Question

Assume that during each second, a job arrives at a webserver with probability 0.03. Use the Poisson distribution to estimate the probability that at least one job will be missed if the server goes down for 57 seconds. Round to at least 6 decimal places, or enter an exact answer using e and exponents.

Answer 1

Answer:

The probability that at lest one job will be missed in 57 second is$=1- e^{-1.71}$ =0.819134

Step-by-step explanation:

Poisson distribution:

A discrete random variable X having the enumerable set {0,1,2,....} as the spectrum, is said to be Poisson distribution.

$P(X=x)=\frac{e^{-\lambda t}({-\lambda t})^x}{x!}$  for x=0,1,2...

λ is the average per unit time

Given that, a job arrives at a web server with the probability 0.03.

Here λ=0.03, t=57 second.

The probability that at lest one job will be missed in 57 second is

=P(X≥1)

=1- P(X<1)

=1- P(X=0)

$=1-\frac{e^{-1.71}(1.71)^0}{0!}$

$=1- e^{-1.71}$

=0.819134