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Math Help - Stats homework

  1. #1
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    Unhappy Stats homework

    I've been trying to figure out how to go about solving this problem and I can't seem to wrap my brain around it. I've made a probability distribution table, but I am not sure I'm doing it correctly. I'll post it below the question.

    Heres the problem:

    An experiment consists of rolling two fair six-sided dice and maximum value observed. Let random variable X represent this maximum value (for instance if on a toss one dice landed on 3 and the other dice landed on 4, the x = max(3,4) = 4.)
    Note that X = {1,2,3,4,5,6}.

    a) Construct a probability distribution for X. Hint: list all the possible outcomes and
    which value of X each corresponds to.
    b) Compute and interpret the expected value of X.



    The table I made looks something like this:
    Dice 1 - 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6
    Dice 2 - 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6
    X - 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6



    Any suggestions or hints would be highly appreciated! Thanks!
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by charles2 View Post
    I've been trying to figure out how to go about solving this problem and I can't seem to wrap my brain around it. I've made a probability distribution table, but I am not sure I'm doing it correctly. I'll post it below the question.

    Heres the problem:

    An experiment consists of rolling two fair six-sided dice and maximum value observed. Let random variable X represent this maximum value (for instance if on a toss one dice landed on 3 and the other dice landed on 4, the x = max(3,4) = 4.)
    Note that X = {1,2,3,4,5,6}.

    a) Construct a probability distribution for X. Hint: list all the possible outcomes and
    which value of X each corresponds to.
    b) Compute and interpret the expected value of X.



    The table I made looks something like this:
    Dice 1 - 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6
    Dice 2 - 1 1 1 1 1 1 2 2 2 2 2 3 3 3 3 4 4 4 5 5 6
    X - 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6



    Any suggestions or hints would be highly appreciated! Thanks!
    Hi, charles2,

    Well, there might be a way to make an abbreviated table, but the most straightforward way is to list all 36 possibilities. This way, there's no chance of making a conceptual error, only a clerical error.

    Even though it's a lot of writing, it's pretty fast on paper, here it is typed

    (1,1) -> 1
    (1,2) -> 2
    (1,3) -> 3
    (1,4) -> 4
    (1,5) -> 5
    (1,6) -> 6
    (2,1) -> 2
    (2,2) -> 2
    (2,3) -> 3
    (2,4) -> 4
    (2,5) -> 5
    (2,6) -> 6
    (3,1) -> 3
    (3,2) -> 3
    (3,3) -> 3
    (3,4) -> 4
    (3,5) -> 5
    (3,6) -> 6
    (4,1) -> 4
    (4,2) -> 4
    (4,3) -> 4
    (4,4) -> 4
    (4,5) -> 5
    (4,6) -> 6
    (5,1) -> 5
    (5,2) -> 5
    (5,3) -> 5
    (5,4) -> 5
    (5,5) -> 5
    (5,6) -> 6
    (6,1) -> 6
    (6,2) -> 6
    (6,3) -> 6
    (6,4) -> 6
    (6,5) -> 6
    (6,6) -> 6

    So reduce it to a table

    Code:
    X   Ways to get X
    =   =============
    1   1
    2   ...
    3   ...
    4   ...
    5   ...
    6   ...
    Then the probability P(X) will be the number "ways to get X" divided by 36.

    For E(X), use the definition to get E(X)=1\cdot P(1)+2\cdot P(2)+\cdots +6\cdot P(6)
    Last edited by undefined; May 12th 2010 at 12:53 AM. Reason: small formatting improvement
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  3. #3
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    I made a table from the number of ways to get X

    Code:
    X     ways to get X
    =     ===========
    1      1
    2      3
    3      5
    4      7
    5      9
    6      11
    # of ways to get X = 36 possibilities

    And then I used this equation to solve E(X)=X*P(X)

    = 1*1/36 + 2*3/36 + 3*5/36 + 4*7/36 + 5*9/36 + 6*11/36 = 4.472222222


    Does this seem correct to anyone?
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by charles2 View Post
    I made a table from the number of ways to get X

    Code:
    X     ways to get X
    =     ===========
    1      1
    2      3
    3      5
    4      7
    5      9
    6      11
    # of ways to get X = 36 possibilities

    And then I used this equation to solve E(X)=X*P(X)

    = 1*1/36 + 2*3/36 + 3*5/36 + 4*7/36 + 5*9/36 + 6*11/36 = 4.472222222


    Does this seem correct to anyone?
    I get the same number, also expressible as 161/36.

    I don't mean to be nitpicky, but your terminology is a bit careless. "E(X)=X*P(X)" is not right; there should be a summation sign , \sum. Of course if you don't know how to type that you can still write E(X)=Sum(X_i*P(X_i)) or some such. Also "# of ways to get X = 36 possibilities" doesn't make sense because X is supposed to stand for a value in {1,2,3,4,5,6}. You could just say that total number of ways to roll dice is 36.

    By the way, I hope you know that the table you wrote down isn't a probability distribution. You would have to divide each entry in the second column by 36, then re-title the column as probability of rolling X, instead of ways to get X.
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  5. #5
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    Quote Originally Posted by undefined View Post
    By the way, I hope you know that the table you wrote down isn't a probability distribution. You would have to divide each entry in the second column by 36, then re-title the column as probability of rolling X, instead of ways to get X.
    Thanks for the information about correct notation. I'm not used to asking math and statistics questions on the internet.

    As far as the probability distribution, is this correct?

    Code:
    X       Probability of rolling X
    =      =========================
    1       1/36
    2       3/36
    3       5/36
    4       7/36
    5       9/36
    6      11/36
    I could also replace "Probability of rolling X" with P(X) right?
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by charles2 View Post
    Thanks for the information about correct notation. I'm not used to asking math and statistics questions on the internet.

    As far as the probability distribution, is this correct?

    Code:
    X       Probability of rolling X
    =      =========================
    1       1/36
    2       3/36
    3       5/36
    4       7/36
    5       9/36
    6      11/36
    I could also replace "Probability of rolling X" with P(X) right?
    Actually, it would not be good to write P(X) because that would imply that X is an event, not a number.

    Let's write it this way.

    Code:
    n      P(X=n)
    =      =====
    1       1/36
    2       3/36
    3       5/36
    4       7/36
    5       9/36
    6      11/36


    EDIT: Ah, I see I introduced the bad notation at the bottom of my first post! Shame on me! Well, capital letters like X often represent events, and I lost track.. happens. It is what might be called "abuse of notation" since people know what you mean, but technically it's not correct. (Or, instead of "abuse of notation" it might simply be called "incorrect." )

    Let's correct what I wrote above and instead write

    E(X) = 1\cdot P(X=1) + 2\cdot P(X=2) + \cdots + 6\cdot P(X=6)

    And by the way when I wrote E(X)=Sum(X_i*P(X_i)) above, I should have written E(X)=Sum(x_i*P(X=x_i)).

    Sorry for the slip!
    Last edited by undefined; May 13th 2010 at 08:48 PM.
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