# Probability Density Function

x = 2, μ (mu) = 5 and σ (sigma) = 3

I am just wanting to confirm may workings for the second part of the probability distribution function as underlined red in the picture below (see Image_1.).

$$e = (2.71828)$$ $$part.two = -\frac{(x - \mu)^2}{2\sigma^2}$$

$$=-\frac{(2 - 5)^2}{2*3^2}$$ $$=-\frac{3^2}{18}$$ $$=e^{-0.5}$$

$$part.two = 0.60653$$

Are my figures correct and have I got the correct answer, otherwise, what have I done wrong?

I do not understand if the fraction $$\frac{(x - \mu)^2}{2\sigma^2}$$ is meant to be $$e^\frac{-(x - \mu)^2}{2\sigma^2}$$ or if it is to be the difference of the constant $$e - \frac{-(x - \mu)^2}{2\sigma^2}$$ Hopefully, I have explained it better and is there anywhere I can test my result next time to ensure I have calculated it right?

Image 1.

– Sycorax
Nov 2, 2018 at 1:53
• Probability density function. Title is correct, that in contents is wrong. should be /(2*sigma^2) or /2/sigma^2. $e^{-0.5}$ , not $e-0.5$. Nov 2, 2018 at 1:57
• Ok, I have updated my question and tried the best to format it correctly. Thanks for having a look. Nov 2, 2018 at 3:33

The formula means $$\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2} \right)$$
After you computed $$-\frac{(x-\mu)^2}{2\sigma^2}$$, call it $$w$$, we then compute $$e^w$$ and then divide by $$\sqrt{2\pi}\sigma$$.
You have yet to finish your computation, you still have to multiply $$e^{-0.5}$$ with $$\frac{1}{\sqrt{2\pi}\sigma}$$.
when we write equality, it means the object on the left and on the right are equal. Avoid stuff like $$-\frac{3^2}{18}=e^{-0.5}$$.
As for how to check your working, one way is open the $$R$$ program, type dnorm(2,5,3). You can do it online as well.