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Math Help - Joint probability!

  1. #1
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    Joint probability!

    A steel manufacturer is testing a new additive for manufacturing an allow of steel. The joint probability mass function of tensile streanth and additive concentrative is

    concentration additive tensile strength 100 150 200
    .02 .05 .06 .11
    .04 .01 .08 .10
    .06 .04 .08 .17
    .08 .04 .14 .12

    I've figured out the marginal probability mass functions but am having trouble with 2 other questions.
    Given that a specimen has an additive concentration of .04, what is the probability that it's strength is 150 or more?
    Find x and y.



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  2. #2
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    Given that a specimen has an additive concentration of .04, what is the probability that it's strength is 150 or more?

    use the formula
    P(A|B) = \frac{ P(A~ and~ B) }{ P(B)}

    where:
    A : the event that it has a strength of 150 or more
    B : the event that it has an addative concentration of 0.04

    You can read P(A and B) from the table, and you can get P(B) from the marginal distribution.



    Find x and y.
    I assume this refers to the population mean of your two variables.

    You said you already have the marginal distributions. Use the marginal distributions for each variable and the formula:
    E(X) = \sum \left( x \times P(X=x) \right)

    E(Y) = \sum \left( y \times P(Y=y)\right)

    Post your solution up to the point you are stuck if you need more help
    Last edited by SpringFan25; March 29th 2011 at 12:51 PM.
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  3. #3
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    for the probability that the strength is 150 or more I got 4+2/2 = 3

    I'm still a little confused on how to use the equations to get the mean population variables

    My marginal probability mass function is P(y) = .91
    P(x) = .091
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  4. #4
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    for the probability that the strength is 150 or more I got 4+2/2 = 3
    All probabilities are between 0 and 1, so both your final answer and the figures you used to calculate it aren't correct.


    My marginal probability mass function is P(y) = .91
    This also doesn't look right im afraid. The probability mass function should show the probability of y taking each of its various values.


    It isn't clear where your misunderstanding is, so here is an explanation of how to do some of this from the gound up. I assume that X is the strength and Y is the concentration


    Given that a specimen has an additive concentration of .04, what is the probability that it's strength is 150 or more?

    The joint probability table shows all possible combinations of X and Y, and the probability of each one.

    To make sure you understand how to read probabilities from the table, lets start by defining two events:
    Event A: strength is 150 or more
    Event B: specimen has a concentration of 0.04

    To find the probability of an event, add up the probabilities for the cells in the table that correspond to the event occuring; as in the examples below.




    The question asks you to find a conditional Probability "what is the probability that it's strength is 150 or more, given that it's concentration is 0.04".

    The standard notation for this is P(A|B), where A and B are as above.

    You should have been taught the formula

    P(A|B) = \frac{P(A~and~B)}{P(B)}

    From the screenshots above, you have:
    P(A~and~B) = 0.18
    P(B) = 0.19

    So
    P(A|B) = \frac{P(A~and~B)}{P(B)} = \frac{0.18}{P(0.19)} = \frac{18}{19}


    Find x
    You need the marginal distribution of X, which looks like this:

    P(X=100) = 0.05 + 0.01 + 0.04 + 0.04 = 0.14
    P(X=150) = 0.06 + 0.08 + 0.08 + 0.14 = 0.36
    P(X=200) = ? (you do it).

    In each case, you find the total probability of X being the value specified, which is the sum of all entries in that column of the table. When you come to do the marginal distribution of Y; youll be looking at rows.


    You should have been taught the formula
    \mu_x = E(X) = \sum \left[x \cdot P(X=x) \right]

    This says that you take every possible value of X, multiply by the probability of it occuring. You add the results of each calculation up to get x

    \mu_x = \left(100 \times P(X=100)\right) + \left(150 \times P(X=150)\right) + \left(200 \times P(X=200)\right)

    You try completing the calculation, and then do x. You'll need to find the marginal distribution of Y.





    Due to a forum limitation the screenshot is shown again below, not sure why:
    Attached Thumbnails Attached Thumbnails Joint probability!-jprob-simpleevents.jpg  
    Last edited by SpringFan25; March 30th 2011 at 03:21 AM.
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  5. #5
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    thank you so much for the breakdown. it helped a ton. My final value for ux= 168 ( seemed high but my calculations seemed right) and uy=.0534
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  6. #6
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    i got the same answers
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