The question goes like this: given the distribution $f(x) = \frac{1}{\sqrt{2\pi}3}\exp^{-\frac{1}{2}(\frac{x}{3})^2}$ what is the probability that $x>1$ ?
What I did: Since this is a normal distribution with $\sigma=3,m=0$: $prob(1<x<\infty)=\int_1^\infty f(x)dx$ and after a lot of handwork I get the probability is $\approx0.84$. I do not know if this answer is correct, however, the reason I come here to ask is if there exists a more direct way to find out this probability or is it mandatory that I solve the integral?
Thanks.
The usual method of calculating these probabilities is by using the fact that if $X\sim N(\mu, \sigma^2)$, thus $$ \frac{X- \mu}{\sigma}\sim N(0,1), $$
and for the cumulative distribution of $N(0,1)$ there are the old fashioned-tables. Thus, $$ \mathbb{P}(X >1) = 1 - \mathbb{P}(X \le 1) = 1 - \mathbb{P}((X-0)/3 \le 1/3) = 1 - \Phi \left(\frac{1}{3} \right). $$ then you can use some software or the aforementioned tables of the standard normal distribution to find that $\Phi(1/3) = 0.63$.