Mean of the Poisson Distribution, $\mu$: 6.4
Standard Deviation of the Poisson Distribution, $\sigma$: 2.5298221281347

P(0) = 0.0016615572731739
P(1) = 0.010633966548313
P(2) = 0.034028692954602
P(3) = 0.072594544969818
P(4) = 0.11615127195171
P(5) = 0.14867362809819
P(6) = 0.15858520330473
P(7) = 0.14499218587861
P(8) = 0.11599374870289

### 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 $$\mu = \lambda = 6.4$$

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 $$\sigma = \sqrt{6.4}$$ Evaluating the expression on the right, we have $$\sigma = 2.5298221281347$$

To complete a poisson distribution table, first identify all of the possible values of X. Since the maximum number of occurences is 8, the values of X range from X = 0 to X = 8.

Next, find each individual poisson probability for each value of X. In this problem, we will be finding 9 probabilities. The sum of all these probabilities will be 1.

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(0) = \frac{{e^{-6.4}} \cdot {6.4^0}}{0!}$$ Remember, 0! is 1. Evaluating the expression, we have $$P(0) = 0.0016615572731739$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(1) = \frac{{e^{-6.4}} \cdot {6.4^1}}{1!}$$ Evaluating the expression, we have $$P(1) = 0.010633966548313$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(2) = \frac{{e^{-6.4}} \cdot {6.4^2}}{2!}$$ Evaluating the expression, we have $$P(2) = 0.034028692954602$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(3) = \frac{{e^{-6.4}} \cdot {6.4^3}}{3!}$$ Evaluating the expression, we have $$P(3) = 0.072594544969818$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(4) = \frac{{e^{-6.4}} \cdot {6.4^4}}{4!}$$ Evaluating the expression, we have $$P(4) = 0.11615127195171$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(5) = \frac{{e^{-6.4}} \cdot {6.4^5}}{5!}$$ Evaluating the expression, we have $$P(5) = 0.14867362809819$$

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

If using a calculator, you can enter $\lambda = 6.4$ 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 $\lambda = 6.4$, we have $$P(6) = \frac{{e^{-6.4}} \cdot {6.4^6}}{6!}$$ Evaluating the expression, we have $$P(6) = 0.15858520330473$$

### $P(7)$ Probability of exactly 7 occurrences

If using a calculator, you can enter $\lambda = 6.4$ and $x = 7$ 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, $x = 7$ and $\lambda = 6.4$, we have $$P(7) = \frac{{e^{-6.4}} \cdot {6.4^7}}{7!}$$ Evaluating the expression, we have $$P(7) = 0.14499218587861$$

### $P(8)$ Probability of exactly 8 occurrences

If using a calculator, you can enter $\lambda = 6.4$ and $x = 8$ 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, $x = 8$ and $\lambda = 6.4$, we have $$P(8) = \frac{{e^{-6.4}} \cdot {6.4^8}}{8!}$$ Evaluating the expression, we have $$P(8) = 0.11599374870289$$