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Math Help - Normal Distribution and CLT

  1. #1
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    Normal Distribution and CLT

    My teacher hasn't taught us how to do this problem but still wants us to do it--

    Suppose that the distribution of the weight of a prepackaged 1-pound bag of carrots is N(1.18,.07^2) and the distribution of the weight of a prepackaged 3-pound bag of carrots is N(3.22,.09^2). Selecting bags at random, find the probability that the sum of three 1-pound bags exceeds the weight of one 3-pound bag. HINT: First determine the distribution of Y, the sum of the three, and then compute P(Y>W), where W is the weight of the 3-pound bag.

    Now, I know Y=X1+X2+X3. Is it also equal to Chi-Square with 3 degrees freedom? I'm completely lost on this problem.
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  2. #2
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    Quote Originally Posted by ShuxRei View Post
    My teacher hasn't taught us how to do this problem but still wants us to do it--

    Suppose that the distribution of the weight of a prepackaged 1-pound bag of carrots is N(1.18,.07^2) and the distribution of the weight of a prepackaged 3-pound bag of carrots is N(3.22,.09^2). Selecting bags at random, find the probability that the sum of three 1-pound bags exceeds the weight of one 3-pound bag. HINT: First determine the distribution of Y, the sum of the three, and then compute P(Y>W), where W is the weight of the 3-pound bag.

    Now, I know Y=X1+X2+X3. Is it also equal to Chi-Square with 3 degrees freedom? Mr F says: No. Your thinking of the sum of the squares of randomly distributed variables.

    I'm completely lost on this problem.
    First of all:

    Let X be the random variable weight of pre-packaged 1-lb bags.
    Let Y be the random variable sum of the weights of three pre-packaged 1-lb bags. Then:

    Y = X + X + X ~ N(1.18 + 1.18 + 1.18, 0.07^2 + 0.07^2 + 0.07^2) = N(3.54, 0.0147) = N(3.54, 0.12124^2).

    Note that 0.12124^2 is the variance.

    See the result in this link.


    Let W be the random variable weight of pre-packaged 3-lb bags.

    You want to calculate Pr(Y - W > 0). So consider the random variable D = Y - W:

    D ~ N(3.54 - 3.22, 0.12124^2 + 0.09^2) = N(0.32, 0.0228) = N(0.32, 0.151^2).

    Note that 0.151^ is the variance.

    See the result in this link.


    Now calculate Pr(D > 0).

    I get 0.983 (correct to three decimal places) as the answer.

    Note: I reserve the right for this reply to contain arithmetic errors.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    First of all:

    Let X be the random variable weight of pre-packaged 1-lb bags.
    Let Y be the random variable sum of the weights of three pre-packaged 1-lb bags. Then:

    Y = X + X + X ~ N(1.18 + 1.18 + 1.18, 0.07^2 + 0.07^2 + 0.07^2) = N(3.54, 0.0147) = N(3.54, 0.12124^2).

    Note that 0.12124^2 is the variance.

    See the result in this link.


    Let W be the random variable weight of pre-packaged 3-lb bags.

    You want to calculate Pr(Y - W > 0). So consider the random variable D = Y - W:

    D ~ N(3.54 - 3.22, 0.12124^2 + 0.09^2) = N(0.32, 0.0228) = N(0.32, 0.151^2).

    Note that 0.151^ is the variance.

    See the result in this link.


    Now calculate Pr(D > 0).

    I get 0.983 (correct to three decimal places) as the answer.

    Note: I reserve the right for this reply to contain arithmetic errors.
    Thank you so, so much. The links are especially helpful for other problems as well. I would have never thought of this.
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  4. #4
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    Just because of this post im making an account lol this helped me a lot thanks
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