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So we have a lognormal random variable Y with mean 10.2 and standard deviation 15.3. If we take 100 random samples with these properties, what is the probability that the average for this sample is less than 9?

Is the central limit theorem relevant here? I'm having trouble because of that "average" part.

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The Central Limit Theorem is very relevant. Let $S$ be the sum of your lognormals. We are interested in the random variable $X=\dfrac{S}{100}$.

Since $S$ is a sum of quite a few nicely behaved independent identically distributed random variables, $S$, and therefore $X$, the sample average (sample mean) has a distribution which is nicely approximated by a normal.

Note that $X$ has mean $10.2$. The variance of $S$ is $(100)(15.3)^2$. Thus the variance of $X$ is $\dfrac{1}{100^2}(100)(15.3)^2=\dfrac{(15.3)^2}{100}$. So the standard deviation of $X$ is $\dfrac{15.3}{10}$. (We could have written this down directly: I wanted to mention the theory.)

Now you want the probability that a normal with known mean $10.2$ and standard deviation $1.53$ is less than $9$. I expect that this is a standard calculation for you.

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Absolutely it is! Thank you so much, I wasn't sure if I should go down this route so thanks for clearing things up!! –  Marcus Nov 21 '12 at 2:08
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