how to find area under normal distribution curve

Question

Find out the area in percentage under standard normal distribution curve of random variable $Z$ within limits from $-3$ to $3$.

my try: probability density function of standard normal distribution is $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
now the area under standard normal distribution curve ($-3\le x\le3$), $$=\frac{1}{\sqrt{2\pi}}\int_{-3}^3e^{-\frac{x^2}{2}}\ dx$$ $$=2\frac{1}{\sqrt{2\pi}}\int_{0}^3e^{-\frac{x^2}{2}}\ dx$$ $$=-\frac{2}{\sqrt{2\pi}}\int_{0}^3e^{-\frac{x^2}{2}}\ dx$$ $$=-\frac{2}{\sqrt{2\pi}}\left(\int_{0}^{\infty}e^{-\frac{x^2}{2}}\ dx-\int_{3}^{\infty}e^{-\frac{x^2}{2}}\ dx\right)$$ $$=-\frac{2}{\sqrt{2\pi}}\left(\frac 12-\int_{3}^{\infty}e^{-\frac{x^2}{2}}\ dx\right)$$

i got stuck here, i don't have any clue to solve above integral. please help me to solve it or give some other method to find the area under the curve.

Answer 1

Hint. There is no elementary antiderivative to evaluate the considered area, but a special function has been designed and studied,

$$ \text{erf}(t):=\frac{2}{\sqrt{\pi}}\int_{0}^te^{-x^2}\ dx,\qquad t \in \mathbb{R}, $$

giving here $$ \frac{1}{\sqrt{2\pi}}\int_{-3}^3e^{-\frac{x^2}{2}}\ dx=\text{erf}\left(\frac{3}{\sqrt{2}}\right) $$ and $\text{erf}(\cdot)$ is implemented in all CAS.

Answer 2

Because the standard normal CDF $\Phi$ cannot be written in closed form, numerical answers depend on numerical integration (printed tables or statistical software).

In R statistical sofware, $\Phi$ is pnorm so pnorm(1) gives $\Phi(1) = P(Z \le 1) = 0.8413,$ where $Z \sim Norm(0,1).$

pnorm(1)
## 0.8413447

The specific answer to your question is $\Phi(3) - \Phi(-3),$ which gives 99.7%.

pnorm(3) - pnorm(-3)
## 0.9973002

Here is a graph, from which you might be able to read some probabilities with 1 or 2-place accuracy.