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Probability Distributions

  1. Jul 4, 2007 #1


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    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.

    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 \\

    Does that look right?

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

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


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  4. Jul 4, 2007 #3

    D H

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    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.
  5. Jul 5, 2007 #4


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    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?
  6. Jul 5, 2007 #5
    Is that between 20 and 45 inclusive? If so, then there are 26 possibilities. If not, then there are only 24 possibilities.
  7. Jul 5, 2007 #6
    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.
  8. Jul 5, 2007 #7


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    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.

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