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:

• Measurements $X_i$ ($X_1,X_2,\dotsc,X_n$)
• “True” value $\mu$
• “Inverse precision” of the measurements (variance) $\sigma$
• Measurement error in the standard scale $Z_i\sim \text{Norm}(0,1)$
• Measurement error in the original scale $\sigma Z_i$

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

$F(x)=\mathbb P(X\leq x)$

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:

• $\mathbb P(0.1\leq Z\leq 0.35)$
\begin{align} \mathbb P(0.10\leq Z\leq 0.35)&=\Phi(0.35)-\Phi(0.10)\\ &=\boxed{0.0970} \end{align}

Note that $\Phi(x)$ can be calculated in R using the pnorm(x) function.

• $\mathbb P(Z\gt 1.25)$
\begin{align} \mathbb P(Z>1.25)&=1-\mathbb P(Z\leq 1.25)\\ &=1-\Phi(1.25)\\ &=\boxed{0.1056} \end{align}
• $\mathbb P (Z\gt -1.20)$
\begin{align} \mathbb P(Z>-1.20)&=1-\mathbb P(Z\leq -1.20)\\ &=1-\Phi(-1.20)\\ &=1-(1-\Phi(1.20))\\ &=\Phi(1.2)\\ &=\boxed{0.8849} \end{align}
• Find such that $\mathbb P(Z\gt c)=0.05$

\begin{align} 1-\Phi(c)&=0.05\\ \Phi(c)&=0.95\\ c&=\boxed{\Phi^{-1}(0.95)} \end{align}

Note that the inverse of standard normal CDF function can be calculated in R using qnorm(0.95)

• Find $c$ such that $\mathbb P(\vert Z\vert <c)=0.95$

\begin{align} \mathbb P(\vert Z\vert >c)&=\mathbb P(-c<Z<c)\\ &=\Phi(c)-\Phi(-c)\\ &=\Phi(c)-(1-\Phi(c))\\ 0.95&=2\Phi(c)-1 \end{align}

Rearrange the terms we can find $\Phi(c)$. Once again, we can use the qnorm(c) function in R to find $c$.

\begin{align} \Phi(c)&=\frac{1.95}{2}=0.975\\ &=\Phi^{-1}(0.975)\\ &=\boxed{1.96} \end{align}

Example:

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

• $\mathbb P(X>4)$
\begin{align} \mathbb P(X>4)&=1-\mathbb P(X<4)\\ &=1-\Phi\left(\frac{4-3}{5}\right)\\ &=1-\Phi(0.2)\\ &-\boxed{0.421} \end{align}
• $\mathbb P(2<X<4)$
\begin{align} \mathbb P(2<X<4)&=F(4)-F(2)\\ &=\Phi\left(\frac{4-3}{5}\right)-\Phi\left(\frac{2-3}{5}\right)\\ &=\Phi(0.20)-\Phi(-0.20)\\ &=2\Phi(0.20)-1\\ &=\boxed{0.159} \end{align}
• $\mathbb P(X<1)$
\begin{align} \mathbb P(X<1)&=F(1)\\ &=\Phi\left(\frac{1-3}{5}\right)\\ &=\Phi(-0.40)\\ &=1-\Phi(0.40)\\ &=\boxed{0.345} \end{align}
• $c$ such that $\mathbb P(X>c)=0.10$

\begin{align} \mathbb P(X>c)&=0.10\\ 1-F(c)&=0.10\\ 1-\Phi\left(\frac{c-3}{5}\right)&=0.10\\ \Phi\left(\frac{c-3}{5}\right)&=0.90\\ \frac{c-3}{5}&=\Phi^{-1}(0.90)\\ c&=\Phi^{-1}(0.90)\times5+3 \end{align}