Question

A random variable X has a Poisson distribution with a mean of 3. What is the probability that X P(1≤X≤3) ?.

Answer 1

P(1≤X≤3) = 0.5974

Explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given interval.

Mean of 3

This means that
`\mu = 3`

P(1≤X≤3) ?

`P(1 \leq X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3)`

So

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 1) = (e^(-3)*3^(1))/((1)!) = 0.1494`

`P(X = 2) = (e^(-3)*3^(2))/((2)!) = 0.2240`

`P(X = 3) = (e^(-3)*3^(3))/((3)!) = 0.2240`

So

`P(1 \leq X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3) = 0.1494 + 0.2240 + 0.2240 = 0.5974`