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[?] Solve for the standard deviation of a normal distribution in W|A?

Bryce Costawong
Posted 4 years ago

Hi folks,

I'm no math or WolframAlpha expert, but I'm taking a probability class and encountered a problem where I have to solve for the standard deviation of a normal distribution. I boiled down the problem into: $$.9 = \int_{117}^{207} \frac{1}{x} \frac{1}{\sqrt{2\pi}} e^{-\frac{(\frac{y - 162}{x})^2}{2}} dy$$ solving for x.
I realize this is super computationally expensive and my TI-84 was making that known, so I plugged it into Wolfy:

.9 = integral(117,207) (1/x * 1/((2*pi)^(1/2)) * (e)^(-((y-162)/x)^2 / 2)dy)

"Wolfram|Alpha doesn't understand your query"

Which is strange because it understands a much simpler command:

.9 = integral(117,207) (x y dy)

I tried just about everything with syntax and could not get it to work, it seems to get stuck when there is (a term with an x) * e^(term with x and y)
My potato calculator eventually pulled through, but I wanted to know what the problem was.

Feedback appreciated, thanks,
Bryce Costawong
POSTED BY: Bryce Costawong
5 Replies
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Solution with Mathematica not WolframAlfa:

INT = Integrate[1/x*1/Sqrt[2*Pi]*Exp[-((y - 162)/x)^2/2], {y, 117, 207}]
(*Erf[45/(Sqrt[2] x)]*)

x /. NSolve[INT == 9/10, x, Reals][[1, 1]]

(* 27.3581 *)
POSTED BY: Mariusz Iwaniuk

By the way, there are other tool sets that parse Context Free Grammars if the Wolfram toolset does not include one.
POSTED BY: David Whitten

Is there a listing of the Extended Backus-Naur Form (EBNF) of the Context Free Grammar (CFG) that the Wolfram Language accepts? ISO/IEC 14977:1996 at https://www.iso.org/standard/26153.html is one format, though it is less readable than others. But even using ISO 14977 is better than none. I think someone told me that there was a Earley Parser as a tool in the Wolfram Toolset.
POSTED BY: David Whitten

Integral (1/x)(1/(2Pi)^(1/2))*e^-((((y-162)/x)^2)/2) dy from y=117 to 207 seems to work
POSTED BY: l van Veen

Yes, however putting ".9 = " does not
POSTED BY: Bryce Costawong
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