### Poisson Distribution Calculator

### Answer:

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 $$ P(8) = 0.11599374870289 $$