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Statistics Continuous Distributions

  1. Feb 2, 2013 #1
    1. If a pair of coils were placed around a homing pigeon and a magnetic field was applied that reverses the earth’s field, it is thought that the bird would be disoriented. Under these circumstances it is just as likely to fly in one direction as in any other. Let θ denote the direction in radians of the bird’s initial flight. θ is uniformly distributed over the interval [0, 2π].
    (a) Find the density for θ.
    (b) Sketch the graph of the density. The uniform distribution is sometimes called the rectangular distribution. Do you see why?
    (c) Shade the area corresponding to the probability that a bird will orient within π/4 radians of home, and find this area using plane geometry.
    (d) Find the probability that a bird will orient within π/4 radians of home by integrating the density over the appropriate region(s), compare your answer to that obtained in part (c).
    (e) If 10 birds are released independently and at least 7 orient within π/4 radians of home, would you suspect that perhaps the coils are not disorienting the birds to the extent expected? Explain based on the probability of this occurring.

    3. I haven't done the other parts because I'm stuck on finding the density for θ:
    Based on the pdf formula, I got this but I don't know what the f(x) dx part should be
    a. f(θ) = ∫2∏0 "f(x) dx?"
     
  2. jcsd
  3. Feb 2, 2013 #2

    HallsofIvy

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    You are told what that density is: "θ is uniformly distributed over the interval [0, 2π]". Do you not understand what "uniform density" means?
     
  4. Feb 2, 2013 #3
    Oh! Just to make sure the equation would be like this:
    f(θ) = 1/(2∏ - 0) = 1/2∏ 0 ≤ θ ≤ 2∏
    0 otherwise
     
  5. Feb 2, 2013 #4

    LCKurtz

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    Yes, that is the density function.
     
  6. Feb 2, 2013 #5
    Thank You!

    For c) Would the red part be the area I shade and calculate?
    pigeon.jpg
     
  7. Feb 2, 2013 #6

    haruspex

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    It's 'within' π/4. So it's up to that error in either direction.
     
  8. Feb 2, 2013 #7
    So these two parts?
    pigeon.jpg
     
  9. Feb 2, 2013 #8

    LCKurtz

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    Have you worked part (b) yet? You need to understand that to work part (c). The area in question is not shown in the pictures you are showing.
     
  10. Feb 3, 2013 #9

    haruspex

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    Hmm.. yes. I hadn't read part (b) - just assumed the area whitehorsey shaded was on the correct graph. Wrong assumption.
     
  11. Feb 3, 2013 #10

    For part b I got this:
    rectangle2.png
    and for why its a rectangle distribution because all values in the range between 0 and 2π are equally likely.

    Oh so I should be shading the attachment above like this:
    rectangle.png
     
  12. Feb 3, 2013 #11

    LCKurtz

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    That's better. To be complete your picture should indicate the "home" direction so you can tell whether or not you are within ##\pi/4## of home.
     
  13. Feb 3, 2013 #12

    haruspex

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    Yes, except that you have fallen back into the range error in your original shading.
     
  14. Feb 3, 2013 #13
    Ah okay! So I put back in the range and as a total I got .25. Is that correct?
     
  15. Feb 3, 2013 #14

    haruspex

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    Yes.
     
  16. Feb 3, 2013 #15
    Thank You!

    As for d, I took the integral of 1/2π from 0 to π/4 and got 1/8 but aren't part c and d's answers suppose to match up?
     
  17. Feb 3, 2013 #16

    haruspex

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    That's the same mistake three times now :frown:
     
  18. Feb 3, 2013 #17
    I'm so sorry!! Lol it's the range! Okay fixed it! THANK YOU!!!

    Lastly e. E(X)= np = 2.5?
     
  19. Feb 3, 2013 #18

    haruspex

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    Yes, but I assume you realise there's more to do.
     
  20. Feb 3, 2013 #19
    Ah yes the explanation. I was wondering how would the 2.5 prove that the coils are not disorienting the birds? I know E(X) stands for the population mean but I can't seem to grasp the concept to how it relates.
     
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