4. Discrete Probability Distributions

Transcription

1 4. Discrete Probabilit Distributions 4.. Random Variables and Their Probabilit Distributions Most of the exeriments we encounter generate outcomes that can be interreted in terms of real numbers, such as heights of children, numbers of voters favoring various candidates, tensile strength of wires, and numbers of accidents at secified intersections. These numerical outcomes, whose values can change from exeriment to exeriment, are called random variables. We will look at an illustrative examle of a random variable before we attemt a more formal definition. A section of an electrical circuit has two relas, numbered and, oerating in arallel. The current will flow when a switch is thrown if either one or both of the relas close. The robabilit that a rela will close roerl is 0.8, and the robabilit is the same for each rela. The relas oerate indeendentl, we assume. Let E i denote the event that rela i closes roerl when the switch is thrown. Then P(E i ) 0.8. When the switch is thrown, a numerical outcome of some interest to the oerator of this sstem is, the number of relas that close roerl. Now, can take on onl three ossible values, because the number of relas that close must be 0,, or. We can find the robabilities associated with these values of b relating them to the underling events E i. Thus, we have P( 0) P( EE ) P( E) P( E). 0.(0.) 0.04 because 0 means that neither rela closes and the relas oerate indeendentl. Similarl, P( ) P( E E E E ) and P( E E P( E ) P( E 0.8(0.) 0,(0.8) 0.3 P( ) P( E E ) P( E E 0.8(0.8) ) ) P( E ) P( E P( E ) P( E 0.64 The values of, along with their robabilities, are more useful for keeing track of the oeration of this sstem than are the underling events E i, because the number of roerl closing relas is the ke to whether the sstem will work. The current will flow if is equal to at least, and this event has robabilit ) ) )

2 P( ) P( P( ) P( ) 0.96 Notice that we have maed the outcomes of an exeriment into a set of three meaningful real numbers and have attached a robabilit to each. Such situations rovide the motivation for Definitions 4. and 4.. or ) Definition 4.. A random variable is a real-valued function whose domain is a samle sace. Random variables will be denoted b uer-case letters, usuall toward the end of the alhabet, such as, Y, and Z. The actual values that random variables can assume will be denoted b lower-case letters, such as x,, and z. We can then talk about the robabilit that takes on the value x, P( x), which is denoted b (x). In the rela examle, the random variable has onl three ossible values, and it is a relativel simle matter to assign robabilities to these values. Such a random variable is called discrete. Definition 4.. A random variable is said to be discrete if it can take on onl a finite number or a countabl infinite number of ossible values x. The robabilit function of, denoted b (x), assigns robabilit to each value x of so that following conditions are satisfied:. P ( x) ( x) 0.. P ( x), where the sum is over all ossible values of x. x The robabilit function is sometimes called the robabilit mass function of, to denote the idea that a mass of robabilit is associated with values for discrete oints. It is often convenient to list the robabilities for a discrete random variable in a table. With defined as the number of closed relas in the roblem just discussed, the table is as follows: x (x) Total.00 This listing constitutes one wa of reresenting the robabilit distribution of. Notice that the robabilit function (x) satisfies two roerties:. 0 ( x) for an x.

3 3. ( x), where the sum is over all ossible values of x. x In general, a function is a robabilit function if and onl if the above two conditions are satisfied. Bar grahs are used to disla the robabilit functions for discrete random variables. The robabilit distribution of the number of closed relas discussed above is shown in Figure 4.. Figure 4.. Grah of a robabilit mass function Functional forms for some robabilit functions that have been useful for modeling real-life data will be given in later sections. We now illustrate another method for arriving at a tabular resentation of a discrete robabilit distribution. Examle 4.: A local video store eriodicall uts its used movies in a bin and offers to sell them to customers at a reduced rice. Twelve coies of a oular movie have just been added to the bin, but three of these are defective. A customer randoml selects two of the coies for gifts. Let be the number of defective movies the customer urchased. Find the robabilit function for and grah the function. Solution: The exeriment consists of two selections, each of which can result in one of two outcomes. Let D i denote the event that the ith movie selected is defective; thus, D i denotes the event that it is good. The robabilit of selecting two good movies ( 0) is

4 4 P ( D ) D P( D on st) P( D on nd D on st) The multilicative law of robabilit is used, and the robabilit for the second selection deends on what haened on the first selection. Other ossibilities for outcomes will result in other values of. These outcomes are convenientl listed on the tree in Figure 3.. The robabilities for the various selections are given on the branches of the tree. Figure 4.. Figure 4.. Outcomes for Examle 4. Clearl, has three ossible outcomes, with robabilities as follows: x (x) Total.00 The robabilities are grahed in the figure below.

5 5 Tr to envision this concet extended to more selections from bins of various structures. We sometimes stud the behavior of random variables b looking at the cumulative robabilities; that is, for an random variable, we ma look at P( b) for an real number b. This is, the cumulative robabilit for evaluated at b. Thus, we can define a function F(b) as F(b) P( b). Definition 4.3. The distribution function F(b) for a random variable is If is discrete, F(b) P( b) F ( b) ( x) b x where (x) is the robabilit function. The distribution function is often called the cumulative distribution function (c.d.f). The random variable, denoting the number of relas that close roerl (as defined at the beginning of this section), has a robabilit distribution given b P( 0) 0.04 P( ) 0.3 P( ) 0.64

6 6 Because ositive robabilit is associated onl for x 0,, or, the distribution function changes values onl at those oints. For values of b at least, but less than, the P( b) P( ). For examle, we can see that P(.5) P(.9) P( ) 0.36 The distribution function for this random variable then has the form 0, x < , 0 x < F ( x) 0.36, x <, x The function is grahed in Figure 4.3. Figure 4.3. Distribution Function Notice that the distribution function is a ste function and is defined for all real numbers. This is true for all discrete random variables. The distribution function is discontinuous at oints of ositive robabilit. Because the outcomes 0,, and have ositive robabilit associated with them, the distribution function is discontinuous at those oints. The change in the value of the function at a oint (the height of the ste) is the robabilit associated with that value x. Since the outcome of is the most robable (() 0.64), the height of the ste at this oint is the largest. Although the function has oints of discontinuit, it is right-hand continuous at all oints. To see this, consider. As we aroach from the left, we have lim F( h) 0.36 F() ; that is, the distribution h 0 function F is right-hand continuous. However, if we aroach from the left, we find lim F( h) F(), giving us that F is not left-hand continuous. Because h 0 a function must be both left- and right-hand continuous to be continuous, F is not continuous at. In general, a distribution function is defined for the whole real line. Ever distribution function must satisf four roerties; similarl, an function satisfing the following four roerties is a distribution function.

7 7. lim F( x) 0 x. lim F( x) x 3. The distribution function is a non-decreasing function; that is, if a < b, F(a) F(b). The distribution function can remain constant, but it cannot decrease, as we increase from a to b. 4. The distribution function is right-hand continuous; that is, lim F( x h) F( x) We have alread seen that, given a robabilit mass function, we can determine the distribution function. For an distribution function, we can also determine the robabilit function. h 0 Examle 4.: A large universit uses some of the student fees to offer free use of its Health Center to all students. Let be the number of times that a randoml selected student visits the Health Center during a semester. Based on historical data, the distribution function of is given below. 0, x < 0 0.6, 0 x < F ( x) 0.8, x < 0.95, x < 3, x 3 For the function above,. Grah F.. Verif that F is a distribution function. 3. Find the robabilit function associated with F. Solution:.

8 8. To verif F is a distribution function, we must confirm that the function satisfies the four conditions of a distribution function. () Because F is zero for all values x less than 0, lim F( x) 0. x () Similarl F is one for all values of x that are 3 or greater; therefore, lim F( x). x (3) The function F is non-decreasing. There are man oints for which it is not increasing, but as x increases, F(x) either remains constant or increases. (4) The function is discontinuous at four oints: 0,,, and 3. At each of these oints, F is right-hand continuous. As an examle, for, lim F( h) 0.95 F(). h 0 Because F satisfies the four conditions, it is a distribution function. 3. The oints of ositive robabilit occur at the oints of discontinuit: 0,,, and 3. Further, the robabilit is the height of the jum at that oint. This gives us the following robabilities. x (x) Exercises 4.. Circuit boards from two assembl lines set u to roduce identical boards are mixed in one storage tra. As insectors examine the boards, the find that it is difficult to determine whether a board comes from line A or line B. A robabilistic assessment of this question is often helful. Suose that the storage tra contains ten circuit boards, of which six came from line A and four from line B. An insector selects two of these identical-looking boards for insection. He is interested in, the number of insected boards from line A. a. Find the robabilit function for. b. Grah the robabilit function of. c. Find the distribution function of. d. Grah the distribution function of. 4.. Among twelve alicants for an oen osition, seven are women and five are men. Suose that three alicants are randoml selected from the alicant ool for final interviews. Let be the number of female alicants among the final three. a. Find the robabilit function for. b. Grah the robabilit function of. c. Find the distribution function of. d. Grah the distribution function of.

9 The median annual income for heads of households in a certain cit is $44,000. Four such heads of household are randoml selected for inclusion in an oinion oll. Let be the number (out of the four) who have annual incomes below $44,000. a. Find the robabilit distribution of. b. Grah the robabilit distribution of. c. Find the distribution function of. d. Grah the distribution function of. e. Is it unusual to see all four below $44,000 in this te of oll? (What is the robabilit of this event?) 4.4. At a miniature golf course, laers record the strokes required to make each hole. If the ball is not in the hole after five strokes, the laer is to ick u the ball and record six strokes. The owner is concerned about the flow of laers at hole 7. (She thinks that laers tend to get backed u at that hole.). She has determined that the distribution function of, the number of strokes a laer takes to comlete hole 7 to be 0, x < 0.05, x < 0.5, x < 3 F ( x) 0.35, 3 x < , 4 x < , 5 x < 6, x 6 a. Grah the distribution function of. b. Find the robabilit function of. c. Grah the robabilit function of. d. Based on (a) through (c), are the owner s concerns substantiated? 4.5. In 005, Derrek Lee led the National Baseball League with a batting average, meaning that he got a hit on 33.5% of his official times at bat. In a tical game, he had three official at bats. a. Find the robabilit distribution for, the number of hits Boggs got in a tical game. b. What assumtions are involved in the answer? Are the assumtions reasonable? c. Is it unusual for a good hitter to go 0 for 3 in one game? 4.6. A commercial building has three entrances, numbered I, II, and III. Four eole enter the building at 9:00 a.m. Let denote the number who select entrance I. Assuming that the eole choose entrances indeendentl and at random, find the robabilit distribution for. Were an additional assumtions necessar for our answer?

10 In 00, 33.9% of all fires were structure fires. Of these, 78% of these were residential fires. The causes of structure fire and the numbers of fires during 00 for each cause are dislaed in the table below. Suose that four indeendent structure fires are reorted in one da, and let denote the number, out of the four, that are caused b cooking. Cause of Fire Number of Fires Cooking 9,706 Chimne Fires 8,638 Incinerator 84 Fuel Burner 3,6 Commercial Comactor 46 Trash/Rubbish 9,906 a. Find the robabilit distribution for, in tabular form. b. Find the robabilit that at least one of the four fires was caused b cooking Observers have noticed that the distribution function of, the number of commercial vehicles that cross a certain toll bridge during a minute is as follows: 0, x < 0 0.0, 0 x < F ( x) 0.50, x < 0.85, x < 4, x 4 a. Grah the distribution function of. b. Find the robabilit function of. c. Grah the robabilit function of Of the eole who enter a blood bank to donate blood, in 3 have te O blood, and in 0 have te O - blood. For the next three eole entering the blood bank, let denote the number with O blood, and let Y denote the number with O - blood. Assume the indeendence among the eole with resect to blood te. a. Find the robabilit distribution for and Y. b. Find the robabilit distribution of Y, the number of eole with te O blood Dail sales records for a car dealershi show that it will sell 0,,, or 3 cars, with robabilities as listed: Number of Sales 0 3 Probabilit

11 a. Find the robabilit distribution for, the number of sales in a two-da eriod, assuming the sales are indeendent from da to da. b. Find the robabilit that at least one sale is made in the next two das. 4.. Four microchis are to be laced in a comuter. Two of the four chis are randoml selected for insection before the comuter is assembled. Let denote the number of defective chis found among the two insected. Find the robabilit distribution for for the following events. a. Two of the microchis were defective. b. One of the four microchis was defective. c. None of the microchis was defective. 4.. When turned on, each of the three switches in the accomaning diagram works roerl with robabilit 0.9. If a switch is working roerl, current can flow through it when it is turned on. Find the robabilit distribution for Y, the number of closed aths from a to b, when all three switches are turned on. 4. Exected Values of Random Variables Because a robabilit can be thought of as the long-run relative frequenc of occurrence for an event, a robabilit distribution can be interreted as showing the long-run relative frequenc of occurrences for numerical outcomes associated with an exeriment. Suose, for examle, that ou and a friend are matching balanced coins. Each of ou flis a coin. If the uer faces match, ou win $.00; if the do not match, ou lose $.00 (our friend wins $.00). The robabilit of a match is 0.5 and, in the long run, ou should win about half of the time. Thus, a relative frequenc distribution of our winnings should look like the one shown in Figure 4.4. The - under the left most bar indicates a loss of $.00 b ou. Figure 4.4. Relative frequenc of winnings

12 On average, how much will ou win er game over the long run? If Figure 4.4 resents a correct disla of our winnings, ou win - (lose a dollar) half of the time and half of the time, for an average of ( ) () 0 This average is sometimes called our exected winnings er game, or the exected value of our winnings. (A game that has an exected value of winnings of 0 is called a fair game.) The general definition of exected value is given in Definition 4.4. Definition 4.4. The exected value of a discrete random variable with robabilit distribution (x) is given b E ( ) x( x) x (The sum is over all values of x for which (x) > 0.) We sometimes use the notation E() μ for this equivalence. Note: We assume absolute convergence when the range of is countable; we talk about an exectation onl when it is assumed to exist... Now ada has arrived, and ou and our friend u the stakes to $0 er game of matching coins. You now win -0 or 0 with equal robabilit. Your exected winnings er game is ( 0) (0) 0 and the game is still fair. The new stakes can be thought of as a function of the old in the sense that, if reresents our winnings er game when ou were laing for $.00, then 0 reresents our winnings er game when ou la for $0.00. Such functions of random variables arise often. The extension of the definition of exected value to cover these cases is given in Theorem 4.. Theorem 4.. If is a discrete random variable with robabilit distribution (x) and if g(x) is an real-valued function of, then E ( g( )) g( x) ( x) x (The roof of this theorem will not be given.) You and our friend decide to comlicate the aoff rules to the coin-matching game b agreeing to let ou win $ if the match is tails and $ if the match is heads. You

13 3 still lose $ if the coins do not match. Quickl ou see that this is not a fair game, because our exected winnings are ( ) () () You comensate for this b agreeing to a our friend $.50 if the coins do not match. Then, our exected winnings er game are (.5) () () and the game is again fair. What is the difference between this game and the original one, in which all aoffs were $? The difference certainl cannot be exlained b the exected value, since both games are fair. You can win more, but also lose more, with the new aoffs, and the difference between the two games can be exlained to some extent b the increased variabilit of our winnings across man games. This increased variabilit can be seen in Figure 4.5, which dislas the relative frequenc for our winnings in the new game; the winnings are more sread out than the were in Figure 4.4. Formall, variation is often measured b the variance and b a related quantit called the standard deviation. Figure 4.5. Relative frequenc of winnings Definition 4.5. The variance of a random variable with exected value μ is given b V ( ) E[ ( μ) ]. We sometimes use the notation for this equivalence. E [( μ ) ] σ Notice that the variance can be thought of as the average squared distance between values of and the exected value μ. Thus, the units associated with σ are the square of the units of measurement for. The smallest value that σ can assume is zero. The variance is zero when all the robabilit is concentrated at a single oint, that is, when takes on a constant value with robabilit. The variance becomes larger as the oints with ositive robabilit sread out more.

14 4 The standard deviation is a measure of variation that maintains the original units of measure, as oosed to the squared units associated with the variance. Definition 4.6. The standard deviation of a random variable is the square root of the variance and is given b σ σ E [( μ) ] is For the game reresented in Figure 4.4, the variance of ou winnings (with μ 0) σ [( ) ] E μ ( ) It follows that σ, as well. For the game reresented in Figure 4.5, the variance of our winnings is σ (.5) and the standard deviation is σ σ Which game would ou rather la? The standard deviation can be thought of as the size of a tical deviation between an observed outcome and the exected value. For the situation dislaed in Figure 4.4, each outcome (- or ) deviates b recisel one standard deviation from the exected value. For the situation described in Figure 4.5, the ositive values average.5 units from the exected value of 0 (as do the negative values), and so.5 units is aroximatel one standard deviation here. The mean and the standard deviation often ield a useful summar of the robabilit distribution for a random variable that can assume man values. An illustration is rovided b the age distribution of the U.S. oulation for 000 and 00 (rojected, as shown in Table 4.). Age is actuall a continuous measurement, but since it is reorted in categories, we can treat it as a discrete random variable for uroses of aroximating its ke function. To move from continuous age intervals to discrete age classes, we assign each interval the value of its midoint (rounded). Thus, the data in Table 4. are interreted as showing that 6.9% of the 000 oulation were around 3 ears of age and that.6% of the 00 oulation is anticiated to be around 45 ears of age. (The oen intervals at the uer end were stoed at 00 for convenience.)

15 5 Table 4.. Age Distribution in 000 and 00 (Projected) Age Interval Age Midoint Under and over *Source: U.S. Census Bureau Interreting the ercentages as robabilities, we see that the mean age for 000 is aroximated b μ x( x) x 3(0.069) 8(0.073) 5(0.44)... 90(0.033) 36.6 (How does this comare with the median age for 000, as aroximated from Table 4..) For 00, the mean age is aroximated b μ x( x) x 3(0.06) 8(0.06) 5(0.8)... 90(0.099) 4.5 Over the rojected eriod, the mean age increases rather markedl (as does the median age). The variations in the two age distributions can be aroximated b the standard deviations. For 000, this is σ ( x μ) ( x) x (3 36.6) (0.069) (8 36.6).6 (0.073) (5 36.6) (0.44)... ( ) (0.033) A similar calculation for the 00 data ields σ 6.3. These results are summarized in Table 4..

16 6 Table 4.. Age Distribution of U.S. Poulation Summar Mean Standard Deviation Not onl is the oulation getting older, on average, but its variabilit is increasing. What are some of the imlications of these trends? We now rovide other examles and extensions of these basic results. Examle 4.3: A deartment suervisor is considering urchasing a hotoco machine. One consideration is how often the machine will need reairs. Let denote the number of reairs during a ear. Based on ast erformance, the distribution of is shown below: Number of reairs, x 0 3 (x) What is the exected number of reairs during a ear?. What is the variance of the number of reairs during a ear? Solution:. From Definition 4.4, we see that E( ) x( x) x 0(0.) (0.3) (0.4) 3(0.).4 The hotocoier will need to be reaired an average of.4 times er ear.. From Definition 4.5, we see that V ( ) ( x μ) ( x) x (0.4) (0.) (.4) (0.3) (.4) (0.4) (3.4) (0.) 0.84 Our work in maniulating exected values can be greatl facilitated b making use of the two results of Theorem 4.. Often, g() is a linear function. When that is the case, the calculations of exected value and variance are eseciall simle.

17 7 Theorem 4.. For an random variable and constants a and b,. E ( a b) ae( ) b. F ( a b) ae( ) b Proof: We sketch a roof of this theorem for a discrete random variable having a robabilit distribution given b (x). B Theorem 4., Notice that x E( a b) x x x a ( ax b) ( x) [( ax) ( x) b( x)] ax( x) x x( x) b ae( ) b x b( x) x ( x) ( x) must equal unit. Also, b Definition 4.5, V ( a b) E[( a b) E( a b)] E[ a b ( ae( ) b)] E[ a ae( )] E a a [ ( x E( )) ] E[ ( E( )) ] a V ( ) An imortant secial case of Theorem 4. involves establishing a standardized variable. If has mean μ and standard deviation σ, then the standardized form of is given b μ Y σ Emloing Theorem 4., one can show that E(Y) 0 and V(Y ). This idea will be used often in later chaters. We illustrate the use of these results in the following examle. Examle 4.4: The deartment suervisor in Examle 4.3 wants to consider the cost of maintenance before urchasing the hotoco machine. The cost of maintenance consists of the

18 8 exense of a service agreement and the cost of reairs. The service agreement can be urchased for $00. With the agreement, the cost of each reair is $50. Find the mean and variance of the annual costs of reair for the hotocoier Solution: Recall that the of Examle 4.3 is the annual number of reairs. The annual cost of the maintenance contract is B Theorem 4., we have E(50 00) 50E( ) 00 50(.4) Thus, the manager could anticiate the average annual cost of maintenance of the hotocoier to be $70. Also, b Theorem 4., V (50 00) 50 V ( ) We will make use of this value in a later examle. (0.84) Determining the variance b Definition 4.5 is not comutationall efficient. Theorem 4. leads us to a more efficient formula for comuting the variance as given in Theorem 4.3. Theorem 4.3. If is a random variable with mean μ, then ( ) V ( ) E μ Proof: Starting with the definition of variance, we have V ( ) E E E E E [( μ) ] ( μ μ ) ( ) E( μ) E( μ ) ( ) μ μ ( ) μ

19 9 Examle 4.5: Use the result of Theorem 4.3 to comute the variance of as given in Examle 4.3. Solution: In Examle 4.3, had a robabilit distribution given b and we found that E().4. Now, E( x 0 3 (x) ) x ( x) x 0 (0.) (0.3) (0.4) 3 (0.).8 B Theorem 4.3, V ( ) E( ) μ.8 (.4) 0.84 We have comuted means and variances for a number of robabilit distributions and noted that these two quantities give us some useful information on the center and sread of the robabilit mass. Now suose that we know onl the mean and the variance for a robabilit distribution. Can we sa anthing secific about robabilities for certain intervals about the mean? The answer is es, and a useful result of the relationshi among mean, standard deviation, and relative frequenc will now be discussed. The inequalit in the statement of the theorem is equivalent to P( μ kσ < < μ kσ ) k To interret this result, let k, for examle. Then the interval from μ - σ to μ σ must contain at least /k ¼ 3/4 of the robabilit mass for the random variable. We consider more secific illustrations in the following two examles.

20 0 Examle 4.6: The dail roduction of electric motors at a certain factor averaged 0, with a standard deviation of 0.. What can be said about the fraction of das on which the roduction level falls between 00 and 40?. Find the shortest interval certain to contain at least 90% of the dail roduction levels. Theorem 4.4: Tchebsheff s Theorem. Let be a random variable with mean μ and variance σ. Then for an ositive k, ) ( k k P < σ μ Proof: We begin with the definition of V() and then make substitutions in the sum defining this quantit. Now, σ μ σ μ σ μ σ μ μ μ μ μ σ k k k k x x x x x x x x V ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( (The first sum stos at the largest value of x smaller than μ kσ, and the third sum begins at the smallest value of x larger than μ kσ; the middle sum collects the remaining terms.) Observe that the middle sum is never negative; and for both of the outside sums, ) ( σ μ k x Eliminating the middle sum and substituting for ( μ) x in the other two, we get σ μ σ μ σ σ σ k k x k x k ) ( ) ( or σ μ σ μ σ σ k k x x k ) ( ) ( or ( ) σ μ σ σ k P k. It follows that ( ) k k P σ μ or ( ) k k P < σ μ

21 Solution:. The interval from 00 to 40 reresents μ - σ to μ σ, with μ 0 and σ 0. Thus, k and 3 k 4 4 At least 75% of all das, therefore, will have a total roduction value that falls in this interval. (This ercentage could be closer to 95% if the dail roduction figures show a mound-shaed, smmetric relative frequenc distribution.). To find k, we must set ( k ) equal to 0.9 and solve for k; that is, 0.9 k 0. k k 0 k 0 The interval or or 3.6 μ 3.6σ to μ3.6σ 0 3.6(0) to 0 3.6(0) 88.4 to 5.6 will then contain at least 90% of the dail roduction levels. Examle 4.7: The annual cost of maintenance for a certain hotoco machine has a mean of $70 and a variance of $00 (see Examle 4.5). The manager wants to budget enough for maintenance that he is unlikel to go over the budgeted amount. He is considering budgeting $400 for maintenance. How often will the maintenance cost exceed this amount? Solution: First, we must find the distance between the mean and 400, in terms of the standard deviation of the distribution of costs. We have

22 400 μ σ Thus, 400 is.84 standard deviations aboe the mean. Letting k.84 in Theorem 4.4, we can find the interval μ.84σ to μ.84σ or 70.84(45.8) to 70.84(45.8) or 40 to 400 must contain at least k (.84) of the robabilit. Because the annual cost is $00 lus $50 for each reair, the annual cost cannot be less than $00. Thus, at most 0. of the robabilit mass can exceed $400; that is, the cost cannot exceed $400 more than % of the time. Examle 4.8: Suose the random variable has the robabilit mass function given in the table below. Evaluate Tchebsheff s inequalit for k. Solution: First, we find the mean of Then and x - 0 (x) /8 3/4 /8 μ x ( x ) ( )( / 8) 0(3 / 4) ( / 8) 0 x E( ) x ( x) ( ) (/ 8) 0 (3/ 4) (/ 8) / 4 Thus, the standard deviation of is x σ E ( ) μ σ σ 4 Now, for, the robabilit is within σ of μ is 4 0 4

23 3 P( μ < σ ) P( μ < ( )) P( μ < ) P( 3 4 0) B Tchebsheff s theorem, the robabilit an random variable is within σ of μ is P( μ < σ ) k Therefore, for this articular random variable, equalit holds in Tchebsheff s theorem. Thus, one cannot imrove on the bounds of the theorem. 3 4 Exercises 4.3. You are to a $.99 to la a game that consists of drawing one ticket at random from a box of unnumbered tickets. You win the amount (in dollars) of the number on the ticket ou draw. The following two boxes of numbered tickets are available. 0,, 0, 0, 0,, 4 I. II. a. Find the exected value and variance of our net gain er la with box I. b. Reeat art (a) for box II. c. Given that ou have decided to la, which box would ou choose, and wh? 4.4. The size distribution of U.S. families is shown in the table below. Number of Persons Percentage 5.7% or more. a. Calculate the mean and the standard deviation of famil size. Are these exact values or aroximations? b. How does the mean famil size comare to the median famil size? 4.5. The table below shows the estimated number of AIDS cases in the United States b age grou.