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.

Quote:

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Originally Posted by ShuxRei

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.

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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.

Quote:

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Originally Posted by mr fantastic

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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Thank you so, so much. The links are especially helpful for other problems as well. I would have never thought of this.

Just because of this post im making an account lol this helped me a lot thanks