# bivariate normal distribution

• May 24th 2009, 02:12 AM
flower3
bivariate normal distribution
if $\displaystyle X,Y \sim BVN$ AND : $\displaystyle \mu \scriptscriptstyle {x} =-3$ $\displaystyle \sigma \scriptscriptstyle {x} =5$ $\displaystyle \mu \scriptscriptstyle {y} =10$ $\displaystyle \sigma \scriptscriptstyle {y} = 3$ $\displaystyle \rho =\frac{3}{5}$ how i can find : $\displaystyle p(-5\prec x \prec 5)$ ???????
• May 24th 2009, 05:10 AM
Moo
Hello,
Quote:

Originally Posted by flower3
if $\displaystyle X,Y \sim BVN$ AND : $\displaystyle \mu \scriptscriptstyle {x} =-3$ $\displaystyle \sigma \scriptscriptstyle {x} =5$ $\displaystyle \mu \scriptscriptstyle {y} =10$ $\displaystyle \sigma \scriptscriptstyle {y} = 3$ $\displaystyle \rho =\frac{3}{5}$ how i can find : $\displaystyle p(-5\prec x \prec 5)$ ???????

If (X,Y) follows a bivariate normal distribution, then in particular, X follows a normal distribution.
Its mean is -3 and its variance is 5^2=25.
$\displaystyle X\sim\mathcal{N}(-3,25)$
--> $\displaystyle \frac{X+3}{5}\sim\mathcal{N}(0,1)$
$\displaystyle \mathbb{P}(-5<X<5)=\mathbb{P}\left(\frac{-5+3}{5}<\frac{X+3}{5}<\frac{5+3}{5}\right)$