Chapter 6 ATE: Random Variables Alternate Examples and Activities


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1 Probability Chapter 6 ATE: Random Variables Alternate Examples and Activities [Page 343] Alternate Example: NHL Goals In 2010, there were 1319 games played in the National Hockey League s regular season. Imagine selecting one of these games at random and then randomly selecting one of the two teams that played in the game. Define the random variable = number of goals scored by a randomly selected team in a randomly selected game. The table below gives the probability distribution of : Goals: Probability: Problem: (a) Show that the probability distribution for is legitimate. (b) Make a histogram of the probability distribution. Describe what you see. (c) What is the probability that the number of goals scored by a randomly selected team in a randomly selected game is at least 6? (a) All of the probabilities are between 0 and 1 and they add to 1, so this is a legitimate probability distribution. (b) The histogram is skewed to the right, which means that the majority of games are relatively low scoring. It is pretty unusual for a team to score 6 or more goals (c) P( 6) = P( = 6) + P( = 7) + P( = 8) + P( = 9) = = Goals 6 8
2 [Page 344] Alternate Example: More Roulette Another wager players can make in Roulette is called a corner bet. To make this bet, a player places his chips on the intersection of four numbered squares on the Roulette table. If one of these numbers comes up on the wheel and the player bet $1, the player gets his $1 back plus $8 more. Otherwise, the casino keeps the original $1 bet. If = net gain from a single $1 corner bet, the possible outcomes are = 1 or = 8. Here is the probability distribution of : Value: $1 $8 Probability: 34/38 4/38 What is the player s average gain? In the long run, the player loses $1 in 34 of every 38 games and gains $8 in 4 of every 38 games. Imagine a hypothetical 38 bets. The player s average gain is: ( 1) 4(8) ( 1) (8) $0.05 If a player were to make $1 corner bets many, many times, the average gain would be about $0.05 per bet. In other words, in the long run, the casino keeps about 5 cents of every dollar bet in roulette. Note: All of the bets in Roulette have the same expected value! However, the bets have different standard deviations, with the red/black bet having the smallest standard deviation and the straightup bet on a single number having the largest standard deviation. [Page 346] Alternate Example: NHL Goals In an earlier alternate example, we defined the random variable to be the number of goals scored by a randomly selected team in a randomly selected game. The table below gives the probability distribution of : Goals: Probability: Problem: Compute the mean of the random variable and interpret this value in context. E ( ) (0)(0.061) (1)(0.154) (9)(0.001) = The mean number of goals for a randomly selected team in a randomly selected game is If you were to repeat the random sampling process over and over again, the mean number of goals scored would be about in the long run.
3 [Page 347] Alternate Example: NHL Goals In the last alternate example, we calculated the mean number of goals for a randomly selected team in a randomly selected game to be = The table below gives the probability distribution for one more time. Goals: Probability: Problem: Compute and interpret the standard deviation of the random variable ( ) (0.061) ( ) (0.154) ( ) (0.001) The standard deviation of is On average, a randomly selected team s number of goals in a randomly selected game will differ from the mean by about 1.63 goals. [Page 349] Alternate Example: Random Normal Many calculators and statistical software packages can generate random numbers from a Normal distribution. For example, on the TI84, pressing RandNorm(0,1) over and over will generate a distribution that can be represented by the standard Normal curve. If Z = the result when RandNorm(0,1) is pressed, the probability that Z is between 1 and 1 is about This is shown in the figure below, where the area under the curve between 1 and 1 is about [Page 351] Alternate example: Weights of ThreeYearOld Females The weights of threeyearold females closely follow a Normal distribution with a mean of = 30.7 pounds and a standard deviation of 3.6 pounds. Randomly choose one threeyearold female and call her weight. Find the probability that the randomly selected threeyearold female weighs at least 30 pounds. State: What s the probability that a randomly selected threeyearold female weighs at least 30 pounds? Plan: The weight of the threeyearold female we choose has the N(30.7, 3.6) distribution. We want to find P( 30). We will standardize this weight and use table A to find the shaded area Do: The standardized value is: z = From Table A, P(Z < 0.19) = , so 3.6 P(Z 0.19) = =
4 Probability Conclude: There is about a 58% chance that the randomly selected threeyearold female will weigh at least 30 pounds. [Page 359] Alternate Example: El Dorado Community College El Dorado Community College considers a student to be fulltime if he or she is taking between 12 and 18 units. The number of units that a randomly selected El Dorado Community College fulltime student is taking in the fall semester has the following distribution. Number of Units () Probability Here is a histogram of the probability distribution along with the mean and standard deviation: Number of Units (0.25) 13(0.10) 18(0.15) ( ) (0.25) ( ) (0.10) ( ) (0.15) At El Dorado Community College, the tuition for fulltime students is $50 per unit. That is, if T = tuition charge for a randomly selected fulltime student, T = 50. Here is the probability distribution for T and a histogram of the probability distribution: Tuition Charge (T) Probability
5 Probability Probability Tuition Charge The mean and standard deviation of this distribution are: 600(0.25) 650(0.10) (900)(0.15) $ T ( ) (0.25) ( ) (0.10) ( ) (0.15) T $103 T Notice that the shapes of both distributions are the same the mean of the distribution of T is 50 times bigger than the mean of : = 50(14.65) the standard deviation of T is 50 times bigger than the standard deviation of : 103 = 50(2.06) [Page 360] Alternate Example: El Dorado Community College In addition to tuition charges, each fulltime student at El Dorado Community College is assessed student fees of $100 per semester. If C = overall cost for a randomly selected fulltime student, C = T. Here is the probability distribution for C and a histogram of the probability distribution: Overall Cost (C) Probability Overall Cost
6 The mean and standard deviation of this distribution are: 700(0.25) 750(0.10) (1000)(0.15) $ C ( ) (0.25) ( ) (0.10) ( ) (0.15) C $103 C Notice that the shapes of both distributions are the same the mean of the distribution of C is $100 larger than the mean of T: = the standard deviation of C is the same as the standard deviation of T [Page 363] Alternate Example: Scaling a Test Problem: In a large introductory statistics class, the distribution of = raw scores on a test was approximately normally distributed with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10. (a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the mean and standard deviation of Y. (b) What is the probability that a randomly selected student has a scaled test score of at least 90? (a) Since Y = , (17.2) 78.8 Y Y 4 4(3.8) 15.2 (b) Since linear transformations do not change the shape, Y has the N(78.8, 15.2) distribution The standardized score for a scaled score of 90 is z According to Table A, 15.2 P(z < 0.74) = Thus, P(Y 90) = = [Page 364] Alternate Example: Speed Dating To save time and money, many single people have decided to try speed dating. At a speed dating event, women sit in a circle and men spend about 5 minutes getting to know a woman before moving on to the next one. What is the probability that the man is taller than the woman in a randomly selected speed dating couple? To answer this question, we need to know about the distribution of men s heights M, the distribution of women s heights W, and what happens when we combine these distributions.
7 [Page 364] Alternate Example: El Dorado Community College El Dorado Community College also has a campus downtown, specializing in just a few fields of study. Fulltime students at the downtown campus only take 3unit classes. Let Y = number of units taken in the fall semester by a randomly selected fulltime student at the downtown campus. Here is the probability distribution of Y: Number of Units (Y) Probability The mean of this distribution is Y = 15 units, the variance is Y = 5.40 units 2 and the standard deviation is Y = 2.3 units. If you were to randomly select 1 fulltime student from the main campus and 1 fulltime student from the downtown campus and add their number of units, the expected value of the sum (S = + Y) would be: S Y [Page 366] Alternate Example: El Dorado Community College Let S = + Y as before. Assume that and Y are independent, which is reasonable since each student was selected at random. Here are all the possible combinations of and Y: P() Y P(Y) S=+Y P(S)=P()P(Y)
8 Here is the probability distribution of S: S P(S) (0.075) 25(0.03) 36(0.045) S ( ) (0.075) ( ) (0.03) ( ) (0.045) S Notice that = Y (29.65 = ) and that (9.63 = ). S S Y [Page 368] Alternate Example: Tuition, Fees, and Books Let B = the amount spent on books in the fall semester for a randomly selected fulltime student at El Dorado Community College. Suppose that B 153and B 32. Recall from earlier that C = overall cost for tuition and fees for a randomly selected fulltime student at El Dorado Community College and C = and C = 103. Problem: Find the mean and standard deviation of the cost of tuition, fees, and books (C + B) for a randomly selected fulltime student at El Dorado Community College. The mean is C B C B = = $ The standard deviation cannot be calculated since the cost for tuition and fees and the cost for books are not independent. Students who take more units will typically have to buy more books. [Page 368] Alternate Example: El Dorado Community College Problem: (a) At the downtown campus, fulltime students pay $55 per unit. Let U = cost of tuition for a randomly selected fulltime student at the downtown campus. Find the mean and standard deviation of U. (b) Calculate the mean and standard deviation of the total amount of tuition for a randomly selected fulltime student at the main campus and for a randomly selected fulltime student at the downtown campus. (a) 55 = 55(15) = $825, 55 = 55(2.3) = $ U Y (b) T U T U = = $ = $163. T U U Y T U T U = = 26570, thus [Page 372] Alternate Example: El Dorado Community College Problem: Suppose we randomly select one fulltime student from each of the two campuses. What are the mean and standard deviation of the difference in tuition charges, D = T U? Interpret each of these values.
9 T U T U = = $ This means that, on average, fulltime students at the main campus pay $92.50 less in tuition than fulltime students at the downtown campus T U T U = = 26570, thus T U = $163. Although the average difference in tuition for the two campuses is $92.50, the difference in tuition for a randomly selected fulltime student from each college will vary from the average difference by about $163, on average. Notice that the standard deviation is the same for the sum of tuition costs and the difference of tuition costs. [Page 373] Alternate Example: A Graphing Calculator Simulation You can imitate the Computer Simulation example on a graphing calculator. For example, in the heading of L1, enter RandNorm(3, 0.9, 500) to randomly generate 500 values from a Normal distribution with a mean of 3 and a standard deviation of 0.9. Likewise, entering RandNorm(1, 1.2) in the heading of L2 will randomly generate 500 values from a Normal distribution with a mean of 1 and a standard deviation of 1.2. In the heading of L3, enter L1+L2 and in the heading of L4, enter L1 L2. Before looking at graphs, review the values in the first row of each list to make sure students understand where each value came from. Now, look at histograms of L1, L2, L3 and L4 and you should see the same things as the Computer Simulation example. [Page 374] Alternate Example: Apples Suppose that the weights of a certain variety of apples have weights that are approximately Normally distributed with a mean of 9 ounces and a standard deviation of 1.5 ounces. If bags of apples are filled by randomly selecting 12 apples, what is the probability that the sum of the weights of the 12 apples is less than 100 ounces? State: What is the probability that a random sample of 12 apples has a total weight less than 100 ounces? Plan: Let = weight of a randomly selected apple. Then 1 = weight of first randomly selected apple, etc. We are interested in the total weight T = Our goal is to find P(T < 100). Do: Since T is a sum of 12 independent Normal random variables, T follows a Normal distribution with mean = = 108 ounces and variance T T = = 27. The standard deviation is 27 = 5.2 ounces. P(T < 100) = normalcdf( 99999, 100, 108, 5.2) = Note: to get full credit on the AP exam when using the calculator command normalcdf, students must clearly identify the shape (Normal), center (mean = 108) and spread (standard deviation = 5.2) somewhere in their work. Conclude: There is about a 6.2% chance that the 12 randomly selected apples will have a total weight of less than 100 ounces. [Page 375] Alternate Example: Speed Dating Suppose that the height M of male speed daters follows a Normal distribution with a mean of 70 inches and a standard deviation of 3.5 inches and the height F of female speed daters follows a T
10 Normal distribution with a mean of 65 inches and a standard deviation of 3 inches. What is the probability that a randomly selected male speed dater is taller than the randomly selected female speed dater with whom he is paired? State: What is the probability that a randomly selected male speed dater is taller than the randomly selected female speed dater with whom he is paired? Plan: We ll define the random variable D = M F to represent the difference between the male s height and the female s height. Our goal is to find P(M > F) or P(D > 0). Do: Since D is the difference of two independent Normal random variables, D follows a Normal distribution with mean D M F = = 5 inches and variance = D M F 3 2 = Thus, the standard deviation is D = 4.61 inches. Thus, P(D > 0) = normalcdf(0, 99999, 5, 4.61) = Note: to get full credit on the AP exam when using the calculator command normalcdf, students must clearly identify the shape (Normal), center (mean = 5) and spread (standard deviation = 4.61) somewhere in their work. Conclude: There is about an 86% chance that a randomly selected male speed dater will be taller than the female he is randomly paired with. Or, in about 86% of speed dating couples, the male will be taller than the female. [Page 384] Alternate Example: Dice, cars, and hoops Problem: Determine whether the random variables below have a binomial distribution. Justify your answer. (a) Roll a fair die 10 times and let =the number of sixes. (b) Shoot a basketball 20 times from various distances on the court. Let Y=number of shots made. (c) Observe the next 100 cars that go by and let C=color. (a) Binary? Yes; success = six, failure = not a six, Independent? Yes; knowing the outcomes of past rolls doesn t provide additional information about the outcomes of future rolls. Number? Yes; there are 10 trials. Success? Yes; the probability of success is always 1/6. This is a binomial setting. is binomial with n = 10 and p = 1/6. (b) Binary? Yes; success = make the shot failure = miss the shot. Independent? Yes; evidence suggests that it is reasonable to assume that making a shot doesn t change the probability of making the next shot. Number? Yes; there are 20 trials. Success? No; the probability of success changes because the shots are taken from various distances. (c) Binary? No. There are more than two possible colors. Also, C is not even a random variable since the outcomes aren t numerical. [Page 386] Alternate Example: Rolling Doubles In many games involving dice, rolling doubles is desirable. Rolling doubles mean the outcomes of two dice are the same, such as 1&1 or 5&5. The probability of rolling doubles when rolling two dice is 6/36 = 1/6. If = the number of doubles in 4 rolls of two dice, then is binomial with n = 4 and p = 1/6.
11 What is P( = 0)? That is, what is the probability that all 4 rolls are not doubles? Since the probability of not getting doubles is 1 1/6 = 5/6, P( = 0) = P(FFFF) = (5/6)(5/6)(5/6)(5/6) = (5/6) 4 = Note: F represents a failure and S represents a Success. What is P( = 1)? There are four different ways to roll doubles once in four tries. For example, the doubles could occur on the first try (SFFF), the second try (FSFF), the third try (FFSF) or the fourth try (FFFS). P(SFFF) = (1/6)(5/6)(5/6)(5/6) = (1/6)(5/6) 3 P(FSFF) = (5/6)(1/6)(5/6)(5/6) = (1/6)(5/6) 3 P(FFSF) = (5/6)(5/6)(1/6)(5/6) = (1/6)(5/6) 3 P(FFFS) = (5/6)(5/6)(5/6)(1/6) = (1/6)(5/6) 3 Thus, the probability of rolling doubles once in 4 attempts is P( = 1) = 4(1/6)(5/6) 3 = [Page 388] Alternate Example: Rolling Doubles Problem: When rolling two dice, the probability of rolling doubles is 1/6. Suppose that a game player rolls the dice 4 times, hoping to roll doubles. (a) Find the probability that the player gets doubles twice in four attempts. (b) Should the player be surprised if he gets doubles more than twice in four attempts? Justify your answer. Let = number of doubles. has a binomial distribution with n = 4 and p = 1/ (a) P( = 2) = 2 = = (b) P( > 2) = P( = 3) + P( = 4) = = = Since there is only a 1.6% chance of getting more than 2 doubles in four rolls, the player should be surprised if this happens. [Page 390] Alternate Example: Rolling Doubles If = number of doubles in 4 attempts, then follows a binomial distribution with n = 4 and p = 1/6. Here is the probability distribution and a histogram of the probability distribution. Value (x) Probability
12 Probability Number of Doubles 3 4 The mean of this distribution is = 0(0.482) + 1(0.386) + + 4(0.001) = 4(1/6) = The variance of this distribution is = ( ) 2 (0.482) + ( ) 2 (0.386) + + ( ) 2 (0.001) = 4(1/6)(5/6) = so the standard deviation is = [Page 392] Alternate Example: Tastes as Good as the Real Thing? The makers of a diet cola claim that its taste is indistinguishable from the full calorie version of the same cola. To investigate, an AP Statistics student named Emily prepared small samples of each type of soda in identical cups. Then, she had volunteers taste each cola in a random order and try to identify which was the diet cola and which was the regular cola. Overall, 23 of the 30 subjects made the correct identification. If we assume that the volunteers really couldn t tell the difference, then each one was guessing with a 1/2 chance of being correct. Let = the number of volunteers who correctly identify the colas. Problem: (a) Explain why is a binomial random variable. (b) Find the mean and the standard deviation of. Interpret each value in context. (c) Of the 30 volunteers, 23 made correct identifications. Does this give convincing evidence that the volunteers can taste the difference between the diet and regular colas? (a) The chance process is each volunteer guessing which sample is the diet cola. Binary? Yes, guesses are either correct or incorrect. Independent? Yes, the results of one volunteer s guess should have no effect on the results of other volunteers. Number? Yes, there are 30 trials. Success? Yes, the probability of guessing correctly is always 50%. Since is counting the number of successful guesses, is a binomial random variable. (b) The mean of is np = 30(0.5) = 15 and the standard deviation of is np(1 p) 30(0.5)(1 0.5) = If this experiment were repeated many times and the
13 volunteers were randomly guessing, the average number of correct guesses would be about 15 and the number of correct guesses would vary from 15 by about 2.74, on average. (c) P( 23) = 1 P( 22) = 1 binomcdf(30, 0.5, 22) = = There is a very small chance that there would be 23 or more correct guesses if the volunteers couldn t tell the difference in the colas. Therefore, we have convincing evidence that the volunteers can taste the difference. [Page 393] Alternate Example: NASCAR Cards and Cereal Boxes In the NASCAR Cards and Cereal Boxes example from section 5.1, we read about a cereal company that put one of 5 different cards into each box of cereal. Each card featured a different driver: Jeff Gordon, Dale Earnhardt, Jr., Tony Stewart, Danica Patrick, or Jimmie Johnson. Suppose that the company printed 20,000 of each card, so there were 100,000 total boxes of cereal with a card inside. If a person bought 6 boxes at random, what is the probability of getting no Danica Patrick cards? Let be the number of Danica Patrick cards obtained from 6 different boxes of cereal. Since we are sampling without replacement, the trials are not independent, so the distribution of is not quite binomial but it is close. If we assume is binomial with n = 6 and p = 0.2, then P( = 0) = (0.2) (0.8) = The actual probability, using the general multiplication rule is: 80,000 79,999 79,998 79,997 79,996 79,995 P(no Danica Patrick cards) = 100,000 99,999 99,998 99,997 99,996 99,995 = These two probabilities are quite close! [Page 394] Alternate Example: Dead Batteries Almost everyone has one a drawer that holds miscellaneous batteries of all sizes. Suppose that your drawer contains 8 AAA batteries but only 6 of them are good. You need to choose 4 for your graphing calculator. If you randomly select 4 batteries, what is the probability that all 4 of the batteries you choose will work? Problem: Explain why the answer isn t P( = 4) = probability is (0.75) (0.25) = The actual Since we are sampling without replacement, the selections of batteries aren t independent. We can ignore this problem if the sample we are selecting is less than 10% of the population. However, in this case we are sampling 50% of the population (4/8), so it is not reasonable to ignore the lack of independence and use the binomial distribution. This explains why the binomial probability is so different from the actual probability.
14 [Page 396] Alternate Example: Teens and Debit Cards In a survey of 506 teenagers ages 1418, subjects were asked a variety of questions about personal finance. (http://www.nclnet.org/personalfinance/66teensandmoney/120nclsurveyteensandfinancialeducation) One question asked teens if they had a debit card. Problem: Suppose that exactly 10% of teens ages have debit cards. Let = the number of teens in a random sample of size 506 that have a debit card. (a) Show that the distribution of is approximately binomial. (b) Check the conditions for using a Normal approximation in this setting. (c) Use a Normal distribution to estimate the probability that 40 or fewer teens in the sample have debit cards. (a) Binary? Yes, teens either have a debit card or they don t. Independent? No, since we are sampling without replacement. However, since the sample size (n = 506) is much less than 10% of the population size (there are millions of teens ages 1416), the responses will be very close to independent. Number? Yes, there is a fixed sample size, n = 506. Success? Yes, the unconditional probability of selecting a teen with a debit card is 10%. (b) Since np = 506(0.10) = 50.6 and n(1 p) = 506(0.90) = are both at least 10, we should be safe using the normal approximation. (c) = np = 506(0.10) = 50.6 and np(1 p) 506(0.1)(0.9) = P( 40) = normalcdf( 9999, 40, 50.6, 6.75) = (The probability using the binomial distribution is 0.064). Note: to get full credit on the AP exam when using the calculator command normalcdf, students must clearly identify the shape (Normal), center (mean = 50.6) and spread (standard deviation = 6.75) somewhere in their work. [Page 398] In the board game Monopoly, one way to get out of jail is to roll doubles. How likely is it that someone in jail would roll doubles on his first, second, or third attempt? If this was the only way to get out of jail, how many turns would it take, on average? [Page 398] Alternate Example: Monopoly The random variable of interest in this example is Y = number of attempts it takes to roll doubles one time. Each attempt is one trial of the chance process. Let s check the conditions for a geometric setting: Binary? Yes, success = doubles, failure = not doubles Independent? Yes, knowing the outcome of previous rolls does not give additional information about future rolls. Trials? Yes, we are counting the number of trials needed to get doubles once. Success? Yes, the probability of success is always 1/6.
15 [Page 398] Alternate Example: Monopoly Problem: Let the random variable Y be defined as in the previous alternate example. (a) Find the probability that it takes 3 turns to roll doubles. (b) Find the probability that it takes more than 3 turns to roll doubles and interpret this value in context. (a) P(Y = 3) = (5/6) 2 (1/6) = (b) Since there are an infinite number of possible values of Y greater than 3, we will use the complement rule. P(Y > 3) = 1 P(Y 3) = 1 P(Y = 3) P(Y = 2) P(Y = 1) = 1 (5/6) 2 (1/6) (5/6) 1 (1/6) (5/6) 0 (1/6) = If a player tried to get out of jail many, many times by trying to roll doubles, about 58% of the time it would take more than 3 attempts. Note: if the first success occurs after the 3 rd attempt, this means that the first three attempts are all failures. Thus, an alternative solution would be P(Y > 3) = P(no doubles and no doubles and no doubles) = (5/6) 3 =
X: 0 1 2 3 4 5 6 7 8 9 Probability: 0.061 0.154 0.228 0.229 0.173 0.094 0.041 0.015 0.004 0.001
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