Question

Suppose that the average number of people log in to a particular web site is 5 per minute.

-What are the mean and variance of the distribution of the number of people log in to this web site?
[Ans.] ok since poisson distribution is 5 (lambda) so is the mean and various are also 5??

-What is the probability that the number of people will log in to this web site is at most 4 in the next minute?
[Ans.] here is my answer but I'm not sure about it.. correct me if I'm wrong
lambda = 5
x~pos(9) and
p(x => 4) = 1-p(x =< 3)
1- 0.265 = 0.735
is this the correct way ?

-What is the probability that the number of people will log in to this web site is between 2 and 5 (inclusive) in the next 2 minutes?
... i still have no idea how to do this

Answer 1

start solving

Answer 2

First part looks right.

2nd part: "At most 4" means "will not exceed 4", meaning \(P(X \le 4)\)

In this part, you have a different time asked. The Poisson distribution in general can be described with the pmf; \[ P(X=x)=\frac{(\lambda t)^xe^{-\lambda t}}{x!}\], where \(\lambda\) is the rate of occurrence, and \(t\) is the time period.
In the 1st and 2nd part, you had \(\lambda=\frac{5}{\text{minute}}, t=1 \text{ minute}, \), so \(\lambda t = 5\)

In this 3rd part, \(\lambda=\frac{5}{\text{minute}}, t=2 \text{ minutes}, \), so \(\lambda t = 10\), so you have a Poisson(10) distribution, and the probability asked is \(P(2 \le X \le 5)\)