Find $c$, $\mu$ and $\sigma^2$ of the normal distribution whose probability function is given by $f\left(x\right) = c e^{- x^2 + 3x}$, $-\infty < X < + \infty$
The standard probability function of a continuous random variable $X$ which follows normal distribution with mean $\mu$ and standard deviation $\sigma$ is
$f \left(x\right) = \dfrac{1}{\sigma \sqrt{2 \pi}} \exp \left[- \dfrac{1}{2} \left(\dfrac{x - \mu}{\sigma}\right)^2\right]$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\text{Given: } f \left(x\right) & = c \exp \left\{- x^2 + 3x\right\} \\\\
& = c \exp \left\{- \left[\left(x^2 - 3x + \dfrac{9}{4}\right) - \dfrac{9}{4}\right] \right\} \\\\
& = c \exp \left\{- \left(x - \dfrac{3}{2}\right)^2 + \dfrac{9}{4} \right\} \\\\
& = c \exp \left\{\dfrac{9}{4}\right\} \times \exp \left\{- \left(x - \dfrac{3}{2}\right)^2 \right\} \\\\
& = c \exp \left\{\dfrac{9}{4} \right\} \times \exp \left\{- \dfrac{1}{2} \left(\dfrac{x - \dfrac{3}{2}}{\dfrac{1}{\sqrt{2}}}\right)^2 \right\} \;\;\; \cdots \; (2)
\end{aligned}$
Comparing equations $(1)$ and $(2)$ we have,
$\mu = \dfrac{3}{2}$
$\sigma = \dfrac{1}{\sqrt{2}}$ $\implies$ $\sigma^2 = \dfrac{1}{2}$
$c \exp \left\{\dfrac{9}{4} \right\} = \dfrac{1}{\sigma \sqrt{2 \pi}}$
$\implies$ $c = \exp \left\{- \dfrac{9}{4} \right\} \times \dfrac{1}{\dfrac{1}{\sqrt{2}} \times \sqrt{2 \pi}}$
i.e. $\;$ $c = \dfrac{\exp \left(-9/4\right)}{\sqrt{\pi}}$