{"id":1845,"date":"2024-05-28T15:44:18","date_gmt":"2024-05-28T19:44:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/?post_type=chapter&#038;p=1845"},"modified":"2024-06-12T14:50:23","modified_gmt":"2024-06-12T18:50:23","slug":"using-the-exponential-distribution-formulas","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/chapter\/using-the-exponential-distribution-formulas\/","title":{"raw":"Using the Exponential Distribution Formulas","rendered":"Using the Exponential Distribution Formulas"},"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\nUse the Exponential Distribution formulas to calculate probabilities of times between events.\r\n\r\n<\/div>\r\n<\/div>\r\nWe were given the following formulas in the <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/part\/exponential-distributions\/\">previous section<\/a>, to calculate the probability of times between events:\r\n<ul>\r\n \t<li>[latex]P(\\text{at most or less than}) = P(X \\le x) = 1 - e^{-\\lambda x}[\/latex]<\/li>\r\n \t<li>[latex]P(\\text{at least or more than}) = P(X \\ge x) = e^{-\\lambda x}[\/latex]<\/li>\r\n<\/ul>\r\nWe will practice using these formulas in the sections below.\r\n<h1>Calculating the probability of AT Most (EXAMPLE)<\/h1>\r\nLet us better understand the Exponential Distribution formula by working through an example.\r\n<h2>Example 37.1<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: On a busy Friday evening, the time between customer arrivals at a supermarket check-out counter follows an Exponential distribution. On average, a customer arrives every 30 seconds.\r\n\r\n<span style=\"color: #003366\"><strong>Question<\/strong><\/span>: What is the probability a customer will arrive in the next minute at the check-out?\r\n\r\n<span style=\"color: #003366\"><strong>Written solution:<\/strong><\/span> Let us use the 'at most' formula and follow the steps below:\r\n<ul>\r\n \t<li><span style=\"color: #003366\">Formula<\/span>: [latex]P(\\text{at most or less than}) = P(X \\le x) = 1 - e^{-\\lambda x}[\/latex]<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li><span style=\"color: #003366\">Time units<\/span>: We will use minutes as the time units.<\/li>\r\n \t<li><span style=\"color: #003366\">Lambda<\/span>: We convert [latex]\\lambda[\/latex] to the number of events per minute: [latex]\\lambda = \\frac{1}{30 \\text{ seconds}} \\times \\frac{60 \\text{ seconds}}{1 \\text{ minute}}=2 \\text{ } \\frac{\\text{customers}}{\\text{per minute}}[\/latex]<\/li>\r\n \t<li><span style=\"color: #003366\">X-values<\/span>: We want the probability of the next customer arriving within the next minute. This gives: [latex]P(x \\le 1)[\/latex].<\/li>\r\n \t<li><span style=\"color: #003366\">e<\/span>: The other letter, [latex]e[\/latex], is a constant. It is the base of the natural log and equal 2.71828.<\/li>\r\n \t<li><span style=\"color: #003366\">Answer<\/span>: Plugging all of this into the formula [latex]P(X \\le x) = 1 - e^{-\\lambda x}[\/latex] gives:\r\n[latex]\r\n\\begin{align}\r\nP(X \\le 1) &amp;= 1 - e^{-2 \\times 1} \\\\\r\n&amp;= 1 - e^{-2} \\\\\r\n&amp;= 1 - 0.1353 \\\\\r\n&amp;= 0.8647\r\n\\end{align}\r\n[\/latex]<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 86.47% chance that a customer will arrive in the next minute at the check-out.\r\n<h1>Calculating the probability of At least (Exercise)<\/h1>\r\nNext, we will use the formulas to solve for the probability of at least a certain amount of time.\r\n<h2>Example 37.2.1<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: The times it takes call center specialist to resolve incoming calls to their call center:\r\n<ul>\r\n \t<li>Follow an exponential distribution.<\/li>\r\n \t<li>Have an average of 2.5 minutes.<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Question:<\/strong><\/span> Specialists will get a warning at 5 minutes to try to wrap up the call (if they can). What is the probability that the specialist receives that 5 minute warning on a call?\r\n\r\n<span style=\"color: #003366\"><strong>You Try:<\/strong><\/span> Fill in the values to answer the question in the exercise below:\r\n\r\n[h5p id=\"104\"]\r\n\r\n<span style=\"color: #003366\"><strong>Answer?<\/strong><\/span>\u00a0Use the values from the above exercise to answer the question. Input your answer in the exercise below:\r\n\r\n[h5p id=\"105\"]\r\n<div>\r\n<h1>Need help with the above exercise? (Click to reveal)<\/h1>\r\n<span style=\"color: #003366\"><strong>Solution:<\/strong><\/span> Let us use the 'at least' formula and follow the steps below:\r\n<ul>\r\n \t<li><span style=\"color: #003366\">Formula<\/span>: [latex]P(\\text{at least or more than}) = P(X \\le x) = e^{-\\lambda x}[\/latex]<\/li>\r\n<\/ul>\r\n<ul>\r\n \t<li><span style=\"color: #003366\">Time units<\/span>: We will use minutes as the time units.<\/li>\r\n \t<li><span style=\"color: #003366\">Lambda<\/span>: We convert [latex]\\lambda[\/latex] to the number of events per minute: [latex]\\lambda = \\frac{1 \\text{ call}}{2.5 \\text{ minutes}} = \\frac{0.4 \\text{ calls}}{\\text{minute}}[\/latex]<\/li>\r\n \t<li><span style=\"color: #003366\">X-values<\/span>: We want the probability of the specialist getting the five minute warning. This means that the call lasts at least 5 minutes. This gives: [latex]P(x \\ge 5)[\/latex].<\/li>\r\n \t<li><span style=\"color: #003366\">e <\/span>= 2.71828.<\/li>\r\n \t<li><span style=\"color: #003366\">Answer<\/span>: Plugging all of this into the formula [latex]P(X \\le x) = 1 - e^{-\\lambda x}[\/latex] gives:\r\n[latex]\r\n\\begin{align}\r\nP(X \\ge 5) &amp;= e^{-0.4 \\times 5} \\\\\r\n&amp;= e^{-2} \\\\\r\n&amp;= 0.1353\r\n\\end{align}\r\n[\/latex]<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 13.53% chance that the call specialist receives the 5 minute warning.\r\n\r\n<\/div>\r\n<h1>Applying the Memoryless Property (Video)<\/h1>\r\nLet us now explore the memoryless property of the exponential distribution in the next example.\r\n<h2>Example 37.2.2<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: We will revisit our previous example...\r\n<ul>\r\n \t<li>The times it takes call center specialist to resolve incoming calls follow an exponential distribution.<\/li>\r\n \t<li>On average, it takes the specialist 2.5 minutes to resolve a call.<\/li>\r\n \t<li>After 5 minutes of being on a call, a warning pops up on the specialist's computer urging them to wrap up the phone call.<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability, after receiving the warning, that it takes at least another 5 minutes to wrap up the call?\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\/Exponential-Probability-Formula-Memoryless-Property.pdf\">Click here to download<\/a> the written solutions shown in the video below:\r\n\r\nhttps:\/\/youtu.be\/ShGZRg5qvig\r\n\r\n<span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is 13.53% chance that the call will take at least another 5 minutes to resolve.\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: Using the Exponential Distribution Formulas<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n[h5p id=\"107\"]\r\n\r\n[h5p id=\"108\"]\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>Use the Exponential Distribution formulas to calculate probabilities of times between events.<\/p>\n<\/div>\n<\/div>\n<p>We were given the following formulas in the <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/part\/exponential-distributions\/\">previous section<\/a>, to calculate the probability of times between events:<\/p>\n<ul>\n<li>[latex]P(\\text{at most or less than}) = P(X \\le x) = 1 - e^{-\\lambda x}[\/latex]<\/li>\n<li>[latex]P(\\text{at least or more than}) = P(X \\ge x) = e^{-\\lambda x}[\/latex]<\/li>\n<\/ul>\n<p>We will practice using these formulas in the sections below.<\/p>\n<h1>Calculating the probability of AT Most (EXAMPLE)<\/h1>\n<p>Let us better understand the Exponential Distribution formula by working through an example.<\/p>\n<h2>Example 37.1<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: On a busy Friday evening, the time between customer arrivals at a supermarket check-out counter follows an Exponential distribution. On average, a customer arrives every 30 seconds.<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: What is the probability a customer will arrive in the next minute at the check-out?<\/p>\n<p><span style=\"color: #003366\"><strong>Written solution:<\/strong><\/span> Let us use the &#8216;at most&#8217; formula and follow the steps below:<\/p>\n<ul>\n<li><span style=\"color: #003366\">Formula<\/span>: [latex]P(\\text{at most or less than}) = P(X \\le x) = 1 - e^{-\\lambda x}[\/latex]<\/li>\n<\/ul>\n<ul>\n<li><span style=\"color: #003366\">Time units<\/span>: We will use minutes as the time units.<\/li>\n<li><span style=\"color: #003366\">Lambda<\/span>: We convert [latex]\\lambda[\/latex] to the number of events per minute: [latex]\\lambda = \\frac{1}{30 \\text{ seconds}} \\times \\frac{60 \\text{ seconds}}{1 \\text{ minute}}=2 \\text{ } \\frac{\\text{customers}}{\\text{per minute}}[\/latex]<\/li>\n<li><span style=\"color: #003366\">X-values<\/span>: We want the probability of the next customer arriving within the next minute. This gives: [latex]P(x \\le 1)[\/latex].<\/li>\n<li><span style=\"color: #003366\">e<\/span>: The other letter, [latex]e[\/latex], is a constant. It is the base of the natural log and equal 2.71828.<\/li>\n<li><span style=\"color: #003366\">Answer<\/span>: Plugging all of this into the formula [latex]P(X \\le x) = 1 - e^{-\\lambda x}[\/latex] gives:<br \/>\n[latex]\\begin{align}  P(X \\le 1) &= 1 - e^{-2 \\times 1} \\\\  &= 1 - e^{-2} \\\\  &= 1 - 0.1353 \\\\  &= 0.8647  \\end{align}[\/latex]<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 86.47% chance that a customer will arrive in the next minute at the check-out.<\/p>\n<h1>Calculating the probability of At least (Exercise)<\/h1>\n<p>Next, we will use the formulas to solve for the probability of at least a certain amount of time.<\/p>\n<h2>Example 37.2.1<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: The times it takes call center specialist to resolve incoming calls to their call center:<\/p>\n<ul>\n<li>Follow an exponential distribution.<\/li>\n<li>Have an average of 2.5 minutes.<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Question:<\/strong><\/span> Specialists will get a warning at 5 minutes to try to wrap up the call (if they can). What is the probability that the specialist receives that 5 minute warning on a call?<\/p>\n<p><span style=\"color: #003366\"><strong>You Try:<\/strong><\/span> Fill in the values to answer the question in the exercise below:<\/p>\n<div id=\"h5p-104\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-104\" class=\"h5p-iframe\" data-content-id=\"104\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Exponential Distribution - Formula Example 37.2\"><\/iframe><\/div>\n<\/div>\n<p><span style=\"color: #003366\"><strong>Answer?<\/strong><\/span>\u00a0Use the values from the above exercise to answer the question. Input your answer in the exercise below:<\/p>\n<div id=\"h5p-105\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-105\" class=\"h5p-iframe\" data-content-id=\"105\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Example 37.2 - At Least - Exponential Distributions - Formula Way Answer\"><\/iframe><\/div>\n<\/div>\n<div>\n<h1>Need help with the above exercise? (Click to reveal)<\/h1>\n<p><span style=\"color: #003366\"><strong>Solution:<\/strong><\/span> Let us use the &#8216;at least&#8217; formula and follow the steps below:<\/p>\n<ul>\n<li><span style=\"color: #003366\">Formula<\/span>: [latex]P(\\text{at least or more than}) = P(X \\le x) = e^{-\\lambda x}[\/latex]<\/li>\n<\/ul>\n<ul>\n<li><span style=\"color: #003366\">Time units<\/span>: We will use minutes as the time units.<\/li>\n<li><span style=\"color: #003366\">Lambda<\/span>: We convert [latex]\\lambda[\/latex] to the number of events per minute: [latex]\\lambda = \\frac{1 \\text{ call}}{2.5 \\text{ minutes}} = \\frac{0.4 \\text{ calls}}{\\text{minute}}[\/latex]<\/li>\n<li><span style=\"color: #003366\">X-values<\/span>: We want the probability of the specialist getting the five minute warning. This means that the call lasts at least 5 minutes. This gives: [latex]P(x \\ge 5)[\/latex].<\/li>\n<li><span style=\"color: #003366\">e <\/span>= 2.71828.<\/li>\n<li><span style=\"color: #003366\">Answer<\/span>: Plugging all of this into the formula [latex]P(X \\le x) = 1 - e^{-\\lambda x}[\/latex] gives:<br \/>\n[latex]\\begin{align}  P(X \\ge 5) &= e^{-0.4 \\times 5} \\\\  &= e^{-2} \\\\  &= 0.1353  \\end{align}[\/latex]<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 13.53% chance that the call specialist receives the 5 minute warning.<\/p>\n<\/div>\n<h1>Applying the Memoryless Property (Video)<\/h1>\n<p>Let us now explore the memoryless property of the exponential distribution in the next example.<\/p>\n<h2>Example 37.2.2<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: We will revisit our previous example&#8230;<\/p>\n<ul>\n<li>The times it takes call center specialist to resolve incoming calls follow an exponential distribution.<\/li>\n<li>On average, it takes the specialist 2.5 minutes to resolve a call.<\/li>\n<li>After 5 minutes of being on a call, a warning pops up on the specialist&#8217;s computer urging them to wrap up the phone call.<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability, after receiving the warning, that it takes at least another 5 minutes to wrap up the call?<\/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\/Exponential-Probability-Formula-Memoryless-Property.pdf\">Click here to download<\/a> the written solutions shown in the video below:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Using the Exponential Distribution Formula and Understanding the Memoryless Property\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/ShGZRg5qvig?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><span style=\"color: #003366\"><strong>Conclusion: <\/strong><\/span>There is 13.53% chance that the call will take at least another 5 minutes to resolve.<\/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: Using the Exponential Distribution Formulas<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"h5p-107\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-107\" class=\"h5p-iframe\" data-content-id=\"107\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Using the Exponential Distribution Formulas Key Takeaways\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-108\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-108\" class=\"h5p-iframe\" data-content-id=\"108\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Using the Exponential Distribution Formulas Key Takeaways 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":1,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1845","chapter","type-chapter","status-publish","hentry"],"part":1820,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1845","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\/1845\/revisions"}],"predecessor-version":[{"id":1965,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1845\/revisions\/1965"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/parts\/1820"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1845\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/media?parent=1845"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=1845"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/contributor?post=1845"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/license?post=1845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}