Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1.2 collisions every four months. Appendix A Statistical Tables (Round your answers to 4 decimal places.) a. What is the probability of having no collisions occur over a four-month period? b. What is the probability of having exactly two collisions in a two-month period? c. What is the probability of having one or fewer collisions in a six-month period?

Accepted Solution

Answer:[tex]a) \simeq 0.3012[/tex]   [tex]b) \simeq 0.0494[/tex] c) [tex]\simeq 0.2438[/tex]Step-by-step explanation:Rate of collision,1.2 collisions every 4 monthsor, [tex]\frac{1.2}{4}[/tex] = 0.3 collisions per  month So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,           P(X =x) = [tex]\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!} [/tex]                                                            for x ∈ N ∪ {0}                        =  0 otherwise --------------------------------------(1)here, [tex]\lambda = 0.3[/tex] collision / monthNo collision over a 4 month period means no collision per month or X =0Putting X = 0 in (1) we get,          P(X = 0) = [tex]\frac{e^{-0.3}\times {\0.3}^{0}}{0!} [/tex]                       [tex]\simeq 0.7408182207[/tex] ------------------------------------(2)Now, since we are calculating  this for 4 months,so, P(No collision in 4 month period)      =[tex]0.7408182207^{4}[/tex]      [tex]\simeq 0.3012[/tex]  -----------------------------------------------------------(3)2 collision in 2 month period means 1 collision per month or X =1Putting X =1 in (1) we get,            P(X =1) = [tex]\frac{e^{-0.3}\times {\0.3}^{1}}{1!} [/tex]                       [tex]\simeq 0.2222454662[/tex] ------------------------------------(4)Now, since we are calculating this for 2 months, so ,P(2 collisions in 2 month period)                 =[tex]0.2222454662^{2}[/tex]                 [tex]\simeq 0.0494[/tex] -----------------------------------------(5)1 collision in 6 months period means                                 [tex]\frac{1}{6}[/tex] collision per monthNow, P(1 collision in 6 months period)= P( X = 1/6]  (which is to be estimated)=[tex]\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}[/tex]= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex] [tex]\simeq 0.6543894283[/tex]-------------------------------------------(6)So,P(1  collision in 6 month period)   =  [tex]0.6543894283^{6}[/tex]    [tex]\simeq 0.0785267444[/tex] ------------------------------------------------(7)So,P(No collision in 6 months period)   = [tex](P(X =0)^{6}[/tex]    [tex]\simeq 0.1652988882[/tex] ---------------------------------(8)so,P(1 or fewer collision in 6 months period)= (8) + (7 ) = 0.0785267444 +0.1652988882 [tex]\simeq  0.2438[/tex] ---------------------------------------------(9)