0
$\begingroup$

I have a restaurant with service time being exponentially distributed. Let's say the food-serving time has a mean of 30 minutes, and the checkout time has a mean of 10 minutes.

Would $\lambda$ in this problem $=\frac{1}{30+10}$ or $=\frac{1}{30} + \frac{1}{10}$?

$\endgroup$
  • $\begingroup$ @orange right so I was looking into that as well but Gamma has too many parameters that it confuses me $\endgroup$ – PTN Mar 28 at 7:36
  • $\begingroup$ math.stackexchange.com/questions/635443/… $\endgroup$ – badatmath Mar 28 at 7:48
  • 1
    $\begingroup$ If you want the mean, you can get it easily: the mean of a sum is the sum of the means, so the mean here is $30+10=40$ minutes. $\endgroup$ – Minus One-Twelfth Mar 28 at 8:05
1
$\begingroup$

Given two independent exponential random variables $X$ and $Y$ with distributions $f_X(t) = \mu e^{-\mu t}$ and $f_Y(t) = \lambda e^{-\lambda t}$, the distribution of the sum $X+Y$ is given by the convolution: \begin{align} f_{X+Y}(t) &= \int_0^t \mu e^{-\mu \tau}\lambda e^{-\lambda (t-\tau)}\;d\tau\\ &=\lambda\mu e^{-\lambda t}\int_0^t e^{-(\mu-\lambda)\tau}\;d\tau\\ &=\frac{\lambda\mu}{\lambda-\mu}\left[e^{-\mu t} - e^{-\lambda t}\right] \end{align}

The mean is given by \begin{align} \frac{\lambda\mu}{\lambda-\mu}\int_0^\infty t\left[e^{-\mu t} - e^{-\lambda t}\right]\;dt&= \frac{\lambda\mu}{\lambda - \mu}\left[\frac{1}{\lambda^2} - \frac{1}{\mu^2}\right]\\ &= \frac{\lambda+\mu}{\lambda\mu} \end{align} which in your case is $40$ minutes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.