Question

Question 7.Consider a Poissonprobability distribution with 2 as the average number of occurrences per time period. a) write the appropriate Poisson probability function. b) what is the average number of occurrences in three time periods? e) write the appropriate Poisson probability function to determine the probability of occurrences in three time periods. d) find the probability of two occurrences in one time period. e) find the probability of six occurrences in three time periods. f) find the probability of five occurrences in two time periods

Answer 1

Solution:

X follows a Poisson distribution with an average number of occurrences 2 per time period.
That is: X ~ Poi( $\lambda$ = 2) per time period.

Part a) Poisson probability distribution function:

Probability mass function of Poisson distribution is:

$P(X=x) = \left\{\begin{matrix} \frac{e^{-\lambda} \times \lambda^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$   

$P(X=x) = \left\{\begin{matrix} \frac{e^{-2} \times 2^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$

Part b) An average number of occurrences per three time period=.........?

We have: an average number of occurrences per time period = 2

then An average number of occurrences per three time period= 2X3=6

That is: X ~ Poi( $\lambda$ = 6) per three time period.

Part c) Poisson probability distribution function for An average number of occurrences per three time period is:

$P(X=x) = \left\{\begin{matrix} \frac{e^{-6} \times 6^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$

Part d) Find: P( 2 occurrences per one time period ) =..............?

That is find:

P(X = 2) =...........?

We use probability distribution stated in part a)

$P(X=x) = \left\{\begin{matrix} \frac{e^{-2} \times 2^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$

$P(X=2) = \frac{e^{-2} \times 2^{2}}{2!}$

$P(X=2) = \frac{0.135335 \times 4}{2 \times 1}$

Part e) Find:

P( 6 occurrences per three time period ) =..............?

That is find:

P(X = 6) =...........?

We use probability distribution stated in part c).

$P(X=x) = \left\{\begin{matrix} \frac{e^{-6} \times 6^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$

$P(X=6) = \frac{e^{-6} \times 6^{6}}{6!}$

$P(X=6) = \frac{0.00247875 \times 46656 }{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$P(X=6) = \frac{0.00247875 \times 46656 }{720}$

Part f) Find :

P( 5 occurrences per two time period ) =..............?

That is find:

P(X = 5) =...........?

We have: an average number of occurrences per time period = 2

thus an average number of occurrences per two time period= 2X2=4

That is: X ~ Poi( $\lambda$ = 4) per two time period.

Thus

Poisson probability distribution function for An average number of occurrences per two time period is:

$P(X=x) = \left\{\begin{matrix} \frac{e^{-4} \times 4^{x}}{x!} & if \: x=0,1,2,3,........... \\ & \\ 0 & otherwise\\ \end{matrix}\right.$

We have to find:

P(X = 5) =...........?

$P(X=5) = \frac{e^{-4} \times 4^{5}}{5!}$

$P(X=5) = \frac{0.01831564 \times 1024 }{5 \times 4 \times 3 \times 2 \times 1}$

$P(X=5) = \frac{0.01831564 \times 1024 }{120}$