1. ## Exponential distribution

Hi

I am really stuck with some parts of this question, and some I think I've got. I can't get my head around this

A bank has a central queue served by four assistants - two of the four assistants are exponentially distributed with mean 4 minutes, the other two exponentially distributed with mean 6 minutes.

When I enter the bank all 4 assistants are busy, but no-one is waiting to be served.

Q1. Find the distribution of the time I have to wait before I can move forward for service, hence show my expected waiting time is 72 seconds

I have got as far as $\displaystyle \sim M(\frac{1}{4} + \frac{1}{4} + \frac{1}{6} + \frac {1}{6})$ or $\displaystyle M(\frac{5}{6})$ but I have no idea how to show the expected waiting time is as above.

Q2. Calculate the probability that I have to wait more than 1 1/2 minutes until an assistant is free to serve me.

I think I've got this one...

$\displaystyle P(X>1.5) = e^{\lambda.x}$ where $\displaystyle \lambda = 5/6$ and $\displaystyle x = 1.5$ giving me 0.287

Any help would be very much appreciated

2. I'm not exactly sure what you did (although you seem to already have the answer), but the distribution of the exponential random variable that is the smallest of the four would be the following:

$\displaystyle P\big(min(X_{1},X_{2},X_{3}, X_{4}) > x \big) = P(X_{i}>x \ \text{for i=1,2,3,4})$

$\displaystyle = P(X_{1}>x) * P(X_{2}>x) *P(X_{3}>x) * P(X_{4}>x)$ (by independence)

$\displaystyle = e^{-x/6}*e^{-x/6}*e^{-x/4}*e^{-x/4}$

$\displaystyle = e^{-5x/6}$

which implies that the cdf is $\displaystyle 1- e^{-5x/6}$, which is the cdf of a exponential distribution with mean 6/5 minutes = 72 seconds