# Homework Help: How do I calculate the average of uncertainties?

1. Sep 10, 2012

### PhizKid

1. The problem statement, all variables and given/known data
Trial 1: 1.05 ± 0.05 m.
Trial 2: 2.0 m.
Trial 3: 1.5 m.
Trial 4: 1.05 ± 0.05 m.

2. Relevant equations
Not sure

3. The attempt at a solution

So to add them up, I just add the "base" measurements" and then tack on the ± and stick in the "uncertainties" like so:

1.05 + 2.0 + 1.5 + 1.05 = 5.6 m.
±
0.05 + 0.05 = 0.1 m.

So the final sum: 5.6 ± 0.1 m.

Now to take the average, I need to the quotient of 5.6±0.1 / 4 but not sure how to do this. The instructions say to divide the averages of the two numbers you are trying to divide (5.6 / 4) and them multiply that quotient by (1.00±δ5.6/5.6 = δ4/4) but 4 doesn't have an average deviation because it's just a number.

So how do I do this?

2. Sep 10, 2012

### Muphrid

The idea is that when you multiply or divide, the percentage uncertainties are added, rather than the absolute uncertainties.

Here, since 4 is a pure number, the percentage uncertainty is just that of your 5.6 measurement.

3. Sep 10, 2012

### PhizKid

So the average of my measurements without the uncertainties is 1.4 m. Would I put down 1.4±0.1 m. as the average of all 4 measurements?

4. Sep 10, 2012

### Muphrid

Is that the same percentage uncertainty that you had before you divided by 4?

5. Sep 10, 2012

### I like Serena

Re: Standard deviation for uncertainties

Hi PhizKid!

Are these different trials for the same measurement?
If so, then if you leave out the ± values, what do you get for the standard deviation?

You should see that the value you get is much higher than 0.05 m.
It means that you can neglect the given uncertainties.
You can take them into account, but that will take more formulas.

6. Sep 10, 2012

### PhizKid

Well, some of the measurements have different uncertainties...for this particular set of experiments, of the 4 trials, only 2 of them have uncertainties, but those two happen to be the same uncertainties. So in the case I have different uncertainties, how would I go about those?

7. Sep 10, 2012

### PhizKid

Re: Standard deviation for uncertainties

Oh...which formulas will it take? Are they far beyond the complications for an introductory mechanics course?

8. Sep 10, 2012

### Muphrid

I'm only talking about the 5.6 number that you had. What was that measurement's percentage uncertainty? I realize it was the sum of 4 trials; that's fine.

9. Sep 10, 2012

### I like Serena

Re: Standard deviation for uncertainties

I think so, but you can judge for yourself.

First off, is this indeed about the average of a number of trials?
That matters.
I'll explain more after you answer that question.

I can say that when adding numbers with an uncertainty, you are supposed to add the squares of the uncertainties.
To find the resulting uncertainty of the sum, you need to take the square root of the sum of the squared uncertainties.

In your case you have 2 sources of uncertainties.
1. The uncertainties in the measurements.
2. The uncertainties that show in the deviation of each measurement to the average.
These need to be handled separately.

10. Sep 10, 2012

### daveb

How can you have an uncertainty for 1 trial of the measurement but not an uncertainty for another trial of the same measurement? That makes no sense.

11. Sep 10, 2012

### I like Serena

The trials have an implicit uncertainty.
The rule is that a half beyond the last digit given is the uncertainty.

So for instance 2.0 is actually 2.0±0.05.

In the case of 1.05 it would be 1.05±0.005.
Since that is not intended, it is explicitly given as 1.05±0.05.

12. Sep 10, 2012

### PhizKid

Re: Standard deviation for uncertainties

Yes, I am trying to take the average of several of the same experiments.

When we took our measurements, we were told to just put the measurement down, and that was it. But when we read a measurement that was in between millimeter lines, we were told to add 0.5 mm to the measurement, then state ±0.5 mm after that measurement. But only if we measured in between the mm lines. Otherwise we didn't have to state the ± uncertainty or whatever.

13. Sep 10, 2012

### I like Serena

Re: Standard deviation for uncertainties

The proper way to do it is to *always* read between the millimeter lines and estimate a number within those lines.
But let's not go into that.
Obviously they want you to take measurements with nice round millimeter numbers which is easier.

To find the uncertainty of "n" trials, the method is as follows:
1. Calculate the average.
2. Calculate the difference of each trial with the average.
3. Square those differences.
4. Add all the squared differences.
5. Divide the sum by (n - 1). In your case this is 3. The result is the so called "variance".
6. Take the square root, which will give you the uncertainty. This is also called the "standard deviation".

This is what your calculator will do for you.
Note that none of the uncertainties of the trials is used for this.
They would be neglected unless they are big enough to matter.
In your case they are not.
Seeing that you are supposed to round to millimeters I don't think you are supposed to take them into account.

Anyway, if you do want to take them into account, it works like this.

To find the resulting uncertainty due to the uncertainties in your trials, you would use the following.
1. Square each uncertainty.
3. Divide by n. In your case this is 4. Again this will give you a "variance".

As it is, "variances" are supposed to be added - not the uncertainties you have.
So you would add the variance due to the average, with the variance due to your trials.
Afterward take the square root and you have your resulting uncertainty.

In your case you will see that the contribution of your trials uncertainties is way less than the uncertainty you get from the average.
Actually, it could also be the other way around.

Last edited: Sep 10, 2012
14. Sep 10, 2012

### Dickfore

When you average measurements with different uncertainties, you need a weighted mean:
$$\bar{x} = \frac{\sum_{i}{\frac{x_i}{\sigma^{2}_{i}}}}{\sum_{i}{\frac{1}{\sigma^{2}_{i}}}}$$
and the uncertainty of the mean is:
$$\sigma_{\bar{x}} = \frac{1}{\sqrt{\sum_{i}{\frac{1}{\sigma^{2}_{i}}}}}$$

15. Sep 11, 2012

### daveb

I am aware of that. I was wondering why he didn't write it for two of the measurements. It seems like he was saying it was an exact measurement