$\Pr(X>1.96)$ where $X \sim \mathcal{N}(0,1)$

Question

I am trying to calculate $\Pr(X>1.96)$ where $X \sim \mathcal{N}(0,1)$. Whether the answer can only be found in normal distribution table?

I took the $1.9$ column and move to the right to reach $0.06$ to get $0.9750$. I know it is the probability of $X<1.96$ (am I right?). So to calculate the $\Pr(X>1.96)$ I just do this : $$\Pr(X>1.96) = 1 - \Pr(X<1.96) \implies \Pr(X>1.96) = 1-0.9750 = 0.025$$

I was wondering if this solution is correct?

Answer 1

yes, you are correct.
Answer can not only be found in normal table, but also in Excel , Matlab, R etc

In excel =1-NORMSDIST(1.96) which gives $0.024997895$
in R 1 -pnorm(1.96,0,1) which gives $0.0249979$

similarly in other software.

Hope this was helpful .

Thanks