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Expected Value and Binomial Random Variable

  1. Jan 21, 2013 #1
    1. In scanning electron microscopy photography, a specimen is placed in a
    vacuum chamber and scanned by an electron beam. Secondary electrons
    emitted from the specimen are collected by a detector and an image is
    displayed on a cathode-ray tube. This image is photographed. In the past
    a 4-  5-inch camera has been used. It is thought that a 35-millimeter
    camera can obtain the same clarity. This type of camera is faster and
    more economical than the 4-  5-inch variety.

    (a) Photographs of 15 specimens are made using each camera system.
    These unmarked photographs are judged for clarity by an impartial judge.
    The judge is asked to select the better of the two photographs from each
    pair. Let X denote the number selected taken by a 35-mm camera. If
    there is really no difference in clarity and the judge is randomly selecting
    photographs, what is the expected value of X?

    (b) Would you be surprised if the judge selected 12 or more photographs
    taken by the 35-mm camera? Explain, based on the probability involved.

    (c) If X ≥ 12, do you think that there is reason to suspect that the judge
    is not selecting the photographs at random?

    3. a) E[X] = np I believe that n = 15 but I'm not sure what p should be.

    As for b) and c), I'm not sure how to approach it.
     
  2. jcsd
  3. Jan 21, 2013 #2

    haruspex

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    The question is not as well worded as it could be. "Random" just means not completely deterministic. But they are trying to indicate the probability that the judge picks the one or the other. What probability do you think they intend?

    For b and c, what do you know about hypothesis testing? Any tests you've been taught?
     
  4. Jan 21, 2013 #3
    If the judge picks one or the other, then the probability would be .5 (chance of picking the 35mm one)?

    Hmm I don't think I have learned hypothesis testing yet.
     
  5. Jan 21, 2013 #4

    HallsofIvy

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    No, that was haruspex's point. "Picking one or the other", even "at random" does NOT necessarily mean "with equal likliehood".
     
  6. Jan 21, 2013 #5

    Ray Vickson

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    Of course this is true in principle, but in practice (especially in introductory prob. and stats courses) the language is used somewhat informally, so that "random" choices really do mean choices made with probabilty 1/r each when there are r possibilities. So, my vote would be that the problem intends p = 1/2 as the meaning of "random choice".
     
  7. Jan 21, 2013 #6

    LCKurtz

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    I agree with Ray. This looks like a problem used to introduce the idea behind hypothesis testing without calling it that, especially since the OP claims not to have studied that yet.
     
  8. Jan 21, 2013 #7
    Okay thanks guys!!

    For b, looking at the solution it says:
    P[X ≥ 12] = 1 - F(11)
    = 1 - .9824
    = .0176

    I think they are doing a cumulative distributive function but I'm not sure why and how they were able to get .9824 out of F(11). I thought F(11) = 0.5^11?
     
  9. Jan 21, 2013 #8

    LCKurtz

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    No, that isn't the value for F(11). F(11) would be f(0)+f(1)+...+f(11) where f is the discrete probability function. It would be easier to calculate f(12)+f(13)+f(14)+f(15). Did you try that? Do you know the discrete probability function for a binomial b(15,1/2) distribution?
     
  10. Jan 22, 2013 #9
    Umm what is the discrete probability function?
     
  11. Jan 22, 2013 #10

    LCKurtz

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  12. Jan 22, 2013 #11
  13. Jan 22, 2013 #12

    Ray Vickson

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    1) Where does the 0.5^15 come from?
    2) You need several such terms (after correcting them!) because you need the probability of a range of values, not just the single value '11'.
     
  14. Jan 23, 2013 #13
    Oh I'm sorry! It was suppose to be 0.5^3.

    Edit: I got it! Thank You everyone!!
     
    Last edited: Jan 23, 2013
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