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
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:
- \[\mathbb P(0.1\leq Z\leq 0.35)\]
\[\begin{align} \mathbb P(0.10\leq Z\leq 0.35)&=\Phi(0.35)-\Phi(0.10)\\ &=\boxed{0.0970} \end{align}\]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
\[\begin{align} \Phi(c)&=\frac{1.95}{2}=0.975\\ &=\Phi^{-1}(0.975)\\ &=\boxed{1.96} \end{align}\]qnorm(c)
function in R to find \(c\).
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 \(\mathbb P(X>c)=0.10\)
\[\begin{align} \mathbb P(X>c)&=0.10\\ 1-F(c)&=0.10\\ 1-\Phi\left(\frac{c-3}{5}\right)&=0.10\\ \Phi\left(\frac{c-3}{5}\right)&=0.90\\ \frac{c-3}{5}&=\Phi^{-1}(0.90)\\ c&=\Phi^{-1}(0.90)\times5+3 \end{align}\]