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Chapter 16Chapter 16Random Random VariablesVariablesRandom Variables and Expected Value

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Betting on Death! Many people in America have life insurance policies.

Although you might not want to think of it this way, when you purchase a life insurance policy, you’re betting that you will die sooner rather than later… Although this is a bet that people really don’t want to win,

it is a bet that they are willing to take just to be sure that their families are financially secure in the event of death.

Most families depend on the income of one or more people in the household. What would happen if that income suddenly disappeared? Life insurance help us handle such disasters.

When you purchase a life insurance policy, it’s in your best interests that the company makes a profit and does well; why do you think that is?

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Betting on Death!Question:Question:

You purchase a policy that charges only $50 a year. If it pays $10,000 for death and $5000 for a permanent disability, is the company likely to make a profit?Actuaries at for the company have determined the

following probabilities in any given year:P (Death) = 1/1000P (Permanently disabled) = 2/1000P (Healthy) = 997/1000

We’ll come back to this problem later on…

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Random Variables and Expected Value A Random VariableRandom Variable is a variable whose values are

numbers that are determined by an outcome of a random event.

Random variables are denoted by capital letters, while the valuesvalues of random variables are denoted with lowercase letters (small letters)

The meanmean of the discrete random variable, X, is also called the expected valueexpected value of X. Notationally, the expected value of X is denoted by E(X). It is what we expect to happen. The formula for expected value is: )()( xPxXEX

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Examples In the experiment of flipping three coins, consider the

outcomes and define the random variable X as the number of heads that appear. The outcomes are {no heads, 1 head, 2 heads, or 3 heads}The outcomes are {no heads, 1 head, 2 heads, or 3 heads} X has values in the set: {0, 1, 2, 3X has values in the set: {0, 1, 2, 3

When rolling two dice and finding the sum, determine the outcomes and the random variable Y. The outcomes are {(1,1), (1,2), (1,3), (1,4), (1,5), etc…}The outcomes are {(1,1), (1,2), (1,3), (1,4), (1,5), etc…} Y has values in the set: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}Y has values in the set: {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

In our life insurance example, what are the outcomes and random variables Z if we define them as the possible payments. The outcomes are {die, disabled, healthy}The outcomes are {die, disabled, healthy} Z has values in the set: {$10,000, $5000, $0}Z has values in the set: {$10,000, $5000, $0}

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Back to Betting on Death So, will the company make a profit for any given

year? How much will they make or lose? These questions are answered by finding the expected

value.

The expected Value is:

PolicyholdePolicyholder Outcomer Outcome

PayoutPayout

xxProbabilityProbability

P(X = x)P(X = x)

Die $10,000 1/1000

Disability $5000 2/1000

Healthy $0 997/1000

20$0$10$10$

1000

9970$

1000

25000$

1000

1000,10$)()(

xPxXEX

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Back to Betting on Death So, what does this mean?

The expected value for the company is a payout, on average, of $20 per customer per year. Since each customer pays $50 Since each customer pays $50 per year, the company expects to make per year, the company expects to make a profit of $30 per customer per year.a profit of $30 per customer per year.

It’s important to note that the insurance company will never really pay anyone $20; it only pays $10,000, $5000, or $0. $20 is the expected expected averageaverage payout payout given a large number of policy holders based on the LLN.

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Labor Costs A car’s air conditioner recently needed to

be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit. What are the outcomes, random variables, and

the probability distribution?

OutcomeOutcome CostCost

xxProbabilityProbability

P(X = x)P(X = x)

Quick fix works $60 ¾ =.75

Replace unit $200 ¼ = .25

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Labor Costs A car’s air conditioner recently needed to

be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit. What is the expected value of the cost of this

repair? 955045

25020075060

$$$

.$.$)()(

xPxXEX

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Labor Costs A car’s air conditioner recently needed to

be repaired at the auto shop. The mechanic said that it could for $60 in 75% of the cases by drawing down and recharging the coolant. If that fails, it will cost an additional $140 to replace the unit.What does this mean in context of this

problem?Car owners with this problem will spend an

average of $95 to get their car fixed at this auto shop.

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Got to Love Those Aces It takes $5 to play a game

From a standard 52 card deck of cards, if you get an ace of hearts, you get $100

If you get any other ace, you get $10 If you get any other heart, you get your $5 back If you get any other card, you lose What is the expect value of this game and is it

worth it to play this game?

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Got to Love Those Aces

OutcomeOutcome X = X = PayoutPayout

Probability: P(X Probability: P(X = x)= x)

Ace of Hearts $95 1/52 = .0192

Other Aces $5 3/52 = .0577

Other Hearts $0 12/52 = .2308

Other Cards -$5 36/52 = .6923

First, you want to determine your possible winnings (let’s include the $5 cost) and probabilities:

Now, we can find the expected value, E(X):

3514630290821

6923523080057750192095

.$.$$.$.$

)(.$)(.$.$.$)()(

xPxXEX

Is this game worth playing?

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Probability Histogram

We can use histograms to display probability distributions as well as distributions of data.

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Discrete Versus ContinuousDiscrete Versus Continuous Earlier this year, we covered the notion of

discrete and continuous, but it needs more exploration now…

A A discrete random variablediscrete random variable has a countable has a countable number of outcomes. In other words, it is number of outcomes. In other words, it is possible for you to count and make a list of possible for you to count and make a list of all of the possible outcomes.all of the possible outcomes.

Discrete random variables take on only integer values. Discrete random variables take on only integer values. Suppose, for example, that we flip a coin and count the Suppose, for example, that we flip a coin and count the number of heads. The number of heads results from a number of heads. The number of heads results from a random process - flipping a coin. And the number of heads random process - flipping a coin. And the number of heads is represented by an is represented by an integerinteger value. value.

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Continuous Random VariableContinuous Random Variable Continuous random variablesContinuous random variables, in contrast,

can take on any value within a range of values. For example, suppose we flip a coin many times and compute the averageaverage number of heads per flip. The average number of heads per flip results from a random process - flipping a coin. And the average number of heads per flip can take on any value between 0 and 1, even a non-integer value. Therefore, the average number of heads per flip is a continuous random variable.

A A continuouscontinuous random variables are not countable. In random variables are not countable. In other words, you cannot list every single possible other words, you cannot list every single possible outcome. outcome. For example, the amount of water can you For example, the amount of water can you put into a 5-gallon container – there are an infinite put into a 5-gallon container – there are an infinite number of possibilities.number of possibilities.

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Example Which of the following is a discrete

random variable?I. The average height of a randomly selected

group of boys. II. The annual number of sweepstakes winners

from New York City. III. The number of presidential elections in the

20th century.

(A) I only (B) II only (C) III only (D) I and II (E) II and III

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Solution The correct answer is B. The annual

number of sweepstakes winners is an integer value and it results from a random process; so it is a discrete random variable. The average height of a group of boys could be a non-integer, so it is not a discrete variable. And the number of presidential elections in the 20th century is an integer, but it does not vary and it does not result from a random process; so it is not a random variable.

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Assignment

Chapter 16

Lesson:Random Variables

and Expected Value

Read:Chapter 16

Problems:1 – 41 (odd)