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Math Help - chaeck for solution

  1. #1
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    chaeck for solution

    Hi,
    I have two probability problems and I need them to be checked please:

    problem-1
    A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

    1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
    My solution:
    lambda=0.03*1=0.03
    f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)


    2- if 50 panesl is randomly sampled from the production process, what is the probability it has no flaws.
    No solution here. I am not sure but it is the same as above but with lambda=0.03*50??

    3- What is the expected number of panels that need to be sampled before flaws are found?
    I am not sure of what I have done for this question:
    lambda=0.03
    (0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66




    problem 2

    the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:

    1-P(X>4)
    My solution: P(X>4)=1-B(4;6,0.6)

    2-P(X=4)
    P(X=4)=b(4;6,0.6)

    3-mean
    mean=n*p=3.6

    4-variance
    Var^2=n*p*(1-p)

    5-P(X=mean)
    Here since mean =3.6, I have problem to find P(X=mean)

    B.
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  2. #2
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    Quote Originally Posted by braddy View Post
    Hi,
    I have two probability problems and I need them to be checked please:

    problem-1
    A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

    1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
    My solution:
    lambda=0.03*1=0.03
    f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)
    Yes, but simplify (or rather calculate the approcimate value of this):

    p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704

    2- if 50 panesl is randomly sampled from the production process, what is the probability it has no flaws.
    No solution here. I am not sure but it is the same as above but with lambda=0.03*50??
    Probability that there are no flaws in a batch of 50 is:

    P(0) = [p(0)]^50 ~= 0.223

    3- What is the expected number of panels that need to be sampled before flaws are found?
    I am not sure of what I have done for this question:
    lambda=0.03
    (0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66
    I'm not quite sure what this is asking for, I might have to think about what it is asking,
    but I expect what they want as an answer is the batch size for which the expected
    number of flawed pannels = 1.

    Let the batch size be N, probability that n are flawed is:

    b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

    and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

    So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

    RonL
    Last edited by CaptainBlack; March 1st 2007 at 09:25 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    Yes, but simplify (or rather calculate the approcimate value of this):

    p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704



    Probability that there are no flaws in a batch of 50 is:

    P(0) = [p(0)]^50 ~= 0.223



    I'm not quite sure what this is asking for, I might have to think about what it is asking,
    but I expect what they want as an answer is the batch size for which the expected
    number of flawed pannels = 1.

    Let the batch size be N, probability that n are flawed is:

    b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

    and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

    So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

    RonL
    there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

    - What is the expected number of panels that need to be sampled before two flaws are found.
    I guess the steps are the same but for E(x)=2.

    CaptainBlack, for the question 2, why did you do p(0)^50?
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  4. #4
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    Quote Originally Posted by braddy View Post
    Hi,
    I have two probability problems and I need them to be checked please:

    problem-1
    A company makes plastic panel used in automobiles. The panel production process is such thast the number of flaws on a panel follows a Poisson process with a mean of 0.03 flaws per panel.

    1- If one panel is randomly selected from the production process, what is the probability it has no flaws.
    My solution:
    lambda=0.03*1=0.03
    f(0,0.03)=(e^(-0.03)*(0.03)^0)/(0!)
    Quote Originally Posted by CaptainBlack View Post
    Yes, but simplify (or rather calculate the approcimate value of this):

    p(0) = f(0,0.03) = (e^(-0.03)*(0.03)^0)/(0!) ~= 0.9704
    Quote Originally Posted by braddy View Post
    3- What is the expected number of panels that need to be sampled before flaws are found?
    I am not sure of what I have done for this question:
    lambda=0.03
    (0.03)*n=1----->n=33.33 for one flaw.so 2 falws-->n=66.66
    Quote Originally Posted by CaptainBlack View Post
    I'm not quite sure what this is asking for, I might have to think about what it is asking,
    but I expect what they want as an answer is the batch size for which the expected
    number of flawed pannels = 1.

    Let the batch size be N, probability that n are flawed is:

    b(n,N,1-p(0)) = N!/(n! (N-n)!) (1-p(0))^n p(0)^(n-1),

    and as this is binomial, the expected number of flawed panels is E(n) = N(1-p(0)) = N 0.02955,

    So if E(n) = 1, the N we require is N=1/0.02955 ~= 33.8.

    RonL
    If probability of at least one flaw in a panel is p = .02955, then the number X of panels sampled before a flawed one is found has a geometric distribution Pr(X=k) = p(1-p)^(k-1) for k = 1,2,3,... The expected number of panels sampled before a flawed panel is found is the mean of this distribution, which is 1/p = 33.8. Same answer but different reasoning.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by braddy View Post
    CaptainBlack, for the question 2, why did you do p(0)^50?
    Because this is the probability that there are no flaws on the first panel, times the probability that there are no flaws on the second panel times ..
    times there are no flaws on the 50th panel = p(0)^50.

    RonL
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  6. #6
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    Quote Originally Posted by JakeD View Post
    If probability of at least one flaw in a panel is p = .02955, then the number X of panels sampled before a flawed one is found has a geometric distribution Pr(X=k) = p(1-p)^(k-1) for k = 1,2,3,... The expected number of panels sampled before a flawed panel is found is the mean of this distribution, which is 1/p = 33.8. Same answer but different reasoning.
    Well that will save me having to think about this some more

    RonL
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  7. #7
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    Thank you

    In the first problem there was a typo. I dont know if you pay attention when I said:[quote]there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

    - What is the expected number of panels that need to be sampled before two flaws are found.

    I guess the steps are the same but for E(x)=2.

    In this case is it still a geometric distribution?

    Now for problem 2:

    the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
    1-P(X>4)
    My solution: P(X>4)=1-B(4;6,0.6)

    2-P(X=4)

    P(X=4)=b(4;6,0.6)

    3-mean
    mean=n*p=3.6

    4-variance
    Var^2=n*p*(1-p)

    5-P(X=mean)
    Here since mean =3.6, I have problem to find P(X=mean)

    Thank you for your time
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  8. #8
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    Quote Originally Posted by braddy View Post
    Thank you

    In the first problem there was a typo. I dont know if you pay attention when I said:
    there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.
    - What is the expected number of panels that need to be sampled before two flaws are found.

    I guess the steps are the same but for E(x)=2.

    In this case is it still a geometric distribution?
    Yes, with p now the probability of 2 or more flaws.
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  9. #9
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    [QUOTE=braddy;41093]Thank you

    In the first problem there was a typo. I dont know if you pay attention when I said:
    there is a mistake in the question .I am sorry they are asking for two flaws ,not 1.

    - What is the expected number of panels that need to be sampled before two flaws are found.

    I guess the steps are the same but for E(x)=2.

    In this case is it still a geometric distribution?

    Now for problem 2:

    the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
    1-P(X>4)
    My solution: P(X>4)=1-B(4;6,0.6)

    2-P(X=4)

    P(X=4)=b(4;6,0.6)

    3-mean
    mean=n*p=3.6

    4-variance
    Var^2=n*p*(1-p)

    5-P(X=mean)
    Here since mean =3.6, I have problem to find P(X=mean)


    Thank you for your time
    my calculator does not give me a approximate value for b(3.6;4,0.6) !
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