Normal Distribution

• Standard Normal
  • Density Function
  • Distribution Function
  • Mean
  • Variance
  • Standard Deviation
• Measurement Error Model
• General Normal Random Variables
  • Mean and Variance
  • Distribution Function
  • Density Function

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

$$\phi (z)=\frac{1}{\sqrt{2\pi }}{e}^{-(\frac{z^2}{2})},-\mathrm{\infty }<z<\mathrm{\infty }$$

Distribution Function

The standard normal distribution function is given by

$$\mathrm{\Phi }(z)={\int }_{-\mathrm{\infty }}^z\phi (t)\mathrm{d}t$$

Note: $\mathrm{\Phi }(z)$ 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

$$\boxed{{\displaystyle \mathrm{\Phi }(z)=1-\mathrm{\Phi }(-z)}}$$

Mean

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

$$\mathbb{E}(Z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}z\phi (z)\mathrm{d}z=0$$

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

Variance

$$\text{Var}(Z)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{z}^2\phi (z)\mathrm{d}z=1$$

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.

$$\mathbb{P}(X<10)\le 0.1$$

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

$$\begin{array}{rl}\mathbb{P}(\frac{x-\mu }{\sigma }<\frac{10-\mu }{\sigma })& \le 0.1\\ \mathbb{P}(z<\frac{10-\mu }{0.1})& \le 0.1\\ ⟹\mathrm{\Phi }(\frac{10-\mu }{0.1})& \le 0.1\\ ⟹\frac{10-\mu }{0.1}& \le {\mathrm{\Phi }}^{-1}(0.1)\\ \mu & \ge 10-0.1{\mathrm{\Phi }}^{-1}(0.1)\end{array}$$ $$\begin{array}{rl}\mathbb{P}(\frac{x-\mu }{\sigma }<\frac{10-\mu }{\sigma })& \le 0.1\\ \mathbb{P}(z<\frac{10-\mu }{0.1})& \le 0.1\\ ⟹\mathrm{\Phi }(\frac{10-\mu }{0.1})& \le 0.1\\ ⟹\frac{10-\mu }{0.1}& \le {\mathrm{\Phi }}^{-1}(0.1)\\ \mu & \ge 10-0.1{\mathrm{\Phi }}^{-1}(0.1)\end{array}$$

Standard Deviation

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

Measurement Error Model

Suppose we have:

• Measurements $X_i$ ($X_1,X_2,\dots ,X_n$)
• “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.

$$\boxed{{\displaystyle X_i=\mu +\sigma Z_i,i=1,2,\dots ,n}}$$

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

$$\boxed{{\displaystyle Z_i=\frac{X_i-\mu }{\sigma },i=1,2,\dots ,n}}$$

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

$$\begin{array}{rl}\mathbb{E}(X)& =\mathbb{E}(\mu +\sigma Z)=\mu +\sigma \underset{0}{\underbrace{\mathbb{E}(Z)}}\\ \mathbb{E}(X)& =\boxed{{\displaystyle \mu }}\end{array}$$ $$\text{Var}(X)=\text{Var}(\mu +\sigma Z)={\sigma }^2\underset{1}{\underbrace{\text{Var}(Z)}}=\boxed{{\displaystyle {\sigma }^2}}$$

Distribution Function

First, start with the definition of distribution function.

$$F(x)=\mathbb{P}(X\le x)$$

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

$$=\mathbb{P}(\frac{X-\mu }{\sigma }\le \frac{x-\mu }{\sigma })$$

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

$$=\mathbb{P}(Z\le \frac{x-\mu }{\sigma })$$

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

$$\boxed{{\displaystyle F(x)=\mathrm{\Phi }\left(\frac{x-\mu }{\sigma }\right)}}$$

Density Function

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

$$\begin{array}{rl}f(x)=F^{\prime }(x)& =\frac{1}{\sigma }\phi \left(\frac{x-\mu }{\sigma }\right)\\ f(x)& =\boxed{{\displaystyle \frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}}}\end{array}$$

Example:

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

• $$\mathbb{P}(0.1\le Z\le 0.35)$$

  $$\begin{array}{rl}\mathbb{P}(0.10\le Z\le 0.35)& =\mathrm{\Phi }(0.35)-\mathrm{\Phi }(0.10)\\ & =\boxed{{\displaystyle 0.0970}}\end{array}$$

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

• $$\mathbb{P}(Z>1.25)$$

  $$\begin{array}{rl}\mathbb{P}(Z>1.25)& =1-\mathbb{P}(Z\le 1.25)\\ & =1-\mathrm{\Phi }(1.25)\\ & =\boxed{{\displaystyle 0.1056}}\end{array}$$

• $$\mathbb{P}(Z>-1.20)$$

  $$\begin{array}{rl}\mathbb{P}(Z>-1.20)& =1-\mathbb{P}(Z\le -1.20)\\ & =1-\mathrm{\Phi }(-1.20)\\ & =1-(1-\mathrm{\Phi }(1.20))\\ & =\mathrm{\Phi }(1.2)\\ & =\boxed{{\displaystyle 0.8849}}\end{array}$$

• Find such that $\mathbb{P}(Z>c)=0.05$

  $$\begin{array}{rl}1-\mathrm{\Phi }(c)& =0.05\\ \mathrm{\Phi }(c)& =0.95\\ c& =\boxed{{\displaystyle {\mathrm{\Phi }}^{-1}(0.95)}}\end{array}$$

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

• Find $c$ such that $\mathbb{P}(|Z|<c)=0.95$

  $$\begin{array}{rl}\mathbb{P}(|Z|>c)& =\mathbb{P}(-c<Z<c)\\ & =\mathrm{\Phi }(c)-\mathrm{\Phi }(-c)\\ & =\mathrm{\Phi }(c)-(1-\mathrm{\Phi }(c))\\ 0.95& =2\mathrm{\Phi }(c)-1\end{array}$$

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

  $$\begin{array}{rl}\mathrm{\Phi }(c)& =\frac{1.95}{2}=0.975\\ & ={\mathrm{\Phi }}^{-1}(0.975)\\ & =\boxed{{\displaystyle 1.96}}\end{array}$$

Example:

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

• $$\mathbb{P}(X>4)$$

  $$\begin{array}{rl}\mathbb{P}(X>4)& =1-\mathbb{P}(X<4)\\ & =1-\mathrm{\Phi }\left(\frac{4-3}{5}\right)\\ & =1-\mathrm{\Phi }(0.2)\\ & -\boxed{{\displaystyle 0.421}}\end{array}$$

• $$\mathbb{P}(2<X<4)$$

  $$\begin{array}{rl}\mathbb{P}(2<X<4)& =F(4)-F(2)\\ & =\mathrm{\Phi }\left(\frac{4-3}{5}\right)-\mathrm{\Phi }\left(\frac{2-3}{5}\right)\\ & =\mathrm{\Phi }(0.20)-\mathrm{\Phi }(-0.20)\\ & =2\mathrm{\Phi }(0.20)-1\\ & =\boxed{{\displaystyle 0.159}}\end{array}$$

• $$\mathbb{P}(X<1)$$

  $$\begin{array}{rl}\mathbb{P}(X<1)& =F(1)\\ & =\mathrm{\Phi }\left(\frac{1-3}{5}\right)\\ & =\mathrm{\Phi }(-0.40)\\ & =1-\mathrm{\Phi }(0.40)\\ & =\boxed{{\displaystyle 0.345}}\end{array}$$

• $c$ such that $\mathbb{P}(X>c)=0.10$

  $$\begin{array}{rl}\mathbb{P}(X>c)& =0.10\\ 1-F(c)& =0.10\\ 1-\mathrm{\Phi }\left(\frac{c-3}{5}\right)& =0.10\\ \mathrm{\Phi }\left(\frac{c-3}{5}\right)& =0.90\\ \frac{c-3}{5}& ={\mathrm{\Phi }}^{-1}(0.90)\\ c& ={\mathrm{\Phi }}^{-1}(0.90)\times 5+3\end{array}$$