A random variable with normal distribution with unknown mean u and variance 2.25. with a suitable value of u. find the greatest value of P(10.9<X<12.1)

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- Jul 4th 2006, 12:25 AMguessProbabilty ans statistics
A random variable with normal distribution with unknown mean u and variance 2.25. with a suitable value of u. find the greatest value of P(10.9<X<12.1)

- Jul 4th 2006, 01:12 AMCaptainBlackQuote:

Originally Posted by**guess**

(10.9+12.1)/2=11.5. Now this seems obviouse to me, but let me know if

you need something more.

RonL - Jul 4th 2006, 01:31 AMguess
ya... u got the ans.. but i does not really know the reason behind it... can you try to explain?

- Jul 4th 2006, 02:19 AMCaptainBlackQuote:

Originally Posted by**guess**

is decreasing as you move away from the mean, which means that if the

mean is not in the centre of the interval (a,b) that P(a<x<b) can be

increased by moving the mean in the direction of the mid point of the interval.

Alternatively you could write:

$\displaystyle

P(a<x<b)=\int_a^b p(x-u) dx

$

where

$\displaystyle p(x-u)=\frac{1}{\sqrt{2\pi}\ \sigma}e^{-(x-u)^2/2\sigma^2}$

Then:

$\displaystyle

P(a<x<b)=\int_a^u p(x-u) dx+\int_u^b p(x-u) dx

$,

changing the variable of integration in each of the integrals:

$\displaystyle

P(a<x<b)=\int_{a-u}^0 p(x) dx+\int_0^{b-u} p(x) dx

$

The rewriting:

$\displaystyle

P(a<x<b)=-\int_0^{a-u} p(x) dx+\int_0^{b-u} p(x) dx

$

Which can now be differentiated (using the fundamental theorem of calculus)

to give:

$\displaystyle

\frac{d}{du} P(a<x<b)=-p(a-u) + p(b-u)

$

So the stationary points of $\displaystyle P(a<x<b)$ as $\displaystyle u$ varies are the solutions of:

$\displaystyle

p(a-u)=p(b-u)

$.

which are solutions of: $\displaystyle (a-u)^2=(b-u)^2$, which if $\displaystyle a \ne b$ is $\displaystyle u=\frac{a+b}{2}$, and this clearly corresponds to a maximum.

Now I could fill in a few more of the details of the argument, but I won't do

that here. I will just observe that the blindingly obvious can have a relativly

complicated proof :eek:

RonL