Question

e number of views of a page on a Web site follows a Poisson per minu (a) Find the probability of no views in 2 minutes, provided that A 3. (b) Find the probability of less than 3 views in 5 minutes, provided that 2 (c) Find λ if you know that the probability of precisely 3 views in a minute is equal to 0.18044.

Answer 1

ANSWER:

The number of views of a page on a Web site follows a Poisson distribution for a mean of lamda per minute. (a) Find the probability of no views in 2 minutes, provide that

For Poisson:

P(x) = e^-$\lambda$ * $\lambda$^x/x!, e = 2.71828

(a) Given,

$\lambda$ = 3/minute, therefore for 2 minutes $\lambda$ = 6 to find P(0)

P(0) = e^-6 * 6⁰/0! = 0.0025

(b) Given, $\lambda$ = 1/minute,

therefore for 5 minutes $\lambda$ = 5.

To find P(X < 3) = P(0) + P(1) + P(2)

P(0) = e^-5 * 5⁰/0! = 0.0067

P(1) = e^-5 * 5¹/1! = 0.0337

P(2) = e^-5 * 5²/2! = 0.0842

Therefore ,

P(X < 3) = 0.0067 + 0.0337 + 0.0842 = 0.1246

(c) P(X = 3) = (4/3) * e^-2 = e^-$\lambda$ * $\lambda$³/3!

Since 4/3 and $\lambda$³/3! are constants,

we can understand from the equation that e^-2 = e^-$\lambda$ .

Therefore $\lambda$ = 2