The Poisson Distribution Calculator will construct a complete poisson distribution, and identify the mean and standard deviation. A poisson probability is the chance of an event occurring in a given time interval. Enter $\lambda$ and the maximum occurrences, then the calculator will find all the poisson probabilities from 0 to max. If you want to find the poisson probability for a specific occurrence only, then have a look at the Poisson Probability Calculator.

Enter a value for $\lambda$ and x. Lambda, $\lambda$, is the average number of occurrences for a given time interval. x is the maximum number of occurrences. The calculator will display the poisson distribution, the mean, and the standard deviation. Then, it will also give you a step by step solution for how to find poisson probabilities.

### Answer:

Mean of the Poisson Distribution, $\mu$: 3.9

Standard Deviation of the Poisson Distribution, $\sigma$: 1.9748417658131

P(0) = 0.020241911445804

P(1) = 0.078943454638637

P(2) = 0.15393973654534

P(3) = 0.20012165750895

P(4) = 0.19511861607122

P(5) = 0.15219252053555

### 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 $$ \mu = \lambda = 3.9 $$

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 $$ \sigma = \sqrt{3.9} $$ Evaluating the expression on the right, we have $$ \sigma = 1.9748417658131 $$

To complete a poisson distribution table, first identify all of the possible values of X. Since the maximum number of occurences is 5, the values of X range from X = 0 to X = 5.

Next, find each individual poisson probability for each value of X. In this problem, we will be finding 6 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 = 3.9 $ 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 = 3.9 $, we have
$$ P(0) = \frac{{e^{-3.9}} \cdot {3.9^0}}{0!} $$
Remember, 0! is 1.
Evaluating the expression, we have
$$ P(0) = 0.020241911445804 $$

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

If using a calculator, you can enter $ \lambda = 3.9 $ 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 = 3.9 $, we have
$$ P(1) = \frac{{e^{-3.9}} \cdot {3.9^1}}{1!} $$
Evaluating the expression, we have
$$ P(1) = 0.078943454638637 $$

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

If using a calculator, you can enter $ \lambda = 3.9 $ 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 = 3.9 $, we have
$$ P(2) = \frac{{e^{-3.9}} \cdot {3.9^2}}{2!} $$
Evaluating the expression, we have
$$ P(2) = 0.15393973654534 $$

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

If using a calculator, you can enter $ \lambda = 3.9 $ 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 = 3.9 $, we have
$$ P(3) = \frac{{e^{-3.9}} \cdot {3.9^3}}{3!} $$
Evaluating the expression, we have
$$ P(3) = 0.20012165750895 $$

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

If using a calculator, you can enter $ \lambda = 3.9 $ 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 = 3.9 $, we have
$$ P(4) = \frac{{e^{-3.9}} \cdot {3.9^4}}{4!} $$
Evaluating the expression, we have
$$ P(4) = 0.19511861607122 $$

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

If using a calculator, you can enter $ \lambda = 3.9 $ 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 = 3.9 $, we have
$$ P(5) = \frac{{e^{-3.9}} \cdot {3.9^5}}{5!} $$
Evaluating the expression, we have
$$ P(5) = 0.15219252053555 $$