Question

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 $f(x)=\frac{2^{x}e^{2}}{x!}$ equation #1 b. what is the expected number of occurrences in three time periods? 6 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 $f(x)=\frac{6^{x}e^{-6}}{x!}$ equation #3 d. compute the probability of two occurrences in one time period (to 4 decimals). 0.2707 e. compute the probability of six occurrences in two time periods (to 4 decimals).

Explanation:

Step1: Recall Poisson formula

The Poisson probability function for one - time period is $f(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$, where $\lambda$ is the average number of occurrences in the time period. If $\lambda = 2$ for one time period, the correct formula is $f(x)=\frac{2^{x}e^{-2}}{x!}$, so the answer to part a is Equation #1.

Step2: Calculate expected value for multiple periods

The expected number of occurrences in one time period is $\lambda$. For $n$ time periods, the expected number of occurrences is $n\lambda$. Given $\lambda = 2$ and $n = 3$, the expected number of occurrences in three time periods is $n\lambda=3\times2 = 6$, which is the answer to part b.

Step3: Determine Poisson formula for multiple periods

For three time periods with $\lambda_{total}=6$, the Poisson probability function is $f(x)=\frac{6^{x}e^{-6}}{x!}$, so the answer to part c is Equation #3.

Step4: Compute probability for one - period case

For one time period with $\lambda = 2$ and $x = 2$, we use $f(x)=\frac{2^{x}e^{-2}}{x!}$. Substitute $x = 2$ into the formula: $f(2)=\frac{2^{2}e^{-2}}{2!}=\frac{4\times e^{-2}}{2}=2e^{-2}\approx2\times0.13534 = 0.2707$ (to 4 decimals), which is the answer to part d.

Step5: Compute probability for multiple - period case

For two time periods, $\lambda=2\times2 = 4$. The Poisson probability function is $f(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$. Substitute $\lambda = 4$ and $x = 6$ into the formula: $f(6)=\frac{4^{6}e^{-4}}{6!}=\frac{4096\times0.01832}{720}\approx\frac{75.9497}{720}\approx0.1055$ (to 4 decimals).

Answer:

a. Equation #1
b. 6
c. Equation #3
d. 0.2707
e. 0.1055