0.112
C) 0.318
D) 0.296
E) 0.211

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The number of traffic accidents per day on a certain section of highway is thought to be Poisson distributed with a mean equal 2.19. Then the probability of no accidents on this section of highway during one day period is approximately:A) 0.457B) 0.112C) 0.318D) 0.296E) 0.211

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Learning Objectives

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Related questions

The number of traffic accidents per day on a certain section of highway is thought to be Poisson distributed with a mean equal 2.19. Then the probability of no accidents on this section of highway during one day period is approximately:

A) 0.457
B) 0.112
C) 0.318
D) 0.296
E) 0.211