# Thread: Difficult Normal Distribution question

1. ## 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)

2. ## Re: Difficult Normal Distribution question

I would rewrite the probability:
$\displaystyle P(-a<X<\mu)=P(X<\mu)-P(X<-a)$

3. ## 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...?

4. ## Re: Difficult Normal Distribution question

Notice $\displaystyle P(X<-a)=P(X>a)=1-P(X<a)$

5. ## 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?

6. ## Re: Difficult Normal Distribution question

Because the graph of normal distribution is symmetric.

7. ## 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...