Question

Suppose that the log-ons to a computer network follow a Poisson process with an average of three counts per minute.

Required:
a. What is the mean time between counts?
b. What is the standard deviation between counts (in minutes)?
c. If it is an average of 3 counts per minute, find the value of x such that P(X < x) = 0.95

Answer 1

a) 3

b)3

c) 8.9872

Explanation:

a) According to the question, mean is 3.

b) Since it is an exponential distribution, standard deviation equals to mean. Therefore, the answer is 3.

c) P[x<X]=0.95

f(x)= λ*e^(-λ*x)

∫f(x)dx, 0 to x

mean = 1/λ =3. So, λ= 1/3.

∫(1/3)*e^(-x/3)dx= 1/3*e^(-x/3)/(-1/3)=-e^(-x/3) 0 to x

-e^(-x/3)+1 = 0.95

-e^(-x/3)=-0.05

x/3=ln(0.05)

x=ln(0.05)*3= 8.9872

Answer 2

a. E(X) = 0.333 minutes

b.
`\mathbf{\sigma_x =0.333 \ minutes}`

c. 0.9986

Explanation:

a.

To find the mean time between counts

Let X be the random variable that illustrates the time between the successive log-ons, then X is exponential with the rate
`\lambda = 3` log-ons per minute.

i.e.

`E(X) = (1)/(\lambda)`

`E(X) = (1)/(3)`

E(X) = 0.333 minutes

b. Since X is attributed to an exponential distribution, The standard deviation between counts is:

`\sigma_x = (1)/(3)`

`\mathbf{\sigma_x =0.333 \ minutes}`

c.

Given that;

The average number of count per minute = 3

The value of X such that P(X < x) = 0.95 can be calculated as;

`\int \limits ^x_0 \lambda e^(-\lambda t)= 0.95`

`3\int \limits ^x_0 \lambda e^(3 t)= 0.95`

`\bigg [ -e^(3t)\bigg ]^x_0 = 0.95`

`e^(3t) = 1-0.95`

-3x = ㏑ (0.05)

-3x = - 2.9957

x = -2.9957/ -3

x = 0.9986

The value of x = 0.9986