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Math Help - Normal distribution.

  1. #1
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    Normal distribution.

    The lenght of a bar, X has a distribution N(10,2^2). The bars are separeted in 3 categories A B and C A for lenght less than 8 B for lenght between 8 and 12 and C for lenght greater than 12
    what is the probability that if we choose 15 bars we have the same number of bars of category A B and C.

    Well tried to solve it that way:
    we have to have 5 bars in each category.
    And i found that the probability of belonging to category A and C is 0.1587 and for B 1-2*0.1587 I think I'm right assuming that.
    then I think the probability we seek is given by:

    Cn(15,5)*(0.1587^10)*(1-2*0.1587)^5

    I can't see why it's not right but the truth is that probably it is wrong since I don't get the result I should.
    Can anyone see where Is my mistake? perhaps I shouldn't have used Combinations and maybe permutations or something...but cant see why...can anyone explain me?
    Thanks for your time
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  2. #2
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    Quote Originally Posted by Mppl View Post
    The lenght of a bar, X has a distribution N(10,2^2). The bars are separeted in 3 categories A B and C A for lenght less than 8 B for lenght between 8 and 12 and C for lenght greater than 12
    what is the probability that if we choose 15 bars we have the same number of bars of category A B and C.

    Well tried to solve it that way:
    we have to have 5 bars in each category.
    And i found that the probability of belonging to category A and C is 0.1587 and for B 1-2*0.1587 I think I'm right assuming that.
    then I think the probability we seek is given by:

    Cn(15,5)*(0.1587^10)*(1-2*0.1587)^5

    I can't see why it's not right but the truth is that probably it is wrong since I don't get the result I should.
    Can anyone see where Is my mistake? perhaps I shouldn't have used Combinations and maybe permutations or something...but cant see why...can anyone explain me?
    Thanks for your time
    is 0,000013 the right answer?

    P= P(Exactly 5 five from the first catagory)*P(Exactly 5 five from the second catagory)*P(Exactly 5 five the third catagory)=0,000013
    P(Exactly 5 five from the first catagory)=nCr(15,5)*(0,15)^5*(1-0,15)^10
    P(Exactly 5 five from the third catagory)=nCr(15,5)*(0,15)^5*(1-0,15)^10
    P(Exactly 5 five from the second catagory)=nCr(15,5)*(0,68)^5*(1-0,68)^10
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  3. #3
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    I think it Can't be that way since those 3 events are not independent, so that multiplication rule can't be applied.
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  4. #4
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    the events are not independent?!! how do you come to that?
    what is the right answer?
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  5. #5
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    Quote Originally Posted by Mppl View Post
    The lenght of a bar, X has a distribution N(10,2^2). The bars are separeted in 3 categories A B and C A for lenght less than 8 B for lenght between 8 and 12 and C for lenght greater than 12
    what is the probability that if we choose 15 bars we have the same number of bars of category A B and C.

    Well tried to solve it that way:
    we have to have 5 bars in each category.
    And i found that the probability of belonging to category A and C is 0.1587 and for B 1-2*0.1587 I think I'm right assuming that.
    then I think the probability we seek is given by:

    Cn(15,5)*(0.1587^10)*(1-2*0.1587)^5

    I can't see why it's not right but the truth is that probably it is wrong since I don't get the result I should.
    Can anyone see where Is my mistake? perhaps I shouldn't have used Combinations and maybe permutations or something...but cant see why...can anyone explain me?
    Thanks for your time
    Calculate the probability p1 of an individual bar being A, from this calculate the probability P1 of exactly 5 from 15 being A. Now calculate the conditional probability p2 of a single bar being B given that it is not A, from this calculate the probability P2 of exactly 5 from 10 are B given that they are not A.

    The final answer is P1.P2

    CB
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  6. #6
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    I didnt get it...why doing it the way you said? Isn't it possible to assume (as I did!) that when choosing 15 it is a binomial distribution? and do it the way I did?

    If it can't be seen the way I saw it why is that so? and can you explain a bit better your resolution?
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  7. #7
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    Quote Originally Posted by Mppl View Post
    I didnt get it...why doing it the way you said?
    If there are exactly 5 A's then there are 10 that are B's and C's, if of these 10 5 are B's the remainder must be C.


    Isn't it possible to assume (as I did!) that when choosing 15 it is a binomial distribution? and do it the way I did?

    If it can't be seen the way I saw it why is that so? and can you explain a bit better your resolution?
    Don't ask me why your method is right or wrong, what argument do you have to justify it? (You are not allowed to just make stuff up or guess)

    CB
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  8. #8
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    Well my method seamed to me so obvious that I thought I didnt have to explain it...
    Anyway here it goes:
    I have 15 places to fill, and I have to do it with 5 A 5 B and 5 C so one way to do it is AAAAABBBBBCCCCC well the probability of that happening is (P(A)*P(C)*P(B))^5 assuming independence, if it cannot be assumed I would be glad if you could explain to me why.
    But going on with my line of thought... the configuration I showed is not the only one that satisfies the criterium there are Cn(15,5) diffrent ways of doing it right? so the final result is, according to that Cn(15,5)*(P(A)*P(C)*P(B))^5
    Thank you for your time.
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  9. #9
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    Quote Originally Posted by Mppl View Post
    Well my method seamed to me so obvious that I thought I didnt have to explain it...
    Anyway here it goes:
    I have 15 places to fill, and I have to do it with 5 A 5 B and 5 C so one way to do it is AAAAABBBBBCCCCC well the probability of that happening is (P(A)*P(C)*P(B))^5 assuming independence, if it cannot be assumed I would be glad if you could explain to me why.
    But going on with my line of thought... the configuration I showed is not the only one that satisfies the criterium there are Cn(15,5) diffrent ways of doing it right? so the final result is, according to that Cn(15,5)*(P(A)*P(C)*P(B))^5
    Thank you for your time.
    The question is what do you think Cn(15,5) means (it is not a notation I am familiar with) and is it the number of permutations of 15 objects 5 of one type 5 of another and 5 of a third type?

    CB
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  10. #10
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    not permutations,combinations. Combinations of 15 5 by 5. is that correct like that?
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  11. #11
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    Quote Originally Posted by Mppl View Post
    not permutations,combinations. Combinations of 15 5 by 5. is that correct like that?
    What is your justification for that?

    Tell us the numerical value that you get.

    I get ~0.001136 by calculation and 0.001146 with SE 0.000034 from simulation

    CB
    Last edited by CaptainBlack; January 27th 2011 at 12:26 AM.
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  12. #12
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    well your resolution seams to be rights, as it is the same value that I should get... but can um give me a more detailed description of why you did things that way?

    I get 0.000005 the justification is here
    I have 15 places to fill, and I have to do it with 5 A 5 B and 5 C so one way to do it is AAAAABBBBBCCCCC well the probability of that happening is (P(A)*P(C)*P(B))^5 assuming independence, if it cannot be assumed I would be glad if you could explain to me why.
    But going on with my line of thought... the configuration I showed is not the only one that satisfies the criterium there are Cn(15,5) diffrent ways of doing it right? so the final result is, according to that Cn(15,5)*(P(A)*P(C)*P(B))^5
    why isn't that right?
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  13. #13
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    Quote Originally Posted by Mppl View Post
    well your resolution seams to be rights, as it is the same value that I should get... but can um give me a more detailed description of why you did things that way?
    We start with the probability that exactly 5 of the 15 are A's P1=BinomialP(5;15,P(A))~=0.0537.

    Now the probability of a B at any position in the 10 that are not A's is P(B| not(A))= P(B)/(1-P(A)). So the probability that exactly 5 of these 10 are B's is P2=BinomialP(5;10,P(B|not(A)) )~=0.0212.

    Then the required probability is P1.P2~=0.001136.

    I get 0.000005 the justification is here why isn't that right?
    There is no need to give a justification of why that is not right, you give no argument about why it should be right (you need the number of permutations of 15 objects comprised of 5 each of three distinct types not a number of combinations)

    CB
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