Difference between the Error function and Normal distribution function?

I have just started reading about the Error function but Distribution theory is not my strong point. So I apologize in advance for asking really simple questions about it.

So the Error function is defined to be $$\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{y=0}^{y=x}e^{-y^2}\mathrm{d}y$$

I have learnt about the Normal Distribution function already which is defined to be $$\displaystyle \Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{z=-\infty}^{z=\infty}e^{-\frac{1}{2}z^2}\mathrm{d}z$$ where $z=\cfrac{x-\mu}{\sigma}$ and that it's graph is

I understand completely that $\Phi(z=-1.14)\approx 0.1271$ is the area under the graph in the left tail shaded $\color{blue}{\mathrm{blue}}$.

But for the Error function graph

if $\mathrm{erf}\left(x=\frac{1}{\sqrt{2}}\right)=0.682689492137086$, what does this value even mean? Is this the value of the area under the curve for $-\infty \le x \le \frac{1}{\sqrt{2}}$?

And finally why do we need the Error function in the first place if we have a Normal distribution function?

• When you integrate, the dummy variable goes away. Your definition of $\Phi$ is wrong. It should be $$\Phi(z) = \frac 1{\sqrt{2\pi}}\int_{-\infty}^z e^{-\frac 12x^2} dx$$ – Tunococ Nov 3 '15 at 1:55
• @Tunococ Thanks that's what I meant it to be, I'll change it now. Any ideas on the rest of the questions? – BLAZE Nov 3 '15 at 2:13

$$\Phi(x)=\frac{1}{2}+\frac{1}{2}\mathrm{erf}(x/\sqrt{2})$$
• Thanks for your answer, it was helpful to see how they were related. Any idea what $\mathrm{erf}\left(\frac{1}{\sqrt{2}}\right)=0.682689492137086$ means? Is it an area? – BLAZE Nov 3 '15 at 2:57
• that is the value of the error function evaluated at $x=1/\sqrt{2}$. It can however be interpreted in a different way if you consider the formulate I gave. Namely it is $2\Phi(1)-1$, which is the area of normal density from $-\infty$ to $1$ multiplied by $2$ and the subtracted by $1$ – Seyhmus Güngören Nov 3 '15 at 3:07