# Percentage point of Normal Distribution.

Let $$X \sim N(65,64)$$ Find the lower $2$% point for $X$; that is, find the value of $x$ such that $Pr(X<x) = 0.02$

i know i need to do something like $\frac{X- 65 }8 = ...$ but not too sure what??

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You are on the right track with doing something like $\dfrac{X-65}{8}$.

Since $X\sim N(65, 64)$, then $\dfrac{X-65}{8} \sim N(0,1)$. It is easy to see that $$Pr(X < x)=Pr\left(\dfrac{X-65}{8}<\dfrac{x-65}{8}\right)=Pr\left(Z<\dfrac{x-65}{8}\right)=0.02$$

where $Z$ is standard normal distributed.

The quantile of a standard normal distribution can be found in many introductory books or some statistical software such as R. You will find out that $Pr(Z<-2.054) = 0.02$. From that, you can easily solve for $x$.

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so $\frac {X-65}8 = -2.054$ hence $x= 48.568$ ??? –  Sam Feb 6 '13 at 0:33
is that correct? –  Sam Feb 6 '13 at 0:40
@jill I think that's correct. –  Patrick Li Feb 6 '13 at 1:18