# What is the physical meaning of the output/ y -value of a normal distribution? (not the area under its curve)

Forgive me for my lack of knowledge regarding math terminology.

I'm learning basic statistics right now, and I can see pretty intuitively that the area under a normal distribution on a certain interval is the probability of a certain range of events happening. For example, if the amount of rainfall tomorrow can be modeled by a normal distribution, and the mean amount of rainfall is 2 inches, I can predict the probability of getting 1-3 inches of rain by finding the integral under the curve between 1 and 3.

Thus the normal distribution doesn't really seem to be important, just its integral. The y-value of its integral (or difference of y-values) is what actually gives us the probabilities we want to find. This means the the y-values of the actual normal distribution are what I can best describe as the rate of change of the probabilities at some point.

So that leads to my question(s). How would you intuitively describe the meaning of the y-values of a normal distribution? The "rate of change of the probabilities" are at a maximum at the mean of the normal distribution; does this mean anything? And thirdly, I'm also considering the possibility that these y-values have no meaning, as my textbook and teacher haven't mentioned them. If they are useless, then why do we even bother looking at the normal distribution? Why don't we just directly look at it's integral?

Thanks, Bryan

-
a physical model related to marbles falling through nails. – Maesumi Mar 2 '13 at 1:47

My interpretation is that you are asking why one should care for the density function of a random variable $X$, when what we use for simple probability calculations is the cumulative distribution function (the integral of the density function).

There are many calculations in which we use the density function. For example, if $X$ has density function $f_X(x)$, then $E(X)=\displaystyle\int_{-\infty}^\infty xf_X(x)\,dx$. More generally, if $Y=g(X)$, then $E(Y)=\displaystyle\int_{-\infty}^\infty g(x)f_X(x)\,dx$.

Remark: Your interpretation of the density function as rate of change of probability is good. Similarly, acceleration is rate of change of velocity. Even if we are only interested in velocity, acceleration is a useful notion.

I would agree with you that cumulative distribution function is more fundamental than density. However, that foes not mean that density of no importance. In the case of the normal, we have the additional fact that while density is an "elementary" function, the cdf is not.

-
I'm going to assume that the integral is for calculating the expected value of that random variable X. In our textbook, we only deal with discrete examples but the concept should be the same for something continuous. In that case the output of fX(x) would be the probability of the x-value occurring. But if I plugged in an x-value into the normal distribution function, my y-value would not be the probability of the x-value occurring. Is there something I'm missing? I'd also agree that acceleration is important, but I have an intuitive understanding of acceleration and I'm missing that here. – Syllabear Mar 2 '13 at 17:52
Discrete probabilities are like weights hung at various places. With continuous distributions, we have a wire of continuously varying density. – André Nicolas Mar 2 '13 at 18:06