I have a bivariate normal distribution, where $ X = (X_1, X_2)^T$. The means are given by $\mu = (0,0)^T$ and the covariance matrix is $ \sum = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix} $.
I've been asked to find $a$ such that the $P(X \in s) = 0.95$ where $ s = \{(x_1,x_2)^T \in \Bbb R : -a \leq x_1 \leq a , -a \leq x_2 \leq a \} $.
Here is my approach: $P(X \in s) = \int ^a _{-a} \int ^a _{-a} f_{X_1,X_2}(x_1,x_2) dx_1 dx_2 = \int ^a _{-a} \int ^a _{-a} f_{X_1}(x_1) f_{X_2}(x_2) dx_1dx_2 = (\int ^a _{-a} \phi (y) dy )^2$ where $\phi(y)$ is the pdf of the standard normal distribution. This is because $X_1, X_2$ are uncorrelated standard normal variables, hence independent.
This equals $(2\Phi(a) -1)^2$. Setting this equal to $0.95$ and solving gives $ a = \Phi ^{-1} (0.98733...) = 2.236$ to (4sf).
Is this correct? The question is only 3 marks which makes me question whether I have overcomplicated this.
Thanks in advance!
(Also, the same question is given but instead where $ s = \{(x_1,x_2)^T \in \Bbb R : x_1^2 + x_2^2 \leq r^2\} $. Do I approach this in the same way? Again this seems overcomplicated for a 3 mark question).