### Poisson Distribution Calculator

### Answer:

Mean of the Poisson Distribution, $\mu$: 4.2

Standard Deviation of the Poisson Distribution, $\sigma$: 2.0493901531919

P(0) = 0.014995576820478

P(1) = 0.062981422646006

P(2) = 0.13226098755661

P(3) = 0.18516538257926

P(4) = 0.19442365170822

P(5) = 0.16331586743491

P(6) = 0.11432110720443

### 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 $$ \mu = \lambda = 4.2 $$

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 $$ \sigma = \sqrt{4.2} $$ Evaluating the expression on the right, we have $$ \sigma = 2.0493901531919 $$

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 = 4.2 $ 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 = 4.2 $, we have $$ P(0) = \frac{{e^{-4.2}} \cdot {4.2^0}}{0!} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.014995576820478 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(1) = \frac{{e^{-4.2}} \cdot {4.2^1}}{1!} $$ Evaluating the expression, we have $$ P(1) = 0.062981422646006 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(2) = \frac{{e^{-4.2}} \cdot {4.2^2}}{2!} $$ Evaluating the expression, we have $$ P(2) = 0.13226098755661 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(3) = \frac{{e^{-4.2}} \cdot {4.2^3}}{3!} $$ Evaluating the expression, we have $$ P(3) = 0.18516538257926 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(4) = \frac{{e^{-4.2}} \cdot {4.2^4}}{4!} $$ Evaluating the expression, we have $$ P(4) = 0.19442365170822 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(5) = \frac{{e^{-4.2}} \cdot {4.2^5}}{5!} $$ Evaluating the expression, we have $$ P(5) = 0.16331586743491 $$

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

If using a calculator, you can enter $ \lambda = 4.2 $ 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 = 4.2 $, we have $$ P(6) = \frac{{e^{-4.2}} \cdot {4.2^6}}{6!} $$ Evaluating the expression, we have $$ P(6) = 0.11432110720443 $$