# Thread: question on limit theorems hw!!!

1. ## question on limit theorems hw!!!

Suppose that 2500 customers subscribe to a telephone exchange. There are 80 trunklines available. Any one customer has the probability of .03 of needing a trunkline on a given call. Consider the situation as 2500 trials with probability of success p=.03, what is the approximate probability that the 2500 customers will tie-up the 80 trunklines at any given time?

2. Originally Posted by lauren2988
Suppose that 2500 customers subscribe to a telephone exchange. There are 80 trunklines available. Any one customer has the probability of .03 of needing a trunkline on a given call. Consider the situation as 2500 trials with probability of success p=.03, what is the approximate probability that the 2500 customers will tie-up the 80 trunklines at any given time?
Let X be the random variable number of customers requiring a trunkline.

X ~ Binomial(n = 2500, p = 0.03)

Calculate $\Pr(X \geq 80) = 1 - \Pr(X \leq 79)$.

(In this golden age of technology the normal approximation to the binomial distribution is not required although its use would be valid).

3. So you are saying we can approx a binomil distribution using a normal distribution?
so we would take mu to be np=2500*.03=75
and sigma^2 to be np(1-p)=2500*.03*.97=72.75
then find th z-score and use a table for the answer??

4. You can use binomial distribution, when the problem satifies the 4 characteristics listed below :

. The experiment consists of n identical trials
. Each trial results in 1 of 2 outcomes called 'success' and 'failure'
. P('success') = p remains constant from trial to trial, and p('failure') = 1-p
. Each trials is independent, i.e. not affected by results of other trials.

This characteristics perfectly suits the problem that you brought up therefore i think it is best to use binominial distribution to solve the problem. If i am not wrong u need to find th z-score and use a table for get the answer. Remember to apply the continuity correction for these kind of problem. Hope it helps.