Find area under the curve of a standard normal distribution Given a standard normal distribution, how can I find the
area under the curve that lies:
1. to the left of z = −1.39;
2. to the right of z = 1.96;
3. between z = −2.16 and z = −0.65;
 A: The areas for one, two, and three standard deviations about the mean are well known, as .68, .95, .997 roughly. But in general, there's no way to find the area under the standard normal distribution without using a computer, calculator, or a chart. 
edit1: most charts will give the area to the left of a given $z$-score. To find other values, use symmetry and complements. For example,
$\operatorname{Pr}(Z > -.5) = 1 - \operatorname{Pr}(Z < -.5)$
and 
$\operatorname{Pr}(Z < - 1) = \operatorname{Pr}(Z > 1)$
edit2: I'm not careful above about whether to use "$\ge$" or "$>$" because the answers will always be the same for a continuous distribution like a normal distribution. Be careful you don't switch them around for discrete distributions because then it will affect the answer.
A: You can use Wolfram alpha. For example, http://www.wolframalpha.com/input/?i=1%2Fsqrt%282pi%29+integral%28e%5E%28-x%5E2%2F2%29%2Cx%2C-infinity%2C-1.39%29 answers your first question. Let me point out that your the meaning of the area under the curve to the left of $-1.39$ is the probability that a standard normal is at most $-1.39$. Make sure that you understand what you're calculating.
As GFauxPas mentions, there is no "nice" expression for the integral of the normal density function (in technical terms, it is not elementary). However, some approximations can be useful:


*

*For small $\epsilon>0$, the probability that a standard normal is between $x$ and $x+\epsilon$ is roughly $\frac{e^{-x^2/2}}{\sqrt{2\pi}} \epsilon$.

*For large $x$, the probability that a standard normal is larger than $x$ is roughly $\frac{e^{-x^2/2}}{\sqrt{2\pi}x}$. (See Mike Spivey's blog.) By symmetry, the same is true for the probability that a standard normal is smaller than $-x$.
