Question

Suppose that phone calls to the President’s office at Lehman College arrive independently and at random, with outside calls arriving at an average rate of 3 in any 10 minute period, and inside calls at an average rate of 4 in any 10 minute period. Compute the probability that 2 or more calls arrive in any 2 minute period.

Answer 1

40.82% probability that 2 or more calls arrive in any 2 minute period.

Explanation:

We have the mean during a time interval, so we use the poisson distribution to solve this question.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given time interval.

10 minutes.

3 outside calls, 4 inside. So mean of 3+4 = 7.

That is, 0.7 calls per minute.

Compute the probability that 2 or more calls arrive in any 2 minute period.

2 minutes, so
`\mu = 2*0.7 = 1.4`

Either less than two calls arrive, or two or more do. The sum of the probabilities of these events is decimal 1. So

`P(X < 2) + P(X \geq 2) = 1`

We want
`P(X \geq 2)`. So

`P(X \geq 2) = 1 - P(X < 2)`

In which

`P(X < 2) = P(X = 0) + P(X = 1)`

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-1.4)*(1.4)^(0))/((0)!) = 0.2466`

`P(X = 1) = (e^(-1.4)*(1.4)^(1))/((1)!) = 0.3452`

`P(X < 2) = P(X = 0) + P(X = 1) = 0.2466 + 0.3452 = 0.5918`

`P(X \geq 2) = 1 - P(X < 2) = 1 - 0.5918 = 0.4082`

40.82% probability that 2 or more calls arrive in any 2 minute period.