# How to remove negative values from a bell curve?

#### lawthesmurf

##### New Member
I have a data set that is looking at the average number of days it takes for a certain action to happen, and the variance is fairly large (~80) and the mean is also around 80. A bell curve gives me data that implies a portion of my data could be negative, but in the case of "Days to action", that's impossible. Is there a way to deal with this? Thank you.

#### ondansetron

##### TS Contributor
I wonder if @lawthesmurf gave us enough information to really know what's being asked and what the data look like. Why not post some histograms of the raw data and try to show us what you're describing?

#### Miner

##### TS Contributor
Time based data is rarely normally distributed. I would try other distributions that commonly model time based data such as weibull or lognormal.

#### GretaGarbo

##### Human
Time based data is rarely normally distributed. I would try other distributions that commonly model time based data such as weibull or lognormal.
Yes.

Or:
...and the variance is fairly large (~80) and the mean is also around 80.
That would rather suggest a Poisson distribution (which can not be negative). But are all data points independent?

#### ondansetron

##### TS Contributor
@GretaGarbo, @Miner, @spunky, I must be the only one wondering if OP has some how standardized values (possibly due to a point-and-click software) and sees negative standardized values which really aren't negative at all. No one else thought this? I'm so obtuse

#### hlsmith

##### Not a robit
I agree with @GretaGarbo on the Poisson dist. That is what I was going to propose, what is it, its mean and variance are equal, plus the positive attribute. And @Miner has a good point with the lognormal.

@ondansetron - I think the OP is just saying if I take +/- 2SD, the data will be negative on the left hand side of the dist, which is impossible.

#### ondansetron

##### TS Contributor
I agree with @GretaGarbo on the Poisson dist. That is what I was going to propose, what is it, its mean and variance are equal, plus the positive attribute. And @Miner has a good point with the lognormal.

@ondansetron - I think the OP is just saying if I take +/- 2SD, the data will be negative on the left hand side of the dist, which is impossible.
I thought of that, but just figured it couldn't hurt to clarify. Otherwise, Poisson and Weibull come to mind right away; the former given mean=variance and the latter given time to something.

#### GretaGarbo

##### Human
But with the mean 80 and variance 80, thus standard deviation about 9, it seems strange to get negative values from 80 +/- 2*9. So Maybe something else happened.

But I guess that we have lost the OP .....

#### ondansetron

##### TS Contributor
But with the mean 80 and variance 80, thus standard deviation about 9, it seems strange to get negative values from 80 +/- 2*9. So Maybe something else happened.

But I guess that we have lost the OP .....
This was part of my initial thinking which is why I tried to understand how the OP could be obtaining "negative values." I give more credence to the end-user-error (in general) than a problem of software or erroneous calculations that are out of the user's control.

#### GretaGarbo

##### Human
I must be the only one wondering if OP has some how standardized values (possibly due to a point-and-click software) and sees negative standardized values which really aren't negative at all.
That is a possibility.

I think the OP is gone. But I wonder how often strange things like this happens "out there". What do people actually draw conclusions from?

#### ondansetron

##### TS Contributor
That is a possibility.

I think the OP is gone. But I wonder how often strange things like this happens "out there". What do people actually draw conclusions from?
I think it's quite often that (lay) people know how to click buttons or copy a script but don't know what they've printed out or how to accurately interpret it.
Quickly reviewing the OP and the OP's other single post, I think this isn't out of the question.