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Math Help - Uniform Distribution question

  1. #1
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    Uniform Distribution question

    Hey all, this is from a past exam but we were not given any solutions.

    The Question:
    Many students find that sitting through an entire lecture is difficult to do without falling asleep. Suppose the number of minutes a single, randomly selected student is asleep during a lecture is uniformly distributed between 7 and 14 minutes.

    1.) Over the course of the semester (say, 50 lectures), what is the probability that a randomly selected student will sleep a total of between 130 and 140 minutes?
    Assume that the minutes slept in any lecture is independent of the number of minutes slept in any other lectures.

    2.) How many lectures must a student attend to be 95% sure that they will have slept at least 28 minutes in total?

    I have done previous parts to the question that ask for E(X) and Std Deviation but I can't work out these 2.

    Any help would be appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by Nguyen View Post
    Hey all, this is from a past exam but we were not given any solutions.

    The Question:
    Many students find that sitting through an entire lecture is difficult to do without falling asleep. Suppose the number of minutes a single, randomly selected student is asleep during a lecture is uniformly distributed between 7 and 14 minutes.

    1.) Over the course of the semester (say, 50 lectures), what is the probability that a randomly selected student will sleep a total of between 130 and 140 minutes?
    Assume that the minutes slept in any lecture is independent of the number of minutes slept in any other lectures.

    2.) How many lectures must a student attend to be 95% sure that they will have slept at least 28 minutes in total?

    I have done previous parts to the question that ask for E(X) and Std Deviation but I can't work out these 2.

    Any help would be appreciated.

    Thanks
    I suggest using the Central Limit Theorem.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    I suggest using the Central Limit Theorem.
    Thanks mr fantastic, but I am getting large numbers and it's not working out for me. Could you show working?

    Thanks.
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  4. #4
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    Quote Originally Posted by Nguyen View Post
    Thanks mr fantastic, but I am getting large numbers and it's not working out for me. Could you show working?

    Thanks.
    You need to show your working.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    You need to show your working.
    oh yes, of course.

    1. \bar{X} \sim N \left( \mu = \frac{21}{2}, \sigma^2 = \frac{7}{50 \cdot 12} \right)

    P(\bar{X} >130) = P \left( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} > \frac{130-(21/2)}{(\sqrt{21}/6)/\sqrt{50}} \right)

    Same goes for P(\bar{X} <140).
    And that is obviously wrong so I am definitely using the wrong values.

    2. Don't know what to do for this one.

    Any help would be nice.

    Thanks
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  6. #6
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    Quote Originally Posted by Nguyen View Post
    oh yes, of course.

    1. \bar{X} \sim N \left( \mu = \frac{21}{2}, \sigma^2 = \frac{7}{50 \cdot 12} \right)

    P(\bar{X} >130) = P \left( \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} > \frac{130-(21/2)}{(\sqrt{21}/6)/\sqrt{50}} \right)

    Same goes for P(\bar{X} <140).
    And that is obviously wrong so I am definitely using the wrong values.

    2. Don't know what to do for this one.

    Any help would be nice.

    Thanks
    The variance of the time spent sleeping in a lecture is 7^2/12 \text{ min}^2, so the variance of the time slept in a total of 50 lectures is 50 (7^2/12)\approx (14.29 \text{ min})^2.

    And the mean times spent sleeping in 50 lectures is 50(10.5) = 525 \text{ min}, so obviously for (1) the answer will be indistinguishable from zero.

    This suggests the possibility of a typo.

    CB
    Last edited by CaptainBlack; May 15th 2011 at 08:00 AM.
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    The variance of the time spent sleeping in a lecture is 7^2/12 \text{ min}^2, so the variance of the time slept in a total of 50 lectures is 50 (7^2/12)\approx (14.29 \text{ min})^2.

    And the mean times spent sleeping in 50 lectures is 50(10.5) = 525 \text{ min}, so obviously for (1) the answer will be indistinguishable from zero.

    This suggests the possibility of a typo.

    CB
    Thanks CaptainBlack.
    That's the problem for me, it is not a typo unfortunately.
    For 1. my tutor said it has to do with working with summations of a random variable and using what we know about how \bar{X} is distributed. But it is not working.
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  8. #8
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    Quote Originally Posted by Nguyen View Post
    Thanks CaptainBlack.
    That's the problem for me, it is not a typo unfortunately.
    For 1. my tutor said it has to do with working with summations of a random variable and using what we know about how \bar{X} is distributed. But it is not working.
    The typo may be in the problem as set. What has been discussed does work with summations of a random variable and using what we know about how \overline{X} is distributed. The typo is probably in the number of lectures in a semester, 15 would seem to make more sense of the problem.

    CB
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  9. #9
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    Hi Huyen Nguyen,

    Maybe you shouldn't lie and just do the STAT1003 assignment yourself. Im fairly certain that ANU stipulates you must not get solutions from other people...including forums. In case you weren't aware, lectures are on monday-wednesday. Go to them and learn something instead of cheating.
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  10. #10
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    This thread is under investigation.

    Thread Closed.
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