Consider a Poisson distribution with a mean of two occurrences per time period.

Question

Consider a Poisson distribution with a mean of two occurrences per time period. a. Which of the following is the appropriate Poisson probability function for one time period? 1. f(x) = 2^x e^-2/x! 2. f(x) = 2^x e^-2/x 3. f(x) = 2^x e^2/x! b. What is the expected number of occurrences in three time periods? c. Select the appropriate Poisson probability function to determine the probability of x occurrences in three time periods. 1. f(x) = 6^x e^6/x! 2. f(x) = 6^x e^-6/x 3. f(x) = 6^x e^-6/x! d. Compute the probability of two occurrences in two time periods (to 4 decimals).

Explanation / Answer

Suppose we conduct a Poisson experiment, in which the average number of successes within a given region is . Then, the Poisson probability is:

f(x) = (e-) (x) / x!

a.

Given, = 2 occurences per time period

So, poisson probability function is,

f(x) = (e-2) (2x) / x!

Hence option 1 is correct.

b.

Expected number of occurences in one time period = 2

Expected number of occurences in three time period = 2 * 3 = 6

c.

Given, = 6 occurences per three time period

So, poisson probability function is,

f(x) = (e-6) (6x) / x!

Hence option 3 is correct.

d)

Expected number of occurences in one time period = 2

Expected number of occurences in two time period = 2 * 2 = 4

So, poisson probability function is,

f(x) = (e-4) (4x) / x!

Probability of two occurrences in two time periods =  (e-4) (42) / 2! = 0.1465

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