{"id":1571,"date":"2024-05-14T09:53:57","date_gmt":"2024-05-14T13:53:57","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/?post_type=chapter&#038;p=1571"},"modified":"2024-06-12T14:46:59","modified_gmt":"2024-06-12T18:46:59","slug":"1571","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/chapter\/1571\/","title":{"raw":"Adjusting the Time Frame (Changing Lambda)","rendered":"Adjusting the Time Frame (Changing Lambda)"},"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\nUnderstand and calculate lambda ([latex]\\lambda[\/latex]) for different length time periods.\r\n\r\n<\/div>\r\n<\/div>\r\n<h2>Changing Lambda ([latex]\\lambda[\/latex])<\/h2>\r\n<ul>\r\n \t<li>If the duration of time given in the question doesn't match the duration given in the average,<\/li>\r\n \t<li>We need to adjust lambda ([latex]\\lambda[\/latex]) to match the duration given in the question<\/li>\r\n \t<li>To do this, we can use: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}[\/latex].<\/li>\r\n<\/ul>\r\n<h1>Changing Lambda ([latex]\\lambda[\/latex]) Example (VIDEO)<\/h1>\r\n<ul>\r\n \t<li>It is important to stress that we must always match the time units.<\/li>\r\n \t<li>In the example from the previous section, the average arrival rate was given per minute.<\/li>\r\n \t<li>The question's time interval was also one minute: \"What is the probability of 5 customers arriving in any given minute?\"<\/li>\r\n \t<li>So, no adjustment of lambda ([latex]\\lambda[\/latex]) was needed since the time intervals matched.<\/li>\r\n \t<li>See the next example to see where an adjustment of [latex]\\lambda[\/latex] is needed.<\/li>\r\n<\/ul>\r\n<h2>Example 31.1<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: The average number of shoppers arriving at a supermarket check-out counter waiting to be served by a cashier is 6 per minute. The arrivals follow a Poisson distribution.\r\n\r\n<span style=\"color: #003366\"><strong>Question<\/strong><\/span>: In any 10-second interval, what is the probability that 2 shoppers will arrive?\r\n\r\n<span style=\"color: #003366\"><strong>Changing Lambda ([latex]\\lambda[\/latex]):<\/strong><\/span>\r\n<ul>\r\n \t<li>There are 60 seconds in one minute. So, we will rewrite our 'old' time ([latex]t_{OLD}[\/latex]) as 60 seconds.<\/li>\r\n \t<li>Our new time interval is 10 seconds ([latex]t_{NEW}[\/latex]).<\/li>\r\n \t<li>This gives a new average of: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}=6\\times \\frac{10}{60}=1[\/latex] arrivals per second.<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Solving for the Probability:<\/strong><\/span>\r\n<ul>\r\n \t<li>Since we want the probability that two shoppers arrive, we use [latex]x=2[\/latex].<\/li>\r\n \t<li>Input [latex]x=2[\/latex] and [latex]\\lambda=1[\/latex] into the formula [latex]P(x) =\\frac{{\\lambda}^x e^{-\\lambda}}{x!}[\/latex]<\/li>\r\n \t<li>This gives: [latex]P(x=2) =\\frac{1^2 e^{-1}}{2!}\u00a0 = \\frac{1\\times 0.367879441}{2} = 0.1839[\/latex]<\/li>\r\n<\/ul>\r\n<span style=\"color: #003366\"><strong>Video Solution<\/strong><\/span>: <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example-31.1.1_WrittenSolutions.pdf\">Click here<\/a>\u00a0to download the written solutions. Also, see the video below:\r\n\r\nhttps:\/\/youtu.be\/PTK8iibwmdA?si=OFWZYomLUBRYTLZi\r\n\r\n<span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 18.39% chance that 2 shoppers will arrive in any 10-second interval.\r\n<h1>Changing Lambda ([latex]\\lambda[\/latex]) Second Example (EXERCISE)<\/h1>\r\nNow, let's practice what you just learnt in the exercise in the next example<span style=\"font-size: 20.5333px\">...<\/span>\r\n<h2>Example 31.2<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: There is an average of 146 crashes per year at Willingdon Avenue and the\u00a0 Highway#1 off-ramp (near BCIT)<span style=\"font-size: 20.5333px\">. Let's assume, for now, that the number of accidents is similar on all days of the week and they follow a Poisson distribution.<\/span>\r\n\r\n<span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability that there are no accidents at this location on any given day that you are at BCIT?\r\n\r\n<strong><span style=\"color: #003366\">You Try<\/span><\/strong>: First, solve for the new average by placing the values in the correct positions below:\r\n\r\n[h5p id=\"81\"]\r\n<div>\r\n<h1>Click here to reveal the solutions:<\/h1>\r\n<ul>\r\n \t<li>There are 365 days in the year. So, we will write our 'old' time ([latex]t_{OLD}[\/latex]) of one year minute as 365 days.<\/li>\r\n \t<li>Our new time interval is 1 day (([latex]t_{NEW}[\/latex])).<\/li>\r\n \t<li>This gives a new average of: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}=146\\times \\frac{1}{365}=0.4[\/latex] accidents per day.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<strong><span style=\"color: #003366\">You Try<\/span><\/strong>: Next, solve for probability by placing values in the correct positions below:\r\n\r\n[h5p id=\"82\"]\r\n<div>\r\n<h1>Click here to reveal the solutions:<\/h1>\r\n<ul>\r\n \t<li>Since we want the probability that no accidents occur, we use [latex]x=0[\/latex].<\/li>\r\n \t<li>Input [latex]x=0[\/latex] and [latex]\\lambda=0.4[\/latex] into the formula [latex]P(x) =\\frac{{\\lambda}^x e^{-\\lambda}}{x!}[\/latex]<\/li>\r\n \t<li>This gives: [latex]P(x=0) =\\frac{0.4^0 e^{-0.4}}{0!}\u00a0 = \\frac{1\\times 0.67032}{1} = 0.67032[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>Changing Lambda in Excel Example (Video)<\/h1>\r\n<ul>\r\n \t<li>Let us practice changing lambda in Excel<\/li>\r\n \t<li>There is no special Excel call - so it's just like doing it by hand<\/li>\r\n \t<li>None-the-less, let's finish this section with a quick video on how to do this Excel<\/li>\r\n<\/ul>\r\n<h2>Example 31.3<\/h2>\r\n<span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: There are an average of 240 accidents per year at 264<sup>th<\/sup> Street and Highway #1<span style=\"font-size: 20.5333px\">. Let's assume, for now, that the number of accidents is similar week-over-week and that they follow a Poisson distribution.<\/span>\r\n\r\n<span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability that there are 10 accidents in any given week at 264<sup>th<\/sup> and Highway 1?\r\n\r\n<span style=\"color: #003366\"><strong>Solution:<\/strong><\/span> Use = POISSON.DIST(10, 240\/52, 0). <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example30-1.xlsx\">Click here<\/a> to download the Excel solutions below:\r\n\r\nhttps:\/\/youtu.be\/sbS3Zy2z07Y\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: Adjusting the Time Frame (Changing Lambda)<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\n[h5p id=\"86\"]\r\n\r\n[h5p id=\"87\"]\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>Understand and calculate lambda ([latex]\\lambda[\/latex]) for different length time periods.<\/p>\n<\/div>\n<\/div>\n<h2>Changing Lambda ([latex]\\lambda[\/latex])<\/h2>\n<ul>\n<li>If the duration of time given in the question doesn&#8217;t match the duration given in the average,<\/li>\n<li>We need to adjust lambda ([latex]\\lambda[\/latex]) to match the duration given in the question<\/li>\n<li>To do this, we can use: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}[\/latex].<\/li>\n<\/ul>\n<h1>Changing Lambda ([latex]\\lambda[\/latex]) Example (VIDEO)<\/h1>\n<ul>\n<li>It is important to stress that we must always match the time units.<\/li>\n<li>In the example from the previous section, the average arrival rate was given per minute.<\/li>\n<li>The question&#8217;s time interval was also one minute: &#8220;What is the probability of 5 customers arriving in any given minute?&#8221;<\/li>\n<li>So, no adjustment of lambda ([latex]\\lambda[\/latex]) was needed since the time intervals matched.<\/li>\n<li>See the next example to see where an adjustment of [latex]\\lambda[\/latex] is needed.<\/li>\n<\/ul>\n<h2>Example 31.1<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: The average number of shoppers arriving at a supermarket check-out counter waiting to be served by a cashier is 6 per minute. The arrivals follow a Poisson distribution.<\/p>\n<p><span style=\"color: #003366\"><strong>Question<\/strong><\/span>: In any 10-second interval, what is the probability that 2 shoppers will arrive?<\/p>\n<p><span style=\"color: #003366\"><strong>Changing Lambda ([latex]\\lambda[\/latex]):<\/strong><\/span><\/p>\n<ul>\n<li>There are 60 seconds in one minute. So, we will rewrite our &#8216;old&#8217; time ([latex]t_{OLD}[\/latex]) as 60 seconds.<\/li>\n<li>Our new time interval is 10 seconds ([latex]t_{NEW}[\/latex]).<\/li>\n<li>This gives a new average of: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}=6\\times \\frac{10}{60}=1[\/latex] arrivals per second.<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Solving for the Probability:<\/strong><\/span><\/p>\n<ul>\n<li>Since we want the probability that two shoppers arrive, we use [latex]x=2[\/latex].<\/li>\n<li>Input [latex]x=2[\/latex] and [latex]\\lambda=1[\/latex] into the formula [latex]P(x) =\\frac{{\\lambda}^x e^{-\\lambda}}{x!}[\/latex]<\/li>\n<li>This gives: [latex]P(x=2) =\\frac{1^2 e^{-1}}{2!}\u00a0 = \\frac{1\\times 0.367879441}{2} = 0.1839[\/latex]<\/li>\n<\/ul>\n<p><span style=\"color: #003366\"><strong>Video Solution<\/strong><\/span>: <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example-31.1.1_WrittenSolutions.pdf\">Click here<\/a>\u00a0to download the written solutions. Also, see the video below:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Converting Lambda (Average Arrival Rate) &amp; Using the Poisson Probability Mass Function\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/PTK8iibwmdA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><span style=\"color: #003366\"><strong>Conclusion:<\/strong><\/span> There is an 18.39% chance that 2 shoppers will arrive in any 10-second interval.<\/p>\n<h1>Changing Lambda ([latex]\\lambda[\/latex]) Second Example (EXERCISE)<\/h1>\n<p>Now, let&#8217;s practice what you just learnt in the exercise in the next example<span style=\"font-size: 20.5333px\">&#8230;<\/span><\/p>\n<h2>Example 31.2<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: There is an average of 146 crashes per year at Willingdon Avenue and the\u00a0 Highway#1 off-ramp (near BCIT)<span style=\"font-size: 20.5333px\">. Let&#8217;s assume, for now, that the number of accidents is similar on all days of the week and they follow a Poisson distribution.<\/span><\/p>\n<p><span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability that there are no accidents at this location on any given day that you are at BCIT?<\/p>\n<p><strong><span style=\"color: #003366\">You Try<\/span><\/strong>: First, solve for the new average by placing the values in the correct positions below:<\/p>\n<div id=\"h5p-81\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-81\" class=\"h5p-iframe\" data-content-id=\"81\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Example 31.2. \u2013 Changing Lambda\"><\/iframe><\/div>\n<\/div>\n<div>\n<h1>Click here to reveal the solutions:<\/h1>\n<ul>\n<li>There are 365 days in the year. So, we will write our &#8216;old&#8217; time ([latex]t_{OLD}[\/latex]) of one year minute as 365 days.<\/li>\n<li>Our new time interval is 1 day (([latex]t_{NEW}[\/latex])).<\/li>\n<li>This gives a new average of: [latex]\\lambda_{NEW}=\\lambda_{OLD}\\times \\frac{t_{NEW}}{t_{OLD}}=146\\times \\frac{1}{365}=0.4[\/latex] accidents per day.<\/li>\n<\/ul>\n<\/div>\n<p><strong><span style=\"color: #003366\">You Try<\/span><\/strong>: Next, solve for probability by placing values in the correct positions below:<\/p>\n<div id=\"h5p-82\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-82\" class=\"h5p-iframe\" data-content-id=\"82\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Example 31.2.2 \u2013 Poisson - Using Formula - Exactly Equal To\"><\/iframe><\/div>\n<\/div>\n<div>\n<h1>Click here to reveal the solutions:<\/h1>\n<ul>\n<li>Since we want the probability that no accidents occur, we use [latex]x=0[\/latex].<\/li>\n<li>Input [latex]x=0[\/latex] and [latex]\\lambda=0.4[\/latex] into the formula [latex]P(x) =\\frac{{\\lambda}^x e^{-\\lambda}}{x!}[\/latex]<\/li>\n<li>This gives: [latex]P(x=0) =\\frac{0.4^0 e^{-0.4}}{0!}\u00a0 = \\frac{1\\times 0.67032}{1} = 0.67032[\/latex]<\/li>\n<\/ul>\n<\/div>\n<h1>Changing Lambda in Excel Example (Video)<\/h1>\n<ul>\n<li>Let us practice changing lambda in Excel<\/li>\n<li>There is no special Excel call &#8211; so it&#8217;s just like doing it by hand<\/li>\n<li>None-the-less, let&#8217;s finish this section with a quick video on how to do this Excel<\/li>\n<\/ul>\n<h2>Example 31.3<\/h2>\n<p><span style=\"color: #003366\"><strong>Problem Setup<\/strong><\/span>: There are an average of 240 accidents per year at 264<sup>th<\/sup> Street and Highway #1<span style=\"font-size: 20.5333px\">. Let&#8217;s assume, for now, that the number of accidents is similar week-over-week and that they follow a Poisson distribution.<\/span><\/p>\n<p><span style=\"color: #003366\"><strong>Question:<\/strong> <\/span>What is the probability that there are 10 accidents in any given week at 264<sup>th<\/sup> and Highway 1?<\/p>\n<p><span style=\"color: #003366\"><strong>Solution:<\/strong><\/span> Use = POISSON.DIST(10, 240\/52, 0). <a href=\"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-content\/uploads\/sites\/2128\/2024\/05\/Example30-1.xlsx\">Click here<\/a> to download the Excel solutions below:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Calculating a probability for a Poisson distribution in Excel and converting the time frame\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/sbS3Zy2z07Y?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/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: Adjusting the Time Frame (Changing Lambda)<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div id=\"h5p-86\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-86\" class=\"h5p-iframe\" data-content-id=\"86\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Adjusting the Time Frame (Changing Lambda) Key Takeaways\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-87\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-87\" class=\"h5p-iframe\" data-content-id=\"87\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Adjusting the Time Frame (Changing Lambda) Key Takeaway 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":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1571","chapter","type-chapter","status-publish","hentry"],"part":237,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1571","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\/1571\/revisions"}],"predecessor-version":[{"id":1958,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1571\/revisions\/1958"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/parts\/237"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapters\/1571\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/media?parent=1571"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/pressbooks\/v2\/chapter-type?post=1571"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/contributor?post=1571"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/1130sandbox\/wp-json\/wp\/v2\/license?post=1571"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}