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

Answer:

P(1≤X≤3) = 0.5974

Step-by-step explanation:

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

So

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