lognormal with very large mean, but small sample value Let $X \sim N(\mu, \sigma^2)$, where $\mu = -800$ and $\sigma = 76$
Let $Y = \exp(X)$, so Y has a lognormal distribution, $E(Y) = \exp(\mu + \sigma^2/2) = \exp(2088)$, which is a very large number.
However, if I first sample from $X$, and then take $\exp()$ of the sampled points, I got all near 0's. This is understandable since $X$ has a big negative mean and 0 is 10 standard deviations away. 
But how should I understand that of $E(Y)$ being a very large number and a sample of all near 0 points? Because a very large variance?
What is the implication of the use of lognormal then?  
EX:
sample <- rnorm(100000, -800, 76)

summary(exp(sample))

 Min.      1st Qu.     Median       Mean    3rd Qu.       Max. 

0.000e+00  0.000e+00   0.000e+00 1.170e-208  0.000e+00 1.169e-203

 A: Note that although the mean is large, the 99.99th percentile is almost 0
 exp(-800+qnorm(0.9999)*76)
 [1] 2.068655e-225

which is to say, only one observation in ten thousand will exceed $2.069\times 10^{-255}$, an extremely small number.
Note that for the log of this random variable, $0$ is $10.526$ standard deviations above the mean. This means that there's only a probability of $3.3\times 10^{-26}$ of the lognormal variable exceeding $1$.
The mean is large because there's a (very) small chance of incredibly large observations. Even though those chances are astronomically small, the values the variable can exceed with those small probabilities are so astoundingly large that the mean is still very big. If you go out far enough, he balance between rarity and size eventually becomes dominated by size, making the mean large.
This the kind of thing you can get with a sufficiently skew variable - even one whose moments all exist!
How big is the (third moment) skewness? Quite large. It's very close to $\exp(8664)$ -- roughly a $5$ with 3762 $0$'s after it. 
