Mean of the Poisson Distribution, $\mu$: 4.5
Standard Deviation of the Poisson Distribution, $\sigma$: 2.1213203435596

P(0) = 0.011108996538242
P(1) = 0.04999048442209
P(2) = 0.1124785899497
P(3) = 0.16871788492456
P(4) = 0.18980762054012
P(5) = 0.17082685848611
P(6) = 0.12812014386458

### Solution:

The mean of the poisson distribution is interpreted as the mean number of occurrences for the distribution. By definition, $\lambda$ is the mean number of successes for a poisson distribution. For this distribution, the mean is $$\mu = \lambda = 4.5$$

The standard deviation of the poisson distribution is interpreted as the standard deviation of the number of occurences for the distribution. To find the standard deviation, use the formula $$\sigma = \sqrt{\lambda}$$ Substituting in the value of $\lambda$ for this problem, we have $$\sigma = \sqrt{4.5}$$ Evaluating the expression on the right, we have $$\sigma = 2.1213203435596$$

To complete a poisson distribution table, first identify all of the possible values of X. Since the maximum number of occurences is 6, the values of X range from X = 0 to X = 6.

Next, find each individual poisson probability for each value of X. In this problem, we will be finding 7 probabilities. The sum of all these probabilities will be 1.

### $P(0)$ Probability of exactly 0 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 0$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 0$ and $\lambda = 4.5$, we have $$P(0) = \frac{{e^{-4.5}} \cdot {4.5^0}}{0!}$$ Remember, 0! is 1. Evaluating the expression, we have $$P(0) = 0.011108996538242$$

### $P(1)$ Probability of exactly 1 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 1$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 1$ and $\lambda = 4.5$, we have $$P(1) = \frac{{e^{-4.5}} \cdot {4.5^1}}{1!}$$ Evaluating the expression, we have $$P(1) = 0.04999048442209$$

### $P(2)$ Probability of exactly 2 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 2$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 2$ and $\lambda = 4.5$, we have $$P(2) = \frac{{e^{-4.5}} \cdot {4.5^2}}{2!}$$ Evaluating the expression, we have $$P(2) = 0.1124785899497$$

### $P(3)$ Probability of exactly 3 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 3$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 3$ and $\lambda = 4.5$, we have $$P(3) = \frac{{e^{-4.5}} \cdot {4.5^3}}{3!}$$ Evaluating the expression, we have $$P(3) = 0.16871788492456$$

### $P(4)$ Probability of exactly 4 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 4$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 4$ and $\lambda = 4.5$, we have $$P(4) = \frac{{e^{-4.5}} \cdot {4.5^4}}{4!}$$ Evaluating the expression, we have $$P(4) = 0.18980762054012$$

### $P(5)$ Probability of exactly 5 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 5$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 5$ and $\lambda = 4.5$, we have $$P(5) = \frac{{e^{-4.5}} \cdot {4.5^5}}{5!}$$ Evaluating the expression, we have $$P(5) = 0.17082685848611$$

### $P(6)$ Probability of exactly 6 occurrences

If using a calculator, you can enter $\lambda = 4.5$ and $x = 6$ into a poisson probability distribution function (poissonPDF). If doing this by hand, apply the poisson probability formula: $$P(x) = \frac{{e^{-\lambda}} \cdot {\lambda^x}}{x!}$$ where $x$ is the number of occurrences, $\lambda$ is the mean number of occurrences, and $e$ is the constant 2.718. Substituting in values for this problem, $x = 6$ and $\lambda = 4.5$, we have $$P(6) = \frac{{e^{-4.5}} \cdot {4.5^6}}{6!}$$ Evaluating the expression, we have $$P(6) = 0.12812014386458$$