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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?

second part of equation Image 1.

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  • $\begingroup$ Please use math typesetting. It's hard to understand what you're asking. More information: math.meta.stackexchange.com/questions/5020/… $\endgroup$
    – Sycorax
    Nov 2, 2018 at 1:53
  • $\begingroup$ Probability density function. Title is correct, that in contents is wrong. should be /(2*sigma^2) or /2/sigma^2. $e^{-0.5}$ , not $e-0.5$. $\endgroup$
    – user158565
    Nov 2, 2018 at 1:57
  • $\begingroup$ Ok, I have updated my question and tried the best to format it correctly. Thanks for having a look. $\endgroup$
    – tcratius
    Nov 2, 2018 at 3:33

1 Answer 1

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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 $\frac{1}{\sqrt{2\pi}\sigma}$.

Comment about your writing:

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.

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