Difficult Normal Distribution question

• Sep 5th 2011, 03:31 AM
Istafa
Difficult Normal Distribution question
X is normally distributed with mean $\displaystyle \mu$ where the mean is greater than zero
If $\displaystyle Pr(a>X)=0.025$
what is the closest decimal approximation for the value
$\displaystyle Pr(-a<X<\mu)$???
a:0.025
b:0.05
c:0.475
d:0.4875
e:-0.025 (obviously wrong choice, just provided for question completeness)
• Sep 5th 2011, 04:37 AM
Siron
Re: Difficult Normal Distribution question
I would rewrite the probability:
$\displaystyle P(-a<X<\mu)=P(X<\mu)-P(X<-a)$
• Sep 5th 2011, 04:45 AM
Istafa
Re: Difficult Normal Distribution question
This looks like a good approach, except how do I find Pr(x<-a)? Pr(X< mu) is obviously 0.5, but this alone doesn't allow me to solve the question...?
• Sep 5th 2011, 04:53 AM
Siron
Re: Difficult Normal Distribution question
Notice $\displaystyle P(X<-a)=P(X>a)=1-P(X<a)$
• Sep 5th 2011, 05:01 AM
Istafa
Re: Difficult Normal Distribution question
This approach gives the answer as c above, but how does P(x<-a)=P(x>a) hold true even when mean is not zero?
• Sep 5th 2011, 05:06 AM
Siron
Re: Difficult Normal Distribution question
Because the graph of normal distribution is symmetric.
• Sep 6th 2011, 03:57 AM
Istafa
Re: Difficult Normal Distribution question
Symmetric about the mean, hence P(X<mu-b) would be equal to P(X>mu+b) for any b but if mu is not zero then P(x<-b) would not equal P(x>b) for an b. Or is there a flaw in this logic? Hmmm...