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Homework Help: Normal Distribution

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data

    If we were to have two normal distributions, say X and Y, which are the scores of students on a test or something, a is the score of a student from X and b is the score of a student from Y, what is the probability that a randomly selected student from X will score more than a randomly selected student from Y, i.ie what's the probability that a>b, or something like a>b+10?
    Both means and SDs given.



    Also, if we have a normal distribution, say for weights of bottles of something, and we know that A is the probability that a randomly chosen bottle will weigh less than X, and we then take a sample of say 6 bottles, and we want to find the probability that 2 or less will weigh less than X, is it just 6C0(1-X)^6 + 6C1(X)(1-X)^5 + 6C2(X)^2(1-X)^4
    or do I have to standardise again for the population of 6 bottles?


    Thanks! :smile:
     
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  3. Jul 28, 2011 #2

    Ray Vickson

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    Is this a homework problem? What have you done on it so far? Where are you having trouble?

    RGV
     
  4. Jul 28, 2011 #3
    It's not homework, just me doing revision before going back to university in september.

    The first one is a problem I came across in a very old exam paper, and don't know where ti begin.

    The second one I just need a yes or no answer.
     
  5. Jul 28, 2011 #4

    I like Serena

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    What kind of distribution is the difference of 2 independent normal distributions?
     
  6. Jul 28, 2011 #5
    Normal with mean u1 - u2 and variance V1 + V2
     
  7. Jul 28, 2011 #6
    Wait, think I get it, find the difference as a normal distribution, and then the probability that this is greater than zero for the probability one is bigger than the other?
     
  8. Jul 28, 2011 #7

    I like Serena

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  9. Jul 28, 2011 #8

    I like Serena

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    As for the second part of your question.

    You have an A which is a probability, and you have an X which is an amount.
    You thought up a formula for a probability, which contains X, but does not contain A.
    Since X is not a probability, isn't it kind of unlikely that you can use it in a formula where a probability is expected? :confused:
     
  10. Jul 28, 2011 #9
    Sorry! Should have been 6C0(1-A)^6 + 6C1(A)(1-A)^5 + 6C2(A)^2(1-A)^4
     
  11. Jul 28, 2011 #10

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    All right. :smile:
    Then I'll just say yes, although I do not understand what you mean by "standardize for the population of 6 bottles".
     
  12. Jul 28, 2011 #11
    I mean do i treat the six bottles as a new normal population, find the z value for X, and a new probability that a bottle chosen from these will be less than X?
     
  13. Jul 28, 2011 #12

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    Ah, now I understand.

    Well, did you open the bottles and mix them?
    Or did you calculate the mean weight of the 6 bottles and compare that?
     
  14. Jul 28, 2011 #13

    Ray Vickson

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    If the 6 bottles are drawn independentally (say, from a large population) you just have a coin-tossing type problem. Each bottle, independent of the others, has a probability 'A' of weighing more than some fixed weight w (you called it X, but I prefer to use the convention that capital letters are random variables, while small letters are given numbers). This is like the problem of getting 3 heads in 6 coin tosses, where each toss has Pr{heads} = A.

    RGV
     
  15. Jul 28, 2011 #14
    I think i'm starting to confuse myself...

    But basically, if i take 6 bottles from a population, and in this population the probability that a randomly chosen bottle weighs less than X is A,
    the probability that 2 of the 6 bottles weighs less than X is the expression i gave above?
     
  16. Jul 28, 2011 #15

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    I already said yes and I'm not changing my answer now! :wink:
    As RGV already said, it's basically a coin-tossing problem with a binomial distribution, which is what you wrote down.
     
  17. Jul 28, 2011 #16
    Thanks to both of you! :smile:
     
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