Probability Distributions

[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.

 

[HallsofIvy]

Why would you use a Gaussian distribution? Or exponential? The Poisson Distribution is most commonly used for "waiting time":
http://en.wikipedia.org/wiki/Poisson_distribution

 

[D H]

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.

 

[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?

 

[daveb]

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

 

[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.

 

[NateTG]

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

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.