# Standard deviation

Gold Member
$$\textit{To get an estimate of your uncertainty, compute the standard deviation. My 4 distances are 150, 120, 100 and 100 parsecs (pc)} The average of my 4 distances is $\bar {x}=\frac{\sum\limits_{i=1}^n {x_i } }{n}$ $\bar {x}=\frac{150pc+120pc+100pc+100pc}{4}=117.5pc$ The average =  120pc \textit{The standard deviation is } $\sigma =\sqrt {\frac{\sum\limits_{i=1}^n {\left( {x_i -\bar {x}} \right)^2} }{n-1}}$ \[ \sigma =\sqrt {\frac{\left( {150-120} \right)^2+\left( {120-120} \right)^2+\left( {100-120} \right)^2+\left( {100-120} \right)^2}{n-1}} =23.805$$

I've computed it but what does it mean? How do I estimate my uncertainty from this number? The book doesn't explain this.

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The standard deviation IS a measure of your uncertainty.

Zurtex
Homework Helper
Standard deviation is sort of the average (or expected) away from the average. If you input a lot of data and after a while you get a stable average and standard deviation, any other point inputted will on average be the standard deviation away from the average.

In other words the standard deviation is how far the points deviate away from the average standardly :tongue:

SpaceTiger
Staff Emeritus
Gold Member
tony873004 said:
I've computed it but what does it mean? How do I estimate my uncertainty from this number? The book doesn't explain this.
Generally, when people report their errors, they report $$\sigma$$, but occasionally you'll see them report $$2\sigma$$ or $$3\sigma$$ (or some other statistics, but $$\sigma$$ is usually standard). These choices for error quoting are fairly arbitrary, it's just a matter of what you're trying to communicate (unfortunately, it can also be used for deception). If the errors are distributed normally, then a 1$$\sigma$$ error bar means that there's about a ~68% chance that the true value lies in that interval, while 2$$\sigma$$ and 3$$\sigma$$ represent about a ~95% and ~99.7% chance that the true value is in the interval. You might quote the latter if you wanted to be conservative with your results. Of course, there are errors on your estimate of the error as well, but let's not get into that.

I don't know what the case is for you, but definitely consider the above before quoting an error.

Gold Member
Thanks for all your replies to a topic not covered by my book!

SpaceTiger, I actually included what you said in my final answer. I got an uncertainty of about 24 parsecs, And there's a 1 pc difference between the closest star and the furthest star in the Pleadies star cluster. I don't know if I should add the intrinsic difference to the 24 or subtract it. I'm guessing subtract. I just concluded it in a sentence to the effect "In the face of a 24 pc uncertainty, the 1 parsec difference in actual star differences is negligable." The teacher liked my answer.

SpaceTiger
Staff Emeritus
Gold Member
tony873004 said:
SpaceTiger, I actually included what you said in my final answer. I got an uncertainty of about 24 parsecs, And there's a 1 pc difference between the closest star and the furthest star in the Pleadies star cluster. I don't know if I should add the intrinsic difference to the 24 or subtract it. I'm guessing subtract. I just concluded it in a sentence to the effect "In the face of a 24 pc uncertainty, the 1 parsec difference in actual star differences is negligable." The teacher liked my answer.
I'm kinda surprised he liked it, cause it's not technically correct to say that the uncertainty on the distance to the center (presumably calculated to be the mean of your data points) is equal to the spread. Actually, the uncertainty on the mean of a set of data points is:

$$\sigma_{\bar{x}}^2=\frac{\sigma^2}{N}$$

where N is the size of the sample. This would mean that your uncertainty is a factor of two smaller than the standard deviation of the sample:

$$\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{4}}$$

This is simply because more measurements will give you more independent data points with which you can estimate the distance, making your final estimate more accurate. The "error" that is given by $$\sigma$$ alone would be the error associated with a single measurement.

Anyway, I'm glad this didn't turn out to be a problem in the end. My apologies for not being clearer about it.

Being that we never covered this in class and the book doesn't cover it either, I don't think the teacher had any choice other than to like my conclusion which is still pretty much correct even though I used 24 instead of 12. 