1. IE 360: Design and Control of Industrial Systems I
Lecture 14
Normal
Random Variables
IE 360: Design and Control of Industrial Systems I
References
Montgomery and Runger
Sections 4-6, 4-7
Copyright  2010 by Joel Greenstein

2. Normal RV
The normal distribution is key in statistics (subject of IE 361, 461)
The normal distribution is also called the Gaussian distribution
The pdf of the normal random variable X with mean μ and variance σ2 is
We use the format “n(x; μ, σ2 )” to emphasize that we need to know the mean and variance to characterize a particular normal rv
X~N(μ, σ2) indicates that X is distributed as a normal rv with parameters μ and σ2

3. Pictures of Normal pdfs
There is an infinite number of normal curves possible
Each different mean and variance induces a different normal pdf
The mean μ sets the center
The variance σ2 sets the width and height
As σ2 increases, the pdf gets flatter and wider

4. Normal RV: Common values to compute
Expected value of a normal rv
Provided without proof
Variance of a normal rv
It can be derived by applying the definition of the variance, but we’ll just look at the answer
Don’t try to integrate the normal pdf by hand – it is IMPOSSIBLE!!!
Instead we use a table of the values of the cdf for the standard normal distribution (Table III in Appendix A of the textbook—bookmark it!)
The standard normal distribution is a normal distribution with mean 0 and variance 1
We generally use Z to denote the standard normal random variable
The standard normal cdf can be used to find the probabilities associated with a normal random variable with any mean and variance by first using a simple transformation
Computing the standard normal transformation
Let X~ N(μ, σ2). Then Z=(X- μ)/σ is distributed as a normal rv with mean 0 and variance 1. (That is, Z is distributed as a standard normal rv.)
For example, if X has μ = -14 and σ2 = 25, then Z = [X-(-14)]/(25)0.5 = (X+14)/5

5. Normal RV Example Assume S~N(10, 25). What is P(S ≤ 10)?
In this case, I’ll also tell you a shortcut.
Any time you need to compute the probability that a rv is less than or equal to its mean, the answer must be 50%
Similarly, the probability that a rv is greater than or equal to its mean must be 50%
Nonetheless, let’s solve this the long way to demonstrate the general technique
We know the variable name, parameters and what to compute
We need to transform this normal rv to a standard normal rv. In general, what we do on one side of an equation or inequality, we need to do on the other.
Now, use the Table III for the cdf
Each row is for the 10ths
The columns are for the 100ths place
So for z=0.00, use the first row, first column of
the second page of the table
So P(S ≤ 10) = P(Z ≤ 0) =

6. More examples Assume S~N(10, 25). What is P(-5 ≤ S ≤ 25)?
Use the same basic technique
Some special things to notice…
The end points are centered on the mean.
Because the normal distribution is symmetric
about the mean…
P(Z ≤ 3) = 1 – P(Z ≤ -3) and P(Z ≤ -3) = P(Z ≥ 3)
P(Z ≤ -3)
P(Z ≥ 3)

7. Inverse Normal Assume S~N(10, 25)
What if I want to know the value of s such that P(S ≤ s) = 15%?
This is called finding the inverse normal value
We know the variable name, parameters and what to compute
We need to transform this to a standard normal rv, but leave s as the unknown
Now, use the inside of the table
Look for the value closest to 0.15
Now, read off the value of z from the table
Here, z≈-1.035
Now, solve for s

8. Baby height example
Assume the height of 6-month old babies is distributed normally with a mean of 27 inches and a standard deviation of 2 inches. “Tall” 6-month old babies are those in the top 5% of all 6-month old babies. What height does a 6-month old baby need to be in order to be considered “tall”?
Step 1: Identify the RV
we know it is normal, let’s call it H = height of 6 mo. babies
Step 2: Identify parameters
Mean (μ) is 27 inches, Variance (σ2) is 4 in2 because the standard deviation is 2 inches.
Step 3: Setup the question
We need h such that P(H > h)=0.05
We know P(H > h)=1-P(H ≤h) so let’s find h such that P(H ≤h) =0.95
Using the table, we see that P(Z ≤1.64) = and P(Z ≤1.65) =
z=1.645 is exactly between them, using linear interpolation
Solve for h
So babies bigger than inches are in the top 5 percent

9. Related reading Montgomery and Runger, Sections 4-6 and 4-7
Some links that might help
Way more info than you probably want
More reasonable
Now you are ready to do HW14