8. In 2010, the average number of pedestrian and bicyclist traffic deaths per mo

Question

8. In 2010, the average number of pedestrian and bicyclist traffic deaths per month was 2.17 in Iowa. Assume that the number traffic deaths of these types of follows a Poisson distribution. (a) What is the probability that no pedestrian or bicyclist traffic deaths will occur during a given month? (b) What is the probability that at least four traffic deaths of these types will occur in a given month? (c) What is the probability that six or more pedestrian or bicyclist traffic deaths will occur in a given month?

Explanation / Answer

Solution :

         We are given that : the average number of pedestrian and bicyclist traffic deaths per month was 2.17 in Iowa and number of pedestrian and bicyclist traffic deaths per month follows Poisson distribution. Thus parameter =2.17

We have to find :

Part a) Probability that no pedestrian or bicyclist traffic death will occur during a given month.

That is we have to find : P( X = 0 ) = .............?

Thus we use following Poisson probability formula :

P(X=x ) = e- * x / x!         x = 0 , 1 , 2 , 3 ,......

Thus

P(X=0 ) = e-2.17 * 2.170 / 0!

P(X=0 ) = 0.114178 * 1 / 1

P(X=0 ) = 0.114178

P(X=0 ) = 0.1142

Part b) Probability that at least four traffic deaths of these types will occur in a given month.

That is we have to find : P( X 4) = ............?

Thus : P( X 4) = 1 - P( X < 4)

         where P( X < 4) = P( X = 0) + P( X = 1) + P( X = 2) + P( X = 3)

Thus P( X < 4) = e-2.17 * 2.170 / 0! + e-2.17 * 2.171 / 1! + e-2.17 * 2.172 / 2! + e-2.17 * 2.173 / 3!

       P( X < 4) = e-2.17 { 2.170 / 0! + 2.171 / 1! +2.172 / 2! + 2.173 / 3! }

       P( X < 4) = 0.114178 { 1 / 1 + 2.17 / 1 + 4.7089 / 2x1 + 10.21831 / 3x2x1 }

      P( X < 4) = 0.114178 { 1 + 2.17 + 4.7089 / 2 + 10.21831 / 6 }

      P( X < 4) = 0.114178 { 1 + 2.17 + 2.35445     + 1.703052    }

      P( X < 4) = 0.114178 { 7.227502 }

       P( X < 4) = 0.825219

Thus P( X 4) = 1 - P( X < 4)

       P( X 4) = 1 - 0.825219

      P( X 4) =0.174781

     P( X 4) =0.1748

Part c) Probability that six or more four traffic deaths of these types will occur in a given month.

That is we have to find : P( X 6) = ............?

Thus : P( X 6) = 1 - P( X < 6)

         where P( X < 6) = P( X = 0) + P( X = 1) + P( X = 2) + P( X = 3)+ P( X = 4)+ P( X = 5)

Thus P( X < 6) = e-2.17 * 2.170 / 0! + e-2.17 * 2.171 / 1! + e-2.17 * 2.172 / 2! + e-2.17 * 2.173 / 3!

                        + e-2.17 * 2.174 / 4! + e-2.17 * 2.175 / 5!

       P( X < 6) = e-2.17 { 2.170 / 0! + 2.171 / 1! +2.172 / 2! + 2.173 / 3! +2.174 / 4! +2.175 / 5! }

       P( X < 6) = 0.114178 { 1 / 1 + 2.17 / 1 + 4.7089 / 2x1 + 10.21831 / 3x2x1

                                      + 22.17374 / 4x3x2x1 + 48.11701 / 5x4x3x2x1 }

      P( X < 6) = 0.114178 { 1 + 2.17 + 4.7089 / 2 + 10.21831 / 6 + 22.17374 / 24 + 48.11701 /120 }

      P( X < 6) = 0.114178 { 1 + 2.17 + 2.35445     + 1.703052 + 0.923906 + 0.400975 }

      P( X < 6) = 0.114178 { 8.552383 }

       P( X < 6) = 0.976491

Thus P( X 6) = 1 - P( X < 6)

       P( X 6) = 1 - 0.976491

      P( X 6) =0.023509

     P( X 6) =0.0235

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