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I have a question about exponential distribution.

"Suppose that during business hours, your business receives a phone call, on average, every $2$ minutes. The distribution for phone calls is given by

$P(t)= \frac{e^{-t/2}}{2}$

What is the probability that you get a phone call in the first $3$ minutes of your day?"

I have solved it by using integrals and the answer is around $22\%$, however, I want to double-check it by using Poisson distribution and can't

I took $\lambda = 1/2$, $t = 3$ and got the equation

$\frac{3}{2} * e ^{-3/2}$, but it gives $0.33$ not $0.22$. It only works if the first number is one and not $3/2$ but why is it so? Please help me find my mistake.

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The waiting time to the first call has an exponential density $\frac{1}{2}e^{-t/2}$ so the probability that you wait less than three minutes is $\int_0^3\frac{1}{2}e^{-t/2}=1-e^{-3/2} \approx 0.78.$ I assume that's what you meant by "taking integrals" and your answer seems numerically correct except that you computed the probability of it not happening rather than it happing.

The number of phone calls that come within three minutes, is, as you suggest, Poisson-distributed with mean $\lambda=3/2,$ so $P(N=n)=\frac{e^{-3/2}(3/2)^n}{n!}.$ So the probability that there are no phone calls in the first three minutes is obtained just by plugging $0$ in for $n$:$$ P(N=0)= e^{-3/2}.$$ The probability that there is one or more phone call in the first three minutes is then $1-e^{-3/2},$ in agreement with what was computed by the exponential distribution.

I'm guessing your error was that you plugged in $n=1$ thinking you wanted the probability of one phone call. But that gives the probability of exactly one phone call, not of one or more.

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  • $\begingroup$ Thanks, spaceisdarkgreen. Now I understand my mistake. I have confused the probability of not receiving calls to receiving 1 call in my Poisson distribution. Thanks again! $\endgroup$ Commented Dec 19, 2019 at 0:11

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