114. Another Normal Density Function Prove that for some constant $k$, $f(x) = ka^{-x^2}$, $a \in (0, \infty)$, is a normal probability density function.
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A normal probability density function has the form: $$ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ where $\mu$ is the mean and $\sigma^2$ is the variance. Show more…
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