Lesson 49 – Symmetry: The language of normal distribution

Hello.

I am the normal distribution.

I am one of the most important density functions in probability. People are so used to me, almost to the point of using me for approximating any data.

You have seen in Lesson 47 and Lesson 48 that I am a good approximation for the distribution function of the sum of independent random variables. The Central Limit Theorem.

My functional form is

 f(x) = \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^{2}}

Did you notice I am merely a symmetric function e^{-x^{2}}, with some constants and parameters?

x can be positive or negative real numbers (continuous random variable)  -\infty < x < \infty .

\mu and \sigma are my control parameters.

Can you tell me what lesson was it where we learn that \mu is the mean or expected value and \sigma is the standard deviation of the distribution?

People use the notation N(\mu, \sigma) to say that I am a normal distribution with mean \mu and standard deviation \sigma.

\mu is my center. It can be positive or negative (-\infty < \mu < \infty), and the function is symmetric to its right and left. \sigma is how spread out I am from the center. It is positive (\sigma > 0).

Look at this example. I am centered on 60 and changing the spread. Narrow and wide flanks. Larger standard deviation results in a wider distribution with more spread around the mean.

In this example, I have the same spread, but I am changing my location (center).

I told you before that I am symmetric around the mean. So the following properties hold for me.

P(X > \mu) = P(X < \mu) = 0.5

P(X > \mu + a) = P(X < \mu - a)

f(\mu + a) = f(\mu-a)

By the way, if I give you the values for \mu and \sigma, do you know how to compute P(X \le x)?

You guessed it correct. Take my probability density function and integrate it from -\infty to x.

P(X \le x) = \int_{-\infty}^{x}\frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^{2}}dx

For example,

Is this the cumulative distribution function  F(x)?

Compute the closed form solution of this integral.

P(X \le x) = \int_{-\infty}^{x}\frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^{2}}dx

You can try change of variables method or the substitution method.

You can try integration by parts.

You can find the anti-derivative and use the fundamental theorem of calculus to solve.

I dare you to compute this integral to a closed form solution by next week.

We will continue our lessons after your tireless effort on this 😉

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