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Before proceeding to the question, bear in english is not my native language and therefore technical terms may be wrong.

So, I'm trying to solve the old exam question, and I have different results from the solutions the teachers gave (they only gave the final answer).

The 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;

They ask, if I pick 3 melons, which store should I choose to maximize the probability of their total weight be above 8Kg.

So, I solved like (for store A) Z = (8 - (2.1 * 3)) / (0.7 * 3) = 0.8095 and then went on the table I found 0.2090 as the answer (did the same thing for store B, 0.2033).

I should choose store A as it has a bigger probability.

BUT, the problem is, the teachers solutions are 0.0804 and 0.0745.

What am I doing wrong? Is it the multiplication of the μ and σ?

I have not study Statistics for 4 years (working), so i'm a little rusty on the subject. A little help would be appreciated.

Thank you very much.

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up vote 2 down vote accepted

I think your standard deviation is wrong, it should be $\sqrt{3*0.7^2}$, in other words $3*0.7^2$ variance because for sums of normal random variables, it's the variances that are additive.

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It adds up. Thank you so much. – Mario Cesar Sep 15 '12 at 19:12
I posted another question related to this one at… It would be amazing if you could help me out in this one too. If not thanks anyway, you have helped me a lot. – Mario Cesar Sep 15 '12 at 21:30

Simply put it is the sum of three independent random variables (melon sizes) selected from A's distribution that you need to compare to the corresponding distribution for three iid melon sizes selected from B's distribution (ignoring finite population considerations). That is why the answer is based on comparing two z scores. The standard deviation that Sam gave is correct for store A. For store B the variance to use is 3 (0.2)$^2$.

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