Normal distribution and extremal value

faruk
Messages
5
Reaction score
0
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
 
Last edited by a moderator:
Physics news on Phys.org
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)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top