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)
The basic idea is that the normal density is symmetric about the mean, andOriginally 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
RonL