# standard deviation of a division

#### michu

##### New Member
Hi,

I have a rather easy question: I have a batch of different dried fruits, let's say n=100, with a diameter of 8±2 cm, 2 being the standard deviation. After throwing them in water the distribution of the batch's diameters is let's say 16±5 cm because of the swelling. This results in a size ratio of 2, but what about the standard deviation of this calculation?

I tried propagation of uncertainty with covariance=0, so:

σ(A/B) ≈A/B*SQRT( (a/A)^2 +(b/B)^2 ). a and b are the standard deviation of A and B, respectively.
Hi,

I have a rather easy question: I have a batch of different dried fruits, let's say n=100, with a diameter of 8±2 cm, 2 being the standard deviation. After throwing them in water the distribution of the batch's diameters is let's say 16±5 cm because of the swelling. This results in a size ratio of 2, but what about the standard deviation of this calculation?

I tried propagation of uncertainty with covariance=0, so:

σ(A/B) ≈A/B*SQRT( (a/A)^2 +(b/B)^2 ). a and b are the standard deviation of A and B, respectively.

When using my example I get 0.8, so is the size ratio of the two population 2±0.8? This seems not too bad given that the extreme ratios 21/6 and 11/10 are 3.5 and 1.1, respectively.

Thanks a lot for the feedback, curious to see whether this is good enough or if there are better means.

#### katxt

##### Active Member
The method you have used works well enough for small a/A and b/B but is only approximate.
You could try it by doing a simulation, either generating A and B using say a normal distribution or using resampling if you have the data.
Is it possible to identify individual fruit before and after soaking?

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#### michu

##### New Member
The method you have used works well enough for small a/A and b/B but is only approximate.
You could try it by doing a simulation, either generating A and B using say a normal distribution (this gives SD about 1.0) or using resampling if you have the data.
Is it possible to identify individual fruit before and after soaking?
hi katxt, thanks a lot for your reply! our a/A and b/B are not that small, 25 and 31 %, respectively, so the approximation might not be great. However, A and B are not that far from having a normal distribution, so your idea seems cool! And I agree that observing the fruits one by one would be better, but this is not possible in that process.

So if you don't mind I would be very happy if you could emphasize a bit more on the method you proposed? Thanks a lot in advance and best regards,
M

#### katxt

##### Active Member
When I read your first post, I only looked at the formula and didn't check your calculations. Now that I've seen your latest post, I realize that you haven't taken the sample size n = 100 into account.
a and b are not the SDs. They are the uncertainties in the sample means (the standard errors) = SD/sqrt(n)
a = 5/sqrt(100) = 0.5 and a/A is 0.5/16 = 3.1% b = 2/sqrt(100) = 0.2 and b/B = 0.2/8 = 2.5% These are small enough that the approximation is fine. Completing the calculations, SD of A/B = turns out to be 0.08 rather than 0.8
The suggestions about resampling are still true, but unnecessary in this case. They would be useful with badly non normal data.
kat

#### katxt

##### Active Member
You have calculation involving uncertain variables and in the formulas a and b are estimates as to how uncertain they are. For your calculation (as I understand your posts) you measure 100 dried fruit and 100 fruit after soaking. You find the means, and divide. So the variables A and B are the sample means, and the a and b are the standard deviation of those means (not of the individual fruit).
The SD of an estimate from a sample is known as the standard error but it is the SD of the mean, not individual fruit.
If you measure a length once and get 120 mm +/- 2 mm then A = 120 and a is 2.
If you measure n = 100 weights and get a mean = 50 g with SD = 5 g then A is 50 and a = SD of A = SE = 5/sqrt(100) = 0.5

#### michu

##### New Member
Hi katxt,

Thanks a lot for that beginners course in statistics, I think all is clear and the standard error of my division of 0.08 makes sense!
Best,
Michu