23.3 Sampling Distributions

Transcription

1 COMMON CORE Locker LESSON Commo Core Math Stadards The studet is expected to: COMMON CORE S-IC.B.4 Use data from a sample survey to estimate a populatio mea or proportio; develop a margi of error through the use of simulatio models for radom samplig. Mathematical Practices COMMON CORE 23.3 Samplig Distributios MP.6 Precisio Name Class Date 23.3 Samplig Distributios Essetial Questio: How is the mea of a samplig distributio related to the correspodig populatio mea or populatio proportio? Explore 1 Developig a Distributio of Sample Meas Resource Locker The tables provide the followig data about the first 50 people to joi a ew gym: member ID umber, age, ad sex. Laguage Objective Work with a parter to compare ad cotrast the stadard error of the mea ad the stadard error of the proportio. ENGAGE Essetial Questio: How is the mea of a samplig distributio related to the correspodig populatio mea or populatio proportio? The mea of the samplig distributio of the sample mea is equal to the populatio mea. Similarly, the mea of the samplig distributio of the sample proportio is equal to the populatio proportio. PREVIEW: LESSON PERFORMANCE TASK View the Egage sectio olie. Discuss the photo ad ote that the U.S. Cesus Bureau gathers data o the umber of people i each household ad the umber of households that have each umber of people, ad that samplig could be used to fid relevat probabilities. The preview the Lesso Performace Task. Image Credits: Juice Images/Shutterstock ID Age Sex ID Age Sex ID Age Sex ID Age Sex ID Age Sex 1 30 M F F M M 2 48 M M M F F 3 52 M F F M F 4 25 F M M M M 5 63 F F M M F 6 50 F M F F F 7 18 F F F F F 8 28 F F M F M 9 72 M M F F F F F M F F Eter the age data ito a graphig calculator ad fid the mea age μ ad stadard deviatio σ for the populatio of the gym s first 50 members. Roud each statistic to the earest teth. µ 41.2; σ 15.3 Use a graphig calculator s radom umber geerator to choose a sample of 5 gym members. Fid the mea age _ x for your sample. Roud to the earest teth. Aswers will vary. Possible aswer: x = 39.5 Module Lesso 3 DO NOT EDIT--Chages must be made through "File ifo" CorrectioKey=NL-A;CA-A Name Class Date 23.3 Samplig Distributios Essetial Questio: How is the mea of a samplig distributio related to the correspodig populatio mea or populatio proportio? Image Credits: Juice Images/Shutterstock S-IC.B.4 Use data from a sample survey to estimate a populatio mea or proportio; develop a margi of error through the use of simulatio models for radom samplig. Explore 1 Developig a Distributio of Sample Meas Resource The tables provide the followig data about the first 50 people to joi a ew gym: member ID umber, age, ad sex. ID Age Sex ID Age Sex ID Age Sex ID Age Sex ID Age Sex 1 30 M F F M M 2 48 M M M F F 3 52 M F F M F 4 25 F M M M M 5 63 F F M M F 6 50 F M F F F 7 18 F F F F F 8 28 F F M F M 9 72 M M F F F F F M F F Eter the age data ito a graphig calculator ad fid the mea age μ ad stadard deviatio σ for the populatio of the gym s first 50 members. Roud each statistic to the earest teth. µ 41.2; σ 15.3 Use a graphig calculator s radom umber geerator to choose a sample of 5 gym members. Fid the mea age _ x for your sample. Roud to the earest teth. Aswers will vary. Possible aswer: x = 39.5 Module Lesso 3 A2_MNLESE385900_U9M23L3.idd /2/14 2:55 PM HARDCOVER PAGES Tur to these pages to fid this lesso i the hardcover studet editio Lesso 23.3

2 C Report your sample mea to your teacher. As other studets report their sample meas, create a class histogram. To do so, shade a square above the appropriate iterval as each sample mea is reported. For sample meas that lie o a iterval boudary, shade a square o the iterval to the right. For istace, if the sample mea is 39.5, shade a square o the iterval from 39.5 to Aswers will vary. A possible Make your ow copy of the class histogram usig the grid show. histogram is show. Frequecy Frequecy Sample Mea D Calculate the mea of the sample meas, μ _ x, ad the stadard deviatio of the sample meas, σ _ x. E F Aswers will vary. Possible aswer: μ _ x = 39.8; σ _ x = 7.9 Now use a graphig calculator s radom umber geerator to choose a sample of 15 gym members. Fid the mea for your sample. Roud to the earest teth. Aswers will vary. Possible aswer: _ x = 40.5 Report your sample mea to your teacher. As other studets report their sample meas, create a class histogram ad make your ow copy of it. Aswers will vary. A possible histogram is show Sample Mea G Calculate the mea of the sample meas, μ x _, ad the stadard deviatio of the sample meas, σ x _. Aswers will vary. Possible aswer: μ x _ = 40.8; σ x _ = 3.4 EXPLORE 1 Developig a Distributio of Sample Meas INTEGRATE TECHNOLOGY Studets will use graphig calculators to first calculate the mea ad stadard deviatio of a populatio, the radomly select samples from that populatio, ad fially calculate statistics for the distributio of sample meas. This simulatio will eable them to see the effect of sample size o how closely the sample meas approximate the populatio mea. QUESTIONING STRATEGIES How ca you chage the way a sample is selected to make the sample mea better match the populatio mea? Icrease the sample size. As the sample size icreases, the sample mea approaches the populatio mea. Module Lesso 3 PROFESSIONAL DEVELOPMENT Learig Progressios I previous lessos, studets costructed probability distributios, ad they explored ormal distributios i detail. I this lesso, studets will costruct distributios of data obtaied from differet samples of the same populatio. Studets should uderstad that the mea ad stadard deviatio for a sample usually will ot match the statistics for the etire populatio or for a differet sample of the same populatio. Studets will lear how to use populatio statistics to determie the likelihood that a sample will have certai characteristics. I the ext lesso they will build o this kowledge to make predictios about populatio parameters from sample statistics. Samplig Distributios 1142

3 EXPLORE 2 Developig a Distributio of Sample Proportios QUESTIONING STRATEGIES What is the differece betwee this samplig distributio ad the oe i the first Explore activity? The first oe was a distributio of sample meas, while this is a distributio of sample proportios. INTEGRATE MATHEMATICAL PRACTICES Focus o Reasoig MP.2 Whe creatig a histogram of sample proportios for a sample size of 5, ask studets why there are oly a few differet values for the sample proportio, eve whe may samples are selected. Studets should recogize that the oly possible values for the proportio are 0, 0.2, 0.4, 0.6, 0.8 ad 1. v Reflect 1. I the class histograms, how does the mea of the sample meas compare with the populatio mea? The mea of the sample meas is close to the populatio mea. 2. What happes to the stadard deviatio of the sample meas as the sample size icreases? 3. What happes to the shape of the histogram as the sample size icreases? Explore 2 Developig a Distributio of Sample Proportios Use the tables of gym membership data from Explore 1. This time you will develop a samplig distributio based o a sample proportio rather tha a sample mea. A B C Fid the proportio p of female gym members i the populatio. Use a graphig calculator s radom umber geerator to choose a sample of 5 gym members. Fid the proportio p of female gym members for your sample. Report your sample proportio to your teacher. As other studets report their sample proportios, create a class histogram ad make your ow copy of it. Frequecy Sample Proportio D Calculate the mea of the sample proportios, μ p, ad the stadard deviatio of the sample proportios, σ p. Roud to the earest hudredth. E The stadard deviatio of the sample meas decreases. The histogram gets closer to the shape of a ormal distributio. p = 0.6 Aswers will vary. All sample proportios will be 0, 0.2, 0.4, 0.6, 0.8, or 1. Aswers will vary. A possible histogram is show. Aswers will vary. Possible aswer: µ p = 0.68; σ p = 0.21 Now use your calculator s radom umber geerator to choose a sample of 10 gym members. Fid the proportio of female members p for your sample. Aswers will vary. All sample proportios will be 0, 0.1, 0.2,, 0.9, or 1. Module Lesso 3 COLLABORATIVE LEARNING Peer-to-Peer Activity Have studets work i pairs. Give each pair the mea ad stadard deviatio for a populatio. Have each studet choose a sample size, determie the stadard error of the mea for that sample size, ad idetify a iterval that cotais either 68%, 95%, or 99.7% of the sample meas. Next, have studets tell their parters the sample size they used ad the upper ad lower bouds of the iterval they foud. Have each studet calculate the percet of sample meas that fall withi the parter s iterval. Have parters check each other s results Lesso 23.3

4 F Report your sample proportio to your teacher. As other studets report their sample proportios, create a class histogram ad make your ow copy of it. G Calculate the mea of the sample proportios, μ p, ad the stadard deviatio of the sample proportios, σ p. Roud to the earest hudredth. v Reflect Aswers will vary. A possible histogram is show. Aswers will vary. Possible aswer: µ p = 0.55; σ p = 0.16 Frequecy Sample Proportio 4. I the class histograms, how does the mea of the sample proportios compare with the populatio proportio? The mea of the sample proportios is close to the populatio proportio. 5. What happes to the stadard deviatio of the sample proportios as the sample size icreases? The stadard deviatio of the sample proportios decreases. Explai 1 Usig the Samplig Distributio of the Sample Mea The histograms that you made i the two Explores are samplig distributios. A samplig distributio shows how a particular statistic varies across all samples of idividuals from the same populatio. I Explore 1, you approximated samplig distributios of the sample mea, _ x, for samples of size 5 ad 15. (The reaso your samplig distributios are approximatios is that you did ot fid all samples of a give size.) The mea of the samplig distributio of the sample mea is deoted µ _ x. The stadard deviatio of the samplig distributio of the sample mea is deoted σ _ x ad is also called the stadard error of the mea. I Explore 1, you may have discovered that µ _ x is close to _ x regardless of the sample size ad that σ _ x decreases as the sample size icreases. You based these observatios o simulatios. Whe you cosider all possible samples of idividuals, you arrive at oe of the major theorems of statistics. Properties of the Samplig Distributio of the Sample Mea If a radom sample of size is selected from a populatio with mea μ ad stadard deviatio σ, the 1. µ _ x = µ, 2. σ _ x = σ_ _, ad 3. The samplig distributio of the sample mea is ormal if the populatio is ormal; for all other populatios, the samplig distributio of the mea approaches a ormal distributio as icreases. The third property stated above is kow as the Cetral Limit Theorem. EXPLAIN 1 Usig the Samplig Distributio of the Sample Mea QUESTIONING STRATEGIES If you create two samplig distributios by radomly selectig samples from the same populatio, with oe distributio havig samples of size 25 ad the other havig samples of size 100, how will the stadard errors of the meas of the two distributios compare? Explai. The stadard error of the mea for samples of 100 will be half as large as the stadard error for samples of 25. Stadard error is iversely proportioal to the square root of the sample size, so quadruplig the sample size reduces the stadard error by a factor of 2. INTEGRATE TECHNOLOGY Review how to use a graphig calculator to fid the probability that a value i a ormal distributio is betwee two specified values. Fid the cumulative distributio fuctio by goig to the DISTR meu ad selectig 2:ormalcdf(, ad eterig the give iformatio i the correct order (smaller value, larger value, mea, stadard deviatio) to fid the probability. To fid the probability that a quatity is less tha a give value, eter 1 EE 99 for the lower value. To fid the probability that a quatity is greater tha a give value, eter 1 EE 99 for the upper value. Module Lesso 3 Samplig Distributios 1144

5 AVOID COMMON ERRORS Remid studets that whe usig a calculator to fid the probability that the mea of a sample is betwee two give values (or above or below a give value), they must first calculate σ _ x, the stadard error of the mea, the use that value as the stadard deviatio that they eter o the calculator. If, istead, they use the stadard deviatio of the populatio, they will obtai a icorrect result. Example 1 Boxes of Crucho cereal have a mea mass of 323 grams with a stadard deviatio of 20 grams. For radom samples of 36 boxes, what iterval cetered o the mea of the samplig distributio captures 95% of the sample meas? Write the give iformatio about the populatio ad a sample. µ = 323 σ = 20 = 36 Fid the mea of the samplig distributio of the sample mea ad the stadard error of the mea. µ x _ = µ = 323 σ x _ = σ = 20 _ = _ The samplig distributio of the sample mea is approximately ormal. I a ormal distributio, 95% of the data fall withi 2 stadard deviatios of the mea. µ _ x - 2 σ _ x = (3.3) = µ _ x + 2 σ _ x = (3.3) = So, for radom samples of 36 boxes, 95% of the sample meas fall betwee grams ad grams. What is the probability that a radom sample of 25 boxes has a mea mass of at most 325 grams? Write the give iformatio about the populatio ad the sample. µ = 323 σ = 20 = 25 Fid the mea of the samplig distributio of the sample mea ad the stadard error of the mea µ x _ = µ = 323 σ x _ = σ = _ = _ = The samplig distributio of the sample mea is approximately ormal. Use a graphig calculator to fid P ( _ x 325). P ( x _ 325) = ormalcdf (-1e99, 325, 323, 4 ) 0.69 So, the probability that the radom sample has a mea mass of at most 325 grams is about Your Tur Boxes of Crucho cereal have a mea mass of 323 grams with a stadard deviatio of 20 grams. 6. For radom samples of 50 boxes, what iterval cetered o the mea of the samplig distributio captures 99.7% of the sample meas? µ x _ = µ = 323 σ _ x = σ = The samplig distributio of the sample mea is approximately ormal. I a ormal distributio, 99.7% of the data fall withi 3 stadard deviatios of the mea. µ _ x -3 σ _ x = (2.8) = µ _ x +3 σ _ x = (2.8) = So, for radom samples of 50 boxes, 99.7% of the sample meas fall betwee grams ad grams. Module Lesso Lesso 23.3

6 7. What is the probability that a radom sample of 100 boxes has a mea mass of at least 320 grams? µ x _ = µ = 323 σ x _ = σ = 20 = = 2 The samplig distributio of the sample mea is approximately ormal. P ( _ x 320) = ormalcdf (320, 1E99, 323, 2) 0.93 So, the probability that the radom sample has a mea mass of at least 320 grams is about Explai 2 Usig the Samplig Distributio of the Sample Proportio Whe you work with the samplig distributio of a sample proportio, p represets the proportio of idividuals i the populatio that have a particular characteristic (that is, the proportio of successes ) ad p is the proportio of successes i a sample. The mea of the samplig distributio of the sample proportio is deoted µ p. The stadard deviatio of the samplig distributio of the sample proportio is deoted σ p ad is also called the stadard error of the proportio. Properties of the Samplig Distributio of the Sample Proportio If a radom sample of size is selected from a populatio with proportio p of successes, the 1. µ p = p, 2. σ p (1 - _ p) p =, ad 3. if both p ad (1 p) are at least 10, the the samplig distributio of the sample proportio is approximately ormal. EXPLAIN 2 Usig the Samplig Distributio of the Sample Proportio QUESTIONING STRATEGIES How are the mea ad stadard deviatio of the samplig distributio of a sample proportio similar to ad differet from the mea ad stadard deviatio of the samplig distributio of a sample mea? I both cases, the mea of the samplig distributio is equal to the correspodig value (mea or proportio) for the populatio. However, the stadard deviatio is calculated somewhat differetly for sample meas ad sample proportios. Example 2 40% of the studets at a uiversity live off campus. Whe samplig from this populatio, cosider successes to be studets who live off campus. For radom samples of 50 studets, what iterval cetered o the mea of the samplig distributio captures 95% of the sample proportios? Write the give iformatio about the populatio ad a sample. p = 0.4 = 50 Fid the mea of the samplig distributio of the sample proportio ad the stadard error of the proportio. µ p = p = 0.4 σ p = p (1 - p) = 0.4 (1-0.4) Check that p ad (1 - p) are both at least 10. p = = 20 (1 - p) = = 30 Sice p ad (1 - p) are both greater tha 10, the samplig distributio of the sample proportio is approximately ormal. I a ormal distributio, 95% of the data fall withi 2 stadard deviatios of the mea. µ p - 2 σ p = (0.069) = µ p + 2 σ p = (0.069) = So, for radom samples of 50 studets, 95% of the sample proportios fall betwee 26.2% ad 53.8%. Module Lesso 3 LANGUAGE SUPPORT Commuicate Math Have studets work with a parter to compare ad cotrast the stadard error of the mea ad the stadard error of the proportio. Moitor studets use of termiology to lear what words ad cocepts are difficult for them to uderstad or to commuicate. INTEGRATE MATHEMATICAL PRACTICES Focus o Critical Thikig MP.3 Discuss with studets the criteria for determiig whether a samplig distributio is approximately ormal. If the populatio is ormally distributed, the samplig distributio of the sample mea will also be ormal. Also, the samplig distributio of a sample mea approaches a ormal distributio as the sample size icreases. The distributio is approximately ormal whe is very large, but we do ot have precise defiitios of approximately ormal or very large. For the samplig distributio of a sample proportio, how close the distributio is to ormal depeds o both ad the proportio p. If p is very large or very small, must be especially large for the distributio to be approximately ormal. We ca use the rule of thumb that says that if both p ad (1 - p) are at least 10, we ca cosider the distributio approximately ormal. Samplig Distributios 1146

7 B What is the probability that a radom sample of 25 studets has a sample proportio of at most 37%? Write the give iformatio about the populatio ad the sample, where a success is a studet who lives off campus. p = 0.4 = 25 Fid the mea of the samplig distributio of the sample proportio ad the stadard error of the proportio. μ p = p = σ p = p (1 - p) = 0.4 (1-0.4 ) 0.4 = Check that p ad (1 - p) are both at least 10. p = = 10 (1 - p) = = 15 Sice p ad (1 - p) are greater tha or equal to 10, the samplig distributio of the sample proportio is approximately ormal. Use a graphig calculator to fid P (p 0.37). P (p 0.37) = ormalcdf (-1e99, 0.37, 0.4, ) So, the probability that the radom sample has a sample proportio of at most 37% is about. YourTur 40% of the studets at a uiversity live off campus. Whe samplig from this populatio, cosider successes to be studets who live off campus. 8. For radom samples of 80 studets, what iterval cetered o the mea of the samplig distributio captures 68% of the sample proportios? µ p = p = 0.4 σ p = = 0.4 (1-0.4) Sice p = 32, (1 - p) = 48, ad both are greater tha 10, the samplig distributio is approximately ormal. I a ormal distributio, 68% of the data fall withi 1 stadard deviatio of the mea. µ p - σ p = = µ p + σ p = = So, for radom samples of 80 studets, 68% of the sample proportios fall betwee 34.5% ad 45.5%. Module Lesso Lesso 23.3

8 9. What is the probability that a radom sample of 60 studets icludes more tha 18 studets who live off campus? µ p = p = 0.4 σ p = = 0.4 (1-0.4) Elaborate Sice p = 24, (1 - p) = 36, ad both are greater tha 10, the samplig distributio is approximately 18_ ormal. p mi = 60 = 0.3 P (p > 0.3) = ormalcdf (0.3, 1e99, 0.4, 0.063) 0.94 So, the probability that the radom sample icludes more tha 18 studets who live off campus is about What is a samplig distributio? A samplig distributio is a distributio that shows how a particular statistic varies across all samples of idividuals from the same populatio. 11. What allows you to coclude that 95% of the sample meas i a samplig distributio are withi 2 stadard deviatios of the populatio mea? The Cetral Limit Theorem says that the samplig distributio of the sample mea is ormal or approximately ormal, so 95% of the sample meas will fall withi 2 stadard deviatios of the mea of the samplig distributio, but the mea of the samplig distributio is equal to the populatio mea, so 95% of the sample meas will fall withi 2 stadard deviatios of the populatio mea. 12. Whe fidig a sample mea or a sample proportio, why is usig the greatest sample size possible (give costraits o the cost ad time of samplig) a desirable thig to do? Icreasig the sample size decreases the variatio i the samplig distributio, which i tur meas that the sample mea or sample proportio will be more accurate because it is more likely to fall closer to the populatio mea or the populatio proportio. 13. Essetial Questio Check-I Whe you repeatedly take radom samples of the same size from a populatio, what does the mea of the samples approximate? The mea of the samples approximates the populatio mea (if the data are umerical) or the populatio proportio (if the data are categorical). ELABORATE INTEGRATE MATHEMATICAL PRACTICES Focus o Commuicatio MP.3 Ask studets to describe differet methods for determiig the probability that a sample mea or proportio falls betwee two give values. They should uderstad that if the give values are multiples of a stadard deviatio away from the mea of the samplig distributio, they ca use the kow percetages of data that fall withi 0, 1, 2, or 3 stadard deviatios of the mea. If ot, they ca calculate z-scores for the give values ad look up the correspodig probabilities o a stadard ormal table. Alteratively, they ca use the cumulative distributio fuctio o a graphig calculator. SUMMARIZE THE LESSON Give the mea ad stadard deviatio of a populatio, how ca you determie the mea ad stadard deviatio of a samplig distributio of sample meas? The mea of the samplig distributio of the sample mea is equal to the mea of the populatio. To fid the stadard deviatio, divide the populatio stadard deviatio by the square root of the sample size. Module Lesso 3 Samplig Distributios 1148

9 EVALUATE Evaluate: Homework ad Practice 1. The geeral maager of a multiplex theater took radom samples of size 10 from the audieces attedig the opeig weeked of a ew movie. From each sample, the maager obtaied the mea age ad the proportio of those who said they liked the movie. The sample meas ad sample proportios are listed i the tables. Olie Homework Hits ad Help Extra Practice ASSIGNMENT GUIDE Cocepts ad Skills Explore 1 Developig a Distributio of Sample Meas Explore 2 Developig a Distributio of Sample Proportios Example 1 Usig the Samplig Distributio of the Sample Mea Example 2 Usig the Samplig Distributio of the Sample Proportio AVOID COMMON ERRORS Practice Exercise 1 Exercise 18 Exercises 2 9 Exercises Whe fidig the probability that a sample has a value withi a give rage, remid studets to pay attetio to whether they are workig with sample meas or sample proportios, so that they use the correct formula for the stadard deviatio of the sample distributio. Sample umber Sample mea (age) Sample proportio (liked the movie) Sample umber Sample mea (age) Sample proportio (liked the movie) a. Based o the 20 samples, what is the best estimate for the mea age of all the people who saw the movie? Explai. The best estimate of the populatio mea is the mea of the sample meas: µ x _ = b. Based o this sample, what is the best estimate for the proportio of all the people who saw the movie ad liked it? Explai. The best estimate of the populatio proportio is the mea of the sample proportios: µ p = c. What could the maager have doe to improve the accuracy of both estimates? The maager could have improved the accuracy of the estimates by usig a sample size greater tha 10. Module Lesso 3 Exercise Depth of Kowledge (D.O.K.) COMMON CORE Mathematical Practices 1 1 Recall of Iformatio MP.5 Usig Tools Skills/Cocepts MP.5 Usig Tools Skills/Cocepts MP.6 Precisio Skills/Cocepts MP.5 Usig Tools 1149 Lesso 23.3

10 O a stadardized sciece test, the seiors at Fillmore High School have a mea score of 425 with a stadard deviatio of For radom samples of 30 seiors, what iterval cetered o the mea of the samplig distributio captures 95% of the mea scores? μ x _ = μ = 425; σ x _ = σ = μ x _ - 2σ x _ = (14.6) = μ _ x + 2σ _ x = (14.6) = So, for radom samples of 30 seiors, 95% of the sample meas fall betwee ad For radom samples of 100 seiors, what iterval cetered o the mea of the samplig distributio captures 68% of the mea scores? μ x _ = μ = 425; σ x _ = σ = 80 = μ x _ - σ x _ = = 417 μ x _ + σ x _ = = 433 So, for radom samples of 100 seiors, 68% of the sample meas fall betwee 417 ad What is the probability that a radom sample of 50 seiors has a mea score of at most 415? μ x _ = μ = 425; σ x _ = σ = _ P ( x 415) = ormalcdf (-1e99, 415, 425, 11.3) 0.19 So, the probability that a radom sample of 50 seiors has a mea score of at most 415 is about VISUAL CUES Whe fidig a iterval that captures a give percetage of the sample meas or proportios, ecourage studets to look back at the figure i the previous lesso that divides the ormal curve ito parts ad idicates what percet of the data i a ormal distributio are cotaied i each part. This visual represetatio may make it easier for visual learers to idetify the required iterval. 5. What is the probability that a radom sample of 25 seiors has a mea score of at least 430? μ x = μ = 425; σ x = σ_ = 80 = _ P ( x 430) = ormalcdf (430, 1e99, 425, 16) 0.38 So, the probability that a radom sample of 25 seiors has a mea score of at least 430 is about For Exercises 6 9, use the followig iformatio: The safety placard o a elevator states that up to 8 people (1200 kilograms) ca ride the elevator at oe time. Suppose the people who work i the office buildig where the elevator is located have a mea mass of 80 kilograms with a stadard deviatio of 25 kilograms. 6. For radom samples of 8 people who work i the office buildig, what iterval cetered o the mea of the samplig distributio captures 95% of the mea masses? μ x = μ = 80; σ_ = 25_ μ x - 2 σ x = 80-2 (8.8) = 62.4 μ x + 2 σ x = (8.8) = 97.6 So, for radom samples of 8 people who work i the office buildig, 95% of the sample meas fall betwee 62.4 kilograms ad 97.6 kilograms. Image Credits: Wested61 GmbH/Alamy Module Lesso 3 Exercise Depth of Kowledge (D.O.K.) COMMON CORE Mathematical Practices Skills/Cocepts MP.6 Precisio 18 2 Skills/Cocepts MP.3 Logic 19 2 Skills/Cocepts MP.3 Logic 20 3 Strategic Thikig MP.6 Precisio 21 3 Strategic Thikig MP.3 Logic Samplig Distributios 1150

11 AVOID COMMON ERRORS Whe usig a graphig calculator to determie the probability that a sample mea or proportio is above or below a give value, studets will obtai icorrect results if they eter the give iformatio i the wrog order. Remid them to read each problem carefully. If the problem asks about meas or proportios that are more tha a certai value, that value should be the first umber etered i the calculator s cumulative distributio fuctio. For questios about meas or proportios that are at most a certai value, that value should be the secod umber etered. 7. For radom samples of 8 people who work i the office buildig, what iterval cetered o the mea of the samplig distributio captures 99.7% of the mea masses? μ x = μ = 80; σ x = σ_ = 25_ μ x - 3 σ x = 80-3 (8.8) = 53.6 μ x + 3 σ x = (8.8) = So, for radom samples of 8 people who work i the office buildig, 99.7% of the sample meas fall betwee 53.6 kilograms ad kilograms. 8. What is the probability that a radom sample of 8 people who work i the office buildig has a mea mass of at most 90 kilograms? μ x = μ = 80; σ x = σ_ = 25_ P ( x 90) = ormalcdf (-1e99, 90, 80, 8.8) 0.87 So, the probability that a radom sample of 8 people who work i the office buildig has a mea mass of at most 90 kilograms is about Based o the elevator s safety placard, what is the maximum mea mass of 8 people who ca ride the elevator at oe time? What is the probability that a radom sample of 8 people who work i the office buildig exceeds this maximum mea mass? The _ maximum mea mass of 8 people who ca ride the elevator at oe time is 1200 = 150 kilograms. 8 μ x = μ = 80; σ x = σ_ = 25_ P ( x > 150) = ormalcdf (150, 1e99, 80, 8.8) So, the probability that a radom sample of 8 people who work i the office buildig has a mea mass that exceeds 150 kilograms is less tha oe quadrillioth A popcor maufacturer puts a prize i 25% of its bags of popcor. Whe samplig from this populatio, cosider successes to be bags of popcor cotaiig a prize. 10. For radom samples of 100 bags of popcor, what iterval cetered o the mea of the samplig distributio captures 95% of the sample proportios? μ p = p = 0.25; σ p = _ = 0.25 (1-0.25) Sice p = 25, (1 - p) = 75, ad both are greater tha 10, the samplig distributio is approximately ormal. μ p - 2 σ p = (0.043) = μ p + 2 σ p = (0.043) = So, for radom samples of 100 bags of popcor, 95% of the sample proportios fall betwee 16.4% ad 33.6%. Module Lesso Lesso 23.3

12 11. For radom samples of 80 bags of popcor, what iterval cetered o the mea of the samplig distributio captures 68% of the sample proportios? μ p = p = 0.25; σ p = _ = 0.25 (1-0.25) Sice p = 20, (1 - p) = 60, ad both are greater tha 10, the samplig distributio is approximately ormal. μ p - σ p = = μ p + σ p = = So, for radom samples of 80 bags of popcor, 68% of the sample proportios fall betwee 20.2% ad 29.8%. 12. What is the probability that a radom sample of 120 bags of popcor has prizes i at most 30% of the bags? μ p = p = 0.25; σ p = _ = 0.25 (1-0.25) Sice p = 30, (1 - p) = 90, ad both are greater tha 10, the samplig distributio is approximately ormal. P (p 0.3) = ormalcdf (-1e99, 0.3, 0.25, 0.04) 0.89 So, the probability that a radom sample of 120 bags of popcor has prizes i at most 30% of the bags is about What is the probability that a radom sample of 60 bags has prizes i more tha 12 bags? μ p = p = 0.25; σ p = _ = 0.25 (1-0.25) Sice p = 15, (1 - p) = 45, ad both are greater tha 10, the samplig distributio is approximately ormal. 12_ p mi = 60 = 0.2 P (p > 0.2) = ormalcdf (0.2, 1e99, 0.25, 0.056) 0.81 So, the probability that a radom sample of 60 bags of popcor has prizes i more tha 12 bags is about Module Lesso 3 Samplig Distributios 1152

13 About 28% of studets at a large school play varsity sports. Whe samplig from this populatio, cosider successes to be studets who play varsity sports. 14. For radom samples of 75 studets, what iterval cetered o the mea of the samplig distributio captures 95% of the sample proportios? μ p = p = 0.28 σ p = _ = 0.28 (1-0.28) Sice p = 21, (1 - p) = 54, ad both are greater tha 10, the samplig distributio is approximately ormal. μ p - 2 σ p = (0.052) = μ p + 2 σ p = (0.052) = So, for radom samples of 75 studets, 95% of the sample proportios fall betwee 17.6% ad 38.4%. 15. For radom samples of 100 studets, what iterval cetered o the mea of the samplig distributio captures 99.7% of the sample proportios? μ p = p = 0.28; σ p = _ = 0.28 (1-0.28) Sice p = 28, (1 - p) = 72, ad both are greater tha 10, the samplig distributio is approximately ormal. μ p - 3 σ p = (0.045) = Image Credits: bikeriderlodo/shutterstock μ p + 3 σ p = (0.045) = So, for radom samples of 100 studets, 99.7% of the sample proportios fall betwee 14.5% ad 41.5%. 16. What is the probability that a radom sample of 45 studets icludes more tha 18 studets who play varsity sports? μ p = p = 0.28; σ p = _ = 0.28 (1-0.28) Sice p = 12.6, (1 - p) = 32.4, ad both are greater tha 10, the samplig 18_ distributio is approximately ormal. p mi = 45 = 0.4 P (p > 0.4) = ormalcdf (0.4, 1e99, 0.28, 0.067) So, the probability that a radom sample of 45 studets icludes more tha 18 studets who play varsity sports is about Module Lesso Lesso 23.3

14 17. What is the probability that a radom sample of 60 studets icludes from 12 to 24 studets who play varsity sports? μ p = p = 0.28; σ p = _ = 0.28 (1-0.28) Sice p = 16.8, (1 - p) = 43.2, ad both are greater tha 10, the samplig 12_ distributio is 24_ approximately ormal. p mi = 60 = 0.2; p max = 60 = 0.4 P (0.2 p 0.4) = ormalcdf (0.2, 0.4, 0.28, 0.058) 0.90 So, the probability that a radom sample of 60 studets icludes from 12 to 24 studets who play varsity sports is about Amog the 450 seiors i a large high school, 306 pla to be i college i the fall followig high school graduatio. Suppose radom samples of 100 seiors are take from this populatio i order to obtai sample proportios of seiors who pla to be college i the fall. Which of the followig are true statemets? Select all that apply. a. Every sample proportio is b. The mea of the samplig distributio of the sample proportio is c. The stadard error of the proportio is about d. The stadard error of the proportio is about e. The samplig distributio of the sample proportio is skewed. f. The samplig distributio of the sample proportio is approximately ormal. _ The populatio proportio is p = 306 = This is also the mea of the samplig 450 distributio of the sample proportio. However, sample proportios obtaied from idividual samples will vary. So, A is false, ad B is true. For radom samples of size 100, the stadard error of the proportio is σ p = 0.68 (1-0.68) So, C is true, ad D is false. 100 Because p = 100 (0.68) = 68, (1 - p) = 100 (0.32) = 32, ad both are greater tha 10, the samplig distributio of the sample proportio is approximately ormal. So, E is false, ad F is true. Aswers: B, C, F INTEGRATE MATHEMATICAL PRACTICES Focus o Reasoig MP.2 To help studets see that the formula for σ p (the stadard error of the proportio) is actually a special case of the formula for σ _ x (the stadard error of the mea), you ca show them the derivatio of a formula for the stadard deviatio of a populatio cosistig of categorical data that either meet or do ot meet a criterio: Let each data value that meets the criterio be 1, ad let each data value that does t meet the criterio be 0. This effectively coverts the categorical data ito umerical data. Assume that a proportio p of the populatio has a value of 1, while a proportio 1 - p has a value of 0. The the mea value of the populatio is p (1) + (1 p) (0) = p. The deviatio of each 1 from the mea is 1 - p, while the deviatio of each 0 from the mea is 0 - p, or -p. The stadard deviatio ca be calculated by multiplyig each fractio of the populatio by its squared deviatio from the mea, the takig the square root: σ = 2 + (1 - p) (-p) 2, which ca be simplified to σ =. The stadard deviatio of the samplig distributio of the mea is, therefore, σ_ = p (1 - p) = p (1 - p), which exactly σ x _ = matches the formula for σ p. Module Lesso 3 Samplig Distributios 1154

15 JOURNAL Have studets create a flow chart showig how to determie the iterval that cotais 95% of the meas of radom samples of size take from a populatio whose mea ad stadard deviatio are kow. H.O.T. Focus o Higher Order Thikig 19. Explai the Error A studet was told that a populatio has a mea of 400 ad a stadard deviatio of 25. The studet was asked to fid the probability that a radom sample of size 45 take from the populatio has a mea of at most 401. The studet etered ormalcdf ( 1e99, 401, 400, 25) o a graphig calculator ad got a probability of about What did the studet do wrog? Show how to fid the correct aswer. The studet used the stadard deviatio of the populatio rather tha the stadard deviatio of the samplig distributio. 25_ σ σ x = = P ( x 401) = ormalcdf (-1e99, 401, 400, 3.73) Draw Coclusios Amada plas to use radom samplig to estimate the percet of people who are truly ambidextrous (that is, they do ot have a domiat right or left had). She suspects that the percet is quite low, perhaps as low as 1%. If she wats the samplig distributio of the sample proportio to be approximately ormal, what miimum sample size should she use? Explai. The samplig distributio of the sample proportio is approximately ormal provided p 10. Substitutig 0.01 for p ad solvig for gives So, the miimum sample size should be Check for Reasoableess Give that about 90% of people are right-haded, you are iterested i kowig what percet of people put their right thumb o top whe they clasp their hads. Havig o other iformatio to go o, you assume that people who put their right thumb o top whe they clasp their hads are those who are also right-haded. You the take a radom sample of 100 people ad fid that 60 put their right thumb o top whe they clasp their hads. Does this result lead you to questio your assumptio? Explai why or why ot. The assumptio that people who put their right thumb o top whe they clasp their hads are those who are also right-haded meas that the populatio proportio p for right thumb o top should equal 0.9. For radom samples of size 100, p = 100(0.9) = 90 ad (1 - p) = 100 (0.1) = 10, so the samplig distributio of the sample proportio should be approximately ormal with mea μ p = 0.9 ad stadard deviatio σ p = 0.9 (1-0.9) = Sice = -10, your radom sample has produced a sample proportio that is 10 stadard deviatios below the mea. This result is so ulikely that you should reject your assumptio. Module Lesso Lesso 23.3

16 Lesso Performace Task Amog the data that the U.S. Cesus Bureau collects are the sizes of households, as show i the table. Number of people i household Number of households 1 31,886, ,635, ,044, ,030, ,940, ,704,873 7 or more 1,749,501 Number of people 31,886,794 77,270,340 54,133,587 60,121,400 34,702,540 16,229,238 12,246,507 a. I the table above, assume that you ca simply use 7 as the umber of people i households with 7 or more people. Complete the third colum of the table. The use that colum to approximate the populatio mea μ (that is, the mea umber of people i a household). Explai your reasoig. Divide the total umber of people by the total umber of households. 31,886, ,270, ,133, ,121, ,702, ,229, ,246,507 31,886, ,635, ,044, ,030, ,940, ,704, ,749,501 = 286,590, ,991,725 So, the approximate populatio mea is about AVOID COMMON ERRORS Recommed that studets orgaize the iformatio they eed for the Lesso Performace Task i a table. Lower boud Upper boud Populatio mea Stadard deviatio This will help studets eter ad track umbers i their correct order. INTEGRATE MATHEMATICAL PRACTICES Focus o Critical Thikig MP.3 Whe usig the ormalcdf(fuctio o a graphig calculator, uder what circumstaces would you eter the desired mea of the sample first? Whe it is the least umber i the rage of data uder cosideratio, eter the desired mea of the sample populatio first. b. Is the actual populatio mea greater tha or less tha the mea that you calculated? Explai. The actual populatio mea is greater because the 7 or more category icludes households with 7 people, 8 people, 9 people, ad so o. So, the actual umber of people livig i those households is greater tha 12,246,507, which makes the umerator of 286,590, ,991,725 greater, which i tur makes the mea greater. c. Give that μ 2.49 ad σ 1.42, fid the probability that a radom sample of 100 households i the Uited States has a mea size of 2.3 people or less. μ x = μ 2.49; σ x = σ_ _ 1.42 = x 2.3) = ormalcdf (-1e99, 2.3, 2.49, 0.142) 0.09 P ( d. Give that μ 2.49 ad σ 1.42, fid the probability that a radom sample of 100 households i the Uited States has a mea size of 2.6 people or more. μ x = μ 2.49; σ x = σ_ = _ 1.42 = x 2.6) = ormalcdf (2.6, 1e99, 2.49, 0.142) 0.22 P ( Module Lesso 3 EXTENSION ACTIVITY Have differet studets choose a sample of either 200, 400, or 600 households; differet rages for the umber of people i the household (for example 1 to 3); ad the fid the probability that, out of a radom sample of that may households i the Uited States, the mea household size will be withi that rage. Have studets compare their results ad how the differet sample sizes affected their results. Scorig Rubric 2 poits: Studet correctly solves the problem ad explais his/her reasoig. 1 poit: Studet shows good uderstadig of the problem but does ot fully solve or explai his/her reasoig. 0 poits: Studet does ot demostrate uderstadig of the problem. Samplig Distributios 1156