Probability square of a normal is in a range; mode of gamma 
  
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*If $X \sim N(1,4)$ find $\mathbb{P}(1<X^2<9)$
  
*If $x=2$ is the unique mode of the $X \sim \Gamma(2,\beta)$ distribution, find the parameter $\beta$.

Well I tried transformation to solve this problem.
Set $Y=X^2$ and I actually found the pdf of $Y$ but when I try to calculate $\mathbb{P}(1<X^2<9)$, I just can't calculate because the pdf of $Y$ is so complicated to integral 1 to 9.
and in number 2. I think I didn't understand the question properly. Does the mode mean that 'the number that came out most'?  
I understand the question as like something that if $X\sim \Gamma(r/2,2)$ it follows $X \sim X^2(r)$.
 A: *

*Don't compute the distribution of $X^2$. Instead (as I already suggested) simply identify all the values of $X$ that satisfy the condition (i.e. work directly with the distribution of $X$). You could do it by computing the distribution of the square but it's effort you don't really need to go to.

*The mode according to the wikipedia article on it: 

The mode of a continuous probability distribution is the value $x$ at which its probability density function has its maximum value.

The density function for a gamma with shape parameter 2 is easy to write down. When finding modes you can safely ignore normalizing constants and work with an unnormalized density, making it easier still. 
As I already suggested, figure out where the maximum of that density function is. It's pretty simple -- there's a unique global maximum (which is also the only local maximum and even the only turning point). The density has properties that make this a particularly simple task.
General guidance: in both cases, draw a picture.
