Question

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?

Answer 1

Answer:

$a) \simeq 0.3012$   $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 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) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}$

                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

         P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}$

                      $\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =$0.7408182207^{4}$

     $\simeq 0.3012$  -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

           P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}$

                      $\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

                =$0.2222454662^{2}$

                $\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

                                $\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6]  (which is to be estimated)

=$\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$-------------------------------------------(6)

So,

P(1  collision in 6 month period)

  =  $0.6543894283^{6}$

   $\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

  = $(P(X =0)^{6}$

   $\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)