Question

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).
0.1055
f. compute the probability of seven occurrences in three time periods (to 4 decimals).
0.1377

Explanation:

Step1: Recall Poisson probability formula

The Poisson probability formula is $f(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$, where $\lambda$ is the average number of occurrences in the given interval and $x$ is the actual number of occurrences.

Step2: Find $\lambda$ for two - time periods

We know from part b that the average number of occurrences in three time - periods is $\lambda_3 = 6$, so the average number of occurrences in one time - period is $\lambda_1=\frac{6}{3}=2$. Then for two time - periods, $\lambda_2 = 2\times2 = 4$.

Step3: Calculate probability for $x = 6$ and $\lambda=4$

Substitute $x = 6$ and $\lambda = 4$ into the Poisson formula $f(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}$. So $f(6)=\frac{4^{6}\times e^{- 4}}{6!}$.
We know that $4^{6}=4096$, $e^{-4}\approx0.018316$, and $6!=720$.
Then $f(6)=\frac{4096\times0.018316}{720}=\frac{74.911336}{720}\approx0.1040$.

Answer:

0.1040