The White Board
Transcription
The White Board
Printer Computer w/ Internet Access Assessment Discussions (4) 0-40 Practice (3) 0-75 Quizzes (4) 0-40 Tests (CST) Test 1 (0-48) TOTAL 0-203 AP Statistics Grading Scale Semester 2 - Unit 1 97 - l00% = A+ 93 - 96% = A 90 - 92% = A87 - 89% = B+ 83 - 86% = B 80 - 82% = B77 - 79% = C+ 73 - 76% = C 70 - 72% = C<69% = INC Name: Teacher: 1 of 70 AP Statistics Sem 2 - Unit 1 Materials List Binomials and Distributions The White Board 04152013 This Page Left Intentionally Blank 2 of 70 AP Statistics Unit Overview: Binomials and Distributions Think about It Page 1 of 4 Unit 1 S2 The following questions will help you study key concepts covered in this unit. 1. You're taking samples of size 10 from a population of 20 marbles and recording the number of blue marbles in each sample. Why isn't this a binomial situation, and why would it be almost binomial if you did the same experiment with a population of 200 marbles? 2. Explain the term continuity correction. Why does it work, and how do you use it? 3. Consider a parent population with mean 75 and standard deviation 7. The population doesn't appear to have extreme skewness or outliers. A. What are the mean and standard deviation of the distribution of sample means for n = 40? B. What's the shape of the distribution? Explain your answer in terms of the central limit theorem. C. What proportion of the sample means of size 40 would you expect to be 77 or less? If you use your calculator, show what you entered. D. Draw a sketch of the probability you found in part C, and label the horizontal axis. 4. Suppose that after several years of offering AP Statistics, a high school finds that final exam scores are normally distributed with mean 78 and standard deviation 6. A. What are the mean, standard deviation, and shape of the distribution of x-bar for n = 50? B. What's the probability a sample of scores will have a mean greater than 80? C. Sketch the distribution curve for part B, showing the area that represents the probability you found. Be sure to label the horizontal axis. 5. In a survey, 600 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 70% said the genders are equal and should have equal opportunities to participate in sports. A. What are the mean, standard deviation, and shape of the distribution of the sample ˆ of parents who say the genders are equal and should have equal proportion p opportunities? Be sure to justify your answer for the shape of the distribution. Use n = 600. B. Using the normal approximation without the continuity correction, sketch the ˆ . Shade equal areas on both probability distribution curve for the distribution of p sides of the mean to show an area that represents a probability of .95, and label the upper and lower bounds of the shaded area as values of p-hat (not z-scores). C. In your sketch from part B, the shaded area shows a .95 probability of what happening? In other words, what does the probability of .95 represent? ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use 3 ofat70www.apexvs.com/TermsOfUse) AP Statistics Unit Overview: Binomials and Distributions Think About It Page 2 of 4 D. Using the normal approximation, what's the probability that a randomly drawn sample of 600 parents will have a sample proportion between 67% and 73%? Draw a sketch of the probability curve, shade the area representing the probability you're finding, and label the z-scores that represent the upper and lower bounds of the probability you're finding. Don't use the continuity correction. E. Now, use the exact binomial calculation to find the probability of getting between, but not including, 67% and 73% of the respondents in a sample of 600 who say the genders are equal and should have equal opportunities. To use the exact binomial, you'll need to convert the proportions to counts by multiplying each proportion by 600. F. Now try it again, but this time find the probability of getting at least 67% but no more than 73%. Use the exact binomial calculation. 6. Annie is a basketball player who makes, on average, 65% of her free throws. Assume each shot is independent and the probability of making any given shot is .65. A. What's the probability Annie will miss three straight free throws before she makes one? (If you use a calculator to get your answer, write your answer in standard notation and show what you do on the calculator as well.) B. During a season, Annie takes 140 free throws. What's the exact binomial probability she'll make at least 100 out of 140 of these throws? C. For part B, are the conditions that permit you to use a normal approximation to the binomial satisfied? Explain. D. Redo part B using a normal approximation, without continuity correction, to the binomial. Draw a sketch of the situation. Then draw the distribution, shade the area representing the probability you're finding, and label the horizontal axis approximately. NOTE: If you use a graphing calculator, show what you do to get the z-scores, and explain your answer in enough detail to show your instructor you understand what you're doing. E. Redo part B using a normal approximation, with continuity correction, to the binomial. NOTE: If you use a graphing calculator, show what you do to get the z-scores on your calculator, and explain your answer in enough detail to show your instructor you understand what you're doing. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use 4 ofat70www.apexvs.com/TermsOfUse) AP Statistics Unit Overview: Binomials and Distributions Think About It Page 3 of 4 Discussion 7. Have you ever been stuck in Jail when playing Monopoly? To get out, you need to roll doubles in three tries or fewer or you have to pay. On average, how many times would people have to roll before getting doubles, and is that number larger than three? This is an average waiting-time question, where you're interested in the number of tries you need on average to get the outcome you want. You'll learn more about situations like this and the probability distributions they produce in the next Tutorial. Meanwhile, this Discussion will help you think about waiting-time situations. Respond to any one (or more) of the following, or respond to another student's posting. As you explain your reasoning, be sure to use what you know about the laws of probability. 1. If you buy a very large bag of candies colored brown, yellow, green, blue, orange, and red, and you start eating them, how many candies would you expect to pick until you got a blue one? Explain your reasoning using what you know about probability. 2. What's the probability you'll get out of Jail in Monopoly without having to pay? In other words, what are the chances you'll roll doubles on two six-sided dice within three tries? 3. You may be familiar with promotional campaigns where a company's products are marked with a letter, under the bottle cap of a soft drink, for example, and you're supposed to spell something to win a prize. In these cases, do you think some game pieces are more common than others? Describe an experiment you could conduct to see if some game pieces are easier to get than others. 4. Say you're playing a game like the one described in topic 3. A soft drink company has a letter printed on each bottle cap and the object is to spell the words I bought a lot of bottles to spell this. You have all the letters you need except p. The company's disclaimer statement says for each bottle you buy there's a 1/200 chance of getting a p. On average, how many bottles would you expect most people to buy in order to get this letter? 5. Create your own question about an average = waiting-time situation, or describe a real one you've seen or participated in. Explain why it's an average = waiting-time situation, and invite other students to answer it. 6. Have you ever won anything in a game like the ones described in topics 3 and 4? Describe the game, and calculate the probability of winning. 8. A sampling distribution is the distribution of a statistic. In other words, if you took the mean of many samples from a population, the set of means would form the distribution of . Some of the questions below have important information you'll need in order to understand the other questions, so before choosing which question to answer please read them all. Respond to one or more of the following or respond to another student's posting. 1. Suppose you had a population of fish whose lengths were normally distributed with a mean of 50 centimeters and a standard deviation of 5 centimeters. You draw a simple random sample of size 10, record the length of each fish, and calculate the mean of the sample lengths. What do you think your sample mean would most likely be? Explain your answer using what you know about random samples and sample means. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use 5 ofat70www.apexvs.com/TermsOfUse) AP Statistics Unit Overview: Binomials and Distributions Think About It Page 4 of 4 2. For the scenario in topic 1, imagine you draw hundreds of samples of size 10 (replacing each fish after you record its length) and calculate the mean of each sample. If you kept doing this until you took every possible sample, you'd get a distribution of sample means called a sampling distribution. What would the shape of the sampling distribution be? Would the standard deviation of the sampling distribution be larger or smaller than the population standard deviation? Explain your answer using what you know about random samples and sample standard deviations. 3. Why might a sampling distribution be helpful if you wanted to estimate the likelihood of getting a particular sample value, say x-bar = 48 centimeters? What's your guess about the likelihood of getting a sample mean of 48 centimeters? Explain your answer using what you know about probability distributions. 4. Building on items 1-3, how might a sampling distribution be helpful if you wanted to estimate a population mean with a sample mean? Outline a scenario for using a sampling distribution for inference. 5. Write a paragraph-long story about a sampling distribution; describe the population, the sampling method, the mean and standard deviation of the population and the sampling distribution. 6. There's a case where a sampling distribution qualifies as a binomial setting. Describe this case, using the criteria for a binomial distribution. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use 6 ofat70www.apexvs.com/TermsOfUse) 1.1 Key Terms inferential statistics 7 of 70 S2 6WDWLVWLFV6WXG\*XLGH ,QWURGXFWLRQWR,QIHUHQWLDO6WDWLVWLFV 3DJHRI 1.1.1 'LUHFWLRQV • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO • .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\ .H\7HUPVDQG&RQFHSWV • QRQH 7XWRULDO6HFWLRQV 3UREDELOLW\DQG&RQILGHQFH,QWHUYDOV 3UREDELOLW\DQG7HVWVRI6LJQLILFDQFH 6WHSVWR,QIHUHQFH BBBBBBBBBBBBBBBBBBB &RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\ XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG 8 of 70 S2 6WDWLVWLFV6WXG\*XLGH ,QWURGXFWLRQWR,QIHUHQWLDO6WDWLVWLFV 3DJHRI &RPPRQ$SSOLFDWLRQV BBBBBBBBBBBBBBBBBBB &RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV $Q\XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG 9 of 70 1.2 Key Terms almost binomial binomcdf on a graphing calculator almost binomial binomcdf on a graphing calculator binomial experiment (setting) binomial probabilities for a range of outcomes binomial probabilities for an exact number of outcomes binompdf on a graphing calculator continuity correction geometric setting normal approximation to the binomial 10 of 70 S2 AP Statistics Study Guide: Binomial Situations (Events) 3DJHRI4 1.2.1 'LUHFWLRQV • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO • .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\ .H\7HUPVDQG&RQFHSWV • • • • DOPRVWELQRPLDO ELQRPFGI ELQRPLDOH[SHULPHQW ELQRPLDOSUREDELOLWLHVIRUDUDQJHRI RXWFRPHV • • • • ELQRPLDOSUREDELOLWLHVIRUDQH[DFW QXPEHURIRXWFRPHV ELQRPLDOVHWWLQJ ELQRPSGI LQIHUHQWLDOVWDWLVWLFV 7XWRULDO6HFWLRQV 7KH'HILQLWLRQRID%LQRPLDO6HWWLQJRU(YHQW &DOFXODWLQJ%LQRPLDO3UREDELOLWLHVIRUDQ([DFW1XPEHURI2XWFRPHV &DOFXODWLQJ%LQRPLDO3UREDELOLWLHVIRUD5DQJHRI2XWFRPHV ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 11 of 70 S2 AP Statistics Study Guide: Binomial Situations (Events) $OPRVW%LQRPLDO ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 12 of 70 Page 2 of 4 AP Statistics Study Guide: Binomial Situations (Events) Page 3 of 4 8VLQJWKH7,/TI-84WR&DOFXODWH%LQRPLDO'LVWULEXWLRQ3UREDELOLWLHV 7R&DOFXODWHD%LQRPLDO3UREDELOLW\IRUD6SHFLILF9DOXHELQRPSGI 7KHSGIVWDQGVIRUSUREDELOLW\GHQVLW\IXQFWLRQ8VHELQRPSGIDQGHQWHUWKHQXPEHUV DQGFRPPDVLQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH % • • • • SUHVVQG >',675@ VFUROOGRZQWR>ELQRPSGI@DQGSUHVV(17(5 .H\LQ<RXUKRPHVFUHHQVKRXOGUHDGELQRPSGI SUHVV(17(5DQG\RXVKRXOGJHW 7R&DOFXODWHD3UREDELOLW\IRUD5DQJHRI9DOXHVELQRPFGI 7KHFGIVWDQGVIRUFXPXODWLYHGHQVLW\IXQFWLRQ(QWHUWKHQXPEHUVDQGFRPPDVLQWKH VDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH% • • • SUHVVQG >',675@ VFUROOGRZQWR>ELQRPFGI@DQGSUHVV(17(5 $VLQELQRPSGIHQWHUWKHQXPEHURIWULDOVWKHSUREDELOLW\IRUHDFKWULDODQGWKH[ YDOXH )RUH[DPSOHIRU%3;LVELQRPFGI ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 13 of 70 AP Statistics Study Guide: Binomial Situations (Events) Page 4 of 4 Using the TI-89 to Calculate Binomial Distribution Probabilities To Calculate a Binomial Probability for a Specific Value: Binomial Pdf f Pdf stands for “probability density function.” Use binomPdf and enter the numbers and commas in the same order as they are given in the probability notation. For example, to calculate B(20,.3,5): 1. From Stats/List Editor, press [F5]:Distr. Scroll down to [B]: Binomial Pdf.... :Press [ENTER]. 2. For Num Trials, n, enter 20. Arrow down one cell. For Prob Success, p, enter .3. Arrow down again. For X Value, enter 5. Press [ENTER] to save your settings, then [ENTER] again to execute the command. You should get .178863 for your answer. 3. You can also access many of the statistics functions by pressing [CATALOG][F3]. This will produce a list of common functions available on the calculator. To Calculate a Binomial Probability for a Range of Values: Binomial Cdf Cdf stands for “cumulative density function.” This function allows you to calculate the probability of a range of possible outcomes from a binomial setting. You will need to identify the lower and upper bounds for that range. To calculate the probability of obtaining five or fewer successes out of 20 trials, with a probability of success of 0.3: 1. From Stats/List Editor, press [F5]:Distr. Scroll down to [C]: Binomial Cdf.... :Press [ENTER]. 2. For Num Trials, n, enter 20. Arrow down one cell. For Prob Success, p, enter .3. Arrow down again. For the Lower Bound, enter 0. Arrow down. For the Upper Bound, enter 5. Press [ENTER] to save your settings, then [ENTER] again to execute the command. 3. You should get .416371 for your answer. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 14 of 70 Page 1 of 3 1.2.3 Why Study Binomial Distributions? Remember that, although statistical inference takes many forms, it often follows these phases: 1. Identify what you want to study. 2. Ensure that your study and your sample are valid and useful for drawing conclusions about what you're studying. This includes understanding how the data were collected (so you know their limitations). 3. Recognize the type of probability distribution that models your situation. The probability distribution tells you how likely or unlikely your findings are. You can also use it to find the range of values your population parameter is likely to have. 4. If you're comparing at least two measures or studying a relationship, conduct significance tests to compare your results against what you'd expect from chance alone. If the study's results are highly unlikely, your results are statistically significant. This Activity focuses on phase 3, the probability distribution, with particular emphasis on the binomial distribution. You'll need to know about a few different types of probability distributions to do inferential statistics. Review Distribution is one of the most important concepts in statistics. Every variable has a distribution, whether the variable is categorical or numerical, continuous or discrete. A variable's distribution consists of the values it takes and how often it takes each of these values. Distributions are often shown in graphs so you can see the relative frequency with which values occur. A binomial distribution (also referred to sometimes as the binomial setting) is a discrete probability distribution of the counts of successes and failures, such as the number of times you get heads when you flip a coin 100 times. Most textbooks list the following four characteristics for the binomial distribution (or setting): In the Binomial Setting, 1. There's a set n of identical trials. 2. The outcome of each trial is either success or failure. 3. The probability of success of each trial is p. (The probability of failure of each trial is q = 1 - p.) 4. Each trial is independent. 15 of 70 S2 Page 2 of 3 Some textbooks list the following as a fifth characteristic, and others use it as an introduction to the previous four: 5. In the binomial setting, we're interested in x, the number of successes observed during the n trials, for x = 0, 1, 2, 3 , … , n. The following are examples of binomial settings: • What's the probability of getting exactly three heads on five flips of a coin? • A coin is flipped 100 times. What's the probability of getting 45 heads? • If the probability of having Rh-negative blood is .4 in a population of 500, what's the probability 210 have type Rh-negative blood? These are all probability settings in which: 1. We know the number of repetitions. 2. The outcome of each trial is either success or failure. 3. We know the probability of success or failure of any trial. 4. The probability doesn't change from trial to trial (the trials are independent.) Other Definitions to Remember Binomial Event: a set of trials within a binomial setting Simple Event or Outcome: a single trial B(n, p): the binomial probability distribution with parameters n, the number of observations, and p, the probability of success on any one observation. If X is a binomial n random variable with B(n, p), B(n, p, x) = P(X = x) = (p)x(1 – p)n–x. x On your calculator, binompdf(n,p,X) calculates the probability of exactly x successes in n trials with a probability of p. With a range of outcomes for X, where X takes on the number of successes 0, 1, 2, 3, …, n, use binomcdf(n,p,X) on your calculator, which calculates the probability of up to and including x successes in n trials with a probability of p for the binomial probability distribution. Remember, the binomial distribution is discrete: Each single value has a probability. This means that, unlike a continuous distribution such as the normal distribution, there's a difference between P(x < 4) and P(x ≤ 4). So for P(x < 4), you add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3). But for P(x ≤ 4) you include the 4, so you add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4). Be careful about your endpoints when using binomcdf. With binomcdf on a calculator, you get the lower-tail probability that includes the upper bound. So, if you enter binomcdf(20,.3,5), you're finding P(x ≤ 5) for B(20, .3). If you need a range with a specific lower and upper bound, remember to subtract like this: (probability below upper bound) – (probability below lower bound). 16 of 70 Page 3 of 3 Be careful about the range you calculate. It may help to draw the range on a number line. For example, if you want to calculate P(4 < x < 8) for B(10, .3), you want the range represented by the underlined numbers: 0 1 2 3 4 5 6 7 8 9 10. So you'd calculate: binomcdf(10,.3,7) – binomcdf(10,.3,4). Note that this will give you the range from 5 to 7. If you wanted to calculate P(4 ≤ x < 8), you include the 4 in your range: 0 1 2 3 4 5 6 7 8 9 10. So you'd calculate: binomcdf(10,.3,7) – binomcdf(10,.3,3). Note that now you're subtracting the 3 (since it's included in binomcdf(10,.3,3)) but not the 4. If you want P(4 ≤ x ≤ 8), you'd include the 8: 0 1 2 3 4 5 6 7 8 9 10. You'd calculate: binomcdf(10,.3,8) – binomcdf(10,.3,3). There are cases where the criteria of the binomial setting might not be satisfied, but which would still be considered binomial. An event is almost binomial if the population size of an experiment is at least 20 times larger than the sample size. (NOTE: 20 is a rule of thumb. Some references use 10 instead of 20.) 17 of 70 Page 1 of 3 1.2.3 Why Study Binomial Distributions? Remember that, although statistical inference takes many forms, it often follows these phases: 1. Identify what you want to study. 2. Ensure that your study and your sample are valid and useful for drawing conclusions about what you're studying. This includes understanding how the data were collected (so you know their limitations). 3. Recognize the type of probability distribution that models your situation. The probability distribution tells you how likely or unlikely your findings are. You can also use it to find the range of values your population parameter is likely to have. 4. If you're comparing at least two measures or studying a relationship, conduct significance tests to compare your results against what you'd expect from chance alone. If the study's results are highly unlikely, your results are statistically significant. This Activity focuses on phase 3, the probability distribution, with particular emphasis on the binomial distribution. You'll need to know about a few different types of probability distributions to do inferential statistics. Review Distribution is one of the most important concepts in statistics. Every variable has a distribution, whether the variable is categorical or numerical, continuous or discrete. A variable's distribution consists of the values it takes and how often it takes each of these values. Distributions are often shown in graphs so you can see the relative frequency with which values occur. A binomial distribution (also referred to sometimes as the binomial setting) is a discrete probability distribution of the counts of successes and failures, such as the number of times you get heads when you flip a coin 100 times. Most textbooks list the following four characteristics for the binomial distribution (or setting): In the Binomial Setting, 1. There's a set n of identical trials. 2. The outcome of each trial is either success or failure. 3. The probability of success of each trial is p. (The probability of failure of each trial is q = 1 - p.) 4. Each trial is independent. Some textbooks list the following as a fifth characteristic, and others use it as an introduction to the previous four: 5. In the binomial setting, we're interested in x, the number of successes observed during the n trials, for x = 0, 1, 2, 3 , … , n. 18 of 70 S1 Page 2 of 3 The following are examples of binomial settings: • What's the probability of getting exactly three heads on five flips of a coin? • A coin is flipped 100 times. What's the probability of getting 45 heads? • If the probability of having Rh-negative blood is .4 in a population of 500, what's the probability 210 have type Rh-negative blood? These are all probability settings in which: 1. We know the number of repetitions. 2. The outcome of each trial is either success or failure. 3. We know the probability of success or failure of any trial. 4. The probability doesn't change from trial to trial (the trials are independent.) Other Definitions to Remember Binomial Event: a set of trials within a binomial setting Simple Event or Outcome: a single trial B(n, p): the binomial probability distribution with parameters n, the number of observations, and p, the probability of success on any one observation. If X is a binomial n random variable with B(n, p), B(n, p, x) = P(X = x) = (p)x(1 – p)n–x. x On your calculator, binompdf(n,p,X) calculates the probability of exactly x successes in n trials with a probability of p. With a range of outcomes for X, use Binomial Cdf on your calculator, which allows you to select lower and upper values for X, and finds the probability of obtaining a count of successes that falls within those values. Remember, the binomial distribution is discrete: Each single value has a probability. This means that, unlike a continuous distribution such as the normal distribution, there's a difference between P(x < 4) and P(x ≤ 4). So for P(x < 4), you add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3). But for P(x ≤ 4) you include the 4, so you add P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4). Be careful about your endpoints when using Binomial Cdf. If you are finding P(X<5), then your lower value will be 0 and your upper value will be 5. However, if you are finding P(X<5), then your values will be 0 and 4. If you need a range with a specific lower and upper value, then be sure to determine whether or not to include those specific values. Generally, words such as ‘at least’, ‘at most’, ‘no greater/ more/fewer/less than’, ‘including’, and ‘inclusive’ indicate the inclusion of an upper or lower value, while words such as ‘less than’, ‘more than’, ‘between’, ‘over/under’, and ‘between’ indicate the omission of that value and the need for the next higher or lower value to be used. 19 of 70 Page 3 of 3 Be careful about the range you calculate. It may help to draw the range on a number line. For example, if you want to calculate P(4 < x < 8) for B(10, .3), you want the range represented by the underlined numbers: 0 1 2 3 4 5 6 7 8 9 10. So you'd calculate: binomcdf(10,.3,7) – binomcdf(10,.3,4). Note that this will give you the range from 5 to 7. If you wanted to calculate P(4 ≤ x < 8), you include the 4 in your range: 0 1 2 3 4 5 6 7 8 9 10. So you'd calculate: binomcdf(10,.3,7) – binomcdf(10,.3,3). Note that now you're subtracting the 3 (since it's included in binomcdf(10,.3,3)) but not the 4. If you want P(4 ≤ x ≤ 8), you'd include the 8: 0 1 2 3 4 5 6 7 8 9 10. You'd calculate: binomcdf(10,.3,8) – binomcdf(10,.3,3). There are cases where the criteria of the binomial setting might not be satisfied, but which would still be considered binomial. An event is almost binomial if the population size of an experiment is at least 20 times larger than the sample size. (NOTE: 20 is a rule of thumb. Some references use 10 instead of 20.) 20 of 70 'LUHFWLRQV 1.2.4 • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKHtutorial. • .HHS\RXUFRPSOHWHGstudy guide in your notebook for later study. .H\7HUPVDQG&RQFHSWV • FRQWLQXLW\FRUUHFWLRQ • QRUPDODSSUR[LPDWLRQWRWKHELQRPLDO 7XWRULDO6HFWLRQV 7KH5HODWLRQVKLS%HWZHHQWKH%LQRPLDO'LVWULEXWLRQDQGWKH1RUPDO 'LVWULEXWLRQ 8VLQJWKH1RUPDO$SSUR[LPDWLRQWR)LQG3UREDELOLWLHV ,QFUHDVH$FFXUDF\E\8VLQJWKH&RQWLQXLW\&RUUHFWLRQ ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 21 of 70 S2 Using the TI-83 or TI-84 to Calculate Binomial Distribution Probabilities 7R&DOFXODWHD%LQRPLDO3UREDELOLW\IRUD6SHFLILF9DOXHELQRPSGI 7KHSGIVWDQGVIRUSUREDELOLW\GHQVLW\IXQFWLRQ8VHELQRPSGIDQGHQWHUWKH QXPEHUVDQGFRPPDVLQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWR FDOFXODWH% • • • • SUHVVQG >',675@ VFUROOGRZQWR>ELQRPSGI@DQGSUHVV(17(5 .H\LQ<RXUKRPHVFUHHQVKRXOGUHDGELQRPSGI SUHVV(17(5DQG\RXVKRXOGJHW 7R&DOFXODWHD3UREDELOLW\IRUD5DQJHRI9DOXHVELQRPFGI 7KHFGIVWDQGVIRUFXPXODWLYHGHQVLW\IXQFWLRQ(QWHUWKHQXPEHUVDQGFRPPDV LQWKHVDPHRUGHUDVWKHSUREDELOLW\QRWDWLRQ)RUH[DPSOHWRFDOFXODWH% • • • SUHVVQG >',675@ VFUROOGRZQWR>ELQRPFGI@DQGSUHVV(17(5 $VLQELQRPSGIHQWHUWKHQXPEHURIWULDOVWKHSUREDELOLW\IRUHDFKWULDODQGWKH [YDOXH )RUH[DPSOHIRU%3;LVELQRPFGI ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 22 of 70 Using the TI-83 or TI-84 to Calculate Normal Distribution Probabilities )LQGLQJ$1RUPDO&XUYH3UREDELOLW\ZLWKQRUPDOFGI 3UHVVQG >',675@7KLVVHOHFWVQRUPDOFGI (QWHU>ORZHUERXQGXSSHUERXQGµDQGσ@)RUWKHELUGHJJH[DPSOHWKHORZHU ERXQGLV]HURWKHXSSHUERXQGLVWKHPHDQLVDQGWKHVWDQGDUGGHYLDWLRQ LV6RSUHVV 3UHVV(17(5WRVHHWKHDQVZHU ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 23 of 70 AP Statistics Study Guide: The Normal Approximation to the Binomial 1.2.4 Directions Use this guide to take notes while you watch the tutorial. Keep your completed study guide in your notebook for later study. Key Terms and Concepts Page 1 of 3 continuity correction normal approximation to the binomial Tutorial Sections The Relationship Between the Binomial Distribution and the Normal Distribution Using the Normal Approximation to Find Probabilities Increase Accuracy by Using the Continuity Correction ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 24 of 70 S2 AP Sta atistics Study Guide: The Normal Approx ximation n to the B Binomial Page 2 of 3 Calculattor Instruc ctions for the TI-89 These instructions assume a thatt you've read d and follow wed the dire ections in the "Getting ator manuall. The steps covered in this guide a are also cov vered Started" chapter of your calcula in the "S Statistics" ch hapter of your calculato or manual. Text in brack kets, like thiis [HOME], gives g the na ame of a bu utton on the calculator. Two sets of brackets b like e [2nd][VAR R-LINK] get you to a co olor-coded ffunction abo ove each key. The diamond d button and brack kets get you u to anotherr color-coded d function sh hown above e each key. The white [A ALPHA] butto on followed by another bracket getts you a lettter. Text in bold, like 3:Regr ressions-> > is a comm and within a menu on tthe calculattor sc creen. Using th he TI-89 to o Calculate e Binomial Distributio on Probabillities To Calcu ulate a Binomial Prob bability for r a Specific c Value: mial Pdf to calculate th Pdf stand ds for “prob bability dens sity function.” Use Binom he probabilitty of observing a specific number of successes in an exercisse that invo olves a fixed d number of trials. Fo or example, to calculate e Binomial Pdf(20,.3,5), P , which is th he probabilitty of seeing g exactly five f successes in 20 tria als when the e probability y of successs is 0.3: 1. 2. 3. 4. 5. 6. 7. Enter List Editor. D / [B]: Binomial B Pdf.... Press [F5]: Distr Enter 20 for Num Trials, n:. 0 for Prob Success, p:. Enter .3 (or 0.3) Enter 5 for X Value: R] to store your y settings, then [EN TER] again to execute the command. Press [ENTER g .178863. You should get ulate a Binomial Prob bability for r a Range o of Values: To Calcu Cdf stands for “cumulative dens sity function n.” Use Bino omial CDF to o calculate tthe probabillity of observ ving success ses that ran nge from a lower bound d to an uppe er bound. . F For example e, to , which is th calculate e Binomial Cdf(20,.3,5) C he probabilitty of seeing g between ze ero and five e successe es in 20 trials when the probability of success is 0.3: 1. 2. 3. 4. 5. 6. Enter List Editor. B Cdff.... Press[F5]: Distr / [C]: Binomial Enter 20 for Num Trials, n:. 0 for Prob Success, p:. Enter .3 (or 0.3) Enter 0 for Lower Value::. U Value::. Enter 5 for Upper _________ ____________ ___________ Copyright © 2011 Apex Learning L Inc. (See ( Terms of Use at www.ap pexvs.com/TerrmsOfUse) 25 of 70 AP Statistics Study Guide: The Normal Approximation to the Binomial Page 3 of 3 7. Press [ENTER] to store your settings, then [ENTER] again to execute the command. 8. You should get .416371. If you want to calculate the probability of seeing between 5 and 10 successes under these same conditions, simply change your Lower Value: to 5 and your Upper Value: to 10. Your answer should be .745347 Using the TI-89 to Calculate Normal Distribution Probabilities Finding a Normal Curve Probability 1. From List Editor, press [F5]: Distr / [4]:Normal Cdf.... 2. You will be prompted for a Lower Bound, an Upper Bound, the mean ( μ ) and the standard deviation ( σ ). For the bird example, enter a Lower Bound of 0. Enter an Upper Bound of 11. Enter a mean of 12. Enter a standard deviation of 0.8. Press [ENTER] to save your settings, then press [ENTER] again to execute the command. 8. Your answer should be .10565. 3. 4. 5. 6. 7. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 26 of 70 6WDWLVWLFV 6WXG\6KHHW %LQRPLDO3UREOHPV 3DJHRI 1.2.6 S2 7KH1RUPDO$SSUR[LPDWLRQWRWKH%LQRPLDO ,QWKLVDFWLYLW\ZH OOORRNDWWZRNLQGVRIGLVWULEXWLRQQRUPDODQGELQRPLDO:KHQWKH QXPEHURIVXFFHVVHVDQGIDLOXUHVIRUDJLYHQVLWXDWLRQDUHHDFKHTXDOWRRUJUHDWHUWKDQ WKHELQRPLDOLVVDLGWRDSSUR[LPDWHWKHQRUPDOGLVWULEXWLRQ0DQ\WH[WERRNVXVH UDWKHUWKDQEXWERWKDUHYDOLG,QRWKHUZRUGVWKHELQRPLDODSSUR[LPDWHVWKH QRUPDOZKHQQS≥DQGQ±S≥1RWHWKDW±SRUWKHSUREDELOLW\RIIDLOXUHLV RIWHQZULWWHQDVTVRQ±SFDQEHZULWWHQDVQT ,QWKHVHFDVHVWKHELQRPLDOUHVHPEOHVWKHQRUPDOVRFORVHO\WKDW\RXFDQXVH]VFRUHV DQGDQRUPDOFXUYHWDEOHWRILQGDUHDVXQGHUWKHFXUYH$VPRVWVLWXDWLRQVLQLQIHUHQWLDO VWDWLVWLFVPHHWWKHUHTXLUHPHQWVIRUWKHELQRPLDODSSUR[LPDWLRQUHVHDUFKHUVUDUHO\QHHG WRILJXUHELQRPLDOSUREDELOLWLHVH[DFWO\WKH\XVXDOO\XVHWKHQRUPDODSSUR[LPDWLRQ LQVWHDG $ELQRPLDOGLVWULEXWLRQWKDWPHHWVWKHUHTXLUHPHQWVIRUWKHQRUPDODSSUR[LPDWLRQZLOO KDYHDPHDQHTXDOWRQSWKHH[SHFWHGYDOXHDQGDVWDQGDUGGHYLDWLRQHTXDOWR QS −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of 70 1.2.7 28 of 70 S2 29 of 70 30 of 70 31 of 70 32 of 70 33 of 70 34 of 70 1.3 Key Terms average waiting time geometcdf on a graphing calculator geometpdf on a graphing calculator geometric distribution 35 of 70 S2 AP Statistics Study: Geometic Probability Distributions 1.3.2 'LUHFWLRQV • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO • .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\ .H\7HUPVDQG&RQFHSWV • DYHUDJHZDLWLQJWLPH • JHRPHWFGI • JHRPHWSGI Page 1 of 3 • JHRPHWULFGLVWULEXWLRQ • JHRPHWULFVHWWLQJ 7XWRULDO6HFWLRQV ,QWURGXFWLRQWR*HRPHWULF3UREDELOLW\'LVWULEXWLRQV 3UREDELOLW\&DOFXODWLRQVLQWKH*HRPHWULF6HWWLQJ $YHUDJH:DLWLQJ7LPH ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 36 of 70 S2 AP Statistics Study: Geometic Probability Distributions Page 2 of 3 TI-83 and TI-84: Finding a 6LQJOH Probability Value for a Geometric Distribution 7RGRWKLVXVHJHRPHWSGIZKLFKVWDQGVIRUJHRPHWULFSUREDELOLW\GHQVLW\IXQFWLRQ • • 3UHVVQG >',675@$/3+$>'@7KLVZLOOSXWWKHFXUVRURQJHRPHWSGI,WZLOODOVRZRUNWR VFUROOGRZQWKHOLVWPDQXDOO\SXWWKHFXUVRURQJHRPHWSGIWKHQSUHVV(17(5 (QWHU>SUREDELOLW\RIVXFFHVVRQHDFKWULDOS@>@>QXPEHURIWULDOVQ@WKHQFORVHWKH SDUHQWKHVLV )RUH[DPSOHLI\RXZDQWWRILQGWKHSUREDELOLW\RIJHWWLQJVXFFHVVLQIRXUWULDOVLIWKH SUREDELOLW\RIVXFFHVVRQHDFKWULDOLV (QWHU ÷ 7KLVWHOOVWKHFDOFXODWRUWKDWLQHDFKWULDOWKHUH VDSUREDELOLW\RI VXFFHVVRIDQGWKDWWKHUHDUHWULDOV7KHQSUHVV(17(5DQG\RXVKRXOGJHW VRPHWKLQJFRQVLVWHQWZLWKQXPEHURIGLJLWVDQGURXQGLQJZLOOYDU\GHSHQGLQJRQ KRZ\RXUFDOFXODWRULVVHW TI-83 and TI-84: Finding a 5DQJH of Probability Values for a Geometric Distribution 7RGRWKLVXVHJHRPHWFGIZKLFKVWDQGVIRUJHRPHWULFSUREDELOLW\GHQVLW\IXQFWLRQ,W ILJXUHVWKHSUREDELOLW\IURPWULDOVXSWRDQGLQFOXGLQJQWULDOV • • 3UHVVQG >',675@$/3+$>(@7KLVZLOOVHOHFWJHRPHWFGI,WZLOODOVRZRUNWRVFUROOGRZQ WKHOLVWPDQXDOO\SXWWKHFXUVRURQJHRPHWFGIWKHQSUHVV(17(5 (QWHU>SUREDELOLW\RIVXFFHVVRQHDFKWULDOS@>@>WKHXSSHUERXQGIRUQXPEHURIWULDOV Q@WKHQFORVHWKHSDUHQWKHVLV )RUH[DPSOHLI\RXZDQWWRILQGWKHSUREDELOLW\RIVXFFHVVLQWKUHHRUIHZHUWULDOVLIWKH SUREDELOLW\RIVXFFHVVRQHDFKWULDOLV (QWHU ÷ 7KLVWHOOVWKHFDOFXODWRUWKDWLQHDFKWULDOWKHUH VDSUREDELOLW\RI DQGWKDWWKHUHDUHWULDOV7KHQSUHVV(17(5DQG\RXVKRXOGJHWQXPEHURIGLJLWV DQGURXQGLQJZLOOYDU\GHSHQGLQJRQKRZ\RXUFDOFXODWRULVVHW TI-83 and TI-84: Generating Random Numbers Using randInt • • • • 3UHVV0$7+ 6FUROODFURVVWR35% 6FUROOGRZQWRUDQG,QWDQGSUHVV(17(5 (QWHUWKH>PD[LPXPQXPEHU@DQG>PLPLPXPQXPEHU@DQGFORVHWKHSDUHQWKHVHV 7KHQSUHVV(17(5HDFKWLPH\RXZDQWDUDQGRPQXPEHU )RUH[DPSOHLI\RXZDQWWRJHQHUDWHUDQGRPQXPEHUVEHWZHHQDQGLQFOXVLYHHQWHU DQGSUHVV(17(5HDFKWLPH\RXZDQWDQXPEHU ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 37 of 70 AP Statistics Study: Geometric Probability Distributions Page 3 of 3 TI-89: Finding a Single Probability Value for a Geometric Distribution To do this, use Geometric Pdf…. 1. Enter List Editor. 2. Press [F5]:Distr / [F]: Geometric Pdf…. 3. You will be prompted for the probability of success (Prob Success, p:) and the trial on which you want the first success to occur (X Value:). 4. For example, if you were rolling a die and wanted to roll a 6, P(roll a 6) = 1/6, and you wanted to find the probability of getting your first 6 on your fourth roll, you would enter 1/6 for Prob Success, p, and 4 for X Value. Press [ENTER] to save your settings, then press [ENTER] again to execute the command. 5. Your answer should be .096451. TI-89: Finding a Range of Probability Values for a Geometric Distribution To do this, use Geometric Cdf. 1. Enter List Editor. 2. Press [F5]:Distr / [G]: Geometric Cdf…. 3. To find the probability of rolling your first 6 within the first four rolls, enter 0 for your Lower Value and 4 for your Upper Value. This will calculate the probability of rolling your first 6 on the first, second, third, or fourth roll. 4. Press [ENTER] to save your settings then press [ENTER] again to execute the command. 5. Your answer should be .517747. If you want to find the probability of rolling your first 6 between the third and fifth rolls, enter 4 for your Lower Value and 5 for your Upper Value. Your answer should be .292567. TI-89: Generating Random Numbers 1. From the home screen, press [2nd][5] and scroll to 7:Probability [ENTER]. 2. Select 4:rand( [ENTER]. 3. This will return you to the command line, where you will enter the upper value for your random integer. Close the parentheses and press [ENTER]. 4. The screen will display a random number between 1 and the number you entered. Press [ENTER] again to generate additional random numbers. ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 38 of 70 AP Statistics Study: Geometric Distribution Problems Page 1 of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− S)Q − S ZKHUHQLVWKHQXPEHURIWULDOVDQGSLVWKHSUREDELOLW\RI VXFFHVVIRUHDFKWULDO 127(0DQ\WH[WERRNVZULWHWKHSUREDELOLW\RIIDLOXUHRU±SDVT6R\RXPD\VHH WKHIRUPXODZULWWHQDV 3 ; = Q = T Q − (S) 7KHSUREDELOLW\\RX OOJHWDVXFFHVVRQRUEHIRUHWKHQWKWULDOLVVLPSO\WKHVXPRIDOO SUREDELOLWLHVIURPRQHWRQ)RULQVWDQFH3[≤ 3[ 3[ 1RWLFHWKDWZH GRQ WLQFOXGH3[ EHFDXVH\RXFDQQHYHUJHWDVXFFHVVZLWKRXWGRLQJDQ\WULDOV 2QDgraphingFDOFXODWRUWKH*HRPHWULF3UREDELOLW\'HQVLW\)XQFWLRQJHRPHWSGI FDOFXODWHVWKHSUREDELOLW\RIWKHILUVWVXFFHVVRQWKHQWKWULDO 7KH*HRPHWULF&XPXODWLYH'HQVLW\)XQFWLRQJHRPHWFGIFDOFXODWHVWKHSUREDELOLW\RI JHWWLQJWKHILUVWVXFFHVVRQRUEHIRUHWKHQWKWULDO,WVXPVWKHSUREDELOLWLHVIRUHDFKRI WKHHYHQWV3ILUVWVXFFHVVRQILUVWWULDO3ILUVWVXFFHVVRQVHFRQGWULDO3ILUVW VXFFHVVRQWKLUGWULDO«3ILUVWVXFFHVVRQ[WKWULDO *HRPHWULFGLVWULEXWLRQVDUHDOZD\VVNHZHGWRWKHULJKWEHFDXVHWKHUH VDVKDUSOLPLWRQ RQHHQG\RXFDQ WKDYHIHZHUWKDQQ DQGDORQJWDLORIGLPLQLVKLQJSUREDELOLWLHVRQ WKHRWKHUHQG ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 39 of 70 AP Statistics Study: Geometric Distribution Problems Page 2 of 2 7KHH[SHFWHGYDOXHRUDYHUDJHYDOXHLQDJHRPHWULFGLVWULEXWLRQLVFDOOHGWKHDYHUDJH ZDLWLQJWLPH,W VDOZD\VS6RLIWKHSUREDELOLW\RIJHWWLQJDIRXUZKHQUROOLQJDVL[ VLGHGGLHLVWKHDYHUDJHZDLWLQJWLPHZRXOGEHWULDOV,WWHOOV\RXWKDWRQDYHUDJH \RX OOKDYHWRUROODGLHVL[WLPHVEHIRUHJHWWLQJ\RXUILUVWIRXU7KLVPD\VHHPFRQIXVLQJ ZKHQ\RXFRQVLGHUWKDWLQDJHRPHWULFSUREDELOLW\GLVWULEXWLRQWKHKLJKHVWIUHTXHQF\LV DWQ n=1 2 3 4 5 6 7 8 9 10 %XWUHFDOOWKDWDQH[SHFWHGYDOXHWDNHVLQWRDFFRXQWDOOSRVVLEOHRXWFRPHV7KHPRVW FRPPRQRXWFRPHPD\EHQ EXWWKHUHDUHSOHQW\RIRWKHUWLPHVZKHUHWKHILUVW VXFFHVVFRPHVDWQ RUPRUH7KHVHKLJKHUYDOXHVSXOOWKHDYHUDJHXSIURPRQHVR WKDWRQDYHUDJH\RXFDQH[SHFWWRUROOWKHGLHVL[WLPHVEHIRUHJHWWLQJ\RXUILUVW VXFFHVV ______________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 40 of 70 1.4 Key Terms central limit theorem (distribution of) mean of the sampling distribution of a sample mean mean of the sampling distribution of a sample proportion sample proportions sampling distribution sampling distribution of a sample mean standard deviation of the sampling distribution of a sample mean standard deviation of the sampling distribution of a sample proportion 41 of 70 S2 6WDWLVWLFV6WXG\*XLGH 6DPSOLQJ'LVWULEXWLRQVDQGWKH&HQWUDO/LPLW7KHRUHP 'LUHFWLRQV 3DJHRI 1.4.2 • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO • .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\ .H\7HUPVDQG&RQFHSWV • • • • &HQWUDO/LPLW7KHRUHP PHDQRIWKHVDPSOLQJGLVWULEXWLRQRID VDPSOHPHDQ VDPSOLQJGLVWULEXWLRQ • VDPSOLQJGLVWULEXWLRQRIDVDPSOH PHDQ VWDQGDUGGHYLDWLRQRIWKHVDPSOLQJ GLVWULEXWLRQRIDVDPSOHPHDQ 7XWRULDO6HFWLRQ 7KH0HDQLQJRID6DPSOLQJ'LVWULEXWLRQ 7KH6DPSOLQJ'LVWULEXWLRQRID6DPSOH0HDQ 7KH&HQWUDO/LPLW7KHRUHP BBBBBBBBBBBBBBBBBBB &RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\ XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG 42 of 70 S2 1.4.2 S2 AP Statistics Page 1 of 4 Study Guide: Sampling Distributions and the Central Limit Theorem Calculator Instructions for the TI-89 These instructions assume that you've read and followed the directions in the "Getting Started" chapter of your calculator manual. The steps covered in this guide are also covered in the "Statistics" chapter of your calculator manual. Text in brackets, like this [HOME], gives the name of a button on the calculator. Two sets of brackets like [2nd][VAR-LINK] get you to a color-coded function above each key. The diamond button and brackets get you to another color-coded function shown above each key. The white [ALPHA] button followed by another bracket gets you a letter. Text in bold, like 3:Regressions-> is a command within a menu on the calculator screen. Section 1: Find the mean of a sampling distribution. Step 1. Clear a list, generate a sample, and store it in list1. 1. Enter List Editor. 2. Clear List 1. 3. Press[2nd][5] / 3:List / 1:seq( [ENTER] [2nd][5] 7:Probability 5:randNorm(20,3), x,1,10) [ENTER] [ENTER]. 4. This generates 10 random numbers from a normal distribution with mean 20 and standard deviation 3. These numbers are automatically stored in list1. Step 2. Find the mean of the sample and store the mean value into a variable. 1. Press [HOME][CATALOG] [ALPHA] M, scroll to mean( , and press [ENTER]. Type the name list1 followed by [)]. [ENTER]. You will see the mean of list1 on the screen. 2. If the command mean(list1) is not highlighted, press[ENTER] to highlight it. Press your right arrow, and your cursor will be placed to the right of your command. 3. Press [STO->]. This places an arrow after your formula, where you can enter a name for the value of the mean of list1. Type a name there (do not use the calculator domain variables like s, y, z, etc., but enter something like a by pressing [ALPHA] a [ENTER]). The mean of list1 is now stored in the variable a. Step 3. Generate nine more random samples. Find and store their means in additional variables. 1. Repeat Steps 1 and 2 to generate additional lists of random values, calculate the new mean of each list, and store each mean in a new variable. You may continue to use list1 to store the list of random values, but be sure to store each new mean before creating the next list of values. 2. For each random sample, store the mean in a new variable. Using the letters a, b, c, d, e, . . . should help you remember where you put them and make it easy to retrieve them. __________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated. 43 of 70 AP Statistics Page 2 of 4 Study Guide: Sampling Distributions and the Central Limit Theorem 3. Continue creating random samples and storing their means in variables until you have 10 random sample means. The table below shows each random sample and the variable letter in which its mean could be stored. Random Sample Variable 1 a 2 b 3 c 4 d 5 e 6 f 7 g 8 h 9 i 10 j Step 4: Find the mean of the sampling distribution. Now that you’ve stored the means of your 10 samples, you need to find the mean of the sampling distribution. 1. Enter List Editor. 2. Clear list2. On line 1 of list2 (list2[1]), enter [ALPHA] a to put the first mean (stored in variable a) in this cell. Then do the same to put the rest of the letters that have the stored means (b, c, d, etc.) on each line of list2. 3. Press [HOME] and enter mean(list2). Press [ENTER]. 4. This will be the mean of your sampling distribution. Section 2: For a population with N(50,5), what proportion of the sample means of size 36 would be expect to be less than 51? Step 1: Solve for P( x < 51) for N(50, (5/√36)). 1. From List Editor, press [F5]:Distr / 4:Normal Cdf…. 2. Enter a Lower Value of , an Upper Value of 51, a mean of 50, and a standard deviation of 5/√(36). 3. Press [ENTER] to save your settings, then [ENTER] again to execute the command. 4. You will see the answer to your question, .88493. Step 2: Graph P( x < 51) for N(50, .83). 1. Press [F1] and uncheck any plots or functions you have active. __________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated. 44 of 70 AP Statistics Page 3 of 4 Study Guide: Sampling Distributions and the Central Limit Theorem 2. Enter List Editor. 3. Press [F5]:Distr / 1:Shade 1: Shade Normal… [ENTER]. Enter a Lower Value of . 4. Enter an Upper Value of 51. 5. Enter a mean of 50 and a standard deviation of 5/6. 6. Make sure the Auto-scale setting is on YES. 7. Press [ENTER] to save your settings, then [ENTER] again to execute. 8. You should see the graph shown below. __________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) TI-89 screens are used with the permission of the publisher. Copyright © 2011, Texas Instruments, Incorporated. 45 of 70 AP Statistics Page 4 of 4 Study Guide: Sampling Distributions and the Central Limit Theorem Directions Use this guide to take notes while you watch the tutorial. Keep your completed study guide in your notebook for later study. Key Terms and Concepts sampling distribution of a sample mean • sampling distribution mean of the sampling distribution of a sample mean • Central Limit Theorem standard deviation of the sampling distribution of a sample mean Tutorial Sections The Meaning of a Sampling Distribution The Sampling Distribution of a Sample Mean The Central Limit Theorem __________________________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) TI-89 screens are used with the permission of the publisher. 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Consider a parent population with mean 75 and a standard deviation 7. The population doesn't appear to have extreme skewness or outliers. A. What are the mean and standard deviation of the distribution of sample means for n = 40? (2 points) B. What's the shape of the distribution? Explain your answer in terms of the Central Limit Theorem. (1 point) C. What proportion of the sample means of size 40 would you expect to be 77 or less? If you use your calculator, show what you entered. (1 point) D. Draw a sketch of the probability you found in part C. (1 point) 2. Suppose over several years of offering AP Statistics, a high school finds that final exam scores are normally distributed with a mean of 78 and a standard deviation of 6. A. What are the mean, standard deviation, and shape of the distribution of x-bar for n = 50? (2 points) B. What's the probability a sample of scores will have a mean greater than 80? (1 point) C. Sketch the distribution curve for part B, showing the area that represents the probability you found. (1 point) 3. Suppose college faculty members with the rank of professor at two-year institutions earn an average of $52,500 per year with a standard deviation of $4,000. In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all two-year institutions in the United States. A. What are the mean and standard deviation of the sampling distribution for n = 60? (2 points) B. What's the shape of the sampling distribution for n = 60? (1 point) C. Calculate the probability the sample mean x-bar is greater than $55,000. (1 point) D. If you drew a random sample with a mean of $55,000, would you consider this sample unusual? What conclusions might you draw? (1 point) 51 of 70 Page 2 of 3 4. A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check the quality of the paper, a random sample of ten pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. The distribution of strengths is known to be normal and the standard deviation, computed from many samples, is known to equal 2 pounds per square inch. The mean is known to be 21 pounds per square inch. A. What are the mean and standard deviation of the sampling distribution for n = 10? (2 points) B. What's the shape of the sampling distribution for n = 10? (1 point) C. Draw a random sample of size 10 from this distribution and compute the mean of the sample, using the randNorm function on your calculator. On a TI-83/84 store the values in L1: MATH [PRB] randNorm(21,2,10) STO [L1]. On a TI-89 Store the values in list1: seq(randNorm(21,2),x,1,10,1) -> list1. On the TI-83/84 you can then get the mean of the values by doing 1–Var stats on the values in L1. Then Press STAT, arrow to calc, select 1-Var stats, Press ENTER, then 2nd [L1] ENTER. On the TI-89 you can then get the mean of the values by doing 1-Var Stats on the values in list1. Enter List Editor, then press F4: 1:1-Var Stats.... Enter list1 as your list and press ENTER twice to do the calculation. Write down all your sample values. What are the mean and sample standard deviation of this sample? Compare them with the mean and standard deviation of the parent population, and explain why they're different or the same. (2 points) D. Using the mean and standard deviation of the sampling distribution (from part A), calculate the probability of getting a sample mean less than what you got. If you use a calculator, show what you entered. (2 points) 5. Suppose a random sample of n = 25 observations is selected from a normal population, with mean 106 and standard deviation 12. A. Find the probability that x-bar exceeds 110. Show the mean and standard deviation you used. (2 points) B. Find the probability that the sample mean deviates from the population mean = 106 by no more than 4. (1 point) C. Sketch the probability distribution from part B, showing the area that represents the probability you found. (1 point) 52 of 70 Page 3 of 3 Acknowledgements Question 3: This question is based on question 7.24 (c & d) from page 261 of Introduction to Probability and Statistics, Tenth Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning Incorporated. Further reproduction is prohibited without permission of the publisher. Question 4: This question is based on question 7.26 from page 262 of Introduction to Probability and Statistics, Tenth Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning Incorporated. Further reproduction is prohibited without permission of the publisher. Question 4: This question is based on question 7.20 from page 261 of Introduction to Probability and Statistics, Tenth Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning Incorporated. Further reproduction is prohibited without permission of the publisher. 53 of 70 6WDWLVWLFV6WXG\*XLGH 6DPSOH3URSRUWLRQV 3DJHRI 1.4.6 'LUHFWLRQV • 8VHWKLVJXLGHWRWDNHQRWHVZKLOH\RXZDWFKWKH7XWRULDO • .HHS\RXUFRPSOHWHG6WXG\*XLGHLQ\RXUQRWHERRNIRUODWHUVWXG\ .H\7HUPVDQG&RQFHSWV • pˆ GLVWULEXWLRQRI • PHDQRIWKHVDPSOLQJGLVWULEXWLRQRIDVDPSOHSURSRUWLRQ • VDPSOHSURSRUWLRQV • VWDQGDUGGHYLDWLRQRIWKHVDPSOLQJGLVWULEXWLRQRIDVDPSOHSURSRUWLRQ 7XWRULDO6HFWLRQV 7KH0HDQLQJRID6DPSOLQJ'LVWULEXWLRQRID6DPSOH3URSRUWLRQ ([HUFLVHV8VLQJWKH6DPSOH3URSRUWLRQ BBBBBBBBBBBBBBBBBBB &RS\ULJKW$3(;2QOLQH/HDUQLQJ,QF$OOULJKWVUHVHUYHG$GYDQFHG3ODFHPHQWDQG$3DUHUHJLVWHUHG WUDGHPDUNVRIWKH&ROOHJH%RDUG7KLVPDWHULDOLVLQWHQGHGIRUWKHH[FOXVLYHXVHRI$3(;HQUROOHGVWXGHQWV$Q\ XQDXWKRUL]HGFRS\LQJUHXVHRUUHGLVWULEXWLRQLVSURKLELWHG 54 of 70 S2 Statistics Assignment Sampling Distribution of p-hat The Sampling Distribution of p-hat Page 1 of 5 1.4.8 S2 ˆ is calculated from a count of successes to failures, the distribution of p ˆ is Because p binomial. You may recall that a binomial distribution of counts has a mean np and a standard deviation np(1 − p) . You may also recall that to turn a count of successes (x) into ˆ ), you divide the count x by the sample size n. Similarly, to turn np and a proportion ( p ˆ , divide each term by np(1 − p) into the mean and standard deviation of the proportion p n: µ pˆ = np = p , σ pˆ = n np(1 − p) n 2 = p(1 − p) . n p(1 − p) . Note n that the term (1 – p), which is the proportion or probability of failure, is written as q in pq many textbooks. So you may see the standard deviation written as . n ˆ is p, and the standard deviation is So, the mean of the distribution of p Using the Normal Approximation When the population is large relative to the sample, and when np ≥ 10 and n(1 – p) ≥ 10, a binomial distribution can be approximated by a normal distribution. This ˆ . When conditions are met, it has a distribution rule applies to the distribution of p N(p, p(1 − p) ). n NOTE: Some textbooks say you can use the normal approximation if np >5 and nq (or n(1p)) is > 5. These are also valid criteria, but for this Assignment use the criteria np ≥ 10 and n(1-p) ≥ 10. The Continuity Correction ˆ uses a continuous distribution to model a The normal approximation to the distribution of p discrete one. Use the continuity correction to offset the error inherent in these situations. To do this, simply proceed as though the interval you're finding also occupies the space .5 below the lower bound and .5 above the upper bound of the interval. ˆ , you need to convert your sample proportions to When using the continuity correction for p counts, and then use the normal approximation on these counts. For example, let's say the proportion of people who will vote for candidate A is .7. You draw a sample of 100 people, and your expected count of successes in a sample (which is the mean of your distribution of counts) is .7(100) = 70. You want the probability that you'd get a sample with between 66 and 68 successes. To use the continuity correction in this scenario, you'd find the normal probability for the interval from 65.5 to 68.5. _____________ © Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of registered users only. No portion of these materials may be70 reproduced or redistributed in any form without the 55 of express written permission of Apex Learning Inc. Statistics Assignment Sampling Distribution of p-hat Page 2 of 5 NOTE: As you advance into doing statistical inference, you may not see the continuity correction very often. In many cases it doesn't make enough of a difference to change your result. Using Your Calculator for the Normal Approximation to p-hat Use normalcdf(lowerbound,upperbound,mean,standard deviation). ˆ > p) = normalcdf(p,E99,µ,σ) • P( p ˆ < p) = normalcdf(–E99,p,µ,σ) Note that you can use any large number in place of • P( p E99. • P(p1 < pˆ < p2) = normalcdf(p1,p2,µ,σ) NOTE: If you don't enter the mean and standard deviation, the calculator assumes a standard normal distribution with mean 0 and standard distribution 1. In that case, you can enter z-scores. On your calculator, that'd be normalcdf(lower z,upper z). _____________ © Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of registered users only. No portion of these materials may be70 reproduced or redistributed in any form without the 56 of express written permission of Apex Learning Inc. Statistics Assignment Sampling Distribution of p-hat Page 3 of 5 Assignment Questions Answer each question completely and send your work back to your instructor. Show all work. Justify your answers clearly and completely with formulas and numerical solutions. Showing your work will allow your instructor to give you partial credit if you make a mistake but show that you still understand the problem. 1. In a survey, 600 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 70% said the genders are equal and should have equal opportunities to participate in sports. A. What are the mean, standard deviation, and shape of the distribution of the ˆ of parents who say the genders are equal and should have sample proportion p equal opportunities? Be sure to justify your answer for the shape of the distribution. Use n = 600. (1 point) B. Using the normal approximation without the continuity correction, sketch the ˆ . Shade equal areas on probability distribution curve for the distribution of p both sides of the mean to show an area that represents a probability of .95, and label the upper and lower bounds of the shaded area as values of p-hat (not z-scores). Show your calculations for the upper and lower bounds. (2 points) C. Considering the sketch in part B, the shaded area shows a .95 probability of what happening? In other words, what does the probability of .95 represent? (2 points) D. Using the normal approximation, what's the probability a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73%? Draw a sketch of the probability curve, shade the area representing the probability you're finding, and label the z-scores that represent the upper and lower bounds of the probability you're finding. Don't use the continuity correction. (2 points) E. Now, use the exact binomial calculation to find the probability of getting between, but not including, 67% and 73% of the respondents in a sample of 600 who say the genders are equal and should have equal opportunities. To use the exact binomial, you'll need to convert the proportions to counts by multiplying each proportion by 600. (1 point) F. Now try it again, but this time find the probability of getting at least 67% but no more than 73%. Use the exact binomial calculation. (1 point) _____________ © Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of registered users only. No portion of these materials may be70 reproduced or redistributed in any form without the 57 of express written permission of Apex Learning Inc. Statistics Assignment Sampling Distribution of p-hat Page 4 of 5 2. Random samples of size n = 500 were selected from a binomial population with p = .1. A. Is it appropriate in this case to use the normal distribution to approximate the ˆ ? What are the mean, standard deviation, and shape of the distribution of p distribution? (1 point) B. Using the normal approximation without the continuity correction, find the ˆ < .12. (1 point) probability that p ˆ < .12). This time use the continuity C. Using the normal approximation, find P( p correction. (3 points) Hint: To use the continuity correction you'll need to use the binomial distribution of counts: • Convert .12 to a count by multiplying it by 500. • Use np and np(1 − p) to find the mean and standard deviation for the normal • • approximation to the binomial for counts. You're looking for the probability of getting a count below a certain number. To use the continuity correction, add .5 to your true upper bound. (You don't need to increase your lower bound because the lower bound is essentially infinity.) ˆ < .12). Compare your answer D. Using the exact binomial calculation, find P( p with the answers you got for parts B and C. Why are they different? (1 point) 3. One of the ways Americans relieve stress is to reward themselves with sweets. Suppose a study claims 52% of Americans admit to overeating sweets when stressed. Suppose also that the 52% figure is correct for the population and that random samples of size n = 100 Americans are selected. ˆ have an approximately normal distribution? If so, A. Does the distribution of p what are its mean and standard deviation? (1 point) ˆ without the continuity correction, what's B. Using the normal approximation of p ˆ greater than .6? (1 point) the probability of getting a sample (n = 100) with p C. Using the normal approximation of the binomial distribution with the continuity ˆ greater correction, what's the probability of getting a sample (n = 100) with p than .6? (3 points) D. Using the exact binomial calculation, what's the probability of getting sample ˆ greater than .6? (1 point) (n = 100) with p _____________ © Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of registered users only. No portion of these materials may be70 reproduced or redistributed in any form without the 58 of express written permission of Apex Learning Inc. Statistics Assignment Sampling Distribution of p-hat Page 5 of 5 4. In 1996 there was a battle in the courts and in the marketplace between Intel and Digital Equipment Corp. about the technology behind Intel's Pentium microprocessing chip. Digital accused Intel of willful infringement on Digital's patents. Although Digital's Alpha microprocessor chip was the fastest on the market at the time, its speed fell victim to Intel's marketing clout. That same year, Intel shipped 76% of the microprocessor market. Suppose a random sample of n = 1,000 personal computer (PC) ˆ be the sales is monitored and the type of microprocessor installed is recorded. Let p proportion of personal computers in the sample with a Pentium microprocessor. ˆ ? (1 A. What are the mean, standard deviation, and shape of the distribution of p point) B. Using the normal approximation without the continuity correction, what's the probability you'd draw a random sample of 1,000 PCs with a proportion of Pentium chips exceeding 80%? (1 point) C. Looking at the answer you got for part B, if you got a simple random sample with ˆ > .8, would you conclude the population proportion may be higher than .76? p Why? (2 points) Acknowledgements Question 1: This question is based on question 7.35 from page 267 of Introduction to Probability and Statistics, Tenth Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning Incorporated. Further reproduction is prohibited without permission of the publisher. Question 3: This question is based on question 7.36 from page 267 of Introduction to Probability and Statistics, Tenth Edition, by W. Mendenhall, R. Beaver, and B. Beaver. Copyright © 1999 by Brooks Cole, division of Thompson Learning Incorporated. Further reproduction is prohibited without permission of the publisher. _____________ © Copyright 2000 Apex Learning Inc. All rights reserved. This material is intended for the exclusive use of registered users only. No portion of these materials may be70 reproduced or redistributed in any form without the 59 of express written permission of Apex Learning Inc. AP Statistics Review: Binomial Situations and Sampling Distributions Page 1 of 11 1.5.2 S2 Key Terms and Concepts Before taking the Quiz, you need to be able to explain the meanings (and recognize symbols in cases where there is an associated symbol) of each of these terms or concepts. You should also know when and how to use them in statistics problems. These terms and concepts are defined in Key Terms. almost binomial average waiting time binomcdf binomial experiment binomial probabilities for a range of outcomes binomial probabilities for an exact number of outcomes binomial setting binompdf Central Limit Theorem continuity correction geometcdf geometpdf geometric distribution geometric setting inferential statistics mean of the sampling distribution of a sample mean mean of the sampling distribution of a sample proportion normal approximation to the binomial ˆ (distribution of) p sample proportions sampling distribution sampling distribution of a sample mean standard deviation of the sampling distribution of a sample mean standard deviation of the sampling distribution of a sample proportion _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 60 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 2 of 11 Objectives, Example Problems, and Study Tips Introduction to Inferential Statistics Objective Define inferential statistics, and give at least two examples. Example 1. Define inferential statistics. Give two examples. Answer 1. Inferential statistics is the process of estimating population parameters from sample statistics. Examples include predicting population proportions, such as the percentage of voters who will vote for a particular candidate, and estimating the mean of a population parameter, such as the mean age of juniors at a high school. Objective Give one reason why probability distributions are important in inferential statistics. Example 1. Give one reason why probability distributions are important in inferential statistics. Tip The answer to this question will become much clearer as you learn more about inferential statistics. For now, it's helpful to have a general idea of why you're learning so much about normal, binomial, and geometric distributions. Answer 1. Probability distributions are used to calculate probabilities so that a researcher can determine the likelihood of an outcome. If a particular outcome is shown to be highly unlikely, then it's statistically significant and something other than chance is probably influencing the results. Objective: Outline the process that an inferential study often takes. Example 1. Outline the process that an inferential investigation often takes. Tip Don't memorize these steps word for word. They're only general guidelines, and you won't have to list them in the Unit Quiz. Once you learn how to do inferential statistics in different settings, you'll find that different cases require different methods. Answer 1. Inference takes many forms, but it often includes the following phases: A. B. C. D. Identify what you want to study. Ensure that the study and the sample data are valid. Use a probability distribution to calculate the likelihood of getting the same results. If you're comparing at least two measures, do significance testing. This involves writing hypotheses and testing them. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 61 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 3 of 11 Binomial Distributions Objective List the four conditions that define a binomial setting, and use those conditions to determine whether a situation is binomial. Examples 1. What are the conditions that define a binomial setting? 2. Suppose we know that 30% of all college students nationwide graduated from private high schools. A researcher wants to find the number of freshmen at a small college in Kansas who graduated from private high schools. The college has 300 freshmen. The researcher randomly selects a student from the freshman class and determines whether the student graduated from a private high school. The researcher repeats this process until 50 different students have been selected. Is this a binomial setting? 3. Consider the situation in the previous question. If the 50 students were sampled from the entire population of 1,250 students at the college, would this be a binomial setting? Tip In a sample-drawing situation, each trial may not be strictly independent. That's because the ratio of the population size to sample size will change each time a sample is drawn. However, we can still consider the setting binomial in certain situations. For instance, the setting is said to be binomial when the population is large compared to the sample. Although we've said in this course that a population 20 times larger than the sample is sufficient, some textbooks say that 10 times is acceptable. Answers 1. A setting is binomial if: A. There are only two possible outcomes for each trial (success or failure). B. Each trial is independent. C. The probability of success for each trial is the same. D. There are a set number of trials. For example, you roll a die 100 times and count the number of times you get a 6. This is a success/failure situation, since the die either lands on 6 or it doesn't. Each trial is independent, and the probability of success on each trial is 1/6. The set number of trials is 100. 2. This situation isn't binomial, since the probability for each trial isn't independent. Within the population of 300 freshmen, the proportion of unsampled students from public and private high schools will change slightly each time a student is drawn. This means that the probability of success for each trial is different. We'd still consider a setting to be binomial, even if each trial isn't independent, if the population is very large compared to the sample size. In this case, the population is only six times as large as the sample size, so we'd probably not consider this to be a binomial setting. 3. Since the population is more than 20 times as large as the sample, we'd consider this a binomial setting. In this course we said that the population must be at least 20 times larger than the sample. Some textbooks say it only needs to be 10 times greater. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 62 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 4 of 11 Objective Solve binomial problems involving at least, at most, and between. Examples 1. You have an octahedral die with faces A, B, C, D, E, F, G, and H. It's a fair diethat is, each face is equally likely to appear on top. You roll the die 75 times. What's the probability of getting exactly 9 A's? 2. For the die in question 1, what's the probability of getting at least 7 A's but less than 12 A's? 3. Suppose we know that 30% of all college students graduated from private high schools. A sample of 50 students is taken from the 1,200 students at a small college, and it's determined whether each student in the sample went to a private high school. Assuming that the 30% proportion applies to your population of interest, what's the probability that at least 18 of the students in the sample went to a private high school. Tips • At least 7 includes 7; less than 12 doesn't include 12. • Remember, the population size should be at least 20 times the sample size to use the binomial distribution. • Although you can solve most (but not all) binomial probability problems using your calculator, you'll have a much better grasp of the concept if you remember the formula: æn ö P( X = x) = çç ÷÷ p x (1 − p) n − x èxø where n = number of trials p = the probability of success for each trial x = number of successes Answers ( )( ) 9 æ 75 ö 7 66 = .139 . You can also use 1. Solve using n = 75, p = 1/8, x = 9. Thus, çç ÷÷ 1 8 è9 ø 8 binompdf(75,1/8,9). 2. This is difficult to do using the formula, but easy on a graphing calculator. If B(n, p, x) represents the probability of getting exactly x successes out of n occurrences of an event that occurs with probability p, then P(7 ≤ X < 12) = B(75, 1/8, 7) + B(75, 1/8, 8) + . . . + B(75, 1/8, 11) = binomcdf(75, 1/8, 11) – binomcdf(75, 1/8, 6), = .62 3. It's okay to use a binomial distribution here, since the population is at least 20 times greater than the sample size. Solve using P(X ≥ 18) = 1 – binomcdf(50,.3,17) = .21. Objective: Use the normal approximation in binomial setting problems. Examples 1. Out of 100 specimens of a certain type of rose bush, 85 will produce flowers annually. Use the normal approximation to calculate the probability that at least 90% of a plot of 250 plants will flower this year. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 63 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 5 of 11 2. Can we use a normal approximation to the binomial in the following situation: Statistics indicate roughly 20% of all adults jog more than twice a week. A sample of 18 adults is asked if they jog more than twice a week. What's the probability only three of these 18 adults jog more than twice a week? Tips • When using the normal approximation, don't forget about the continuity correction. • There are some rules of thumb for deciding if you can use a normal approximation. Some textbooks say that the expected numbers of successes and failures should be at least 5, or np ≥ 5 and n(1 – p) ≥ 5. But the Tutorials in this course use 10 instead of 5, or np ≥ 10 and n(1 – p) ≥ 10. Either rule will work, though in this course it's preferable to use 10. Answers 1. The answer is: µ x = 250(.85) = 212.5, σ x = 250(.85)(.15) = 5.65 . Thus .9(250) = 225, 224.5 − 212.5 = 2.12 . 5.65 So P(z > 2.12) = .017. Note that the exact answer is given by 1 – binomcdf(250, .85, 224) = .013. so now we find P(X ≥ 225), which is z225 = 2. You can't use a normal approximation here. Some texts will tell you that, to use a normal approximation, both np and n(1 – p) must be at least 10 (though some books say 5). Here np = 18(.2) = 3.6 < 10. Note: Your selections from the Mendenhall textbook may tell you that np and n(1 – p) must be at least 5, but in this course (and on the AP Exam) it's safer to use the rule that they should be at least 10. Geometric Distributions Objective List the four conditions that define a geometric setting. Use these conditions to determine whether a situation is geometric. Examples 1. Describe a geometric setting and explain how, if at all, it differs from a binomial setting. 2. An event occurs with probability p. On average, how long will we have to wait for the first success of the event? 3. A sack contains eight red dice and nine green dice. We remove one die at a time (and do not replace it) until we get a red die. Is this a geometric situation? Tip The expected value of X in a geometric setting is the average waiting time and is always 1/p. Answers 1. A random variable is binomial if the only two possible outcomes are success or failure, if it occurs a set number of times with each occurrence being independent of the others, and if each occurrence has the same probability of success. In a geometric setting, there _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 64 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 6 of 11 aren't a set number of trials. Rather, our interest is in the number of trials until the first success (or in some cases, a certain number of successes) occurs. However, like the binomial, each occurrence must be independent of the others and occur with the same probability each time. 2. The answer is: 1/p. 3. No, this isn't geometric. The probability of getting a red die when the first marble is drawn is 8/17. But the probability of getting a red die on the second draw is either 7/16 or 8/16 depending on the color of the die drawn first. Objective Solve problems involving the geometric distribution. Examples 1. There are 10 different prizes in boxes of Googily-Snaps. Prize #4 is the most valuable to collectors. What's the probability that you'll get prize #4 without having to buy more than four boxes of Googily-Snaps? (Assume that all prizes are equally likely in each box.) 2. Given the situation in question 1, suppose you're unlucky and don't get your prize in one the first four boxes. On average, how long will you have to wait to get prize #4? Tips While you can solve most geometric probability problems using your calculator, you'll have a much better grasp of the concept if you remember the formula: P( X = n) = (1 − p ) ⋅p where n = the number of the trial that yields the first success p = the probability of success for each trial n −1 Some geometric probability problems require you to know and understand the formula. Don't assume you can do it all on your calculator. Geometric distributions are skewed right. In other words, the probability of getting your first success decreases with each trial. The expected value for the number of trials is the average waiting time, or 1/p, where p is the probability of success for each trial. Answers 1. The answer is: P(X ≤ 4) = G(1) + G(2) + G(3) + G(4) = geometcdf(.1,4) = .3439. 2. This is asking for the average waiting time, which is 1/p. Thus 1/.1 = 10 boxes. Sampling Distributions: Means and Proportions Objective Define sampling distribution and use the definition to answer questions about sampling distributions. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 65 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 7 of 11 Examples 1. True or False: A sampling distribution of the mean retirement age in North America can consist of as few as 100 sample values. 2. Consider drawing samples of size 2 from {A,B,C,D,E} and computing the mean of each sample. The sampling distribution would consist of how many values? Tips If you're not sure what a sampling distribution is, look it up in the Glossary. Remember, a distribution is composed of all possible values rather than a subset of all possible values. In most cases, all possible values is a theoretical limit, and so the distribution is a theoretical construct rather than an actual set of values. Answers 1. False. A sampling distribution is composed of all possible values of a sample statistic using samples of the same size. Since the population of retirees in North America is composed of millions of individuals, 100 sample values is only a simulation of a distribution, it isn't the actual distribution. While 1,000 values may give a pretty good idea of what the distribution looks like, it's still just a simulation or an approximation unless it's composed of all possible values. 2. The sampling distribution would consist of 10 values, since there are 10 ways to select a æ5 ö sample of size 2 from a set with 5 elements ( çç ÷÷ = 10 ). Even though there are only 10 è2ø values in this distribution, it's a sampling distribution, since it's composed of all possible samples of a given size drawn from the population. Objective: Determine the shape, mean, and standard deviation of a sampling distribution. Examples 1. Samples of size 10 are drawn from a large (N > 10,000), symmetric population with a mean of 45 and a standard deviation of 9. What are the mean and standard deviation of the sampling distribution of the mean for samples of size 10, and what's the shape of the distribution? 2. A population has proportion .35 of some characteristic of interest. What are the mean ˆ for samples of size 50? What's and standard deviation of the sampling distribution of p ˆ the shape of the sampling distribution of p ? 3. A distribution is strongly skewed to the left. Samples of size n are drawn and the sampling distribution of the sample mean is constructed. What's the shape of the sampling distribution of x if n = 10, if n = 25, and if n = 100? Tips • The shape of the sampling distribution tends to resemble the shape of the parent population for small samples. ˆ , we must have np ≥ • To use the normal approximation to the sampling distribution of p 10 and n(1 – p) ≥ 10. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 66 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 8 of 11 • Remember that the Central Limit Theorem doesn't apply until the sample size (n) is large. In many cases, a sample size of 30 is large enough to make the sampling distribution close to normal. Answers 1. The answer is: µ x = 45, σ x = 9 = 2.85 . The sample size is small, so the shape will 10 resemble that of the parent population, that is, it will be symmetric about its mean. (.35)(1 − .35) = .067 . Samples of size 50 are relatively 50 large, so we'd expect the shape of the sampling distribution to be approximately normal. (Note that the assumptions we must make to be able to use the normal approximation ˆ are satisfied in this case.) to p 2. The answer is: µ pˆ = .35, σ pˆ = 3. For n = 10, we'd expect the sampling distribution to resemble the parent population. That is, it would still be skewed to the left, but would bunch more about the population mean than the original population. For n = 25, the distribution would be more normal than the original, but the extent to which it would resemble the original population depends on how severe the skewness is. We'd usually expect samples of size 25 to produce sampling distributions that are approximately normal. If n = 100, we'd certainly expect an approximately normal sampling distribution regardless of the shape of the original population. Objective Summarize the Central Limit Theorem and use elements of the theorem to answer questions about sampling distributions. Examples 1. True or False: The Central Limit Theorem tells us that the sampling distribution of a sample mean will be approximately normal regardless of the shape of the parent population. 2. We create a sampling distribution of x by taking samples of size 70 from a population whose mean is known to be 5 and whose standard deviation is 1. What can we say about the sampling distribution of x ? Tips Remember that the Central Limit Theorem doesn't apply until the sample size (n) is large. In many cases, a sample size of 30 is large enough to make the sampling distribution close to normal. For symmetric distributions with no outliers, samples smaller than 30 may be sufficient. If you can't easily summarize the Central Limit Theorem in your own words, look it up in the Glossary and review the Tutorial. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 67 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 9 of 11 Answers 1. False. The Central Limit Theorem tells us that the shape of the sampling distribution of a sample mean will be approximately normal for large samples. Just how large is large enough depends on the shape of the parent population, but samples of size 30 or greater are often large enough. For symmetric distributions with no outliers, even small sample sizes will yield approximately normal sampling distributions. 2. Because the sample size is large, the Central Limit Theorem applies, and you can say that the shape of the sampling distribution of x will be approximately normal even though you're given no information about the shape of the parent population. You also 1 = .12 . Note: The last two facts would be true know that µ x = 5 and that σ x = 70 regardless of the shape of the original distribution or the sample size. Objective Solve problems involving the sampling distribution of a sample mean or of a sample proportion. Examples 1. A popular soda comes in 12-oz cans. However, the actual volume of soda in the can varies normally with a mean of 11.9 oz and a standard deviation of .3 oz. What's the probability that the mean amount of soda in a six-pack is less than 12 oz? 2. A popular soda comes in 12-oz cans. However, the actual volume of soda in the can varies normally with a mean of 11.9 oz and a standard deviation of .3 oz. What's the probability that the mean amount of soda in a six-pack is between 11.7 oz and 12 oz? 3. The probability of winning at roulette is about .474. Suppose you bet 50 times. What's your probability of being even or ahead after 50 bets and after 1,000 bets? Tips • When you calculate the standard deviation of the sampling distribution, don't forget to divide by the square root of the sample size. Answers 1. You need to find the probability of getting a mean less than or equal to 12. 12 − 11.9 = .82 è Area = .793. P( x ≤ 12) = z12 = .3 6 _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 68 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 10 of 11 2. You need to find the probability of getting a mean between 11.7 and 12, inclusive. ) = .742. P(11.7 ≤ x ≤ 12) = normalcdf(11.7,12,11.9, .3 6 3. You need to find the probability of getting a proportion greater than or equal to .5, which ˆ ≥ .5). Because 50(.474) ≥ 10 and 50(1 – .474) ≥ 10, we can use the normal is P( p approximation to the sampling distribution of a sample proportion. This would be as follows: (.474)(1 − .474) For n = 50: µ pˆ = .474, σ pˆ = = .071 , normalcdf(.5,10,.474,.071) 50 = .36. For n = 1000: σ pˆ = (.474)(1 − .474) = .016 , normalcdf(.5,10,.474,.016) 1000 = .05. ___________ TI-83 screens are used with the permission of the publisher. Copyright ã 1996, Texas Instruments, Incorporated. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 69 of 70 AP Statistics Review: Binomial Situations and Sampling Distributions Page 11 of 11 About the Unit Quiz What to Bring • Scratch paper • Calculator • Approved formula sheet • Approved tables You can't have any reference materials other than those specifically mentioned above. You won't be able to ask for help during the Quiz. Hints and Tips for the Free-Response Portion • Show your work. The test corrector won't assume you used proper set up and methods if you reach the correct answer. It's up to you to communicate the methods that you used. Answers alone, without appropriate justification, will receive no credit. • Take your time reading the question. Since we want to see how well you can apply your knowledge to new and somewhat unfamiliar situations, take some time to think about the question. If you don't understand the question, you're unlikely to find the right answer. Read the entire question before beginning to answer. • Most questions will be given in several parts. The answers from one section will often be used in subsequent sections. Missing points in an early section does not mean you'll lose points in subsequent sections. Again, read the entire question to see how the different sections connect to each other. • The calculator. As in the AP Exam, this Quiz will test you on how well you know statistics, not on how well you can use your calculator. Be sure you understand the concepts behind the calculator operations. Don't use "calculator-speak" in your answer— the instructor doesn't want to read a set of steps for your graphing calculator! Use your calculator for doing the mechanics, but be sure to clearly communicate your process for solving the problem. • Use Units. If units are given in the problem, make sure that you give them in your answer. • Answer the Question. Finally, be very careful to answer the question asked. Before you move on, read over your answer to make sure you're providing exactly what the question asks for. Generally, an answer to a question you weren't asked will receive no credit. _____________ Copyright © 2011 Apex Learning Inc. (See Terms of Use at www.apexvs.com/TermsOfUse) 70 of 70