Answer:
Step-by-step explanation:
Given that :
A web page is accessed at an average of 20 times an hour.
Therefore:
a. he rate parameter λ of the distribution of the time until the first hit = 20
b. What is the expected time between hits?
Let consider E(Y) to be the expected time between the hits; Then :
E(Y) = 1/λ
E(Y) = 1/20
E(Y) = 0.05 hours
E(Y) = 3 minutes
(c.) What is the probability that there will be less than 5 hits in the first hour?
Let consider X which follows Poisson Distribution; Then,
P(X<5) [tex]\sim[/tex] G(∝=5, λ = 20)
For 5 hits ; the expected time will be :
Let 5 hits be X
E(X) = ∝/λ
E(X) = 5/20
E(X) =1/4
E(X) = 0.25 hour
E(X) = 15 minutes
From above ; we will see that it took 15 minutes to get 5 hits; then
[tex]P(\tau \geq 0.25) = \int\limits^{\alpha}_{0.25} \dfrac{\lambda^{\alpha}}{\ulcorner^{\alpha}} t^{a\pha-1} \ e^{-\lambda t} \, dt[/tex]
[tex]P(\tau \geq 0.25) = \int\limits^{5}_{0.25} \dfrac{20^{5}}{\ulcorner^{5}} t^{5-1} \ e^{-20 t} \, dt[/tex]
[tex]\mathbf{P(\tau \geq 0.25) =0.4405}[/tex]
