# Parameters of normal distribution: why are they so different from other distributions?

In all, at least well-known probabilities, the parameters are not related to mean and variance. In other words, usually we provide a specification on how a single events occur (e.g., p in Bernoulli trial, frequency in Poisson / exponential). Then, we can calculate the expected value and variance. In contrast, in Normal distribution we provide mean (which is expected value) and variance (standard deviation). So, this looks as though this is not an information about single events.

Why is this so different for Normal distribution?

Thanks

• It is probably not as different as you are making it out to be. For the exponential distribution, you could think of specifying the mean survival time (of, say, a lightbulb), and for the Poisson distribution you might think of the mean number of occurrences in an interval of time (or space), e.g. the number of phone calls you get in a day. In both cases you provide the mean parameter. – snar Dec 23 '18 at 18:18
• Thanks for the answer. This sounds reasonable. – John Dec 24 '18 at 6:37

## 1 Answer

Normal distribution is a continuous distribution, while the others you mentioned (Bernoulli, Poisson) are discrete.

For continuous distributions, the probability of a single event is $$0$$. The discussion will always be about events within a range.

• But exponential distribution is continuous distribution, but it is still defined by lamda (i.e., frequency of events), but not by the summary statistics as normal distribution. – John Dec 23 '18 at 17:27
• I am not sure what distinction you are making between normal distribution defined by two parameters (mean and variance) or the exponential defined by one parameter (lambda, which is the reciprocal of the mean). – herb steinberg Dec 23 '18 at 17:43
• But put it other way, why there are no distributions with variance as a parameter? – John Dec 23 '18 at 18:04
• Technically, there are. The Gamma distribution has two parameters, which together determine the mean and variance. Neither parameter alone determines the variance, but that is just a question of parametrization. For instance, you can think of an affine function as $f(x) = ax + b$ for parameters $a, b$, or $f(x) = c(x - h)$. – snar Dec 23 '18 at 18:22