Yes, for standard normal distribution, the formula is $f(x)=\frac{e^{-x^{2}/2}}{\sqrt{2\pi}}$. If you want to find the likelihood of something falling between -.5 and .8 standard deviations away, you would want to compute a complicated integral $\int_{-.5}^{.8}{\frac{e^{-x^{2}/2}}{\sqrt{2\pi}}dx}$. On the calculator, you can use fnInt( like this:

`fnInt(e^(-.5X^2)/√(2π),X,-.5,.8`

But that is the tough way to do it :P You might want to use

normalcdf( and

normalpdf( since those are more accurate, faster, and designed for this. The above would simply be:

invNorm( might also be of interest.