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.