The equation for a normal distribution with a mean of 20 and a variance of 30 is
$f(x)=\frac{1}{\sqrt{30 \pi}} e^{\frac{-(x-20)^{2}}{60}}$

Question

The equation for a normal distribution with a mean of 20 and a variance of 30 is
 $f(x)=\frac{1}{\sqrt{30 \pi}} e^{\frac{-(x-20)^{2}}{60}}$

Quizplus · Accepted Answer

The correct equation for a normal distribution with a mean ($\mu$) of 20 and a variance ($\sigma^2$) of 30 is $f(x) = \frac{1}{\sqrt{2\pi \cdot 30}} e^{-\frac{(x-20)^2}{2 \cdot 30}}$, because the variance should be in the denominator inside the exponential as $2\sigma^2$, and the square root in the denominator outside the exponential should be $\sqrt{2\pi\sigma^2}$.

The Equation for a Normal Distribution with a Mean of 20