{"id":1463,"date":"2024-05-10T17:44:21","date_gmt":"2024-05-10T21:44:21","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/?post_type=chapter&#038;p=1463"},"modified":"2024-06-12T14:45:29","modified_gmt":"2024-06-12T18:45:29","slug":"binomial-distributions-at-least","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/chapter\/binomial-distributions-at-least\/","title":{"raw":"Binomial Distributions - At Least","rendered":"Binomial Distributions &#8211; At Least"},"content":{"raw":"<div class=\"textbox textbox--learning-objectives\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Learning Objectives<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nCalculate the probability of at least [latex] x[\/latex] successes or, [latex]P(X \\ge x)[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\nCalculating [latex]P(X \\ge x)[\/latex] is similar to calculating [latex]P(X \\gt\u00a0 x)[\/latex] with only one difference:\r\n<ul>\r\n \t<li>[latex]P(X \\ge x)[\/latex], the probability of <a href=\"https:\/\/dictionary.cambridge.org\/dictionary\/english\/at-least\">at least<\/a> [latex] x[\/latex] successes, includes the [latex] x[\/latex] value.<\/li>\r\n \t<li>[latex]P(X \\gt\u00a0 x)[\/latex], the probability of <a href=\"https:\/\/dictionary.cambridge.org\/dictionary\/english\/more-than\">more than<\/a> [latex] x[\/latex] successes, does not include the [latex] x[\/latex] value.<\/li>\r\n<\/ul>\r\nIn other words:\r\n<ul>\r\n \t<li>[latex] P(X \\gt x) = 1- P(X \\le x) = 1-\\text{BINOM.DIST}(x, n, p, 1) [\/latex]<\/li>\r\n \t<li>[latex] P(X \\ge x) = 1- P(X \\le x-1) = 1-\\text{BINOM.DIST}(x-1, n, p, 1) [\/latex]<\/li>\r\n<\/ul>\r\nWhy do we use [latex]x-1[\/latex] in the above formula?\r\n<ul>\r\n \t<li>It has to do with what it means to take a complement.<\/li>\r\n \t<li>When taking a complement, we take all values outside of that sample space.<\/li>\r\n \t<li>Since [latex]x[\/latex] is included in the range, we 'stop' at [latex]x-1[\/latex] when taking the complement: [latex] P(X \\ge x) = 1 - [P(X=0)+P(X=1)+...+P(X=x-1)] [\/latex]<\/li>\r\n<\/ul>\r\n<h1>Using Formulas to Calculate At least (Example)<\/h1>\r\nLet us revisit the hotel example from the previous section to highlight the similarities and differences between 'more than' and 'at least' calculations.\r\n<h2>Example 28.1 \u2013 Using the Formula<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: A hotel's records indicate that 65% of its guests are visitors from Canada.\r\n\r\n<span style=\"color: #003366\"><strong>Question<\/strong><\/span>: From a random sample of 12 guests, what is the probability that at least 10 of them are from Canada?\r\n\r\n<span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: We know the following:\r\n<ul>\r\n \t<li>[latex]P(\\text{at least 10}) = P(x\\ge 10) = P(x=10)+P(x=11) + P(x=12) [\/latex]<\/li>\r\n \t<li>[latex]n=12[\/latex] and [latex]p=0.65[\/latex].<\/li>\r\n<\/ul>\r\nThis gives:\r\n<ul>\r\n \t<li>[latex]P(x=10) = {}_{12}C_{10} \\cdot (0.65)^{10} \\cdot (1 - 0.65)^{12-10} = \\frac{12!}{10!2!} \\cdot (0.65)^{10} \\cdot (0.35)^2 =66(0.01346)(0.1225) = 0.10885 [\/latex]<\/li>\r\n \t<li>[latex]P(x=11) = {}_{12}C_{11} \\cdot (0.65)^{11} \\cdot (1 - 0.65)^{12-11} = \\frac{12!}{11!1!} \\cdot (0.65)^{11} \\cdot (0.35)^1 =12(0.00875)(0.35) = 0.03675 [\/latex]<\/li>\r\n \t<li>[latex]P(x=12) = {}_{12}C_{12} \\cdot (0.65)^{12} \\cdot (1 - 0.65)^{12-12} = \\frac{12!}{12!0!} \\cdot (0.65)^{12} \\cdot (0.35)^0 =1(0.00569)(1) = 0.00569 [\/latex]<\/li>\r\n \t<li>[latex]P(x\\ge 10) =\u00a0 P(x=10)+P(x=11)+P(x=12) = 0.10885+0.03675+ 0.00569 = 0.1513[\/latex]<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is an 15.13% chance that at least 10 of them are from Canada.\r\n<h1>Using Excel to Calculate At least (VIDEO)<\/h1>\r\n<ul>\r\n \t<li>Using Excel's <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c#:~:text=Returns%20the%20individual%20term%20binomial,is%20constant%20throughout%20the%20experiment.\">BINOM.DIST()<\/a> function is much quicker than using the formula shown in the previous section<\/li>\r\n \t<li>Again, just be careful of which [latex]x[\/latex] value to include and don't forget to take the complement.<\/li>\r\n \t<li>We will try this out in the next example.<\/li>\r\n<\/ul>\r\n<h2>Example 28.2 \u2013 Using Excel and a Complement<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: Let us now revisit example 28.1 but we will use Excel.\r\n\r\n<span style=\"color: #003366\"><strong>Question<\/strong><\/span>: Can you use Excel's <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c#:~:text=Returns%20the%20individual%20term%20binomial,is%20constant%20throughout%20the%20experiment.\">BINOM.DIST<\/a> to calculate the probability of at least 10 guests being from Canada?\r\n\r\n<span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example28-1.xlsx\">Click here<\/a>\u00a0to download the Excel solution file. Also, see the video below:\r\n\r\nhttps:\/\/youtu.be\/50bhgrlVn1Y\r\n\r\n<span style=\"color: #003366\"><strong>Conclusion<\/strong><\/span>: Again, here is an 15.13% chance that at least 10 of them are from Canada.\r\n<h1>Key Takeaways (EXERCISE)<\/h1>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways: Binomial Distributions - At Least<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n[h5p id=\"75\"]\r\n\r\n[h5p id=\"76\"]\r\n\r\n<\/div>\r\n<\/div>\r\n<h1>Your Own Notes (EXERCISE)<\/h1>\r\n<ul>\r\n \t<li>Are there any notes you want to take from this section? Is there anything you'd like to copy and paste below?<\/li>\r\n \t<li>These notes are for you only (they will not be stored anywhere)<\/li>\r\n \t<li>Make sure to download them at the end to use as a reference<\/li>\r\n<\/ul>\r\n[h5p id=\"16\"]","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Learning Objectives<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Calculate the probability of at least [latex]x[\/latex] successes or, [latex]P(X \\ge x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<p>Calculating [latex]P(X \\ge x)[\/latex] is similar to calculating [latex]P(X \\gt\u00a0 x)[\/latex] with only one difference:<\/p>\n<ul>\n<li>[latex]P(X \\ge x)[\/latex], the probability of <a href=\"https:\/\/dictionary.cambridge.org\/dictionary\/english\/at-least\">at least<\/a> [latex]x[\/latex] successes, includes the [latex]x[\/latex] value.<\/li>\n<li>[latex]P(X \\gt\u00a0 x)[\/latex], the probability of <a href=\"https:\/\/dictionary.cambridge.org\/dictionary\/english\/more-than\">more than<\/a> [latex]x[\/latex] successes, does not include the [latex]x[\/latex] value.<\/li>\n<\/ul>\n<p>In other words:<\/p>\n<ul>\n<li>[latex]P(X \\gt x) = 1- P(X \\le x) = 1-\\text{BINOM.DIST}(x, n, p, 1)[\/latex]<\/li>\n<li>[latex]P(X \\ge x) = 1- P(X \\le x-1) = 1-\\text{BINOM.DIST}(x-1, n, p, 1)[\/latex]<\/li>\n<\/ul>\n<p>Why do we use [latex]x-1[\/latex] in the above formula?<\/p>\n<ul>\n<li>It has to do with what it means to take a complement.<\/li>\n<li>When taking a complement, we take all values outside of that sample space.<\/li>\n<li>Since [latex]x[\/latex] is included in the range, we &#8216;stop&#8217; at [latex]x-1[\/latex] when taking the complement: [latex]P(X \\ge x) = 1 - [P(X=0)+P(X=1)+...+P(X=x-1)][\/latex]<\/li>\n<\/ul>\n<h1>Using Formulas to Calculate At least (Example)<\/h1>\n<p>Let us revisit the hotel example from the previous section to highlight the similarities and differences between &#8216;more than&#8217; and &#8216;at least&#8217; calculations.<\/p>\n<h2>Example 28.1 \u2013 Using the Formula<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: A hotel&#8217;s records indicate that 65% of its guests are visitors from Canada.<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: From a random sample of 12 guests, what is the probability that at least 10 of them are from Canada?<\/p>\n<p><span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: We know the following:<\/p>\n<ul>\n<li>[latex]P(\\text{at least 10}) = P(x\\ge 10) = P(x=10)+P(x=11) + P(x=12)[\/latex]<\/li>\n<li>[latex]n=12[\/latex] and [latex]p=0.65[\/latex].<\/li>\n<\/ul>\n<p>This gives:<\/p>\n<ul>\n<li>[latex]P(x=10) = {}_{12}C_{10} \\cdot (0.65)^{10} \\cdot (1 - 0.65)^{12-10} = \\frac{12!}{10!2!} \\cdot (0.65)^{10} \\cdot (0.35)^2 =66(0.01346)(0.1225) = 0.10885[\/latex]<\/li>\n<li>[latex]P(x=11) = {}_{12}C_{11} \\cdot (0.65)^{11} \\cdot (1 - 0.65)^{12-11} = \\frac{12!}{11!1!} \\cdot (0.65)^{11} \\cdot (0.35)^1 =12(0.00875)(0.35) = 0.03675[\/latex]<\/li>\n<li>[latex]P(x=12) = {}_{12}C_{12} \\cdot (0.65)^{12} \\cdot (1 - 0.65)^{12-12} = \\frac{12!}{12!0!} \\cdot (0.65)^{12} \\cdot (0.35)^0 =1(0.00569)(1) = 0.00569[\/latex]<\/li>\n<li>[latex]P(x\\ge 10) =\u00a0 P(x=10)+P(x=11)+P(x=12) = 0.10885+0.03675+ 0.00569 = 0.1513[\/latex]<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is an 15.13% chance that at least 10 of them are from Canada.<\/p>\n<h1>Using Excel to Calculate At least (VIDEO)<\/h1>\n<ul>\n<li>Using Excel&#8217;s <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c#:~:text=Returns%20the%20individual%20term%20binomial,is%20constant%20throughout%20the%20experiment.\">BINOM.DIST()<\/a> function is much quicker than using the formula shown in the previous section<\/li>\n<li>Again, just be careful of which [latex]x[\/latex] value to include and don&#8217;t forget to take the complement.<\/li>\n<li>We will try this out in the next example.<\/li>\n<\/ul>\n<h2>Example 28.2 \u2013 Using Excel and a Complement<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: Let us now revisit example 28.1 but we will use Excel.<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: Can you use Excel&#8217;s <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c#:~:text=Returns%20the%20individual%20term%20binomial,is%20constant%20throughout%20the%20experiment.\">BINOM.DIST<\/a> to calculate the probability of at least 10 guests being from Canada?<\/p>\n<p><span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example28-1.xlsx\">Click here<\/a>\u00a0to download the Excel solution file. Also, see the video below:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Using Excel&#39;s BINOM.DIST to calculate the probability of at least &#39;x&#39; successes\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/50bhgrlVn1Y?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><span style=\"color: #003366\"><strong>Conclusion<\/strong><\/span>: Again, here is an 15.13% chance that at least 10 of them are from Canada.<\/p>\n<h1>Key Takeaways (EXERCISE)<\/h1>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways: Binomial Distributions &#8211; At Least<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"h5p-75\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-75\" class=\"h5p-iframe\" data-content-id=\"75\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key Takeaways for Binomial Distributions - At Least\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-76\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-76\" class=\"h5p-iframe\" data-content-id=\"76\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key Takeaways for Binomial Distributions - At Least - Solutions\"><\/iframe><\/div>\n<\/div>\n<\/div>\n<\/div>\n<h1>Your Own Notes (EXERCISE)<\/h1>\n<ul>\n<li>Are there any notes you want to take from this section? Is there anything you&#8217;d like to copy and paste below?<\/li>\n<li>These notes are for you only (they will not be stored anywhere)<\/li>\n<li>Make sure to download them at the end to use as a reference<\/li>\n<\/ul>\n<div id=\"h5p-16\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-16\" class=\"h5p-iframe\" data-content-id=\"16\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key takeaways, notes and comments from this section document tool.\"><\/iframe><\/div>\n<\/div>\n","protected":false},"author":865,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1463","chapter","type-chapter","status-publish","hentry"],"part":231,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1463","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/users\/865"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1463\/revisions"}],"predecessor-version":[{"id":1956,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1463\/revisions\/1956"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/parts\/231"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1463\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/media?parent=1463"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=1463"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/contributor?post=1463"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/license?post=1463"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}