Chapter 7.2Chapter 7.2

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

Chapter 7.2 Means Chapter 7.2 Means and Variances of and Variances of Random VariablesRandom Variables

Notations Notations

The mean, The mean, x-bar, x-bar, of a set of observations of a set of observations is their ordinary average.is their ordinary average.

The mean The mean μμ is the average of the is the average of the possible values of the random variable X. possible values of the random variable X. Also called the expected value of X. Also called the expected value of X.

Since we’ll be finding means for different Since we’ll be finding means for different random variables, we’ll use random variables, we’ll use μμxx for the for the

mean of the random variable X.mean of the random variable X.

Finding the Mean of a Finding the Mean of a Discrete Random Discrete Random Variable.Variable.

Write the distribution like this:Write the distribution like this:

To find To find μμ,, multiply each possible value, x, by its multiply each possible value, x, by its probability, then sum the products.probability, then sum the products.

μμ may not be a value of x. may not be a value of x.

1 1 2 2X i ik kx p x p x p x p

Value of X: Value of X: xx11 XX22 ……………….... xxkk

Probability: Probability: pp11 pp22 ……………….... ppkk

Finding the Variance of a Finding the Variance of a Discrete Random Discrete Random Variable.Variable.

Write the distribution like this:Write the distribution like this:

Variance is found by:Variance is found by:σσ22

x x =(x=(x11 – – μμ))22●●pp11 + (x + (x22 – – μμ))22●●pp22 +…+ (x +…+ (xkk - - μμ))22●●ppkk

To find the standard deviation, you must To find the standard deviation, you must square root the variance. square root the variance.

Value of X: Value of X: xx11 XX22 ……………….... xxkk

Probability: Probability: pp11 pp22 ……………….... ppkk

Law of Large numbers:Law of Large numbers:

Draw independent observations at Draw independent observations at random from any population with finite random from any population with finite mean mean μμ. As the number of observations . As the number of observations drawn increases, the mean x-bar of the drawn increases, the mean x-bar of the observed values eventually approaches observed values eventually approaches the mean the mean μμ of the population. of the population.

Rules for Means:Rules for Means:

Rule 1. If X is a random variable and a Rule 1. If X is a random variable and a and b are fixed numbers, then:and b are fixed numbers, then:

μμa+bxa+bx = a + b = a + bμμxx

Rule 2. If X and Y are random variables, Rule 2. If X and Y are random variables, then:then:

μμx+yx+y = = μμxx + + μμyy

Rules for Variances:Rules for Variances:

Rule 1. If X is a random variable and a Rule 1. If X is a random variable and a and b are fixed numbers, then:and b are fixed numbers, then:

σσ22a+bxa+bx = b = b22σσ22

xx

Rule 2. If X and Y are independent Rule 2. If X and Y are independent random variables, then:random variables, then:

σσ22x+yx+y = = σσ22

xx + + σσ22yy

σσ22x-yx-y = = σσ22

xx + + σσ22yy

Rules for VarianceRules for Variance

If X and Y have a correlation If X and Y have a correlation ρρ, then , then

σσ22x+yx+y = = σσ22

xx + + σσ22y y + 2+ 2ρσρσxxσσy y

σσ22x-yx-y = = σσ22

xx + + σσ22y y - 2- 2ρσρσxxσσy y

Chapter 7.2Chapter 7.2

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

Warm Up 12/16 and 12/17Warm Up 12/16 and 12/17

At a large university students have either a final exam At a large university students have either a final exam or a final paper at the end of a course. The table below or a final paper at the end of a course. The table below lists the distribution of the number of final exams that lists the distribution of the number of final exams that students at the university will take, and their associated students at the university will take, and their associated probabilities. What are the mean and standard probabilities. What are the mean and standard deviation of this distribution (remember to show ALL deviation of this distribution (remember to show ALL work)?work)?

XX 00 11 22 33

P(X)P(X) 0.050.05 0.250.25 0.400.40 0.300.30

Extra Examples: #1Extra Examples: #1

Suppose you buy a raffle ticket at the fair. Suppose you buy a raffle ticket at the fair. The ticket costs $1 and the prize is a The ticket costs $1 and the prize is a new truck. The new truck is valued at new truck. The new truck is valued at approximately $22,000. Second prize is approximately $22,000. Second prize is a $100 gift certificate to Southpoint Mall. a $100 gift certificate to Southpoint Mall. The fair sells 50,000 raffle tickets. What The fair sells 50,000 raffle tickets. What are your average winnings per ticket? are your average winnings per ticket?

Make a probability distribution table to Make a probability distribution table to represent, X, the payoff. represent, X, the payoff.

The probability The probability distribution of X is :distribution of X is :

PayoffX:PayoffX: $0$0 $100$100$22,000$22,000

Probability: 49998/50000Probability: 49998/50000 1/50000 1/50000 1/500001/50000

(.99996)(.99996) (.00002)(.00002)(.00002)(.00002)

So, the average long-run payoff is So, the average long-run payoff is ($0*.99996)+($100*.00002)+($0*.99996)+($100*.00002)+($22,000*.00002) = $0.442($22,000*.00002) = $0.442

Does this make sense? Does this make sense?

It does if you consider the long run. Also you It does if you consider the long run. Also you could think about it from the fair’s point of view. could think about it from the fair’s point of view. From their side you can see that they pay out From their side you can see that they pay out an average of 44 cents per raffle ticket sold. an average of 44 cents per raffle ticket sold.

Further, you could figure their cost to be .442 * Further, you could figure their cost to be .442 * $1(50000 tickets sold) = $22,100$1(50000 tickets sold) = $22,100

Further, you could figure their profit to be Further, you could figure their profit to be

(1-.442) * $1(50000 tickets sold) = $27,900(1-.442) * $1(50000 tickets sold) = $27,900

Another Example #2 Another Example #2 Finding an Expected Value.Finding an Expected Value.

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

Solution: Let event A represent having an automobile Solution: Let event A represent having an automobile accident. Then event A’ represents not having an accident. Then event A’ represents not having an automobile accident. In 1993, the probability of having automobile accident. In 1993, the probability of having an automobile accident was:an automobile accident was:

P(A) = P(A) =

Which means P(A’) 0.884. You can calculate the Which means P(A’) 0.884. You can calculate the payoffs for A and A’ as follows. If the person had an payoffs for A and A’ as follows. If the person had an accident, he or she paid $638 and received $1796, so accident, he or she paid $638 and received $1796, so the person received $1796 - $638 = $1158. the person received $1796 - $638 = $1158.

14,100,0000.116

121,100,000

If the person didn’t have an accident, he or she If the person didn’t have an accident, he or she paid $638. Thus the expected value waspaid $638. Thus the expected value was

The $429.66 represents an average amount The $429.66 represents an average amount paid by each policy-holder to cover insurance paid by each policy-holder to cover insurance company expenses (and produce a profit for company expenses (and produce a profit for the insurance companies). the insurance companies).

( ) (1158) ( ' ) ( 638)

0.116(1158) 0.884(638) $429.66

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