The probability that a randomly chosen grain weighs less than the mean grain weight

Question

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.

Answer 1

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.

Answer 2

This is the question in mathmatical statistics 7th edition. Q4.183.

But first, this qustion has a big mistake about the mean value. $E(Y)=e^{\mu+\sigma^2/2}$

The big hint is in Q4.182: A random variable Y is said to have a log-normal distribution if X=ln(Y) has a normal distribution.

Thus, $P(Y<e^{11})=P(X<11)$ and X has a normal distribution with $\mu=3$ and $\sigma=4$

$P(X<11)=P(X<\mu+2\sigma)=0.5+0.9545/2 = 0.97725$

However, the answer in the book is 0.8020 which I believe is incorrect