I am trying to calculate the following probability
P(X>a) a is the lower bound of integration
where X is following normal distribution with parammter $\mu$ and $\sigma$. I tried the integartion of probability density function of normal distribution but I did something wrong I guess
$\int_{a}^{}$ f(x) dx
Thank you in advance
If I properly understand, what you want is to compute, for given $(a,\mu,\sigma)$ $$P(a)=\frac{1}{\sqrt{2 \pi \sigma ^2}}\int_a^\infty e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}\,dx$$ Using $\sigma >0$, you have $$P(a)=\frac{1}{2} \text{erfc}\left(\frac{a-\mu }{\sigma\sqrt{2} }\right)=\frac 12\left( 1-\text{erf}\left(\frac{a-\mu }{\sigma\sqrt{2} }\right)\right)$$ This involves non elementary functions but you can get quite good approximations using for example $$\mathrm{erf}\!\left(t\right)\sim \sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha \,t^2}{1+\beta\, t^2}\,t^2 \Big)}$$ where $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi }\qquad \text{and} \qquad \beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$
Have a look here.
Edit
In a quite recent paper is given the very nice approximation $$\frac{1}{2} \left(1+\text{erf}\left(\frac{t}{\sqrt{2}}\right)\right)\sim 2^{-22^{1-41^{t/10}}}$$