First off all, my general understanding of gaussians is not very good, and I'm having issues getting my head around this because I cannot find an explanation of them I can understand.
I'm working with probability density functions and gaussian distribution and essentially I'm trying to sketch some gaussians given the variance and the mean.
I understand the variance controls the width of the guassian, and the mean corresponds to the peak.
Someone please correct me if I'm wrong, but I am under the impression that the end points of the gaussian are given by $ \mu$ $ ^{+}_{-} \frac{1}{\sqrt{2\piσ^2}}$ where $\mu$ is the mean.
And the x-coordinate of the peak is equal to the mean.
What I don't understand is how to determine the y-coordinate of the peak.
An example given is mean = 0, variance = 1. The end points are approximately 4 and -4 (or rather 0.3989...), and the peak has x-coordinate (0,0.4)
Where has 0.4 come from?
The Gaussian PDF is maximized at the mean (this is called a mode of the distribution). The Gaussian has full support (i.e. the PDF is non-zero on the whole real line), so the "end points" idea is wrong.
You can find the value of the Gaussian PDF at the peak by plugging into the Gaussian density: $f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-(x-\mu)^2/(2 \sigma^2)}$ to see that the peak value of the Gaussian pdf (which occurs at $x=\mu$) is $\frac{1}{\sqrt{2 \pi \sigma^2}}$.