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Homework Help: Probability Homework Help

  1. Aug 30, 2011 #1
    The following table gives the probability distribution of X, the number of defective products in a production line in a day.
    x P(X=x)

    0 0.04

    1 0.19

    2 0.35

    3 0.26

    4 0.16

    a. Evaluate P(|X - expected value| >= standard dev)
    b. The cost of maintaining the machine at the production line is given by C = 50 + 3x + 2x^2. Find the mean maintaining cost.
     
    Last edited: Aug 30, 2011
  2. jcsd
  3. Aug 30, 2011 #2

    Ray Vickson

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    Show your work. What have you done so far? Where are you stuck?

    RGV
     
  4. Aug 30, 2011 #3
    I only managed this much:

    Expected value = 2.31
    Standard deviation = 1.074

    P( |X - 2.31| >= 1.074)
    =P( -1.074 >= (X - 2.31) >= 1.074 )
    =P( 1.236 >= X >= 3.384)

    I'm stuck here. For question b, I assume I have to calculate the expected value of C by substituting the variables x and x^2 with their respective expected values?
     
  5. Aug 31, 2011 #4

    vela

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    This doesn't make sense since 1.236 clearly isn't less than or equal to 3.384.

    The condition |X - 2.31| ≥ 1.074 is equivalent to (X-2.31 ≥ 1.074) or (X-2.31 ≤ -1.074), so
    [tex]P(\lvert X-2.31 \rvert \ge 1.074) = P(X-2.31 \ge 1.074) + P(X-2.31 \le -1.074)[/tex]
    Yes, because E(C) = E(50 + 3x + 2x^2) = E(50) + E(3x) + E(2x2) = 50 + 3 E(x) +2 E(x2)
     
  6. Aug 31, 2011 #5

    Ray Vickson

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    |X - 2.31| >= 1.074 <--> either X - 2.31 >= 1.074 or 2.31 - X >= 1.074.

    RGV
     
  7. Aug 31, 2011 #6
    So is this how it goes, based on the given values of x and P(X=x):

    P(|X-2.31| >= 1.074)
    = P( X-2.31 >= 1.074) + P( 2.31-X >= 1.074)
    = P( X >= 3.384) + P( X <= 1.236)
    = 0.16 + 0.19 + 0.04
    = 0.39
     
  8. Sep 1, 2011 #7

    vela

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    Yes, that's right.
     
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