# The average rate of job submissions in a computer center is 2 per minute.

Full Question: The average rate of job submissions in a computer center is 2 per minute. If it can be assumed that the number of submissions per minute has a Poisson distribution, calculate the probability that

(a) more than two jobs will arrive in a minute

$P(X>2) = 1- P(X≤ 2) = 1- [P(X=0)+P(X=1)+P(X=2)]$

and I got 0.3233235 as my answer. Is that correct?

Example: For $P(X=1),$ I did $((2^1)(e^-2))/1!$

(b) at least 30 sec will elapse between any two jobs

This is where I'm confused. A similar problem was answered using Exponential distribution in the book even though the question said this was a Poisson distribution problem.

I tried with Exponential and this is what I got:

Since 30 seconds is $1/2$ of 1 minute, $(1/2) * λ$ or $(1/2)*2 = 1$

$P(X ≥ 1/2) = 1 - P(x < 1/2) = 1 - (1- e^-1) = 0.3678794$

Please tell me if I used the correct distribution functions for both a and b, and if I answered them correctly. Thank you.

• It looks right to me. Nov 30, 2016 at 9:57
• See Question 1986790 on this site for one explanation of the connection between Poisson and exponential distributions. Nov 30, 2016 at 23:58
• In the second part, why is it 1 - (1− e−1)? Sep 9, 2018 at 13:26

(a) Counting Poisson events: Let $X \sim Pois(\lambda = 2).$ Then $P(X > 2) = 0.3233236.$ From R statistical software:
1 - ppois(2, 2)   $'ppois' is Poisson CDF ## 0.3233236  (b) Exponential interarrival times: The waiting time$W$for the next job is$W \sim Exp(rate = \lambda = 2).$By the no-memory property, the starting time is not relevant, and$W$is also the wait between job arrivals (interarrival time). Then$P(W > .5) = 0.3678794.$[Note that some books and software (including R) parameterize the exponential distribution with the rate$\lambda,$while others use the mean$\mu = 1/\lambda.$Please check the definition and the PDF and CDF in your text.] 1 - pexp(.5, 2) # 'pexp' is exponential CDF (rate parameter) ## 0.3678794  This also amounts to saying no Poisson event occurs in a time interval of length 1/2, for which the arrival rate is$\lambda^\prime = 1.$Then$P(X^\prime = 0) = 0.3678794.\$
 dpois(0, 1)     # 'dpois' is Poisson PDF