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Additional tasks describe the sampling distribution of song mean lengths, then estimate the probability that total music time for 40 songs exceeds available airtime given show scheduling constraints, followed by provided answer references.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/ap-exam-review-sampling-distributions-answer-key/314013/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/ap-exam-review-sampling-distributions-answer-key/314013.png","ImageObject",300,407,{"name":42,"@type":43},"Arica Lee","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-10-21","2026-09-20",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",9,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How do you decide whether a sample mean greater than 10 is more likely for n = 15 or n = 50?","Question",{"text":63,"@type":64},"Compare how sample size affects variability of the sample mean. The larger sample size typically produces a more concentrated sampling distribution around the population mean, so the probability of exceeding 10 changes accordingly.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How is the probability P(sample mean \u003C 7.5) for n = 50 computed?",{"text":68,"@type":64},"Use the normal distribution for the sampling distribution of the sample mean. Convert 7.5 to a z-score using the given standard deviation (0.3 inch) and then evaluate the corresponding left-tail probability.",{"name":70,"@type":61,"acceptedAnswer":71},"What is the sampling distribution of the sample mean song lengths for samples of 40 songs?",{"text":72,"@type":64},"It is centered at the population mean (3.9 minutes) with a variability determined by the population standard deviation (1.1 minutes) and the sample size. 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Big Town Fisheries recently stocked a new lake in a city park with 2,000 fish of various sizes. The distribution of the lengths of these fish is approximately normal.  \n(a) Big Town Fisheries claims that the mean length of the fish is 8 inches . If the claim is true, which of the following would be more likely?  \n• A random sample of 15 fish having a mean length that is greater than 10 inches or  \n• A random sample of 50 fish having a mean length that is greater than 10 inches Justify your answer.  \n(b) Suppose the standard deviation of the sampling distribution of the sample mean for random samples of size 50 is 0.3 inch. If the mean length of the fish is 8 inches, use the normal distribution to compute the probability that a random sample of 50 fish will have a mean length less than 7.5 inches .  \nAPExam Review Page 1 of 6 Sampling Distributions  \n2010 AP® STATISTICS FREE-RESPONSE QUESTIONS  \n2. A local radio station plays 40 rock-and-roll songs during each 4-hour show. The program director at the station needs to know the total amount of airtime for the 40 songs so that time can also be programmed during the show for news and advertisements. The distribution of the lengths of rock-and-roll songs, in minutes, is roughly symmetric with a mean length of 3.9 minutes and a standard deviation of 1.1 minutes.  \n(a) Describe the sampling distribution of the sample mean song lengths for random samples of 40 rock-and-roll songs.  \n(b) If the program manager schedules 80 minutes of news and advertisements for the 4-hour (240-minute) show, only 160 minutes are available for music. Approximately what is the probability that the total amount of time needed to play 40 randomly selected rock-and-roll songs exceeds the available airtime?  \nAPExam Review Page 2 of 6 Sampling Distributions  \n2009 AP® STATISTICS FREE-RESPONSE QUESTIONS  \nAPExam Review Page 3 of 6 Sampling Distributions  \nANSWERS: AP Statistics Exam FR: Sampling Distributions  \n2007 \\#3 (Big Town Fisheries)  \nAPExam Review Page 4 of 6 Sampling Distributions  \n2010 \\#2 (Radio Station)  \nAPExam Review Page 5 of 6 Sampling Distributions  \n2009 \\#2 (Tire Manufacturer)  \nAPExam Review Page 6 of 6 Sampling Distributions","cbCaimfbdpOB77L0","https://ap.wps.com/l/cbCaimfbdpOB77L0","pdf",458182,"English","# AP Statistics Free-Response Questions\n## 2007 #3 Big Town Fisheries\n## 2010 #2 Radio Station\n## 2009 #2 Tire Manufacturer\n## Answers and References","[{\"question\":\"How do you decide whether a sample mean greater than 10 is more likely for n = 15 or n = 50?\",\"answer\":\"Compare how sample size affects variability of the sample mean. The larger sample size typically produces a more concentrated sampling distribution around the population mean, so the probability of exceeding 10 changes accordingly.\"},{\"question\":\"How is the probability P(sample mean \\u003c 7.5) for n = 50 computed?\",\"answer\":\"Use the normal distribution for the sampling distribution of the sample mean. Convert 7.5 to a z-score using the given standard deviation (0.3 inch) and then evaluate the corresponding left-tail probability.\"},{\"question\":\"What is the sampling distribution of the sample mean song lengths for samples of 40 songs?\",\"answer\":\"It is centered at the population mean (3.9 minutes) with a variability determined by the population standard deviation (1.1 minutes) and the sample size. The distribution shape is described based on the given roughly symmetric population assumption.\"}]","AP Exam Review - Sampling Distributions - Answer Key | PDF",1789899846,15]