The probability that a randomly chosen grain weighs less than the mean grain weight If Y has a log-normal distribution with parameters $\mu$ and $\sigma^2$, it can be shown that
$E(Y)=e^\frac{\mu + \sigma^2}{2}$ and
$V(Y)=e^{2\mu +\sigma^2}(e^{\sigma^2}-1)$.
The grains composing polyerstalline metals tend to have weights that follow a log-normal distribution. For a type of aluminum, grain weights have a log-normal distribution with $\mu=3$ and $\sigma^2=4$ (in units of $10^{-4}$g).
Part A: Find the mean and variance of the grain weights.
Answer: $E(Y)=e^\frac{19}{2}$ and $V(Y)=e^{22}(e^{16}-1)$ 
Part B: Find an interval in which at least 75% of the grain weights should lie.
Answer: (0, $e^\frac{19}{2}+2e^{11}(e^{16}-1)^{1/2}$)
Part C: Find the probability that a randomly chosen grain weighs less than the mean grain weight.
I am unsure what to do for this part. Can someone sketch the procedures for me?
As of 12/28/13 there still instead an answer to this question.

It can also be shown that for a log-normal distribution, the cumulative distribution function is 
$$F(x)=\Phi \left( \frac{ \log x - \mu} {\sigma}\right).$$
Here, $\Phi$ is the cumulative distribution function of the standard normal. So the answer is
$$
\Phi \left(\frac{\frac{19}{2} -e^{19/2} }{e^{11}\sqrt{e^{16} -1}}   \right),
$$
very close to 0.5, as you might expect.
