Mathematics: 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.

BStep-by-step explanation:Just took the quiz...

Suppose that phone calls to the President’s office

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23.01.2022

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

Step-by-step 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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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)

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.4}*(1.4)^{0}}{(0)!} = 0.2466$

$P(X = 1) = \frac{e^{-1.4}*(1.4)^{1}}{(1)!} = 0.3452$

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.

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