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Math Help - question regarding binomial and exponential distributions

  1. #1
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    question regarding binomial and exponential distributions

    Hi

    i need help with the attached question

    I have attempted the question but i am stuck on parts d) to h)

    My answers so far are

    a) i substituted in z=3 hence my answer was 0.112

    b) P(X>2\mid X<3)=\frac{0.184}{0.888}=0.207

    c) X\sim Bin(6,\:0.112) P(X=2)=\dbinom{6}{2}(0.112)^{2}(1-0.112)^{4}=0.117

    my guess to part d was
    \dbinom{6}{0}(0.112)^{0}(0.888)^{6}=0.4903

    Probability of first three not failing is (0.888)^{3}=0.7 hence answer is \frac{0.4903}{0.7}=0.7

    any help is appreichated

    thanks
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  2. #2
    Moo
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    Hello,

    I haven't looked at the other questions yet, but the first one is false

    a) i substituted in z=3 hence my answer was 0.112
    You're working with a probability density function. The probability of getting a value equal to 3, while working with a rv with pdf, is 0.
    What you're asked is the probability that the man fails to do it in 3 minutes. Which means it's the probability that it takes more than 3 minutes to do the task.

    So you're looking for P(Z>3), which is \int_3^\infty f_Z(z) ~dz, and not f_Z(3)

    I guess you made the same mistake for question b)
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  3. #3
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    is the answers for part a) 1-[1-e^{-(0.5).(3)}]=0.223
    part b) P(X>2\mid X<3)=\frac{0.368}{0.778}=0.473

    part c) X\sim Bin(6,\:0.223) P(X=2)=\dbinom{6}{2}(0.223)^{2}(1-0.223)^{4}=0.272
    Last edited by cooltowns; January 24th 2010 at 03:28 AM.
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