{"id":1334,"date":"2024-05-09T16:22:57","date_gmt":"2024-05-09T20:22:57","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/?post_type=chapter&#038;p=1334"},"modified":"2024-06-12T14:42:22","modified_gmt":"2024-06-12T18:42:22","slug":"binomial-distributions-at-least-or-at-most","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/chapter\/binomial-distributions-at-least-or-at-most\/","title":{"raw":"Binomial Distributions - At Most","rendered":"Binomial Distributions &#8211; At Most"},"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 most [latex] x[\/latex] number of successes.\r\n\r\n<\/div>\r\n<\/div>\r\nIf we want to calculate the probability of a range of x values using the formula:\r\n<ul>\r\n \t<li>We need to add all possible probabilities together.<\/li>\r\n \t<li>Ex: [latex] P(\\text{at most 2 sales})=P(x\u22642)=P(x=0)+P(x=1)+P(x=2) [\/latex]<\/li>\r\n<\/ul>\r\nIf we want to calculate the probability of a range of x values using <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>:\r\n<ul>\r\n \t<li>Set 'cumulative' = TRUE (or 1)<\/li>\r\n \t<li>Ex: P(at most 2 sales) = P(x \u2264 2) = BINOM.DIST(2, [latex]n[\/latex], p, TRUE)<\/li>\r\n<\/ul>\r\n<h1>total of All Possible Outcomes (Example)<\/h1>\r\nLet us re-examine the salesperson problem:\r\n<ul>\r\n \t<li>There are ten calls made in the day.<\/li>\r\n \t<li>This means that there are between 0 and 10 possible sales made in a day.<\/li>\r\n \t<li>What if we add up all of those possible probabilities?<\/li>\r\n \t<li>Let us examine this in the next example.<\/li>\r\n<\/ul>\r\n<h2>Example 26.1.1<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: Again, let a salesperson call 10 clients in a day. Let the odds of the salesperson making a sale with any one of the clients be [latex] p[\/latex]=0.3\r\n\r\n<span style=\"color: #003366\"><strong>Question<\/strong><\/span>: What is the probability that the salesperson makes between 0 and 10 sales in a day?\r\n\r\n<span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: In the previous section, we found the probabilities for all possible number of sales:\r\n<p style=\"padding-left: 40px\"><img class=\"alignnone wp-image-1381\" src=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/SalesPerson_AllProbs.jpg\" alt=\"screenshot of all probabilities for x values between 0 and 10\" width=\"132\" height=\"329\" \/><\/p>\r\nTo calculate the probability of between 0 and 10 sales in the day, add up all these probabilities:\r\n\r\n[latex]\r\n\r\n\\begin{align*}\r\n\r\nP(0\\le x\\le10) &amp;= P(x=0)+P(x=1) + P(x=2) + P(x=3) + ... + P(x=9) + P(x=10) \\\\\r\n\r\n&amp;= 0.02825+ 0.12106+0.23347+0.26683+0.20012+0.10292+0.03676+0.009+0.00145+0.00014+0.00001 \\\\\r\n\r\n&amp;= 1\r\n\r\n\\end{align*}\r\n\r\n[\/latex]\r\n\r\n<span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is a 100% chance that the salesperson makes between 0 and 10 sales in a day. Why is this? Because that range encompasses all the possible outcomes that can happen and we know that the sum of all possible probabilities in a probability distribution must equal to 1.\r\n<h1>total of All Possible Outcomes in excel (Video)<\/h1>\r\n<ul>\r\n \t<li>How would we do the calculation in the previous example in Excel?<\/li>\r\n \t<li>We would use <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">Cumulative<\/a> = TRUE or 1.<\/li>\r\n \t<li>Let us try this in the next example.<\/li>\r\n<\/ul>\r\n<h2>Example 26.1.2.<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: We will use Excel's <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">binom.dist<\/a> call to calculate the probability of between 0 and 10.\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\/Example26-1-2.xlsx\">Click here<\/a> to download the Excel solutions. Also, see the video below:\r\n\r\nhttps:\/\/youtu.be\/nNSSi1purfY\r\n<h1>Using the Formula to calculate 'At most' (example)<\/h1>\r\n<ul>\r\n \t<li>Let us now consider the probability of at most a certain number of successes occurring.<\/li>\r\n \t<li>We denote 'at most' with a '<b>\u2264'<\/b> symbol.<\/li>\r\n \t<li>It can also be stated as 'less than or equal to.'<\/li>\r\n<\/ul>\r\n<h2>Example 26.2.1 \u2013 Using the Formula<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: A hotel's records indicate that 5% of its guests are visitors from the U.S.A.\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 most one of them is from the U.S.A.?\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 most one}) = P(x\\le 1) = P(x=0) + P(x=1) [\/latex]<\/li>\r\n \t<li>[latex]n=12[\/latex] and [latex]p=0.05[\/latex].<\/li>\r\n<\/ul>\r\nThis gives:\r\n<ul>\r\n \t<li>[latex] P(x=0) = {}_{12}C_0 \\cdot (0.05)^0 \\cdot (1 - 0.05)^{12-0} = \\frac{12!}{0!12!} \\cdot (0.05)^0 \\cdot (0.95)^12 =1(1)(0.5404) = 0.5404 [\/latex]<\/li>\r\n \t<li>[latex] P(x=1) = {}_{12}C_1 \\cdot (0.05)^1 \\cdot (1 - 0.05)^{12-1} = \\frac{12!}{1!11!} \\cdot (0.05)^1 \\cdot (0.95)^11 =12(0.05)(0.5688) = 0.34128 [\/latex]<\/li>\r\n \t<li>[latex] P(x\\le1) =\u00a0 P(x=0)+P(x=1) = 0.5404+ 0.34128 = 0.8816 [\/latex]<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is an 88.16% chance that at most one of their guests is from the U.S.A.\r\n<h1>Using Excel to calculate 'At Most' (EXample)<\/h1>\r\nIn 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, when cumulative is set to TRUE (or 1):\r\n<ul>\r\n \t<li>Excel calculates the probability of having up to and including [latex]x[\/latex] ('<span style=\"color: #003366\"><a style=\"color: #003366\" href=\"https:\/\/support.microsoft.com\/en-us\/office\/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c#:~:text=If%20cumulative%20is%20TRUE%2C%20then,that%20there%20are%20number_s%20successes.\">number_s<\/a>'<\/span>) successes.<\/li>\r\n \t<li>BINOM.DIST([latex]x[\/latex], [latex]n[\/latex], [latex]p[\/latex], 1) = P([latex]X[\/latex] \u2264 [latex]x[\/latex])<\/li>\r\n \t<li>Let us revisit example 26.2 to better understand this concept.<\/li>\r\n<\/ul>\r\n<h2>Example 26.2.2 \u2013 Using BINOM.DIST()<\/h2>\r\nIf we wanted to solve example 26.2 using Excel's <span style=\"color: #003366\"><a style=\"color: #003366\" href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">BINOM.DIST()<\/a><\/span> function:\r\n<p style=\"padding-left: 40px\">P(x \u2264 1) = BINOM.DIST(1, 12, 0.05, 1) = BINOM.DIST(1, 12, 0.05, TRUE) = 0.8816<\/p>\r\n<a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example26-2-1.xlsx\">Click here<\/a> to download the Excel solution file for the above problem.\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 Most<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n[h5p id=\"71\"]\r\n\r\n[h5p id=\"72\"]\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 most [latex]x[\/latex] number of successes.<\/p>\n<\/div>\n<\/div>\n<p>If we want to calculate the probability of a range of x values using the formula:<\/p>\n<ul>\n<li>We need to add all possible probabilities together.<\/li>\n<li>Ex: [latex]P(\\text{at most 2 sales})=P(x\u22642)=P(x=0)+P(x=1)+P(x=2)[\/latex]<\/li>\n<\/ul>\n<p>If we want to calculate the probability of a range of x values using <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>:<\/p>\n<ul>\n<li>Set &#8216;cumulative&#8217; = TRUE (or 1)<\/li>\n<li>Ex: P(at most 2 sales) = P(x \u2264 2) = BINOM.DIST(2, [latex]n[\/latex], p, TRUE)<\/li>\n<\/ul>\n<h1>total of All Possible Outcomes (Example)<\/h1>\n<p>Let us re-examine the salesperson problem:<\/p>\n<ul>\n<li>There are ten calls made in the day.<\/li>\n<li>This means that there are between 0 and 10 possible sales made in a day.<\/li>\n<li>What if we add up all of those possible probabilities?<\/li>\n<li>Let us examine this in the next example.<\/li>\n<\/ul>\n<h2>Example 26.1.1<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: Again, let a salesperson call 10 clients in a day. Let the odds of the salesperson making a sale with any one of the clients be [latex]p[\/latex]=0.3<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: What is the probability that the salesperson makes between 0 and 10 sales in a day?<\/p>\n<p><span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: In the previous section, we found the probabilities for all possible number of sales:<\/p>\n<p style=\"padding-left: 40px\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-1381\" src=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/SalesPerson_AllProbs.jpg\" alt=\"screenshot of all probabilities for x values between 0 and 10\" width=\"132\" height=\"329\" srcset=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/SalesPerson_AllProbs.jpg 184w, https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/SalesPerson_AllProbs-65x162.jpg 65w\" sizes=\"auto, (max-width: 132px) 100vw, 132px\" \/><\/p>\n<p>To calculate the probability of between 0 and 10 sales in the day, add up all these probabilities:<\/p>\n<p>[latex]\\begin{align*}    P(0\\le x\\le10) &= P(x=0)+P(x=1) + P(x=2) + P(x=3) + ... + P(x=9) + P(x=10) \\\\    &= 0.02825+ 0.12106+0.23347+0.26683+0.20012+0.10292+0.03676+0.009+0.00145+0.00014+0.00001 \\\\    &= 1    \\end{align*}[\/latex]<\/p>\n<p><span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is a 100% chance that the salesperson makes between 0 and 10 sales in a day. Why is this? Because that range encompasses all the possible outcomes that can happen and we know that the sum of all possible probabilities in a probability distribution must equal to 1.<\/p>\n<h1>total of All Possible Outcomes in excel (Video)<\/h1>\n<ul>\n<li>How would we do the calculation in the previous example in Excel?<\/li>\n<li>We would use <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">Cumulative<\/a> = TRUE or 1.<\/li>\n<li>Let us try this in the next example.<\/li>\n<\/ul>\n<h2>Example 26.1.2.<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: We will use Excel&#8217;s <a href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">binom.dist<\/a> call to calculate the probability of between 0 and 10.<\/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\/Example26-1-2.xlsx\">Click here<\/a> to download the Excel solutions. Also, see the video below:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Cumulative Probabilities in Excel&#39;s Binom.dist() Function\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/nNSSi1purfY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h1>Using the Formula to calculate &#8216;At most&#8217; (example)<\/h1>\n<ul>\n<li>Let us now consider the probability of at most a certain number of successes occurring.<\/li>\n<li>We denote &#8216;at most&#8217; with a &#8216;<b>\u2264&#8217;<\/b> symbol.<\/li>\n<li>It can also be stated as &#8216;less than or equal to.&#8217;<\/li>\n<\/ul>\n<h2>Example 26.2.1 \u2013 Using the Formula<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: A hotel&#8217;s records indicate that 5% of its guests are visitors from the U.S.A.<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: From a random sample of 12 guests, what is the probability that at most one of them is from the U.S.A.?<\/p>\n<p><span style=\"color: #003366\"><strong>Solution<\/strong><\/span>: We know the following:<\/p>\n<ul>\n<li>[latex]P(\\text{at most one}) = P(x\\le 1) = P(x=0) + P(x=1)[\/latex]<\/li>\n<li>[latex]n=12[\/latex] and [latex]p=0.05[\/latex].<\/li>\n<\/ul>\n<p>This gives:<\/p>\n<ul>\n<li>[latex]P(x=0) = {}_{12}C_0 \\cdot (0.05)^0 \\cdot (1 - 0.05)^{12-0} = \\frac{12!}{0!12!} \\cdot (0.05)^0 \\cdot (0.95)^12 =1(1)(0.5404) = 0.5404[\/latex]<\/li>\n<li>[latex]P(x=1) = {}_{12}C_1 \\cdot (0.05)^1 \\cdot (1 - 0.05)^{12-1} = \\frac{12!}{1!11!} \\cdot (0.05)^1 \\cdot (0.95)^11 =12(0.05)(0.5688) = 0.34128[\/latex]<\/li>\n<li>[latex]P(x\\le1) =\u00a0 P(x=0)+P(x=1) = 0.5404+ 0.34128 = 0.8816[\/latex]<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is an 88.16% chance that at most one of their guests is from the U.S.A.<\/p>\n<h1>Using Excel to calculate &#8216;At Most&#8217; (EXample)<\/h1>\n<p>In 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, when cumulative is set to TRUE (or 1):<\/p>\n<ul>\n<li>Excel calculates the probability of having up to and including [latex]x[\/latex] (&#8216;<span style=\"color: #003366\"><a style=\"color: #003366\" href=\"https:\/\/support.microsoft.com\/en-us\/office\/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c#:~:text=If%20cumulative%20is%20TRUE%2C%20then,that%20there%20are%20number_s%20successes.\">number_s<\/a>&#8216;<\/span>) successes.<\/li>\n<li>BINOM.DIST([latex]x[\/latex], [latex]n[\/latex], [latex]p[\/latex], 1) = P([latex]X[\/latex] \u2264 [latex]x[\/latex])<\/li>\n<li>Let us revisit example 26.2 to better understand this concept.<\/li>\n<\/ul>\n<h2>Example 26.2.2 \u2013 Using BINOM.DIST()<\/h2>\n<p>If we wanted to solve example 26.2 using Excel&#8217;s <span style=\"color: #003366\"><a style=\"color: #003366\" href=\"https:\/\/support.microsoft.com\/en-us\/office\/binomdist-function-506a663e-c4ca-428d-b9a8-05583d68789c\">BINOM.DIST()<\/a><\/span> function:<\/p>\n<p style=\"padding-left: 40px\">P(x \u2264 1) = BINOM.DIST(1, 12, 0.05, 1) = BINOM.DIST(1, 12, 0.05, TRUE) = 0.8816<\/p>\n<p><a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example26-2-1.xlsx\">Click here<\/a> to download the Excel solution file for the above problem.<\/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 Most<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"h5p-71\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-71\" class=\"h5p-iframe\" data-content-id=\"71\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key Takeaways for Binomial Distributions - At Most\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-72\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-72\" class=\"h5p-iframe\" data-content-id=\"72\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Key Takeaways for Binomial Distributions - At Most - 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":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1334","chapter","type-chapter","status-publish","hentry"],"part":231,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1334","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\/1334\/revisions"}],"predecessor-version":[{"id":1954,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1334\/revisions\/1954"}],"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\/1334\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/media?parent=1334"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=1334"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/contributor?post=1334"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/license?post=1334"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}