Question 1<\/h3>\r\nRecall what you learned in Preview Assignment 8.F. What did you discover about the role that\u00a0[latex] n [\/latex] plays in the shape of the binomial distribution?\r\n\r\n<\/div>\r\nFor a fixed [latex] p [\/latex], as the number of trials, [latex] n [\/latex], in a binomial experiment increases, the probability distribution of the random variable\u00a0[latex] X [\/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex] np \\geq 10 \\mbox{ and } n(1-p) \\geq 10 [\/latex], the probability distribution will be approximately symmetric and bell shaped.\r\n\r\nIn other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex] np [\/latex], and the expected number of failures, [latex] n(1-p) [\/latex], are both at least 10.\r\n\r\nQuestion 2<\/h3>\r\nRecall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?\r\n\r\n<\/div>\r\nIn addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.\r\n\r\nIn probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex] X [\/latex], is represented in the normal distribution by the interval from [latex] X - 0.5 \\mbox{ to } X + 0.5[\/latex].\r\n\r\nWhy is this necessary? Consider the current example:\r\n\r\nThe probability that George successfully makes exactly 270 free throws is [latex] P(X= 270) = 0.0766 [\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex] P(X= 270.2) [\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.\r\n\r\nNow suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex] X [\/latex] is a continuous random variable.\r\n\r\nThe probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:\r\n\r\n[latex] -P(X<a), P(X>a), \\mbox{ or } P(a<X<b) [\/latex]\r\n\r\nFor a continuous random variable, [latex] P(X = a) = 0 [\/latex] because there are an infinite number of possible numbers on any interval. Also, [latex] P( X\\leq a) = P(X<a) [\/latex] for a [\/latex]continuous random variable.\r\n\r\nThus, we will use [latex] P(269.5 <X < 270.5) [\/latex] to approximate [latex] P(X = 270)[\/latex].\r\n\r\nQuestion 3<\/h3>\r\n\r\n\r\n\r\n\r\nExact Probability\u00a0 Using Binomial<\/td>\r\n Approximate\r\n\r\nProbability Using\r\n\r\nNormal<\/td>\r\n Graphical Depiction<\/td>\r\n<\/tr>\r\n \r\n[latex] P(X = 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X \\leq 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X > 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] (260 < X < 280) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nQuestion 4<\/h3>\r\n4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.\r\n\r\n \t- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\r\n \t
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\r\n \t
- Find the probability that George will make more than 285 free throws.<\/li>\r\n \t
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\r\n \t
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n \r\n\r\n ","rendered":"
In a previous in-class activity, you learned about the role that [latex]p[\/latex] plays in the shape of the binomial distribution. The binomial probability distribution is skewed right if [latex]p < 0.5[\/latex], symmetric and approximately bell shaped if [latex]p= 0.5[\/latex], and skewed left if [latex]p > 0.5[\/latex]. Let\u2019s discuss the role that [latex]n[\/latex] plays in its shape.<\/p>\n
![\"\"](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5738\/2022\/08\/18174835\/8F-InClass-1-300x193.png\")
<\/p>\n
Credit: iStock\/skynesher<\/p>\n
\nQuestion 1<\/h3>\n
Recall what you learned in Preview Assignment 8.F. What did you discover about the role that\u00a0[latex]n[\/latex] plays in the shape of the binomial distribution?<\/p>\n<\/div>\n
For a fixed [latex]p[\/latex], as the number of trials, [latex]n[\/latex], in a binomial experiment increases, the probability distribution of the random variable\u00a0[latex]X[\/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex]np \\geq 10 \\mbox{ and } n(1-p) \\geq 10[\/latex], the probability distribution will be approximately symmetric and bell shaped.<\/p>\n
In other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex]np[\/latex], and the expected number of failures, [latex]n(1-p)[\/latex], are both at least 10.<\/p>\n
\nQuestion 2<\/h3>\n
Recall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?<\/p>\n<\/div>\n
In addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\n
In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[\/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \\mbox{ to } X + 0.5[\/latex].<\/p>\n
Why is this necessary? Consider the current example:<\/p>\n
The probability that George successfully makes exactly 270 free throws is [latex]P(X= 270) = 0.0766[\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex]P(X= 270.2)[\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.<\/p>\n
Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[\/latex] is a continuous random variable.<\/p>\n
The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:<\/p>\n
[latex]-P(Xa), \\mbox{ or } P(a\nQuestion 3<\/h3>\n\n\n\n\nExact Probability\u00a0 Using Binomial<\/td>\n Approximate<\/p>\nProbability Using<\/p>\n
Normal<\/td>\n
Graphical Depiction<\/td>\n<\/tr>\n \n[latex]P(X = 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X \\leq 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X > 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex](260 < X < 280)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n <\/p>\n
\nQuestion 4<\/h3>\n
4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.<\/p>\n
\n- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\n
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\n
- Find the probability that George will make more than 285 free throws.<\/li>\n
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\n
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\n<\/ol>\n<\/div>\n
<\/p>\n
<\/p>\n
<\/p>\n","protected":false},"author":574340,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5159","chapter","type-chapter","status-publish","hentry"],"part":4997,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/574340"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions"}],"predecessor-version":[{"id":5166,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions\/5166"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/4997"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5159"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5159"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5159"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Question 2<\/h3>\r\nRecall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?\r\n\r\n<\/div>\r\nIn addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.\r\n\r\nIn probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex] X [\/latex], is represented in the normal distribution by the interval from [latex] X - 0.5 \\mbox{ to } X + 0.5[\/latex].\r\n\r\nWhy is this necessary? Consider the current example:\r\n\r\nThe probability that George successfully makes exactly 270 free throws is [latex] P(X= 270) = 0.0766 [\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex] P(X= 270.2) [\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.\r\n\r\nNow suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex] X [\/latex] is a continuous random variable.\r\n\r\nThe probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:\r\n\r\n[latex] -P(X<a), P(X>a), \\mbox{ or } P(a<X<b) [\/latex]\r\n\r\nFor a continuous random variable, [latex] P(X = a) = 0 [\/latex] because there are an infinite number of possible numbers on any interval. Also, [latex] P( X\\leq a) = P(X<a) [\/latex] for a [\/latex]continuous random variable.\r\n\r\nThus, we will use [latex] P(269.5 <X < 270.5) [\/latex] to approximate [latex] P(X = 270)[\/latex].\r\n\r\nQuestion 3<\/h3>\r\n\r\n\r\n\r\n\r\nExact Probability\u00a0 Using Binomial<\/td>\r\n Approximate\r\n\r\nProbability Using\r\n\r\nNormal<\/td>\r\n Graphical Depiction<\/td>\r\n<\/tr>\r\n \r\n[latex] P(X = 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X \\leq 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X > 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] (260 < X < 280) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nQuestion 4<\/h3>\r\n4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.\r\n\r\n \t- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\r\n \t
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\r\n \t
- Find the probability that George will make more than 285 free throws.<\/li>\r\n \t
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\r\n \t
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n \r\n\r\n ","rendered":"
In a previous in-class activity, you learned about the role that [latex]p[\/latex] plays in the shape of the binomial distribution. The binomial probability distribution is skewed right if [latex]p < 0.5[\/latex], symmetric and approximately bell shaped if [latex]p= 0.5[\/latex], and skewed left if [latex]p > 0.5[\/latex]. Let\u2019s discuss the role that [latex]n[\/latex] plays in its shape.<\/p>\n
<\/p>\n
Credit: iStock\/skynesher<\/p>\n
\nQuestion 1<\/h3>\n
Recall what you learned in Preview Assignment 8.F. What did you discover about the role that\u00a0[latex]n[\/latex] plays in the shape of the binomial distribution?<\/p>\n<\/div>\n
For a fixed [latex]p[\/latex], as the number of trials, [latex]n[\/latex], in a binomial experiment increases, the probability distribution of the random variable\u00a0[latex]X[\/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex]np \\geq 10 \\mbox{ and } n(1-p) \\geq 10[\/latex], the probability distribution will be approximately symmetric and bell shaped.<\/p>\n
In other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex]np[\/latex], and the expected number of failures, [latex]n(1-p)[\/latex], are both at least 10.<\/p>\n
\nQuestion 2<\/h3>\n
Recall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?<\/p>\n<\/div>\n
In addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\n
In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[\/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \\mbox{ to } X + 0.5[\/latex].<\/p>\n
Why is this necessary? Consider the current example:<\/p>\n
The probability that George successfully makes exactly 270 free throws is [latex]P(X= 270) = 0.0766[\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex]P(X= 270.2)[\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.<\/p>\n
Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[\/latex] is a continuous random variable.<\/p>\n
The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:<\/p>\n
[latex]-P(Xa), \\mbox{ or } P(a\nQuestion 3<\/h3>\n\n\n\n\nExact Probability\u00a0 Using Binomial<\/td>\n Approximate<\/p>\nProbability Using<\/p>\n
Normal<\/td>\n
Graphical Depiction<\/td>\n<\/tr>\n \n[latex]P(X = 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X \\leq 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X > 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex](260 < X < 280)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n <\/p>\n
\nQuestion 4<\/h3>\n
4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.<\/p>\n
\n- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\n
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\n
- Find the probability that George will make more than 285 free throws.<\/li>\n
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\n
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\n<\/ol>\n<\/div>\n
<\/p>\n
<\/p>\n
<\/p>\n","protected":false},"author":574340,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5159","chapter","type-chapter","status-publish","hentry"],"part":4997,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/574340"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions"}],"predecessor-version":[{"id":5166,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions\/5166"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/4997"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5159"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5159"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5159"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Question 3<\/h3>\r\n\r\n\r\n\r\n\r\nExact Probability\u00a0 Using Binomial<\/td>\r\n Approximate\r\n\r\nProbability Using\r\n\r\nNormal<\/td>\r\n Graphical Depiction<\/td>\r\n<\/tr>\r\n \r\n[latex] P(X = 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X \\leq 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] P(X > 270) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n \r\n[latex] (260 < X < 280) [\/latex]<\/td>\r\n <\/td>\r\n <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nQuestion 4<\/h3>\r\n4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.\r\n\r\n \t- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\r\n \t
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\r\n \t
- Find the probability that George will make more than 285 free throws.<\/li>\r\n \t
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\r\n \t
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n \r\n\r\n ","rendered":"
In a previous in-class activity, you learned about the role that [latex]p[\/latex] plays in the shape of the binomial distribution. The binomial probability distribution is skewed right if [latex]p < 0.5[\/latex], symmetric and approximately bell shaped if [latex]p= 0.5[\/latex], and skewed left if [latex]p > 0.5[\/latex]. Let\u2019s discuss the role that [latex]n[\/latex] plays in its shape.<\/p>\n
<\/p>\n
Credit: iStock\/skynesher<\/p>\n
\nQuestion 1<\/h3>\n
Recall what you learned in Preview Assignment 8.F. What did you discover about the role that\u00a0[latex]n[\/latex] plays in the shape of the binomial distribution?<\/p>\n<\/div>\n
For a fixed [latex]p[\/latex], as the number of trials, [latex]n[\/latex], in a binomial experiment increases, the probability distribution of the random variable\u00a0[latex]X[\/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex]np \\geq 10 \\mbox{ and } n(1-p) \\geq 10[\/latex], the probability distribution will be approximately symmetric and bell shaped.<\/p>\n
In other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex]np[\/latex], and the expected number of failures, [latex]n(1-p)[\/latex], are both at least 10.<\/p>\n
\nQuestion 2<\/h3>\n
Recall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?<\/p>\n<\/div>\n
In addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\n
In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[\/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \\mbox{ to } X + 0.5[\/latex].<\/p>\n
Why is this necessary? Consider the current example:<\/p>\n
The probability that George successfully makes exactly 270 free throws is [latex]P(X= 270) = 0.0766[\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex]P(X= 270.2)[\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.<\/p>\n
Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[\/latex] is a continuous random variable.<\/p>\n
The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:<\/p>\n
[latex]-P(Xa), \\mbox{ or } P(a\nQuestion 3<\/h3>\n\n\n\n\nExact Probability\u00a0 Using Binomial<\/td>\n Approximate<\/p>\nProbability Using<\/p>\n
Normal<\/td>\n
Graphical Depiction<\/td>\n<\/tr>\n \n[latex]P(X = 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X \\leq 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex]P(X > 270)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n \n[latex](260 < X < 280)[\/latex]<\/td>\n <\/td>\n <\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n <\/p>\n
\nQuestion 4<\/h3>\n
4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.<\/p>\n
\n- Find the probability that George will make at least 268 free throws for the\u00a0 season.<\/li>\n
- Find the probability that George will make between 256 and 284 free throws,\u00a0 inclusive.<\/li>\n
- Find the probability that George will make more than 285 free throws.<\/li>\n
- Would it be surprising or unusual if George were to only make 230 free\u00a0 throws for the season? Explain.<\/li>\n
- How does the answer in Part A compare to the exact binomial probability you\u00a0 derived in Question 2, Part E of the preview assignment?<\/li>\n<\/ol>\n<\/div>\n
<\/p>\n
<\/p>\n
<\/p>\n","protected":false},"author":574340,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5159","chapter","type-chapter","status-publish","hentry"],"part":4997,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/574340"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions"}],"predecessor-version":[{"id":5166,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions\/5166"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/4997"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5159"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5159"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5159"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
| Exact Probability\u00a0 Using Binomial<\/td>\r\n | Approximate\r\n\r\nProbability Using\r\n\r\nNormal<\/td>\r\n | Graphical Depiction<\/td>\r\n<\/tr>\r\n | |||||||||||||||
| [latex] P(X = 270) [\/latex]<\/td>\r\n | <\/td>\r\n | <\/td>\r\n<\/tr>\r\n | |||||||||||||||
| [latex] P(X \\leq 270) [\/latex]<\/td>\r\n | <\/td>\r\n | <\/td>\r\n<\/tr>\r\n | |||||||||||||||
| [latex] P(X > 270) [\/latex]<\/td>\r\n | <\/td>\r\n | <\/td>\r\n<\/tr>\r\n | |||||||||||||||
| [latex] (260 < X < 280) [\/latex]<\/td>\r\n | <\/td>\r\n | <\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n \r\n \r\n Question 4<\/h3>\r\n4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.\r\nIn a previous in-class activity, you learned about the role that [latex]p[\/latex] plays in the shape of the binomial distribution. The binomial probability distribution is skewed right if [latex]p < 0.5[\/latex], symmetric and approximately bell shaped if [latex]p= 0.5[\/latex], and skewed left if [latex]p > 0.5[\/latex]. Let\u2019s discuss the role that [latex]n[\/latex] plays in its shape.<\/p>\n
Credit: iStock\/skynesher<\/p>\n \n Question 1<\/h3>\nRecall what you learned in Preview Assignment 8.F. What did you discover about the role that\u00a0[latex]n[\/latex] plays in the shape of the binomial distribution?<\/p>\n<\/div>\n For a fixed [latex]p[\/latex], as the number of trials, [latex]n[\/latex], in a binomial experiment increases, the probability distribution of the random variable\u00a0[latex]X[\/latex] becomes nearly symmetric and bell shaped. As a rule of thumb, if [latex]np \\geq 10 \\mbox{ and } n(1-p) \\geq 10[\/latex], the probability distribution will be approximately symmetric and bell shaped.<\/p>\n In other words, the binomial distribution can be approximated well by the normal distribution when n is large enough so that the expected number of successes, [latex]np[\/latex], and the expected number of failures, [latex]n(1-p)[\/latex], are both at least 10.<\/p>\n \n Question 2<\/h3>\nRecall the free throw scenario in Preview Assignment 8.F. Suppose that over the course of a season, Paul George (a top free-throw shooter) will shoot 300 free throws. The top free-throw shooters in the league have a probability of about 0.90 of making any given free throw. Can this binomial distribution be approximated by the normal distribution?<\/p>\n<\/div>\n In addition to verifying that both the number of successes and the number of failures exceed 10, we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\n In probability theory, a continuity correction is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[\/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \\mbox{ to } X + 0.5[\/latex].<\/p>\n Why is this necessary? Consider the current example:<\/p>\n The probability that George successfully makes exactly 270 free throws is [latex]P(X= 270) = 0.0766[\/latex]. So, there is a 7.66% chance that he will make exactly 270 free throws out of 300. The [latex]P(X= 270.2)[\/latex] free throws is 0 because he cannot make 0.2 of a free throw. The number of free throws made is a discrete random variable.<\/p>\n Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[\/latex] is a continuous random variable.<\/p>\n The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:<\/p>\n [latex]-P(X Probability Using<\/p>\n Normal<\/td>\n <\/p>\n 4) Next, let\u2019s use the normal approximation with a continuity correction to approximate\u00a0 the following binomial probabilities. Use the DCMP Normal Distribution tool to\u00a0 calculate these probabilities.<\/p>\n <\/p>\n <\/p>\n <\/p>\n","protected":false},"author":574340,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5159","chapter","type-chapter","status-publish","hentry"],"part":4997,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/574340"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions"}],"predecessor-version":[{"id":5166,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/revisions\/5166"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/4997"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5159\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5159"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5159"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5159"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} |