Normal distribution and extremal value

faruk
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Normal distribution.

What is the value of sigma (dispersion) for maximal probability P(1<x<2) ?

Excel calculation: sigma is about 1.471. But what would be an analytical solution?

http://img500.imageshack.us/img500/558/normdistrib19ql.gif
 
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HINT: Calculate the derivative with respect to \sigma of the integral of the distribution function over the given interval.
 
Tide said:
HINT: Calculate the derivative with respect to sigma of the integral of the distribution function over the given interval.

That's exact the problem.
exp(-x²) belongs to the unintegratable functions. It's the cause we use the table of the normal distribution probability.

I hope to be wrong. Please help me.
 
faruk said:
That's exact the problem.
exp(-x²) belongs to the unintegratable functions. It's the cause we use the table of the normal distribution probability.

I hope to be wrong. Please help me.

Your precise wording is wrong. exp(-x2) is integrable- it's integral just doesn't happen to be an elementary function. (Actually, its integral is the error function because that's how the error function is defined!)

But you don't need to know the function itself you only need to know its derivative. What is this derivative:
\frac{d}{dx}\left(\int_a^x e^{-t^2}dt\right)

Hint: What is this derivative:
\frac{d}{dx}\left(\int_a^x f(t)dt\right)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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