Evaluating normal distribution integral

Question

How one can show that $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{2.92}e^{-x^2/2}dx=0.99825$$

Answer 1

Note that the Gaussian integral

$$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-x^2 / 2} dx = 1$$

can be computed directly, so it's a little easier to try approximating

$$\frac{1}{\sqrt{2\pi}} \int_{2.92}^{\infty} e^{-x^2/2} dx$$

Due to the very fast decay of the exponential function, it's a pretty good estimate to just compute

$$\frac{1}{\sqrt{2\pi}} \int_{2.92}^{10} e^{-x^2/2} dx$$

which can be approximated as $0.00175$ (using your favorite numerical method, e.g. Simpsons' rule). To estimate the error involved in this approximation, note that

$$\int_{10}^{\infty} e^{-x^2/2} dx < \int_{10}^{\infty} e^{-5x} dx \sim 10^{-22}$$

where we have used that $x^2 > 10x$ for $x > 10$.