Any experiment that has characteristics two and three and where n<\/em> = 1 is called a Bernoulli Trial<\/span> (named after Jacob Bernoulli who, in the late 1600s, studied them extensively). A binomial experiment takes place when the number of successes is counted in one or more Bernoulli Trials.<\/p>\r\n\r\n The state health board is concerned about the amount of fruit available in school lunches. Forty-eight percent of schools in the state offer fruit in their lunches every day. This implies that 52% do not. What would a \"success\" be in this case?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n A trainer is teaching a dolphin to do tricks. The probability that the dolphin successfully performs the trick is 35%, and the probability that the dolphin does not successfully perform the trick is 65%. Out of 20 attempts, you want to find the probability that the dolphin succeeds 12 times. State the probability question mathematically.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n A fair, six-sided die is rolled ten times. Each roll is independent. You want to find the probability of rolling a one more than three times. State the probability question mathematically.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n a. This is a binomial problem because there is only a success or a __________, there are a fixed number of trials, and the probability of a success is 0.70 for each trial.<\/p>\r\n\r\n<\/div>\r\n b. If we are interested in the number of students who do their homework on time, then how do we define X<\/em>?<\/p>\r\n\r\n<\/div>\r\n b. X<\/em> = the number of statistics students who do their homework on time<\/p>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n c. What values does x<\/em> take on?<\/p>\r\n\r\n<\/div>\r\n c. 0, 1, 2, \u2026, 50<\/p>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n d. What is a \"failure,\" in words?<\/p>\r\n\r\n<\/div>\r\n The probability of a success is p<\/em> = 0.70. The number of trials is n<\/em> = 50.<\/p>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n e. If p<\/em> + q<\/em> = 1, then what is q<\/em>?<\/p>\r\n\r\n<\/div>\r\n e. q<\/em> = 0.30<\/p>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n Sixty-five percent of people pass the state driver\u2019s exam on the first try. A group of 50 individuals who have taken the driver\u2019s exam is randomly selected. Give two reasons why this is a binomial problem.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n It has been stated that about 41% of adult workers have a high school diploma but do not pursue any further education. If 20 adult workers are randomly selected, find the probability that at most 12 of them have a high school diploma but do not pursue any further education. How many adult workers do you expect to have a high school diploma but do not pursue any further education?<\/p>\r\nLet X<\/em> = the number of workers who have a high school diploma but do not pursue any further education.\r\n X<\/em> takes on the values 0, 1, 2, ..., 20 where n<\/em> = 20, p<\/em> = 0.41, and q<\/em> = 1 \u2013 0.41 = 0.59. X<\/em> ~ B<\/em>(20, 0.41)<\/p>\r\n Find P<\/em>(x<\/em> \u2264 12). P<\/em>(x<\/em> \u2264 12) = 0.9738. (calculator or computer)<\/p>\r\n\r\n Go into 2nd<\/sup> DISTR. The syntax for the instructions are as follows:<\/p>\r\n To calculate (x<\/em> = value): binompdf(n<\/em>, p<\/em>, number)<\/strong> if \"number\" is left out, the result is the binomial probability table. \r\n<\/span>To calculate P<\/em>(x<\/em> \u2264 value): binomcdf(n<\/em>, p<\/em>, number)<\/strong> if \"number\" is left out, the result is the cumulative binomial probability table. \r\n<\/span>For this problem: After you are in 2nd<\/sup> DISTR, arrow down to binomcdf. Press ENTER. Enter 20,0.41,12). The result is P<\/em>(x<\/em> \u2264 12) = 0.9738.<\/strong><\/p>\r\n\r\n<\/div>\r\n If you want to find P<\/em>(x<\/em> = 12), use the pdf (binompdf). If you want to find P<\/em>(x<\/em> > 12), use 1 - binomcdf(20,0.41,12).<\/p>\r\n\r\n<\/div>\r\nThe probability that at most 12 workers have a high school diploma but do not pursue any further education is 0.9738.\r\n The graph of X<\/em> ~ B<\/em>(20, 0.41) is as follows:<\/p>\r\n\r\n The y<\/em>-axis contains the probability of x<\/em>, where X<\/em> = the number of workers who have only a high school diploma.<\/p>\r\nThe number of adult workers that you expect to have a high school diploma but not pursue any further education is the mean, \u03bc<\/em> = np<\/em> = (20)(0.41) = 8.2.\r\n\r\nThe formula for the variance is \u03c32<\/sup> = npq<\/em>. The standard deviation is \u03c3<\/em> = \\(\\sqrt{npq}\\). \r\n<\/span>\u03c3<\/em> = \\(\\sqrt{\\left(20\\right)\\left(0.41\\right)\\left(0.59\\right)}\\) = 2.20.\r\n\r\n<\/div>\r\n About 32% of students participate in a community volunteer program outside of school. If 30 students are selected at random, find the probability that at most 14 of them participate in a community volunteer program outside of school. Use the TI-83+ or TI-84 calculator to find the answer.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n In the 2013 Jerry\u2019s Artarama<\/em> art supplies catalog, there are 560 pages. Eight of the pages feature signature artists. Suppose we randomly sample 100 pages. Let X<\/em> = the number of pages that feature signature artists.<\/p>\r\n\r\n The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Suppose we randomly sample 200 people. Let X<\/em> = the number of people who will develop pancreatic cancer.<\/p>\r\n\r\n During the 2013 regular NBA season, DeAndre Jordan of the Los Angeles Clippers had the highest field goal completion rate in the league. DeAndre scored with 61.3% of his shots. Suppose you choose a random sample of 80 shots made by DeAndre during the 2013 season. Let X<\/em> = the number of shots that scored points.<\/p>\r\n\r\nNotation for the Binomial: B<\/em> = Binomial Probability Distribution Function<\/h3>\r\nX<\/em> ~ B<\/em>(n<\/em>, p<\/em>)\r\n\r\nRead this as \"X<\/em> is a random variable with a binomial distribution.\" The parameters are n<\/em> and p<\/em>; n<\/em> = number of trials, p<\/em> = probability of a success on each trial.\r\n\r\n<\/div>\r\n
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