# the variance of a gaussian PDF?

A problem is this: The probability density function of the univariate Gaussian with mean $$μ$$ and variance $$σ2, N(μ,σ2)$$:

$$f_x(x) = \frac{1}{\sqrt(2*pi*σ2)} * e^-(x-μ)^2/(2*σ2)$$

The pdf of a Gaussian random variable X is given by: $$f_x(x) = \frac{n}{(3*\sqrt(2*pi))} * exp(-(n^2(x-2)^2)/18)$$.

What is the mean and variance of X?

I got the mean right: 2, but the variance wrong:9.

Why is 9 not the right answer here? I thought it would be $$3^2$$?

You omit $$n$$ in the expression. Your PDF can be written as $$f_X(x)=\frac{1}{\sqrt{2\pi}(3/n)}\exp(-(x-2)^2/(2(3/n)^2))$$ where $$\sigma_2=3/n$$.