# Statistical Problem (part 2)

Following my question I found another problem.

Having the same data from the other question:

There are 2 melon stores. The melon weights follow a normal distribution.

• Store A -> μ = 2.1Kg, σ = 0.7Kg;
• Store B -> μ = 2.5Kg, σ = 0.2Kg;

But now what I can't figure out is:

If I buy 6 melons from each store, what is the probability of the sum of the weight of the melons from A being greater than the sum of the weight of the melons from the store B.

I'm really, really rusty at this stuff, and my exame is in a week...

So far I have that μA = 6*2.1 and σA = sqrt(6*0.7^2) and μB = 6*2.5 and σB = sqrt(6*0.2^2)

Thank you guys.

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## 2 Answers

So if we call $W_A$ the weight of 6 melons from store $A$, and $W_B$ the weight of 6 melons from store $B$, you are interested in the distribution of $W_A-W_B$. In particular, we want to find $P(W_A-W_B>0)$. We know the sum of two Gaussians is a Gaussian, so how does the difference of two Gaussians behave? Well, try to write down the distribution of $-W_B$. If you are still stuck, give a shout.

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Well, i tried, but I am completely lost. The WA−WB > 0 would have a μ = μA - μB and a σ^2 = σA^2 + σB^2 ? The forms we have state that f(x) is not integrable, so how can i write the distribution function?... man, I am really lost. – Mario Cesar Sep 16 '12 at 0:15
So in your other question, you knew how to find $P(Gaussian<x)$ by referring to a table. In this question, you need to find $P(Gaussian>x)$, which can be rewritten to make your table useful. It sounds like your book is reminding you that the gaussian density $f(x)=\frac{1}{\sqrt{2\pi \sigma^2}}e^{-x^2/(2\sigma^2)}$ is not integrable by any simple calculus means, so it's just reiterating to use your Z table to help you. – Alex R. Sep 16 '12 at 0:22
Well, this question is clearly not for me. Thank you anyway, i'll accept your answer because it is (most likely) correct. – Mario Cesar Sep 16 '12 at 0:31
Naw, don't give up. Which detail is still bothering you? – Alex R. Sep 16 '12 at 0:37
Well, it's a little late here and maybe is me spending the whole day studying... I'll give it another try tomorrow, right now I can't even solve the other question i solved before. Thanks. – Mario Cesar Sep 16 '12 at 0:43

I think the problem here is that you haven't stated the complete question and Sam guessed at it. The question Sam answered is:

If you choose 6 melons from one of the stores what is the probability that the total weight from Store A would be greater than what you would have gotten if you selected 6 from store B.

But your previous question asked

If you choose 3 melons from either store A or store B which selection has the higher probability that the total weight is greater than 8 kg?

Is it this second question with 6 substituted for 3 and possibly a different weight replacing 8?

If it is the latter than just as your instructors did for the previous question you need to compute the 2 z scores for the sums of 6 iid melon weights from store A and store B. Use the standard deviations and means as you correctly described them above in your question and compare the probabilities to 8 kg or whatever value you need to substitute for it.

Based on your comment Sam was giving you the right approach. Let me just fill in somemore detail so that you can finish it off yourself. As he pointed out if X and Y are two independent normal ranom variance the sum X-Y is normal and has mean equal to the difference of the two means and variance equal to the sum of the two variances. So in this probem Take W$_A$-W$_B$ and divide it by sqrt(6*0.7^2+6*0.2^2) to get the Standard normal random variable. Then the z score is [μA - μB)]/sqrt(6*0.7^2+6*0.2^2). So look at the standard normal table to see what the probability is that a standard normal is larger than this z score. (In all these problem if the sample standard deviation is used t distribution should be used. But in this case since the variances are different the Welch test would be used instead of the t test that pools the variance estimates.)

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No. This question only have the mean and variance in common with the first one. The question is really "what is the probability of the 6 melons from store A weight more than the 6 melons from store B". The answer in the solutions is 0.0892, and I have no idea how to get there. – Mario Cesar Sep 16 '12 at 14:15
I have amended my answer to give you additional help. – Michael Chernick Sep 16 '12 at 14:45
but, what would WA-WB be? the difference between the means? – Mario Cesar Sep 16 '12 at 14:54
[W$_A$-W$_B$-(μ$_A$ - μ$_B$)]/sqrt(6*0.7^2+6*0.2^2) is standard normal. Therefore you need P{[W$_A$-W$_B$-(μ$_A$ - μ$_B$)]/sqrt(6*0.7^2+6*0.2^2) >0}. – Michael Chernick Sep 16 '12 at 16:21
Sorry for confusion. you want P[W$_A$-W$_B$>0] W$_A$-W$_B$ is a random variable with a known normal distribution. divide both sides of the inequality by sqrt(6*0.7^2+6*0.2^2) then subtract[μA - μB)]/sqrt(6*0.7^2+6*0.2^2) from both sides. The left side is now a standard normal random variable and the right side is a number you can compute. So look up the probability that a standard normal is greater than the number you get on the right side of the inequality. – Michael Chernick Sep 16 '12 at 16:35