Respuesta :
Answer:
[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}= e^{-4}= 0.0183[/tex]
Step-by-step explanation:
Previous concepts
Let X the random variable that represent the number of new visitors to a website in one hour. We know that [tex]X \sim Poisson(\lambda=4)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
Solution to the problem
On this case we are interested on the probability of having no visitors to the website:.
Using the pmf we can find this probability like this:
[tex]P(X=0)=\frac{e^{-\lambda} \lambda^0}{0!}=e^{-\lambda}= e^{-4}= 0.0183[/tex]
