Can you calculate the standard Deviation from this data???

T

Tin_Whisker

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Jul 25, 2014
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I have a set of weights of cases of apples. Each case contains 6 apples. the weights I have are net weights so only the weight of the apples is counted. I would like to estimate the st. deviation of the apples.

Calculating the st. dev. of the cases is easy, but that is not what I am after. Since it is the variance in individual apple weights that comprise the variance in the case weights, there should be a way to calc this assuming the apple weights are normally distributed in the cases. I know engineers use a sum of squares approach for relating tolerances on interacting dimensions, which seems related but I can not gronk how to turn that into what i need.


Here are the case weights in lb. for reference:

1.86
1.89
1.87
1.88
1.91
1.87
1.91
1.89
1.93
1.90
1.91
1.90
1.89
1.89
1.89
1.89
1.90
1.89
1.87
1.88
1.87
1.87
1.85
1.87
1.87
1.88
1.89
1.87
1.88
1.88
1.88
1.88
1.89
1.90
1.91
1.89
1.87
1.88
1.90
1.90
1.92
1.87
1.88
1.89
1.90
1.87
1.87
1.86
1.87
1.89
1.88
1.88
1.89
1.88
1.90
1.89
1.92
1.91
1.90
1.89
1.88
1.90
1.90
1.89
1.89
1.88
1.88
1.88
1.88
1.87
1.87
1.86
1.88
1.88
1.92
1.88
1.89
1.87
1.88
1.92
1.91
1.89
1.90
1.88
1.89
1.88
1.89
1.88
1.89
1.86
1.88
1.88
1.86
1.87
1.86
1.86

 
Tin_Whisker said:
Surely there is someone out there that can help me with this.
Click to expand...
Strange, I thought I had answered this. The answer has to do with the Central Limit Theorem which basically says that if you take N sets of M random samples from an underlying population then, as N and M become large, the means (arithmetic averages) of those sets will approach a normal distribution with a mean of the underlying distribution and a variance of the underlying distribution divided by M.

Practically, this means that if you take the average weight of the apples in each of your crates, xj, and average those together you will have a number 'close to' the actual average weight, \(\displaystyle \mu\), of the apples. If you also compute the variance [average of the squared difference xj-\(\displaystyle \mu\)] and multiply by 6, you will have a number 'close to' the variance of the underlying population.

Oh, almost forgot, you might want to look at
https://www.khanacademy.org/math/pr...sampling_distribution/v/central-limit-theorem
 
Last edited:
I actually understand (most of) that and it not only makes sense I think I can use it to solve this. Thanks!
 
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