# normal vs nonnormal distribution??

#### fiddlestix327

##### New Member
Hi, we just learned about the nonnormal distribution in class today, but I'm afraid that I'm still a little confused.
I'm stuck on this problem:

What is the probability that a randomly selected observation exceeds the
a. Mean of a normal distribution?
b. Median of a normal distribution?
c. Mean of a nonnormal distribution?
d. Median of a nonnormal distribution?

So far, I believe that to parts b and d, for the median of both normal and nonnormal, it would be .5, since the median is the 50th percentile of the distribution. I also think that it would be .5 for the mean of a normal distribution, since, given that its normal, should be about the same value as the median. But what about the mean of a nonnormal distribution?? I want to say that it depends on a distribution's skewness, but I"m not sure.

Any sort of help would be appreciated. Thanks!

#### Dragan

##### Super Moderator
Hi, we just learned about the nonnormal distribution in class today, but I'm afraid that I'm still a little confused.
I'm stuck on this problem:

What is the probability that a randomly selected observation exceeds the
a. Mean of a normal distribution?
b. Median of a normal distribution?
c. Mean of a nonnormal distribution?
d. Median of a nonnormal distribution?

Well, what kind of non-normal distribution did you learn about in class?

That is, if it was a non-normal symmetric distribution (a uniform or rectangular distribution), then the answer would also be 0.50 i.e because skewness is zero i.e. the mean is equal to the median.

If, on the other hand, it was an asymmetric distribution , skewness non-zero, then the probability of randomly selecting a value greater than the mean is going to depend on the shape (or moments) associated with that particular distribution.