How many significant figures in the average area?

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Anshul23
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Homework Statement


I've been given a list of lengths of the side of a small metal sheet in the shape of a square. The length of the side of this square sheet varies with temperature. I have to calculate areas for each length and then calculate the average area. What should be the uncertainity and number of significant figures in the average area if the length with least sig figs has 8 significant figures. Is the uncertainity in average area related to the standard deviation in any way.

Homework Equations


Area of a square = length^2

The Attempt at a Solution


Calculated the average area using the calculator but it gives over 14 sig figs whereas the least sig figs in a length entry are just 8.
 
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If the only calculation being done is multiplying/dividing then I believe you need to go with the least number of sig figs in that particular calculation.
In the case of addition/subtraction, the main thing that matters is how many places digits are after the decimal.

(If you need more help with sig figs I will try to find a webpage which explains them well.
I think uncertainty has to do with the smallest place digit but do not know about standard deviation.)
 
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Thank you so much for your answer. My professor has assigned us an activity in which we have to calculate the average area. Some of my friends used average area +- the standard deviation which makes no sense to me. I believe in the same process that you've mentioned.
 
Here is the link to a webpage with understandable explanation of significant figures and how to use them in calculations:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html

I hope this helps, sorry I can't help with the standard deviation except to recommend searching "what is standard deviation" if you haven't already. I just did so and one of the first links which came up, https://www.mathsisfun.com/data/standard-deviation.html appears to explain it clearly and with easy-to-follow examples (though I do not know if what you are doing is more complicated whether it would help).
 
Anshul23 said:
Some of my friends used average area +- the standard deviation which makes no sense to me
Agreed. The average of a set of exactly known numbers would also be exactly known. The distribution of the numbers creates no source of error.