Probability from Chi-Squared distribution If I have a random variable called $A$ and $A \sim \mathcal{N}(0,1)$, how can I get the value of $X$ if $P(X<5A^2<6.52)=0.5310$. It seems that the answer is $2.4109$, but I don't know how to reach that value.
Thanks!

Although $A^2 \sim \chi^2(1)$, but $5A^2$ is not, since it can't be expressed as sum of i.i.d. standard normal r.v.s.
$P(5A^2<6.52)=P(A^2<\dfrac{6.52}{5})$ which could be found with pchisq(1, 6.52/5) in R, e.g.
Also, by change of variable (Jacobian), we can find the the pdf of the r.v. $Y=5A^2$.
$f_Y(y)=f_A(a)\left|\dfrac{da}{ay}\right|=\dfrac{e^{-a^2/2}}{\sqrt{2\pi}}.\dfrac{1}{10a}=\dfrac{e^{-y/10}}{\sqrt{2\pi}.2\sqrt{5y}}$
Now, $F_Y(6.52) = P(Y\leq 6.52) = \int_{0}^{6.52}f_Y(y)dy=\int_{0}^{6.52}\dfrac{e^{-y/10}}{\sqrt{2\pi}.2\sqrt{5y}}dy$, since $Y \geq 0$
With numerical integration, e.g., with R, we can compute the probability.
integrate(function(y) exp(-y/10)/sqrt(y), 0, 6.52)$value / (2*sqrt(5)*sqrt(2*pi))

