muchen 牧辰

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


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



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.


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

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,


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.

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


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 \(c\).

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


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 \(\mathbb P(X>c)=0.10\)

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