# What does it mean if the cumulative normal = 1?

What is the significance of cumulative normal = 1?

I am trying to program a game, however I have inherited code from another programmer. He is using the cumulative norm. his code is outputting a value of one for a certain value x, mean and standard deviation.

I have a product value which lets say is = 2.5. The standard deviation is 0.625. The x value was generated iteratively and eventually comes to 10.931456.

So are you saying that X or the mean is the 2.5, x is the 10.93, and the sd = 0.625?

Also that the probability of 2.5 being at the point x = 10.93 is 1?

-

If I understand your question correctly, you have a normally distributed random variable $X$, and the program says that the probability that $X\le x$ is $1$. Now for a normally distributed random variable $X$, the probability that $X\le x$ can never be exactly equal to $1$, it must always be less than $1$.

However, the tails decay very rapidly, so we may have $P(X\le x)$ for all practical purposes. So for example we could easily have $P(X\le x)=1-10^{-12}$. For instance, the probability that a standard normal $Z$ is $\le 6$ is about $1-2\times 10^{-9}$.

The programmer may have decided to round such a number to $1$. Or else, because of precision limitations, there is automatic rounding.

Deciding to round is very sensible. When you are modelling a real phenomenon using the normal distribution, that distribution will never model the phenomenon exactly. Sort of close is the best one can hope for.

Added: For your edited question, I do not know what the $2.5$ means. But the range from $2.5$ to $10.93$ is $8.43$, which is $8.43/0.625$ standard deviation units, about $13.5$ standard deviation units. That is an astoundingly large range, really quite inconsistent with a normal distribution. For most practical (or even impractical) purposes, in a normal there is virtually nothing $4$ standard deviation units or more away from the mean.

-
Yes it is assumed one since the value gets closer and closer to 1. And yes because of precision limits. Can you see my edit above? – user1064028 Feb 4 '12 at 17:18
So is my edit above (in the original question) correct? – user1064028 Feb 4 '12 at 17:34
@user1064028: I added some stuff to the answer. I cannot know what the numbers stand for, except for the standard deviation. It makes sense that the $x$ is $10.93$. The number $2.5$ cannot be "at" $10.93$, that doesn't make sense. The random variable $X$ (the numerical result of the experiment) can in principle be $\ge 10.93$, though with a normal with SD $0.625$, and mean $2.5$, or even a mean as large as $8$, this can basically never happen. That is consistent with the computed cumulative distribution function being virtually $1$ at $10.93$. – André Nicolas Feb 4 '12 at 17:51
Ok thank you it is making sense now – user1064028 Feb 4 '12 at 17:58