$X$ is $N(10,1)$. Find $f(x|(x-10)^2 < 4)$

This is a homework question. I can only figure out that X is normally distributed with mean 10 and variance 1.

Can you please explain what is meant to be found here.


  • 3
    $\begingroup$ I would start by first analyzing the condition. Under what conditions is $(x-10)^2<4$ $\endgroup$ Commented May 1, 2013 at 4:33
  • $\begingroup$ Thank you very much. So it means I have to find the probability that x lies between 8 and 12 in the given normal distribution? $\endgroup$
    – PasanW
    Commented May 1, 2013 at 4:37
  • 1
    $\begingroup$ that is correct. This is a case of substitution. The expression on the right hand side is basically telling you what you need to do to x in this "given" situation. SInce the variables are the same, it is simplys substitution. Again, intuitively, if I told you that f(x,y)=2xy and I told you to find f(x,y) given x=4, then it is simply our task to find f(4,y). Here we are happy that we only have one variable and so all we need to do is find f(x|8<x<12). Be careful though here, because you didn't ask for probability. Your problem says find f(x|(x-10)^2<4), not P(X..) $\endgroup$ Commented May 1, 2013 at 4:54

1 Answer 1


It is best to rewrite $\{x\;|\;(x-10)^2<4\}$ as $\{x\;|\;-2<x-10<2\}$. In terms of the normalized quantity $z=(x-\mu)/\sigma$ this is $\{z\;|\;-2<z<2\}$.

I don't know what $f$ is doing in this problem. If what you really meant was $P\{x\;|\;(x-10)^2<4\}$, then a table of probabilities for $z$-scores is all you need.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .