# Normally distributed random variables

• Sep 11th 2011, 03:05 AM
gbooker
Normally distributed random variables
Not sure about this question, i think the use of letters instead of numbers confuses me. If anyone could provide me with some help, it would be much appreciated

X is a normally distributed random variable. If a<µ<b and Pr(X>a)=p and Pr(X<b)=q, find:
a) Pr(a<X<b)
b) Pr(X<a/X<b)
• Sep 11th 2011, 03:22 AM
Plato
Re: Normally distributed random variables
Quote:

Originally Posted by gbooker
X is a normally distributed random variable. If a<µ<b and Pr(X>a)=p and Pr(X<b)=q, find:
a) Pr(a<X<b)
b) Pr(X<a/X<b)

For part a), $\displaystyle \mathcal{P}(X\le a)=1-\mathcal{P}(X>a)$.
Do you know how to rewrite $\displaystyle \mathcal{P}(a<X<b)~?$

For part b, is that notation suppose to be $\displaystyle \mathcal{P}(X<a|X<B)~?$ OR you you mean $\displaystyle \mathcal{P}(X\color{red}>a|X<B)~?$
• Sep 11th 2011, 04:13 AM
gbooker
Re: Normally distributed random variables
[QUOTE=Plato;679274]For part a), $\displaystyle \mathcal{P}(X\le a)=1-\mathcal{P}(X>a)$.
Do you know how to rewrite $\displaystyle \mathcal{P}(a<X<b)~?$

no I can't think of it, I might have learned it, but I just can't remember it at the moment

For part b, is that notation suppose to be $\displaystyle \mathcal{P}(X<a|X<B)~?$ OR you you mean $\displaystyle \mathcal{P}(X\color{red}>a|X<B)~?$[/QUOTE

sorry about the confusion, i meant the first one
• Sep 11th 2011, 04:19 AM
Plato
Re: Normally distributed random variables
Quote:

Originally Posted by gbooker
Quote:

Originally Posted by Plato
For part a), $\displaystyle \mathcal{P}(X\le a)=1-\mathcal{P}(X>a)$.
Do you know how to rewrite $\displaystyle \mathcal{P}(a<X<b)~?$

no I can't think of it, I might have learned it, but I just can't remember it at the moment

For part b, is that notation suppose to be $\displaystyle \mathcal{P}(X<a|X<B)~?$ OR you you mean $\displaystyle \mathcal{P}(X\color{red}>a|X<B)~?$

sorry about the confusion, i meant the first one

In that case note that $\displaystyle (-\infty,a]\cap(-\infty,b]=(-\infty,a]$