1. Chapter 7.2  Warm Up Do #’s 36,39,42  Homework Due 12/16 th and 12/19 th  # 37,38, 44, 45, 46, 55, 56, 57, 61, 68  POP Mini Quiz Fri/Mon  Review Worksheet over Break  Chapter 7 Quiz on Jan 4 th and Jan 5 th

2. Chapter 7.2 Means and Variances of Random Variables

3. Notations  The mean, x-bar, of a set of observations is their ordinary average.  The mean μ is the average of the possible values of the random variable X. Also called the expected value of X.  Since we’ll be finding means for different random variables, we’ll use μ x for the mean of the random variable X.

4. Finding the Mean of a Discrete Random Variable.  Write the distribution like this:  To find μ, multiply each possible value, x, by its probability, then sum the products.  μ may not be a value of x. Value of X: x1x1x1x1 X2X2X2X2……….. xkxkxkxkProbability: p1p1p1p1 p2p2p2p2……….. pkpkpkpk

5. Finding the Variance of a Discrete Random Variable.  Write the distribution like this:  Variance is found by: σ 2 x =(x 1 – μ) 2 ●p 1 + (x 2 – μ) 2 ●p 2 +…+ (x k - μ) 2 ●p k  To find the standard deviation, you must square root the variance. Value of X: x1x1x1x1 X2X2X2X2……….. xkxkxkxk Probability: p1p1p1p1 p2p2p2p2……….. pkpkpkpk

6. Law of Large numbers:  Draw independent observations at random from any population with finite mean μ. As the number of observations drawn increases, the mean x-bar of the observed values eventually approaches the mean μ of the population.

7. Rules for Means:  Rule 1. If X is a random variable and a and b are fixed numbers, then:  μ a+bx = a + bμ x  Rule 2. If X and Y are random variables, then:  μ x+y = μ x + μ y

8. Rules for Variances:  Rule 1. If X is a random variable and a and b are fixed numbers, then:  σ 2 a+bx = b 2 σ 2 x  Rule 2. If X and Y are independent random variables, then:  σ 2 x+y = σ 2 x + σ 2 y  σ 2 x-y = σ 2 x + σ 2 y

9. Rules for Variance  If X and Y have a correlation ρ, then  σ 2 x+y = σ 2 x + σ 2 y + 2ρσ x σ y  σ 2 x-y = σ 2 x + σ 2 y - 2ρσ x σ y

10. Chapter 7.2  Warm Up Do #’s 36,39,42  Homework Due 12/16 th and 12/17 th  # 37,38, 44, 45, 46, 55, 56, 57, 61, 68  POP Mini Quiz Fri/Mon  Review Worksheet over Break  Chapter 7 Quiz on Jan 4 th and Jan 5 th

11. Warm Up 12/16 and 12/17  At a large university students have either a final exam or a final paper at the end of a course. The table below lists the distribution of the number of final exams that students at the university will take, and their associated probabilities. What are the mean and standard deviation of this distribution (remember to show ALL work)? X0123 P(X)0.050.250.400.30

12. Extra Examples: #1  Suppose you buy a raffle ticket at the fair. The ticket costs $1 and the prize is a new truck. The new truck is valued at approximately $22,000. Second prize is a $100 gift certificate to Southpoint Mall. The fair sells 50,000 raffle tickets. What are your average winnings per ticket?  Make a probability distribution table to represent, X, the payoff.

13. The probability distribution of X is :  PayoffX:$0$100$22,000  Probability: 49998/500001/50000 1/50000 (.99996)(.00002)(.00002) (.99996)(.00002)(.00002)  So, the average long-run payoff is ($0*.99996)+($100*.00002)+($22,000*.0 0002) = $0.442

14. Does this make sense?  It does if you consider the long run. Also you could think about it from the fair’s point of view. From their side you can see that they pay out an average of 44 cents per raffle ticket sold.  Further, you could figure their cost to be.442 * $1(50000 tickets sold) = $22,100  Further, you could figure their profit to be (1-.442) * $1(50000 tickets sold) = $27,900 (1-.442) * $1(50000 tickets sold) = $27,900

15. Another Example #2 Finding an Expected Value.  In 1993, there were 121,100,000 cars in use in the United States. That year, there were 14,100,000 automobile accidents. The average premium paid in 1993 for automobile collision insurance was $638, and the average automobile collision claim paid by insurance companies was $1796. What was the expected value of the insurance coverage for a person who paid $638 for collision insurance in 1993?

16.  Solution: Let event A represent having an automobile accident. Then event A’ represents not having an automobile accident. In 1993, the probability of having an automobile accident was:  P(A) =  Which means P(A’) 0.884. You can calculate the payoffs for A and A’ as follows. If the person had an accident, he or she paid $638 and received $1796, so the person received $1796 - $638 = $1158.

17.  If the person didn’t have an accident, he or she paid $638. Thus the expected value was  The $429.66 represents an average amount paid by each policy-holder to cover insurance company expenses (and produce a profit for the insurance companies).