Question

If we assume the number of visits to a blog has a Poisson distribution, with average of 99 visits per second.

Let X be the waiting time until one visit to the blog:

a) What distribution does X follow? State the probability density function of x and give an appropriate graphing rage of x for f(x)

b) What is the probability that at least 50 seconds are needed until one visit occur?

Answer 1

a) As we are given here the mean visits per second as 99. Therefore the mean waiting time for 1 visit would be given as 1/99 sec

For exponential distribution, the parameter is reciprocal of its mean. Therefore the distribution of waiting here is given as:

$X \sim exp(99)$

The PDF here is therefore given as:

$f_x(x) = \lambda e^{-\lambda x}$

This is the required PDF here.

The appropriate range here is given as:
X ranges from 0 to Infinity here that is [0, $\infty$ ) here.

b) The probability that at least 50 seconds are needed until one visit occur is computed here as:

= Probability that waiting time is more than 50 seconds

$P(X \geq 50) = \int_{50}^{\infty }99e^{-99x} \ dx = e^{-99*50} = 0$

Therefore 0 is the required probability here.