Poisson Distribution Calculator
Answer:
Mean of the Poisson Distribution, $\mu$: 5.9
Standard Deviation of the Poisson Distribution, $\sigma$: 2.4289915602982
P(0) = 0.0027394448187684
P(1) = 0.016162724430733
P(2) = 0.047680037070663
P(3) = 0.093770739572305
P(4) = 0.13831184086915
P(5) = 0.1632079722256
P(6) = 0.16048783935517
P(7) = 0.13526832174221
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 $$ \mu = \lambda = 5.9 $$
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 $$ \sigma = \sqrt{5.9} $$ Evaluating the expression on the right, we have $$ \sigma = 2.4289915602982 $$
To complete a poisson distribution table, first identify all of the possible values of X. Since the maximum number of occurences is 7, the values of X range from X = 0 to X = 7.
Next, find each individual poisson probability for each value of X. In this problem, we will be finding 8 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 = 5.9 $ 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 = 5.9 $, we have $$ P(0) = \frac{{e^{-5.9}} \cdot {5.9^0}}{0!} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.0027394448187684 $$
$P(1)$ Probability of exactly 1 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(1) = \frac{{e^{-5.9}} \cdot {5.9^1}}{1!} $$ Evaluating the expression, we have $$ P(1) = 0.016162724430733 $$
$P(2)$ Probability of exactly 2 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(2) = \frac{{e^{-5.9}} \cdot {5.9^2}}{2!} $$ Evaluating the expression, we have $$ P(2) = 0.047680037070663 $$
$P(3)$ Probability of exactly 3 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(3) = \frac{{e^{-5.9}} \cdot {5.9^3}}{3!} $$ Evaluating the expression, we have $$ P(3) = 0.093770739572305 $$
$P(4)$ Probability of exactly 4 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(4) = \frac{{e^{-5.9}} \cdot {5.9^4}}{4!} $$ Evaluating the expression, we have $$ P(4) = 0.13831184086915 $$
$P(5)$ Probability of exactly 5 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(5) = \frac{{e^{-5.9}} \cdot {5.9^5}}{5!} $$ Evaluating the expression, we have $$ P(5) = 0.1632079722256 $$
$P(6)$ Probability of exactly 6 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(6) = \frac{{e^{-5.9}} \cdot {5.9^6}}{6!} $$ Evaluating the expression, we have $$ P(6) = 0.16048783935517 $$
$P(7)$ Probability of exactly 7 occurrences
If using a calculator, you can enter $ \lambda = 5.9 $ 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 = 5.9 $, we have $$ P(7) = \frac{{e^{-5.9}} \cdot {5.9^7}}{7!} $$ Evaluating the expression, we have $$ P(7) = 0.13526832174221 $$