The Standard Normal Distribution

Why Use the Standard Normal Distribution

One reason to use the standard normal distribution is to solve probability questions that could otherwise be difficult for a calculator to handle.

Example: $Xsim N(1000000,122500)$. What is $mathbb{P}(999500leq Xleq 1001000)$?

$begin{aligned}&mathbb{P}(999500leq Xleq 1001000)=[1.2em]&mathbb{P}left(dfrac{999500-mu}{sigma}leq Zleq dfrac{1001000-mu}{sigma}right)=[1.2em]&mathbb{P}left(dfrac{999500-1000000}{sqrt{122500}}leq Zleq dfrac{1001000-1000000}{sqrt{122500}}right)[1.2em]&=mathbb{P}left(dfrac{-10}{7}leq Zleq dfrac{20}{7}right)[1.2em]&=0.9213end{aligned}$

Another reason to use the standard normal distribution is that you will be provided with a table of key values for it – called a percentage points table.

The percentage points table is most useful for finding $Z$ values from probabilities.

Example: $Xsim N(10,9)$ and $mathbb{P}(X<x)=0.95$

Find $x$.

$mathbb{P}(X<x)=0.95$

$mathbb{P}left(Z<dfrac{x-mu}{sigma}right)=0.95$

$mathbb{P}left(Z<dfrac{x-10}{3}right)=0.95$

From the percentage points table we find:

$dfrac{x-10}{3}=1.6449$

$x-10=4.9347$

$x=14.9347$

Transform to Standard Normal to Find an Unknown

The biggest reason to use the standard normal distribution is that doing so can help us find $mu$ and $sigma$ if one of them is unknown.

Example: Suppose $Xsim N(mu,4)$ and $mathbb{P}(Xleq 31)=0.9$. Find $mu$.

$mathbb{P}(Xleq 31)=0.9$

$mathbb{P}left(Zleq dfrac{31-mu}{2}right)=0.9$

From the percentage points table:

$dfrac{31-mu}{2}=1.2816$

$31-mu=2.5632$

$mu=31-2.5632$

$mu=28.4368$

If $sigma$ is unknown instead of $mu$ we can use the same method to get an equation for it.

Simultaneous Equations from Standard Normal

If we are given two probabilities, we can convert to the standard normal distribution and use the percentage points table to get two equations, so can solve for two unknowns. This means that we can find both $mu$ and $sigma$.

Example: Suppose $Xsim N(mu,sigma^{2})$ and $mathbb{P}(X<0.2)=0.05,mathbb{P}(X<0.9)=0.8$. Find $mu$ and $sigma$.

$mathbb{P}(X<0.2)=0.05$

$mathbb{P}left(Z<dfrac{0.2-mu}{sigma}right)=0.05$

Use percentage points table:

$dfrac{0.2-mu}{sigma}=-1.6449$

$0.2-mu=-1.6449sigma;;;(1)$

$mathbb{P}(X<0.9)=0.8$

$mathbb{P}left(Z<dfrac{0.9-mu}{sigma}right)=0.8$

Use percentage points table:

$dfrac{0.9-mu}{sigma}=0.8416$

$0.9-mu=0.8416sigma;;;(2)$

$0.7=2.4865sigma;;;(2)-(1)$

$sigma=0.2815$

$0.9-mu=0.8416times 0.2815$

$0.9-mu=0.2369$

$mu=0.6631$