Guassian Probability density function

AI Thread Summary
The discussion focuses on solving a homework problem involving the Gaussian probability density function (PDF) given by p_x(x)=1/(C√(2π)) e^(-(x-4)²/18). To find the constant C, it is determined that C equals 3, as derived from the relationship between the exponent and standard deviation. The probability of x being greater than or equal to 2 is calculated by substituting x=2 into the PDF, yielding a value of approximately 0.1065. Participants also discuss using the Q(x) function for probability calculations and share tips on formatting LaTeX code for better presentation. The conversation highlights the importance of understanding the steps involved in solving Gaussian distribution problems.
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Homework Statement



The PDF (probability density function) of a Gaussian variable x is given by.

$$p_x(x)=\frac{1}{C \sqrt{2 \pi}} e^{\frac{-(x-4)^2}{18}}$$

a) Find C
b)find the probability of x≥2 --> ##P(x≥2)##

Homework Equations



$$ \frac{dF_X(x)}{dx} x=P(x<X≤x+Δx)$$

The Attempt at a Solution



So i get stuck on how to solve the above for C. I have an example of a similar problem that my professor did in class but it skips a lot of steps that I need to see to fully understand. It seems like he started with taking the integral of the signal by using an integral table?

In my textbook I do see that the above is a standard of a gaussian or normal probability density. It looks something like this.

$$p_X(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^2}{2}$$
$$F_X(x)=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x} e^{\frac{-x^2}{2}}dx$$

Any hints on where to start?

Any help is much appreciated! Thank you!
 
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The parameter C is the standard deviation.
The denominator of the power of the exponent is equal to 2(C^2).
Hence C = 3, as 2 times 9 is 18.

p_x(x)=\frac{1}{3 \sqrt{2\pi}}e^{\frac{-(x-4)^2}{18}}

My approach to this was to look at the formula given on this wikipedia page:
https://en.wikipedia.org/wiki/Normal_distribution

To find the probability of x=2 I think maybe we could substitute.

p_x(2)=\frac{1}{3 \sqrt{2\pi}}e^{\frac{-(2-4)^2}{18}}
 
Last edited:
x^2\sqrt{x}[\tex]<br /> was trying some latex here. semi-success.
 
Substituted the value for x=2 in my Casio and I get 0.1064826685.
Will try and plot in Mathematica for confirmation.
 
Have plotted them :)

The files are attached to this post. Need to work out how to get them to flash up here.
 

Attachments

Plotted :)

Not sure quite how to embed the image so that it appears in this post. But its attached & the hand-calculated value looks reasonable :)

GaussianPhysForum.gif


GaussianPhysForumzoom.gif


yes! Think this is how its done.
 
Wow it was really that simple! I had that equation written down right on the scratch pad where I was working this problem and didnt see that I guess.

To find the probability I followed that other example that simply used the ##Q(x)## function. And then take the result and look up the probability in the table that goes along with that function.

For reference the function looks like this (from my text) ## Q(x)= \frac{1}{x \sqrt{2 \pi}} e^{\frac{-x^2}{2}}##

Thanks for the help AugustCrawl!
 
Oh btw a little note. Your latex code looks ok. If you use those tage be sure to use "[\itex]" i think your just leaving out the i. Or you can use two dollar signs $$ before and after for a separate line of code or two hash tags ## for code to be on the same line.
 
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