The formula for pdf of normal distribution is
$$ f(x)= \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$
and I have a pdf that is: $$ke^{-x^2-7x}$$
But I face a paradox in calculating $\mu$
From the formula:
$ {2\sigma^2} = 1 $ ______So_____ ${-x^2-7x} = {-(x-\mu)^2}$ and this gives $\mu = 0$ and then there won't be any -7x
can somebody declare it for me?
On
Completing the square yields $$ -(x^2+7x) = -\left(x+\frac{7}{2}\right)^2+\frac{49}{4} $$ so $\mu$ can only be $\mu = -7/2$. With that in hand, $$ k e^{-(x^2+7x)} = k e^{-\left(x+\frac{7}{2}\right)^2+\frac{49}{4}} = (ke^{\frac{49}{4}}) e^{-\left(x+\frac{7}{2}\right)^2} = k' e^{-\left(x+\frac{7}{2}\right)^2} $$ where $k' = ke^{\frac{49}{4}}$. Can you finish from there?
$ke^{-x^{2}-7x}=ke^{49/4} e^{-(x+\frac 7 2)^{2}}$. So $\mu =-\frac 7 2,\sigma=1/{\sqrt 2}$ and, for the given function to be a pdf, $k$ must be such that $ke^{49/4}=\frac 1 {\sqrt {\pi }}$.