# Probability Distributions

1. Jul 4, 2007

### danago

A group of students wish to determine how long, on average, customers are waiting in line at a supermarket before being served.

The students conduct trials and record the times taken. They found that they were kept waiting for an average of 7 minutes.

If a customer goes to that same supermarket, what is the probability they will be waiting more than 7 minutes in line before being served?

Originally, i would have thought to model this situation using a Gaussian distribution, but i am not given a standard deviation to work with, only a mean.

My next thought was to use an exponential distribution function.

$$\begin{array}{c} P(x > 7) = \int_7^\infty {\frac{{e^{ - \frac{x}{7}} }}{7}} dx \\ = \mathop {\lim }\limits_{a \to \infty } \int_7^a {\frac{{e^{ - x/7} }}{7}} dx \\ = \mathop {\lim }\limits_{a \to \infty } \left[ { - e^{ - x/7} } \right]_7^a \\ = \mathop {\lim }\limits_{a \to \infty } (e^{ - 1} - e^{ - a/7}) \\ = e^{ - 1} \approx 0.368 \\ \end{array}$$

Does that look right?

Also, if i was given more information, would a gaussian distribution have been suitable?

Thanks,
Dan.

Last edited: Jul 4, 2007
2. Jul 4, 2007

### HallsofIvy

Staff Emeritus
3. Jul 4, 2007

### D H

Staff Emeritus
Strictly speaking, a Gaussian distribution is never appropriate as a model of a queueing process. How can one wait a negative amount of time in a line?

Queueing models often are modeled as Poisson processes with an underlying exponential distribution. Your assumption was a good one. Moreover, you were given but one statistic. This fits well with the exponential distribution, which takes only one parameter.

4. Jul 5, 2007

### danago

Alright thanks for the replies :)

I have another question i needed help with:

A spinner has 100 equal segments marked out on it, numbered from 1 to 100. What is the probability that the dial lands on a number between 20 and 45?

I would have just thought the probability would have been (45-20)/100=0.25

Is it as straight forward as that, or am i missing something?

5. Jul 5, 2007

### daveb

Is that between 20 and 45 inclusive? If so, then there are 26 possibilities. If not, then there are only 24 possibilities.

6. Jul 5, 2007

### ZioX

Well, think about it logically. If the probability is uniform, that is, each segment has an equal chance of being hit, then it would be that simple.

7. Jul 5, 2007

### NateTG

Well, as soon as the problem defines 'average' in a useful fashion...

You assume that 7 minutes is the mean, but...
If 7 minutes is the median, the problem becomes very easy.
If 7 minutes is the mode, then things get a bit more interesting.