Poisson Distribution Calculator
Answer:
Mean of the Poisson Distribution, $\mu$: 5.2
Standard Deviation of the Poisson Distribution, $\sigma$: 2.2803508501983
P(0) = 0.0055165644207608
P(1) = 0.028686134987956
P(2) = 0.074583950968686
P(3) = 0.12927884834572
P(4) = 0.16806250284944
P(5) = 0.17478500296342
P(6) = 0.15148033590163
P(7) = 0.11252824952692
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 $$ \mu = \lambda = 5.2 $$
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 $$ \sigma = \sqrt{5.2} $$ Evaluating the expression on the right, we have $$ \sigma = 2.2803508501983 $$
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 = 5.2 $ 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 = 5.2 $, we have $$ P(0) = \frac{{e^{-5.2}} \cdot {5.2^0}}{0!} $$ Remember, 0! is 1. Evaluating the expression, we have $$ P(0) = 0.0055165644207608 $$
$P(1)$ Probability of exactly 1 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(1) = \frac{{e^{-5.2}} \cdot {5.2^1}}{1!} $$ Evaluating the expression, we have $$ P(1) = 0.028686134987956 $$
$P(2)$ Probability of exactly 2 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(2) = \frac{{e^{-5.2}} \cdot {5.2^2}}{2!} $$ Evaluating the expression, we have $$ P(2) = 0.074583950968686 $$
$P(3)$ Probability of exactly 3 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(3) = \frac{{e^{-5.2}} \cdot {5.2^3}}{3!} $$ Evaluating the expression, we have $$ P(3) = 0.12927884834572 $$
$P(4)$ Probability of exactly 4 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(4) = \frac{{e^{-5.2}} \cdot {5.2^4}}{4!} $$ Evaluating the expression, we have $$ P(4) = 0.16806250284944 $$
$P(5)$ Probability of exactly 5 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(5) = \frac{{e^{-5.2}} \cdot {5.2^5}}{5!} $$ Evaluating the expression, we have $$ P(5) = 0.17478500296342 $$
$P(6)$ Probability of exactly 6 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(6) = \frac{{e^{-5.2}} \cdot {5.2^6}}{6!} $$ Evaluating the expression, we have $$ P(6) = 0.15148033590163 $$
$P(7)$ Probability of exactly 7 occurrences
If using a calculator, you can enter $ \lambda = 5.2 $ 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 = 5.2 $, we have $$ P(7) = \frac{{e^{-5.2}} \cdot {5.2^7}}{7!} $$ Evaluating the expression, we have $$ P(7) = 0.11252824952692 $$