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16-1

Chapter 16 Random Variables What’s It About?

Chapter 16 introduces random variables and describes them with probability models. We look at expected values and standard deviations, and examine the effects of shifting and scaling on mean and variance. We introduce the all-important concept that when adding or subtracting independent random variables, we add their variances.

Comments

We’ve already dealt with many of these concepts back in Part I, but there we were describing data—the distribution, center, and spread. Now we turn our attention to probability models. The ideas are the same, but rather than examining actual data we examine what might happen in theory. Instead of a distribution of observed results, we have the probabilities of the possible outcomes. Instead of the mean of observed values, we have the mean of the probability model – the expected value. The standard deviation of the data measured the spread of the data. Now it measures the spread of the probability distribution. And the effects of shifting and scaling are the same: Both affect the center, but only scaling affects the spread.

The notion that we add the variances of sums or differences of independent random variables may seem complicated and counterintuitive at first (especially differences), but there’s nothing in this chapter that’s more important in terms of setting the stage for really understanding other probability models, sampling distribution models, and the otherwise scary-looking formulas for inference that lie ahead. Be sure students leave this chapter with that idea well established. Calling this the Pythagorean Theorem of Statistics helps them understand and remember it.

Looking Ahead

We discuss the properties of random variables in general here, but the ideas form the basis for much of the rest of the course. We’ll be looking at some special cases in Chapter 17: the Geometric and Binomial models. The really important applications begin to arrive in Chapter 18, when we consider models for the sampling distributions of two sample statistics: sample proportions and sample means. Those, in turn, form the basis for the inference methods found in Chapter 19 and beyond.

The all-important “variances add” mantra pays both immediate and long-term benefits:

• In Chapter 17 we find the standard deviation for a Binomial model.

• In Chapter 18 we find the standard deviation for the sampling models for sample proportions and for sample means.

• In Chapter 22 we find the standard deviation for the difference of two sample proportions.

• In Chapter 24 we find the standard deviation for the difference of two sample means.

Class Do’s

Continue to emphasize the importance of vocabulary and notation. Random variables have a probability model; actual sample data have a distribution. The model’s mean is the expected value of the variable – the long-run average – and can be written µ or E(X), but not x. The model’s

variance is Var(X) or σ 2, not s2 .

Copyright 2010 Pearson Education, Inc.

16-2 Part IV Randomness and Probability

Help students understand that the expected value may very well be impossible on any given trial. The chapter’s opening example about death and disability insurance, for instance, results in an expected value of $20. The policy could pay off $10,000, or $5000, or nothing at all. No individual policy will pay off $20. That’s just the expected average payout per policy if there are a very large number of policies. This is a good time to revisit the Law of Large Numbers.

The LLN also explains how folks who run gambling games and lotteries can make such juicy profits despite offering a multi-million dollar prize. They would be foolish to offer such a payoff if there were only a few players. Even though the probability is very low, someone might win – a large downside risk. But the huge prize generates headlines (especially when someone does win) and entices a large number of people to play. Sure, one or two may win megabucks, but the LLN ensures that so many others will lose that running the game will be a very lucrative venture.

You can make sense of the formula for the variance by looking at the formula for mean. In a sample, to “average” something we add individual values and divide by a count. For random variables the “add-and-divide” routine is simply replaced by “multiply by the probabilities and add.” We find the mean by multiplying each possible value by the corresponding probability and then summing. We do the same to find the variance, this time multiplying squared deviations from the mean by the probabilities and summing. (Ultimately, a calculator or computer will do the heavy lifting.)

Students find it difficult to tell when they’re multiplying by a constant (2X) and when they are summing separate random variables ( X

1+ X

2) . These aren’t the same. Doubling the roll of a

single die produces more variability in the result than summing the two dice. Help them Think about this issue as they attack each problem. You can read a good discussion in Peter Flanagan-Hyde’s article “Random Variables vs. Algebraic Variables” at AP Central.

Spend time with the “variances add” idea. There are three important issues:

• First, notice that it’s the variance that actually deals with deviations from the mean, though they are squared. We can add on the variability from another random variable only at this stage – not after we have taken the square root. If we want the total variability, we need to add variances. Only then can we take the square root to get our units in line with the original units (“square dollars” aren’t helpful). Students know from the Pythagorean Theorem that a right triangle with legs of 3 and 4 does not have a hypotenuse of 7. Just as Geometry’s Pythagorean Theorem is about areas, the Pythagorean Theorem of Statistics is about variances – squared deviations from the mean.

• Most confusing to students is the fact that we still add the variances even if we are interested in the difference of the random variables. Think of variance as a measure of uncertainty. In any situation, the more variability there is in the possible outcomes, the less certain we can be of the results. Every time another random outcome is introduced, another layer of uncertainty is added to the final outcome.

• All of this discussion is true only if the variables are independent. If one outcome affects what will happen next, then the uncertainty (read “variance”) is diminished, so we should not add the variances.

You can read a more in-depth analysis of these issues in “Why Variances Add – And Why It Matters” at AP Central.


Chapter 16 Random Variables 16-3

The Importance of What You Don’t Say

Don’t think that the stuff about adding variances isn’t very important. Don’t be tempted to shortchange or even skip the discussion about means and standard deviations of the sum and difference of random variables. Many of the otherwise scary-looking formulas for inference – and the all-important decisions about when they can be applied – rest on this idea. Not only that, adding variances has been the key to correct solutions on at least two of the AP’s free response questions, and is sure to be found among the multiple choice problems as well.

Class Examples

1. Play Greedy Pig. All the students stand up. You start rolling a die. Each student gets the number of points on the die. Students then may keep their points by sitting down, or remain standing for another roll. With each roll, points accumulate for those who choose to remain standing. BUT . . . if you roll a 5, all those still standing lose their points, ending the round with a score of 0. Play a couple of practice rounds to be sure they understand the game, then play 5 rounds to see who can amass the highest total score. (Play with your door closed – this gets loud . . . .) Afterward, ask what strategies they used. Some will have a points-based rule: Sit down when you have 12 points or more. Others will count rolls: Sit down after 4 rolls. Some will have a combination strategy: Stand for 6 rolls, but sit early if you get a lot of points. Tell them there is an optimal strategy, and let this sit for a couple of days. Some will play (or simulate) the game, and all will be interested to get back to it later.

After students understand expected value, they can determine the strategy by noting that if they have P points now and decide to remain standing, there is 1 chance in 6 that the next roll will give them P + 1, P + 2, P + 3, P + 4, 0, or P + 6 points. Their expected score if they remain standing is therefore:

P + 1( ) 16 + P + 2( ) 1

6 + P + 3( ) 16 + P + 4( ) 1

6 + 0 ⋅ 16 + P + 6( ) 1

6 =5P +16

6

That’s greater than their current score P if P < 16, so in the long run the optimal strategy is to sit down whenever your point total hits 16 (or more).

2. To introduce the concept of random variable and expected value, propose a game. A player pays you $5 and draws a card from a deck. If he draws the ace of hearts, you pay him $100. For any other ace, you pay $10, and for any other heart, you pay $5. If he draws anything else, he loses. See if anyone would be willing to play. Would they play for a top prize of $200? Point out that such “games” are common. Obviously those include casino games and lotteries, but sports betting, the stock market, and the entire insurance industry also work this way.

List the values of the random variable, payoff: 95, 5, 0, and –5. List the corresponding probabilities to create the probability model. Appeal to the idea that this is a model (i.e., theoretical perfection) to find the expected value (or average) outcome. In a perfect world we’d expect 52 plays of this game to yield one $100 winner, three $5 winners, twelve $0 winners, and 36 people who lose their $5. Write the average as:

E( X )=

95(1) + 5(3) + 0(12) + (−5)(36)

52= −$1.35

Discuss the meaning of the result. Obviously no one ever loses a $1.35 playing this game. Get the students to explain what it really means, and be sure to bring in the Law of Large Numbers.



Now rewrite the calculation by distributing the common denominator: 3 361 12

52 52 52 52( ) 95( ) 5( ) 0( ) ( 5)( ) $1.35E X = + + + − = −

Written in this form, it is easy to see how the expected value is computed directly from the probability model. Have students check to see if they would play for a top prize of $200.

The variance is the expected value of squared deviations from the mean, so:

( ) ( )( ) ( )

2 2 3152 52

2 2 361252 52

( ) 95 ( 1.35) ( ) 5 ( 1.35) ( )

0 ( 1.35) ( ) ( 5) ( 1.35) ( )

( ) $13.80

Var X

SD X

= − − + − −

+ − − + − − −=

Be sure students can explain why the standard deviation is so much larger than the mean in this situation.

3. Talk about combining independent random variables.

• Consider a dice game: no points for rolling a 1, 2, or 3; 5 points for a 4 or 5; 50 points for a 6. Find the expected value and the standard deviation.

Solution:

Think

I want to find the mean and expected value of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 5 points, and 6 is worth 50 points.

Let Y = the number of points.

Roll 1,2,3 4,5 6 Y 0 5 50

P(Y) 36 2

6 16

Show

3 2 1( ) 0 5 50 10

6 6 6E Yµ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 2 2 23 2 1( ) (0 10) (5 10) (50 10) 325

6 6 6

( ) ( ) 325 18.03

Var Y

SD Y Var Y

σ

σ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = − + − + − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= = = ≈

Tell

I expect to win 10 points per roll in the long run. The standard deviation is 18.03 points, suggesting a highly variable game.



• Imagine doubling the points awarded. What are the new mean and standard deviation?

Think

I want to find the mean and expected value of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 10 points, and 6 is worth 100 points. This is double the payout of the original dice game.

Show

(2 ) 2 ( ) 20E Y E Y= = 2(2 ) 2 ( ) 4(325) 1300Var Y Var Y= = = , so (2 ) 36.06SD Y =

Tell

To calculate the mean and standard deviation of a game in which the rewards are doubled, just double the mean and standard deviation of the original game. I expect to win 20 points per roll in the long run. The standard deviation is 36.06 points, suggesting a highly variable game.

• Now imagine just playing the game twice. What are the mean and the standard deviation of your total points?

Think

I want to find the mean and expected value of the total of two plays of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 10 points, and 6 is worth 100 points.

Show and Tell

To calculate the mean of the sum of two random variables, add the means.

1 2 1 2( ) ( ) ( ) 10 10 20E Y Y E Y E Y+ = + = + = .

I expect to win 20 points total for every two rolls in the long run.

To find the standard deviation of the sum of independent random variables, use the Pythagorean Theorem of Statistics: add the variances, and then take the square root.

1 2 1 2 1 2( ) ( ) ( ) ( ) 325 325 650 25.50SD Y Y Var Y Y Var Y Var Yσ = + = + = + = + = ≈

Notice that this is not the same situation as playing once for twice as much. Think about tossing a coin once—heads you win $50. That’s not the same as tossing a coin five times at $10 a head. Each outcome is a potentially different value of the random variable, and they vary independently. For independent random variables, the variances add.

• Suppose you and a friend both play the dice game. What are the mean and standard deviation of the difference in your winnings?

Think

I want to find the mean and expected value of the difference of two plays of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 10 points, and 6 is worth 100 points.

Show and Tell

To calculate the mean of the difference of two random variables, subtract the means.

1 2 1 2( ) ( ) ( ) 10 10 0E Y Y E Y E Y− = − = − = .



The difference in points won is expected to be 0 points in the long run.

To find the standard deviation of the difference of independent random variables, use the Pythagorean Theorem of Statistics: add the variances, and then take the square root.

1 2 1 2 1 2( ) ( ) ( ) ( ) 325 325 650 25.50SD Y Y Var Y Y Var Y Var Yσ = − = − = + = + = ≈

Point out that there is as much potential variability (looking at the range of values) in the difference (–50 to 50) as in the sum (0 to 100). Variances add, even when you are subtracting random values.

4. Some justifications of the “variances add” idea:

• You mix a quart of lemonade. There should be 32 fluid ounces, but your measuring process isn’t perfectly accurate so, while 32 ounces is the expected amount, there remains some variability. You pour a 12-ounce glass. Of course, that measurement is not perfect either, so it’s actually somewhere around 12, give or take a little. How much is left in the pitcher? Should be around 20 ounces, of course, but given the uncertainty about the initial amount and the variability in how far you filled the glass, you can’t say exactly how much is left. Indeed, you are less sure about the amount left because you removed an unknown amount. Subtracting some lemonade has increased the variability in the amount remaining. Variances add.

• We have two piles of fruit: grapefruit weighing from 16 to 24 ounces and oranges weighing from 9 to 13 ounces. We pick one of each at random. The sum of the weights could be as high as 24 + 13 = 37 ounces or as low as 9 + 16 = 25 ounces, a range of 12 ounces. The difference in the weights has a max of 24 – 9 = 15 ounces and a min of 16 – 13 = 3 ounces, again a range of 12 ounces. Okay, this is range, not variances, but both measure spread – and it’s the same for sums as for differences (in both cases the sum of the ranges of the two variables, 8 +4).

• Imagine a survey of the class that asked how many hours a student slept between noon yesterday and noon today. The second question is how many hours the student was awake during that period of time. We could find the mean and standard deviation for each set of answers. What are the mean and standard deviation of the total? Certainly the mean = 24 hours and standard deviation = 0. We don’t add the variances this time because the answers are not independent.

5. 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.)

Solution:

Think – We want to know the mean and standard deviation of the total time spent preparing a car. We believe that the two phases of the process are independent, so variances add.

Let M = mechanical time, A = appearance time, T = total time



Show – T = M + A

( ) ( ) ( ) 90 60 150E T E M E A= + = + = minutes.

2 2( ) ( ) ( ) 15 5 250

( ) 250 15.8 minutes

Var T Var M Var A

SD T

= + = + =

= =

Tell – We expect the total preparation time to take an average of 150 minutes (2.5 hours), with a standard deviation of 15.8 minutes.

• What are the mean and standard deviation of the total expense to prepare a car?

Solution:

Think – 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. We believe that the two phases of the process are independent, so variances add.

Let M = mechanical hours ( ) 1.5E M = ( ) 0.25SD M = A = appearance hours ( ) 1.0E A = ( ) 0.10SD A = T = total expenses

Show – 50 6T M A= + ( ) (50 6 ) (50 ) (6 ) 50 ( ) 6 ( ) 50(1.5) 6(1.0) $81E T E M A E M E A E M E A= + = + = + = + =

2 2

2 2

( ) (50 6 ) (50 ) (6 ) 50 ( ) 6 ( )

2500(.25 ) 36(0.10 ) 156.50

Var T Var M A Var M Var A Var M Var A= + = + = += + =

( ) 156.50 $12.51SD T = =

Tell – We expect the total preparation expense to average $81, with a standard deviation of $12.51.

• What are the mean and standard deviation of the difference in costs for the two phases of the operation?

Solution:

Think – We want to know the mean and standard deviation of the difference in expenses 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. We believe that the two phases of the process are independent, so variances add.

Show – 50 6D M A= − ( ) (50 6 ) (50 ) (6 ) 50 ( ) 6 ( ) 50(1.5) 6(1.0) $69E D E M A E M E A E M E A= − = − = − = − =

2 2

2 2

( ) (50 6 ) (50 ) (6 ) 50 ( ) 6 ( )

2500(.25 ) 36(0.10 ) 156.50

Var D Var M A Var M Var A Var M Var A= + = + = += + =

( ) 156.50 $12.51SD D = =

Tell – We expect the difference in preparation expenses to average $69, with a standard deviation of $12.51.



• 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.)

Solution:

Think – To calculate the probability that it will take longer to do the appearance prep than the mechanical checkup, we need to know the mean and standard deviation of the difference in time spent on each. If the difference in prep times in less than 0, then the appearance prep took longer. The two phases of the process are independent, so variances add. We will assume that each of the processes varies according to a Normal model. Let D M A= − .

Show – ( ) ( ) ( ) ( ) 90 60 30E D E M A E M E A= − = − = − = minutes. 2 2( ) ( ) ( ) ( ) 15 5 250 15.8 minutes.SD D SD M A Var M Var A= − = + = + = ≈

The difference in prep times varies according to the Normal model N(30, 15.8).

Tell – We expect the appearance prep to take less time than the mechanical checkup about 2.88% of the time.

6. Have students design their own games of chance. They need to describe the rules, the payoffs, and the cost of playing. Games could be based on cards, dice, coins, spinners, random numbers, and so on. The objective is to create a game that looks appealing, so many people would be eager to play, but with an expected value and standard deviation that would make it likely that the person running the game would realize a profit.

Resources

ActivStats

• Chapter 16—Simulations in this chapter generalize the Randomness Tool to continuous random variables. Students will use this version of the tool later to understand the Central Limit Theorem (CLT) and inference, so it is worthwhile to introduce it here.

Activity-Based Statistics

• The Lazy Student

Web Links

• “Random Variables vs. Algebraic Variables” and “Why Variances Add – And Why It Matters”, both on the Statistics page at apcentral.collegeboard.com.

Assignments

Split the reading into two sections, assigning about five problems a day as you cover the chapter in three days. Continue to assign a couple of these problems for a few extra days along with work in the next chapter to be sure students understand the concepts. Three chapter quizzes are provided.

z = y − µσ

z = 0 − 3015.8

z ≈ −1.899


16-9

AP Statistics Quiz A – Chapter 16 Name 1. A biology professor responds to some student questions by e-mail. The probability model

below describes the number of e-mails that the professor may receive from students during a day.

e-mails received 0 1 2 3 4 5 Probability 0.05 0.10 0.20 0.25 0.30 0.10

a. How many e-mails should the professor expect to receive each day?

b. What is the standard deviation?

c. If it takes the professor an average of ten minutes to respond to each e-mail, how much time should the professor expect to spend responding to student e-mails each day?

2. The American Veterinary Association claims that the annual cost of medical care for dogs

averages $100 with a standard deviation of $30, and for cats averages $120 with a standard deviation of $35. a. Find the expected value for the annual cost of medical care for a person who has one dog

and one cat.

b. Find the standard deviation for the annual cost of medical care for a person who has one dog and one cat.

c. Suppose that a couple owns four dogs. i. Find the expected value for the annual cost of medical care for the couple’s dogs.

ii. Find the standard deviation for the annual cost of medical care for the couple’s dogs.


16-10

AP Statistics Quiz A – Chapter 16 – Key 1. A biology professor responds to some student questions by e-mail. The probability model

below describes the number of e-mails that the professor may receive from students during a day.

e-mails received 0 1 2 3 4 5 Probability 0.05 0.10 0.20 0.25 0.30 0.10

a. How many e-mails should the professor expect to receive each day?

Let X = number of e-mails received.

( ) ( ) ( ) ( ) ( ) ( ) ( )0 0.05 1 0.10 2 0.20 3 0.25 4 0.30 5 0.10 2.95E X = + + + + + = e-mails per day

b. What is the standard deviation?

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2 2

2 2 2

0 2.95 0.05 1 2.95 0.10 2 2.95 0.20

3 2.95 0.25 4 2.95 0.30 5 2.95 0.10

1.7475

Var X = − + − + −

+ − + − + −=

( ) 1.7475 1.32SD X = = e-mails per day

c. If it takes the professor an average of ten minutes to respond to each e-mail, how much

time should the professor expect to spend responding to student e-mails each day? Let Y = amount of time spent responding to e-mails

1 2 10...Y X X X= + + +

( ) ( ) ( ) ( ) ( )1 2 10 1 2 10... ... 2.95 2.95 ... 2.95 29.5E Y E X X X E X E X E X= + + + = + + + = + + + =

minutes per day

2. The American Veterinary Association claims that the annual cost of medical care for dogs averages $100 with a standard deviation of $30, and for cats averages $120 with a standard deviation of $35. a. Find the expected value for the annual cost of medical care for a person who has one dog

and one cat. ( ) ( ) ( ) $100 $120 $220E D C E D E C+ = + = + =

b. Find the standard deviation for the annual cost of medical care for a person who has one

dog and one cat. ( ) ( ) ( ) 2 230 35 2125Var D C Var D Var C+ = + = + = , so ( ) 2125 $46.10SD D C+ = =

c. Suppose that a couple owns four dogs.

i. Find the expected value for the annual cost of medical care for the couple’s dogs. ( )1 2 3 4 $100 $100 $100 $100 $400E D D D D+ + + = + + + =

ii. Find the standard deviation for the annual cost of medical care for the couple’s dogs.

( ) 2 2 2 21 2 3 4 30 30 30 30 3600Var D D D D+ + + = + + + = , so ( )1 2 3 4 3600 $60SD D D D D+ + + = =


16-11

AP Statistics Quiz B – Chapter 16 Name A fast food restaurant just leased a new freezer and food fryer for three years. The service contract for the freezer offers unlimited repairs for a fee of $125 a year plus a $35 service charge for each repair needed. The restaurant’s research suggested that during a given year 80% of these freezers need no repairs, 11% needed to be serviced once, 5% twice, 4% three times, and none required more than three repairs. 1. Find the expected number of repairs this kind of freezer is expected to need each year. Show

your work.

2. Find the standard deviation of the number of repairs each year.

3. What are the mean and standard deviation of the restaurant’s annual expense for the service contract?

4. How many times should the restaurant expect to have to get this freezer repaired over the three-year term of the lease?

5. What is the standard deviation of the number of repairs that may be required during the three-year term of the lease? On what assumption does your calculation rest? Do you think this assumption is reasonable?

6. The yearly service contract for the food fryer estimates a mean annual cost of $140 with a standard deviation of $40. What is the expected value and standard deviation of the total cost for the service contracts for the freezer and the food fryer?

7. Which service contract should the restaurant expect to cost more each year? How much more? With what standard deviation?


16-12

AP Statistics Quiz B – Chapter 16 – Key A fast food restaurant just leased a new freezer and food fryer for three years. The service contract for the freezer offers unlimited repairs for a fee of $125 a year plus a $35 service charge for each repair needed. The restaurant’s research suggested that during a given year 80% of these freezers need no repairs, 11% needed to be serviced once, 5% twice, 4% three times, and none required more than three repairs. 1. Find the expected number of repairs this kind of freezer is expected to need each year. Show

your work. E(X) = 0(0.80) + 1(0.11) + 2(0.05) + 3(0.04) = 0.33 repairs


Var(X) = (0 – 0. 33)2(0.80) + (2 – 0. 33)2 (0.05) + (3 – 0. 33)2(0.04) = 0.561

Standard deviation = 0.561 = 0.749

3. What are the mean and standard deviation of the restaurant’s annual expense for the service contract?

Let C = $125 + $35X; E(C) = $125 + $35(0.33) = $136.55 Standard deviation(C) = $35(0.749) = $26.22

4. How many times should the restaurant expect to have to get this freezer repaired over the three-year term of the lease?

E(X1 + X2 + X3) = 0.33 + 0.33 + 0.33 = 0.99 repairs

5. What is the standard deviation of the number of repairs that may be required during the three-year term of the lease? On what assumption does your calculation rest? Do you think this assumption is reasonable?

Var(X1 + X2 + X3) = 0.561 + 0.561 + 0.561 = 1.683, so standard deviation(C) = 1.297 The assumption is that the number of repairs is independent from year to year. This might be incorrect because some freezers might need more service than others.

6. The yearly service contract for the food fryer estimates a mean annual cost of $140 with a standard deviation of $40. What is the expected value and standard deviation of the total cost for the service contracts for the freezer and the food fryer?

E(freezer + fryer) = $136.55 + $140 = $276.55 Var(freezer + fryer) = ($26.22)2 + ($40)2 = 2287.49, so standard deviation = $47.83

7. Which service contract should the restaurant expect to cost more each year? How much more?

With what standard deviation? The food fryer’s service contract is expected to cost more. E(fryer – freezer) = $140 - $136.55 = $3.45 more Var(fryer – freezer) = $47.83 (same as the sum in problem 6)


16-13

AP Statistics Quiz C – Chapter 16 Name

A small business just leased a new computer and color laser printer for three years. The service contract for the computer offers unlimited repairs for a fee of $100 a year plus a $25 service charge for each repair needed. The company’s research suggested that during a given year 86% of these computers needed no repairs, 9% needed to be repaired once, 4% twice, 1% three times, and none required more than three repairs.

1. Find the expected number of repairs this kind of computer is expected to need each year. Show your work.

__________


__________

3. What are the mean and standard deviation of the company’s annual expense for the service contract?

__________

__________

4. How many times should the company expect to have to get this computer repaired over the three-year term of the lease?

__________

5. What is the standard deviation of the number of repairs that may be required during the three-year lease period? On what assumption does your calculation rest? Do you think this assumption is reasonable? Explain.

__________

6. The service contract for the printer estimates a mean annual cost of $120 with standard deviation of $30. What is the expected value and standard deviation of the total cost for the service contracts on computer and printer?

__________

__________

7. Which service contract should the company expect to cost more each year? How much more? With what standard deviation?

__________

__________


16-14

AP Statistics Quiz C – Chapter 16 Name

A small business just leased a new computer and color laser printer for three years. The service contract for the computer offers unlimited repairs for a fee of $100 a year plus a $25 service charge for each repair needed. The company’s research suggested that during a given year 86% of these computers needed no repairs, 9% needed to be repaired once, 4% twice, 1% three times, and none required more than three repairs.

1. Find the expected number of repairs this kind of computer is expected to need each year. Show your work.

E(X) = 0(0.86) + 1(0.09) + 2(0.04) + 3(0.01) = 0.20 repairs


Var( X) = (0 − 0.2)2 (0.86) + (1− 0.2)2(0.09) + (2 − 0.2)2(0.04) + (3 − 0.2)2(0.01) Var(X) = 0.30 SD(X) = 0.55 repairs

3. What are the mean and standard deviation of the company’s annual expense for the service

contract?

Let C = 100 + 25X; E(C) = 100 + 25(0.20) = $105; SD(C) = 25(0.55) = $13.69 4. How many times should the company expect to have to get this computer repaired over the

three-year term of the lease?

1 2 3( ) 0.20 0.20 0.20 0.60E X X X+ + = + + = repairs.

5. What is the standard deviation of the number of repairs that may be required during the three-

year lease period? On what assumption does your calculation rest? Do you think this assumption is reasonable? Explain.

We must assume that the number of repairs is independent from year to year. This may be risky as some computers may be more trouble-prone than others.

1 2 3( ) 0.30 0.30 0.30 0.90Var X X X+ + = + + = , so SD(X) = 0.95 repairs.

6. The service contract for the printer estimates a mean annual cost of $120 with standard

deviation of $30. What is the expected value and standard deviation of the total cost for the service contracts on computer and printer?

E(C + P) = 105 + 120 = $225; 2 2( ) (13.69) 30 1087Var C P+ = + = , so SD(C + P) = $32.98

7. Which service contract should the company expect to cost more each year? How much more?

With what standard deviation?

The printer; E(P – C) = 120 - 105 = $15 more; SD(P - C) = $32.98 (same as the sum)


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