Question

Suppose X has a Poisson distribution with a mean of .4. How would I find the probability when X=0, X=4, X $\leqslant$2 and when X=8?

Answer 1

Hint: Poisson distribution gives the method for determining probability in cases in which probability of occurrence is small and constant. It is usually used in cases like mean activity for a radioactive sample is given then probability for any activity in a given interval can be found.
Formula used:
The formula for probability by Poisson’s distribution is given as:
$P(X = n) = {{{\mu ^n}} \over {n!}}{e^{ - \mu }}$.
X is a random variable, $\mu$ is the mean of the variable for the distribution and n is the value of X at which or up to which we are determining the probability.

Complete answer:
We are given that the mean of the variable for the distribution is $\mu$=0.4. Now, substituting the value of n one by one we get:
(i) $P(X = 0) = {{{\mu ^0}} \over {0!}}{e^{ - \mu }} = {e^{ - (0.4)}} = 0.67$
(ii) $P(X = 4) = {{{{(0.4)}^4}} \over {4!}}{e^{ - (0.4)}} = 7.15 \times {10^{ - 4}}$
(iii) $P(X = 8) = {{{{(0.4)}^8}} \over {8!}}{e^{ - (0.4)}} = 1.089 \times {10^{ - 8}}$
(iv) Up to X=2, we have X =0, X =1 and then X=2. Adding these three will give us X $\leqslant$2 probability.
So,
$\eqalign{
  & P(X = 1) = \dfrac{{{{(0.4)}^1}}}{{1!}}{e^{ - (0.4)}} = 0.268 \cr
  & P(X = 2) = \dfrac{{{{(0.4)}^2}}}{{2!}}{e^{ - (0.4)}} = 0.0536 \cr
  & \therefore P(X \leqslant 2) = 0.67 + 0.268 + 0.0536 = 0.9916 \cr} $

Note:
Poisson distribution is a discrete distribution so up to 2, we could simply add the previous probabilities. There are also continuous distributions like normal distribution in which simple addition does not give the probability. In such cases integration has to be performed. A random variable could be anything that we define like X could be the number of times heads appear in n throws of a die. Mean is generally obtained by adding all products of the variables and their respective probabilities also, mean of a quantity is the average of that quantity. It might not be an actual random variable number for our distribution, it is just a statistical tool at our dispense.