ELEC 321

Normal Distribution

Updated 2017-10-11

Standard Normal

Standard Normal Random Variable is denoted by \(Z\). The notation \(Z\) ~ \(N(0,1)\) means that “\(Z\) is a normal random variable mean of 0 and variance of 1”.

Density Function

The standard normal density is given by

\[\varphi(z)=\frac{1}{\sqrt{2\pi}}e^{-(\frac{z^2}{2})},\quad-\infty<z<\infty\]

Distribution Function

The standard normal distribution function is given by

\[\Phi(z)=\int_{-\infty}^z\varphi(t)\mathrm dt\]

Note: \(\Phi(z)\) cannot be calculated in close form

Therefore, it is usually better to use the standard normal table or the function pnorm(z).

Due to the symmetry of the distribution function

\[\boxed{\Phi(z)=1-\Phi(-z)}\]

Mean

The mean, or expected value is given by (as always):

\[\mathbb E(Z)=\int_{-\infty}^\infty z\varphi(z)\mathrm dz=0\]

Notice the expected value for standard normal is at 0 since the standard normal centers around 0.

Variance

\[\text{Var}(Z)=\int_{-\infty}^\infty z^2\varphi (z)\mathrm dz=1\]

Example: concrete mix

A machine fills 10-pound bags of dry concrete mix. The actual weight of the mix put into the bag is a normal random variable with standard deviation \(\sigma=0.1\) pound. The mean can be set by the machine operator

a. is the mean at which the machine should be set if at most 10% of the bags can be underweight?

Let \(X\sim \text{Norm}(\mu, \sigma^2)\) where \(X\) is the actual weight. Thus we can express the following.

\[\mathbb P(X<10)\leq 0.1\]

Which means the probability of weight less than 10 pounds is 0.1.

\[\begin{align} \mathbb P(\frac{x-\mu}{\sigma}<\frac{10-\mu}{\sigma})&\leq0.1\\ \mathbb P(z<\frac{10-\mu}{0.1})&\leq0.1\\ \implies\Phi(\frac{10-\mu}{0.1})&\leq0.1\\ \implies \frac{10-\mu}{0.1}&\leq \Phi^{-1}(0.1)\\ \mu&\geq 10-0.1\Phi^{-1}(0.1) \end{align}\] \[\begin{align} \mathbb P(\frac{x-\mu}{\sigma}<\frac{10-\mu}{\sigma})&\leq0.1\\ \mathbb P(z<\frac{10-\mu}{0.1})&\leq0.1\\ \implies\Phi(\frac{10-\mu}{0.1})&\leq0.1\\ \implies \frac{10-\mu}{0.1}&\leq \Phi^{-1}(0.1)\\ \mu&\geq 10-0.1\Phi^{-1}(0.1) \end{align}\]

Standard Deviation

Since the variance equals to 1, standard deviation also equals to 1: \(\sigma=1\).

Measurement Error Model

Suppose we have:

Then we can model the errors as follows.

\[\boxed{X_i=\mu+\sigma Z_i,\quad i=1,2,\dots,n}\]

Using this equation, we can find the error of the individual measurement to be

\[\boxed{Z_i=\frac{X_i - \mu}{\sigma},\quad i=1,2,\dotsc,n}\]

General Normal Random Variables

This applies to any normal random variables that aren’t standardized. These random variables are denoted as \(X\sim \text{Norm}(\mu,\sigma^2)\), which stands for “X is a normal random variable with a mean of \(\mu\) and a variance of \(\sigma^2\)”.

Manipulating the mean (\(\mu\)) shifts the distribution left and right. Manipulating the variance (\(\sigma^2\)) changes the amplitude and thickness of the distribution.

Mean and Variance

Recall that \(X=\mu+\sigma Z\) and \(Z\sim\text{Norm}(0,1)\iff Z=\frac{X-\mu}{\sigma}\) , we can substitute \(Z\) into \(X\) and find the expected value and variance functions.

\[\begin{align} \mathbb E(X)&=\mathbb E(\mu+\sigma Z)=\mu+\sigma\underbrace{\mathbb E(Z)}_0\\ \mathbb E(X)&=\boxed{\mu} \end{align}\] \[\text{Var}(X)=\text{Var}(\mu+\sigma Z)=\sigma^2\underbrace{\text{Var}(Z)}_1=\boxed{\sigma^2}\]

Distribution Function

First, start with the definition of distribution function.

\[F(x)=\mathbb P(X\leq x)\]

Next, we subtract \(\mu\) and divide \(\sigma\) on both sides of the inner inequality.

\[=\mathbb P\left(\frac{X-\mu}{\sigma}\leq\frac{x-\mu}{\sigma}\right)\]

Recall that \(Z=\frac{X-\mu}{\sigma}\), we plug it in.

\[=\mathbb P\left(Z\leq \frac{x-\mu}{\sigma}\right)\]

Notice that this is the standard normal distribution function. Thus,

\[\boxed{F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)}\]

Density Function

Recall that \(F'(X)=f(x)\)and \(\Phi'(z)=\varphi(z)=\frac{1}{\sqrt{2\pi}}e^{-\frac12z^2}\), the density function is simply as follows.

\[\begin{align} f(x)=F'(x)&=\frac{1}{\sigma}\varphi\left(\frac{x-\mu}{\sigma}\right)\\ f(x)&=\boxed{\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac12\left(\frac{x-\mu}{\sigma}\right)^2}} \end{align}\]

Example:

Let \(Z\sim\text{Norm}(0,1)\), calculate:

Example:

Let \(X\sim \text{Norm}(3, 25)\), calculate: