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 $f(x)=\frac{2^{x}e^{-2}}{x!}$
2 $f(x)=\frac{2^{x}e^{-2}}{x}$
3 $f(x)=\frac{2^{x}e^{2}}{x!}$

• select your answer -

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 $f(x)=\frac{6^{x}e^{6}}{x!}$
2 $f(x)=\frac{6^{x}e^{-6}}{x}$
3 $f(x)=\frac{6^{x}e^{-6}}{x!}$

• select your answer -

d. compute the probability of two occurrences in one time period (to 4 decimals).

Explanation:

Step1: Recall Poisson probability formula

The Poisson probability function is $f(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$, where $\lambda$ is the mean number of occurrences in the given time - period and $x$ is the actual number of occurrences. Given $\lambda = 2$ for one time - period.

Step2: Answer part a

Substitute $\lambda = 2$ into the Poisson formula. The correct Poisson probability function for one time - period is $f(x)=\frac{2^{x}e^{-2}}{x!}$. So the answer for part a is 1.

Step3: Answer part b

If the mean number of occurrences per time - period is $\lambda_1=2$, for $n = 3$ time - periods, the expected number of occurrences $\lambda=n\lambda_1$. So $\lambda=3\times2 = 6$.

Step4: Answer part c

For three time - periods, $\lambda = 6$. The Poisson probability function is $f(x)=\frac{6^{x}e^{-6}}{x!}$. So the answer for part c is 3.

Step5: Answer part d

For one time - period, $\lambda = 2$ and $x = 2$. Substitute into the Poisson formula: $f(2)=\frac{2^{2}e^{-2}}{2!}=\frac{4\times e^{-2}}{2}$. Since $e^{-2}\approx0.13534$, then $f(2)=\frac{4\times0.13534}{2}=0.2707$.

Answer:

a. 1. $f(x)=\frac{2^{x}e^{-2}}{x!}$
b. 6
c. 3. $f(x)=\frac{6^{x}e^{-6}}{x!}$
d. 0.2707