[Math] Exponential distribution question!

probability distributions

Suppose that the time between calls from your best friend has an exponential
distribution with a mean time of $3$ days.

(a) If you just received a call from her, what is the probability that you will receive the next call within the next $2$ days?

(b) You realize that you have not received a call in at least $2$ days. What is the probability that she will not call you within the next $2$ days?

My attempt:

if $\mu=3$ then $m=1/\mu = {1/3}$ hence $X \sim \exp(1/3)$

(a) $P(X\leq2) = 1-e^{-1/3*2} = 0.4866$

(b) $P(X > 2) = 1-P(X\leq2) = 1-(1-e^{-1/3*2}) = e^{-2/3} = 0.5134$

I think my answers are incorrect. I also tried $X \sim \exp(3)$ but those answer make even less sense than the ones above.

Will anyone please point me in the correct direction if my approach is incorrect please?

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Your answers are correct. Perhaps tell us why you thought they were incorrect, because it's important to improve on the intuition that led you to think this.

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[Math] exponential distribution – get some meaning

It seems obvious that the more time it passes by without an earthquake happening, the more likely it is than an earthquake will happen.

As explained in the comments, to use an exponential distribution is to assume the opposite: that whatever time passed by without an earthquake happening, still as likely as ever it is than an earthquake will happen.

So, if one thinks this rate should not be constant, how to model the time $X$ until the first earthquake happens? The hypothesis is that, for every $t\geqslant0$, when $s\to0^+$, $$ P[X\leqslant t+s\mid X\geqslant t]=r(t)s+o(s), $$ for some function $r$. Thus, the function $\bar F:t\mapsto P[X\geqslant t]$ is such that $$ \bar F(t+s)=\bar F(t)\cdot(1-r(t)s+o(s)), $$ for every $t\geqslant0$, that is, $$ \bar F'(t)=-r(t)\bar F(t). $$ Since $\bar F(0)=1$ by definition, one gets $$ \bar F(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right), $$ and finally, $X$ has PDF $$ f(t)=\exp\left(-\int_0^tr(u)\,\mathrm du\right)r(t)\,\mathbf 1_{t\geqslant0}. $$ One sees that every density $f$ can be written like that, since one recovers the instantaneous rate $r$ through the formula $$ r(t)=\frac{f(t)}{\bar F(t)}=\frac{f(t)}{1-F(t)}. $$

[Math] Reverse Poisson Distribution problem

I believe you are being asked to focus on a typical 5-minute period of time. Then the average number $X$ of incoming calls within 5 minutes has the distribution $X \sim \mathsf{Pois}(\lambda = 4.5).$ A bar plot of this distribution is shown below:

Poisson Model: Computations with the Poisson PDF $P(X = i) = e^{-\lambda}\frac{\lambda^i}{i!},$ where $\lambda = 4.5,$ show that $P(X \le 6) = 0.8311 < 0.9$ and $P(X \le 7) = 0.9134.$ Thus seven agents would suffice to serve incoming customers at least 90% of the time.

Trial and error with Poisson CDF: Computations using R statistical software (where the CDF of a Poisson distribution is denoted ppois) are as follows:

ppois(6, 4.5)
## 0.8310506
ppois(7, 4.5)
## 0.9134135

Using a Poisson quantile function: If you have such software available, you can use the inverse CDF or quantile function qpois to get to the answer without exploration.

qpois(.9, 4.5)
## 7

Note: Without knowing the context of this problem in your course, I suppose something like this is the approach you are expected to take. However, there are some unstated assumptions involved. A complete analysis of such a problem would involve finding the required number of servers $k$ required in an M/M/k queue (at steady state) that are sufficient to keep the average number of customers in the system below $k$ 90% of the time. This would involve knowing the exponential arrival rate $\lambda = 4.5$ per five minutes or $0.9$ per minute, and knowing the exponential rate (often denoted $\mu)$ at which each server can finish serving customers.

For starters: here are a few unstated assumptions: (a) No agents are busy at the beginning of the 5-minute period. (b) Each agent handles only one call within this period. (c) We are not concerned whether agents are still handling calls from this 5-minute period ends and the next begins.

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Related Solutions

[Math] exponential distribution – get some meaning

[Math] Reverse Poisson Distribution problem

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