Talk:Standard deviation

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I noticed the External Links section was recently removed. I am the editor who updated the LaTeX notation in the 'Rapid calculation methods' section earlier today to improve the article's technical clarity. Regarding the external link I provided (Firgelli), it is a live, high-precision tool that supports Bessel's correction and weighted calculations—both core themes of this article.

I replaced the previous link (standard-deviation-calculator.net) because it was a low-utility site. If an 'External Links' section is to be maintained, I believe it should point to tools that handle the advanced cases discussed in the text, such as step-by-step variance derivations. I'd appreciate feedback on whether this resource meets the community's standards for utility. ProjectAutomate (talk) 20:38, 27 March 2026 (UTC)Reply

Wikipedia isn't for supplying calculators. Sure, if there is one out there with many citations and/or is otherwise notable and reliable, we sometimes would include it. The calculator you are adding was posted March 8, 2026 according to itself. That alone would seem to indicate that it is not reliable as a source. I could be wrong, but if were forced to make a guess with no additional information, I would guess that this case is instead more like low-level WP:CITESPAM. —Quantling (talk | contribs) 21:21, 27 March 2026 (UTC)Reply

Fair point, Quantling. I saw the External Links section was a bit of a mess with that old .net site and thought a high-precision tool would be a better fit, but I'll defer to the community standards on that. My primary interest is just cleaning up the LaTeX notation and technical clarity in these articles. Thanks for the heads-up on the policy. ProjectAutomate (talk) 21:33, 27 March 2026 (UTC)Reply

It was not good before, but there was a link to expected value (from the word average, which was not ideal). Is that link still needed somewhere? Jens Ahlström (talk) 01:49, 30 March 2026 (UTC)Reply

I am pretty sure that the biases given for n=3 and n=9 are wrong. Using the given formula I get in R

> for (n in c(3,9))

print(c(n,1/((sqrt(2/(n-1))*gamma(n/2)/gamma((n-1)/2)))))

[1] 3.000000 1.128379

[1] 9.000000 1.031661

which corresponds to 13% (not 1.3%) and 3% (not smaller than 0.1%). That these values are correct can be seen by:

> s=c(); for (i in 1:10000) s=c(s,sd(rnorm(3)))

> 1/mean(s)

[1] 1.128619 ~2026-41396-52 (talk) 13:28, 24 July 2026 (UTC)Reply

I have corrected them. DSNice (talk) 13:51, 24 July 2026 (UTC)Reply