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Question: At ABC insurance company, suppose the patient insurance inquiries arrive at mean rate of $2.2$ calls per minute. Compute the probability of waiting more than $30$ seconds for the next call.

I am confused here. When I first read the question, I thought of modelling the situation using Poisson distribution because of $2.2$ calls per minute. However, the question is asking for time between two calls, which is not discrete.

If I follow my initial thought to solve the question, then define $X$ to be the number of calls per $30$ seconds, then from the question, $\lambda = 2.2 /2 = 1.1.$ Then I think we need to calculate $\mathbb{P}(X = 1).$

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  • $\begingroup$ The Poisson model is appropriate. Under this model, the time until the next call follows an exponential distribution with mean $\frac{1}{2.2}$. $\endgroup$
    – Yuta
    Sep 6, 2018 at 5:11
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    $\begingroup$ Waiting more than $30$ seconds for the next call means there is no calls in the first $30$-second interval. So the required probability should be $\mathbb{P}(X=0)$. $\endgroup$
    – Yuta
    Sep 6, 2018 at 5:17

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I believe you can solve using either.

Using Poisson

Let $X$ be the number of calls in $30$ seconds. Then $X \sim Poisson(\lambda = 1.1)$ and $P(X=0) = e^{-1.1}$

Using Exponential

Let $Y$ be the time until the first call. The average time to the first call is $60/2.2 = 27.27$ seconds.

That means $E[Y] = 27.27$ which means $\frac{1}{\lambda} = 27.27$ so $\lambda = .0366$ and then $P(Y > 30) = 1-P(Y \le 30) = 1-(1-e^{-30 *.0366}) = e^{-1.1}$

I think the key thing is that $\lambda$ is a rate parameter so if someone says $2.2$ every minute you can chop it up to suit your needs... I am learning this stuff as well so let me know if you have any doubts.

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