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Math Help - Distribution Function

  1. #1
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    Distribution Function

    I am trying to solve this question
    Let X be the number of years a particular type of machine is in good working condition and need no replacement. Assume that X has the probability mass function

    P(1) = 0.4, P(2) = 0.3, P(3) = 0.2, P(4) = 0.08, P(5) = 0.02.

    Find the distribution function F. Sketch F. Find the probability that the machine needs no replacement during the first 3 years.
    I have been able to get the distribution function. the correct anwer given to me is that P( machine needs no replacement during the first 3 years) = 0.3.
    However, from the distribution function F sketch, I get P(machine needs no replacement during the first 3 years) = 0.7.

    what is the right answer for this? if 0.3, why is it so?

    Thank you.

    regards,

    tommy
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  2. #2
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    Quote Originally Posted by tommyhakinen View Post
    I am trying to solve this question
    I have been able to get the distribution function. the correct anwer given to me is that P( machine needs no replacement during the first 3 years) = 0.3.
    However, from the distribution function F sketch, I get P(machine needs no replacement during the first 3 years) = 0.7.

    what is the right answer for this? if 0.3, why is it so?

    Thank you.

    regards,

    tommy
    P(1) is the probability that the machine needs replacent after 1 year

    P(2) is the probability that the machine needs replacent after 2 years

    etc.

    So the probability that the machine needs to be replaces in 3 or fewer years is P(1)+P(2)+P(3)=0.7

    So the probability that the machine needs no replacement in 3 years is 1 minus the probability that it needs replacement in 3 or less years.

    CB
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  3. #3
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    thanks for the explanation. but i still don't get it. it is mentioned that X is the no. of years a particular type of machine is in good condition and need no replacement. the question asks the probability that machine need no replacement during the first 3 years, which is P(X < 3).

    P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.7

    Does that already answer the question? why do we need to do 1 - P(<3)? or is there possible that something wrong with this question? eg. it should be asking "probability that the machine needs replacement during the first 3 years?".

    Thanks.

    tommy
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  4. #4
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    Quote Originally Posted by tommyhakinen View Post
    thanks for the explanation. but i still don't get it. it is mentioned that X is the no. of years a particular type of machine is in good condition and need no replacement. the question asks the probability that machine need no replacement during the first 3 years, which is P(X < 3).

    P(X < 3) = P(X=0) + P(X=1) + P(X=2) = 0.7

    Does that already answer the question? why do we need to do 1 - P(<3)? or is there possible that something wrong with this question? eg. it should be asking "probability that the machine needs replacement during the first 3 years?".

    Thanks.

    tommy
    You are not reading the question carefully enough:

    P(1) is the probability of satisfactory operation for 1 year meaning it is Q(2) the probability of failure in the second year, and so on.

    CB
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  5. #5
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    i see. got it now. thanks alot
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