Poisson Distribution Calculator
Answer:
Mean of the Poisson Distribution, $\mu$: 6.1
Standard Deviation of the Poisson Distribution, $\sigma$: 2.4698178070457
P(0) = 0.0022428677194858
P(1) = 0.013681493088863
P(2) = 0.041728553921033
P(3) = 0.084848059639435
P(4) = 0.12939329095014
P(5) = 0.15785981495917
P(6) = 0.16049081187515
P(7) = 0.13985627891978
P(8) = 0.10664041267633
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 $$ \mu = \lambda = 6.1 $$
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 $$ \sigma = \sqrt{6.1} $$ Evaluating the expression on the right, we have $$ \sigma = 2.4698178070457 $$
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 = 6.1 $ 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 = 6.1 $, we have $$ P(0) = \frac{{e^{-6.1}} \cdot {6.1^0}}{0!} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.0022428677194858 $$
$P(1)$ Probability of exactly 1 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(1) = \frac{{e^{-6.1}} \cdot {6.1^1}}{1!} $$ Evaluating the expression, we have $$ P(1) = 0.013681493088863 $$
$P(2)$ Probability of exactly 2 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(2) = \frac{{e^{-6.1}} \cdot {6.1^2}}{2!} $$ Evaluating the expression, we have $$ P(2) = 0.041728553921033 $$
$P(3)$ Probability of exactly 3 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(3) = \frac{{e^{-6.1}} \cdot {6.1^3}}{3!} $$ Evaluating the expression, we have $$ P(3) = 0.084848059639435 $$
$P(4)$ Probability of exactly 4 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(4) = \frac{{e^{-6.1}} \cdot {6.1^4}}{4!} $$ Evaluating the expression, we have $$ P(4) = 0.12939329095014 $$
$P(5)$ Probability of exactly 5 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(5) = \frac{{e^{-6.1}} \cdot {6.1^5}}{5!} $$ Evaluating the expression, we have $$ P(5) = 0.15785981495917 $$
$P(6)$ Probability of exactly 6 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(6) = \frac{{e^{-6.1}} \cdot {6.1^6}}{6!} $$ Evaluating the expression, we have $$ P(6) = 0.16049081187515 $$
$P(7)$ Probability of exactly 7 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(7) = \frac{{e^{-6.1}} \cdot {6.1^7}}{7!} $$ Evaluating the expression, we have $$ P(7) = 0.13985627891978 $$
$P(8)$ Probability of exactly 8 occurrences
If using a calculator, you can enter $ \lambda = 6.1 $ 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 = 6.1 $, we have $$ P(8) = \frac{{e^{-6.1}} \cdot {6.1^8}}{8!} $$ Evaluating the expression, we have $$ P(8) = 0.10664041267633 $$