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Math Help - Can any one pls help with this Probibality question with formulas?

  1. #1
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    Can any one pls help with this Probibality question with formulas?

    XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

    Number of
    Breakdowns------------- Probability
    0------------------------------- 0.3
    1 -------------------------------0.4
    2------------------------------- 0.1
    3------------------------------- 0.2
    More than 3----------------- 0.0

    (a) What is the expected number of breakdowns in any given day?
    (b) What is the variance for this distribution?
    (c) What is the probability that there will be at least 2 breakdowns in a day?
    Last edited by stoorrey; December 19th 2009 at 08:33 AM. Reason: Can you solve the whole question Pls
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  2. #2
    MHF Contributor matheagle's Avatar
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    E(X)=(0)(.3)+(1)(.4)+(2)(.1)+(3)(.2)

    Likewise E(X^2)=(0)^2(.3)+(1)^2(.4)+(2)^2(.1)+(3)^2(.2)

    and V(X)=E(X^2)-[E(X)]^2

    P(X\ge 2)=.1+.2
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  3. #3
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    Can you solve the whole question pls?

    Will be highly appreciated
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  4. #4
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    Quote Originally Posted by stoorrey View Post
    Will be highly appreciated
    Please show some effort. What have you done? Where are you still stuck? What part of matheagle's reply do you not understand?
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    XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

    Number of
    Breakdowns------------- Probability
    0------------------------------- 0.3
    1 -------------------------------0.4
    2------------------------------- 0.1
    3------------------------------- 0.2
    More than 3----------------- 0.0

    (a) What is the expected number of breakdowns in any given day?
    (b) What is the variance for this distribution?
    (c) What is the probability that there will be at least 2 breakdowns in a day?

    Answers:
    a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
    c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by malaynsamani View Post
    XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

    Number of
    Breakdowns------------- Probability
    0------------------------------- 0.3
    1 -------------------------------0.4
    2------------------------------- 0.1
    3------------------------------- 0.2
    More than 3----------------- 0.0

    (a) What is the expected number of breakdowns in any given day?
    (b) What is the variance for this distribution?
    (c) What is the probability that there will be at least 2 breakdowns in a day?

    Answers:
    a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
    c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
    a) is incorrect, that's not E(X)
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  7. #7
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    Quote Originally Posted by malaynsamani View Post
    Answers:
    a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
    c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
    A. You should know by definition why this isn't right. Expected value isn't (necessarily) a percentage, unless the random value is a percentage (but then it still wouldn't mean what it normally means. . .). The expected value is just that the value you would expect to get. In this case, since our random variable is a discreet number, referring to number of breakdowns, then we should get a specific number (and not necessarily an integer). To calculate expected value we take the sum of all random variables, multiplied by their probabilities: 0*P(X=0)+1P(X=1)+2P(X=2)+3P(X=3). Why does this work? Remember what the expected value is - the mean. Imagine you have a perfect sample of 100 days. Among this sample I would expect to have 30% have 0 breakdowns, 40% have 1 breakdown, 10% have 2 breakdowns, 20% have 3 breakdowns, and 0% to have more than 3 breakdowns. That means I would EXPECT to have, among a perfect sample of 100: 30 0's, 40 1's, 10 2's, 20 3's and 0' "More than 3's". Now how would I go about averaging those numbers? And that is your expected value.
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  8. #8
    MHF Contributor matheagle's Avatar
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    and I thought I answered this a week ago
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  9. #9
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    Oh man you are right. And the OP isn't the fella who I responded to. I'm all mixed up here.
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  10. #10
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    Hello Matheagle

    Quote Originally Posted by matheagle View Post
    a) is incorrect, that's not E(X)

    If (a) is not E(X) the would you mind tell me the correct answer or formula if you know it. thnaks
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  11. #11
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    Quote Originally Posted by stoorrey View Post
    If (a) is not E(X) the would you mind tell me the correct answer or formula if you know it. thnaks
    He was responding to someone else. Question A is asking for the expected value of X, and was answered.
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  12. #12
    MHF Contributor matheagle's Avatar
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    so answer this, who's on first?
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