1. Application of Random Variables

2. The Sum of the Variance
The sum of variance is a concept that is very important. It has been the subject of many AP questions and it is an idea that you should understand. (NOTE: This is for independent random variables only!)
Looking at Rules 2 and 3, is the following true?
Given that Var(X) = Var(Y), then is Var(X+Y) = Var(2X)?
Surprisingly, the answer is NO!!!
Rules that work in Algebra do not necessarily work with random variables!!!
Assume that Var(X) = 50 and Var(Y) = 50, then according to our rules:
Rule 2: Var(2X) = 22(50) = 4(50) = 200
Rule 3: Var(X + Y) = (50) + (50) = 100
It’s obvious that these are not the same!!!
If you think about it, Var(X +X) ≠ Var(2X) since these are demonstrating two different scenarios:
Var(X + X) represents playing the game twice, whereas Var(2X) represents doubling all the possible values.
For example, let X = 0, 7, -5, E(X) = -1.33, and Var(X) = 18.9, then
For Var(X + X): Var(X) = = 37.8
For Var(2X), we have to change our random variable to 2X = 0, 14, -10 (which obviously has a greater spread).
Var(2X) = 22(18.9) = 4(18.9) = 75.6

3. The Sum of the Variance
What is going on here?! If the variances are the same for each random variable, then why don’t the rules match up?
The main reason is because the two random variable are independent. Since the random variables are independent, each one reduces the variability.
Consider the following example: You know that you have a 40% chance of winning a game. The game pays out $100 for every $1 that you bet. You have $5. What is the best strategy to win? Should you bet all $5 at once or should you play 5 games at $1 per game? Assume that each game is independent.

4. The Sum of the Variance
It’s in your best interest to spread it out. Why?
Because there is less variance if you spread it out or rather you have a lower chance of losing. If you play 5 games you increase you chances of winning at least one game.
P(win one game) = .4
P(win at least once out of 5 games) = .9222
If you play 5 games, you have a 92% chance of winning at least once whereas if you played just one game, you only have a 40% chance of winning.

5. The Sum of the Variance
This concept is based on independence. With independent random variables, we don’t expect to lose all five games. Because we spread the risk across five games, we should expect our standard deviation and variance to be smaller.
This is the fundamental principle that insurance companies, lotteries, and casinos use to make all their money.

6. Example 1: Die
You wish to play a game with a die. If you roll a 1, 2, or 3, you get 0 points; if you roll 4 or 5, you get 5 points; and if you roll a 6, you get 50 points. Find the expected value, the standard deviation, and interpret you results.
Outcome
Points
P(X = x)
1, 2, 3
1/2
4, 5 5
1/3 6 50
1/6
E(X) = 0(1/2) + 5(1/3) +50(1/6) = 10
Var(X) = (0 – 10)2(1/2) + (5 – 10)2(1/3) + (50 – 10)2(1/6)
Var(X) = 325
SD(X) =

7. Example 1: Die
You wish to play a game with a die. If you roll a 1, 2, or 3, you get 0 points; if you roll 4 or 5, you get 5 points; and if you roll a 6, you get 50 points. Find the expected value, the standard deviation, and interpret you results.
So, what does E(X) = 10 and SD(X) = mean?
I expect to win 10 points per roll, on average, in the long run. The standard deviation is points, suggesting a highly variable game.

8. Example 1: Die Variations
If you played the game twice, what would be your new expected value and standard deviation?
E(X + Y) = E(X) + E(Y) = = 20
Var(X + Y) = Var(X) + Var(Y) = = 650
SD(X + Y) =
If you doubled the points, what would be your new expected value and standard deviation?
E(2X) = 2E(X) = 2(10) = 20
Var(2X) = a2Var(X) = 22(325) = 4(325) = 1300
SD(2X) =

9. Example 1: Die Variations
OK, one more variation of the die game. Let’s assume that you and your friend are going to play this game. Your total score will be the difference in your individual scores (X – Y). Find the expected value and the standard deviation.
E(X – Y) = 10 – 10 = 0
Var(X – Y) = = 650
SD(X – Y) =
Don’t forget!!! Variances ADD!!!

10. Example 2: The Auto Dealer
Suppose a used car dealer runs autos through a two-stage process to get them ready to sell. The mechanical checkup costs $50 per hour and takes an average of 90 minutes, with a standard deviation of 15 minutes. The appearance prep (wash, polish, etc.) costs $6 per hour and takes an average of 60 minutes, with a standard deviation of 5 minutes.
What are the mean and standard deviation of the total time spent preparing a car? (Note that we cannot find the standard deviation if we do not believe that the two phases of the process are independent, an important assumption to check.)

11. Example 2: The Auto Dealer
Let’s start by defining the what we are working with: Let M = mechanical time, A = appearance time, T = total time, where T = M + A
E(T) = E(M) + E(A) = = 150 minutes.
Var(T) = Var(M) + Var(A) = = 250
SD(T) =
We expect the total preparation time to take an average of 150 minutes (2.5 hours), with a standard deviation of 15.8 minutes.

12. Example 2: The Auto Dealer
What are the mean and standard deviation of the total expense to prepare a car?
We want to know the mean and standard deviation of the total expense of preparing a car. We can find mean and standard deviation of expenses for each part of the preparation by multiplying the mean and standard deviation of the times by the rate of $50/hr for mechanical time and $6/hr for appearance time. We believe that the two phases of the process are independent, so variances add.

13. Example 2: The Auto Dealer
Let M = mechanical hours E(M) = SD(M) = 0.25
A = appearance hours E(A) = SD(A) = 0.08
T = total expenses
T = 50M + 6A
E(T ) = E(50M + 6A) = E(50M) + E(6A)
= 50E(M) + 6E(A) = 50(1.5) + 6(1.0) = $81
Var(T ) = Var(50M + 6A) = Var(50M) +Var(6A)
= (502)Var(M) + (62)Var(A)
= 2500(.252) + 36(.082) =
SD(T ) =
We expect the total preparation expense to average $81, with a standard deviation of $12.51.

14. Example 2: The Auto Dealer
What is the probability that it will take longer to do the appearance prep than the mechanical checkup? (Note that we cannot answer this question unless we believe each phase of the process can be described by a Normal model.)

15. Example 2: The Auto Dealer
We want to know the probability that it will take longer to do the appearance prep than the mechanical checkup. In order to calculate this, we need to know the mean and standard deviation of the difference in time spent on each type of preparation. Let D = M – A. Since we are subtracting the amount of time for Appearance from the time for Mechanical work, if the difference in prep times in less than 0, that means the appearance prep took longer. We believe that the two phases of the process are independent, so variances add.
In order to calculate a probability, we will assume that each of the processes varies according to a Normal model. Therefore, the difference in times will vary Normally as well.

16. Example 2: The Auto Dealer
E(D) = E(M − A) = E(M) − E(A) = 90 − 60 = 30minutes.
Var(D) = Var(M − A) = Var(M) + Var(A)
SD(D) =
The difference in prep times varies according to the Normal model N(30, 15.8).
Remember that N(μ, σ) represents the normal curve with a mean of μ and a standard deviation of σ
Find normalcdf(-E99, , 0, 1) =
Or use normalcdf(-E99, 0, 30, 15.8) =
We expect the appearance prep to take more time than the mechanical checkup about 2.88% of the time.

17. Assignment Chapter 16 Lesson: Random Variables and Expected Value
Read:
Problems:
1 – 41 (odd)