I think this is an easy one but I'm unfortunately a bit stumped. In short, why are standard deviations of actual values different than the standard deviations of those values expressed as a percentage of a total?

Say I sell 30 total widgets on day 1 of three different types (A-20, B-7, C-3). Day 2 sells 30 again, and the breakdown is (A-24, B-2, C-4).

For widget A, the 2 day data set calculates a mean of 22.0 and standard deviation of 2.8. For widget B, the 2 day data set calculates a mean of 4.5 and standard deviation of 3.5. For widget C, the 2 day data set calculates a mean of 3.5 and standard deviation of 0.7.

A new day of widgets sold comes in. Day 3 also sells a total of 30 widgets , and the breakdown is (A-27, B-2, C-1). When I calculate Day 3's sales standard deviations from the mean I get (27-22.0)/2.8 = 1.77. I apply the same math for B and C to calculate their respective day 3 sales in terms of standard deviations from the mean.

Now when I convert widget A's sales to percentages of the total widgets sold for each day (day 1 = 20/30 = 66.67%, day 2 = 24/80 = 80%, and day 3 = 27/30 = 90%), and calculate Day 3's mean of 73.3% and sd of 9.43%, I'm able to calculate Day's 3 standard deviations from the mean (90%-73.3%) / 9.43% = 1.77.

Good stuff - it ties out! The below is the actual part I'm stumped on.

Say we change day 2 to reflect 25 total widgets sold, with A selling 24 widgets. The SD from the mean stands at 1.77 but this certainly changes the percentages. The previous 66.7%, 80%, and 90% breakdown with mean of 73.3% now changes to 66.7%, 96% (=24/25), and 90% with mean of 81.33%. The previous SD of 9.43% changes to 20.74%. New SD from mean for percentages is at 0.42 = (90%-81.3%)/20.74%.

Why isn't it still 1.77?

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  • $\begingroup$ Welcome to Cross Validated! It'd be more useful to explain &/or illustrate your calculations than to show us a lot of numbers. Anyway, if 5 out of 10 widgets sold were A one day, & 5 out of 100 the next day, what results would you get for the mean & standard deviation of the counts & the proportions? $\endgroup$ Feb 4, 2015 at 18:27
  • $\begingroup$ Thanks! And good call. Will update my question shortly. $\endgroup$
    – Jeff
    Feb 4, 2015 at 18:44

1 Answer 1


There's no reason to expect the counts to be linear functions of the percentages except when the total is constant. Suppose 5 out of 10 widgets sold were A one day, & 5 out of 100 the next day. The counts have mean 5 & standard deviation 0; the percentages (which have gone from 50% down to 5%), mean 27.5% & standard deviation 31.8%. So the count isn't varying at all from day to day; the percentage is varying quite a bit.


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