Poisson Distribution Calculator
Answer:
Mean of the Poisson Distribution, $\mu$: 3.5
Standard Deviation of the Poisson Distribution, $\sigma$: 1.870828693387
P(0) = 0.030197383422319
P(1) = 0.10569084197811
P(2) = 0.1849589734617
P(3) = 0.21578546903865
P(4) = 0.18881228540882
P(5) = 0.13216859978617
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 $$ \mu = \lambda = 3.5 $$
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 $$ \sigma = \sqrt{3.5} $$ Evaluating the expression on the right, we have $$ \sigma = 1.870828693387 $$
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 = 3.5 $ 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 = 3.5 $, we have $$ P(0) = \frac{{e^{-3.5}} \cdot {3.5^0}}{0!} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.030197383422319 $$
$P(1)$ Probability of exactly 1 occurrences
If using a calculator, you can enter $ \lambda = 3.5 $ 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 = 3.5 $, we have $$ P(1) = \frac{{e^{-3.5}} \cdot {3.5^1}}{1!} $$ Evaluating the expression, we have $$ P(1) = 0.10569084197811 $$
$P(2)$ Probability of exactly 2 occurrences
If using a calculator, you can enter $ \lambda = 3.5 $ 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 = 3.5 $, we have $$ P(2) = \frac{{e^{-3.5}} \cdot {3.5^2}}{2!} $$ Evaluating the expression, we have $$ P(2) = 0.1849589734617 $$
$P(3)$ Probability of exactly 3 occurrences
If using a calculator, you can enter $ \lambda = 3.5 $ 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 = 3.5 $, we have $$ P(3) = \frac{{e^{-3.5}} \cdot {3.5^3}}{3!} $$ Evaluating the expression, we have $$ P(3) = 0.21578546903865 $$
$P(4)$ Probability of exactly 4 occurrences
If using a calculator, you can enter $ \lambda = 3.5 $ 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 = 3.5 $, we have $$ P(4) = \frac{{e^{-3.5}} \cdot {3.5^4}}{4!} $$ Evaluating the expression, we have $$ P(4) = 0.18881228540882 $$
$P(5)$ Probability of exactly 5 occurrences
If using a calculator, you can enter $ \lambda = 3.5 $ 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 = 3.5 $, we have $$ P(5) = \frac{{e^{-3.5}} \cdot {3.5^5}}{5!} $$ Evaluating the expression, we have $$ P(5) = 0.13216859978617 $$