{"id":253,"date":"2021-07-14T15:58:59","date_gmt":"2021-07-14T15:58:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-exponential-distribution\/"},"modified":"2023-12-05T09:16:03","modified_gmt":"2023-12-05T09:16:03","slug":"the-exponential-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-exponential-distribution\/","title":{"raw":"What is an Exponential Distribution?","rendered":"What is an Exponential Distribution?"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Given the mean of an exponential distribution, create a model for the distribution<\/li>\r\n \t<li>Calculate probabilities for an exponential distribution<\/li>\r\n \t<li>Calculate percentiles for an exponential distribution<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Exponents<\/h3>\r\nExponents are used to represent repeated multiplication of the same number. For example,\r\n<p style=\"text-align: center;\">[latex]x^3 = x \\cdot x \\cdot x[\/latex]<\/p>\r\nso, multiplying [latex]x[\/latex] by iteself [latex]3[\/latex] times.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224353\/CNX_BMath_Figure_10_02_013_img.png\" alt=\"On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.\" \/>\r\nThis is read [latex]a[\/latex] to the [latex]{m}^{\\mathrm{th}}[\/latex] power.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: [latex]e[\/latex]<\/h3>\r\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\r\n\r\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">as n increases without bound<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\r\n\r\n<\/div>\r\nThe\u00a0<strong>exponential distribution<\/strong> is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.\r\n\r\nValues for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.\r\n\r\nThe exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Negative Exponents<\/h3>\r\n<h4>The Negative Rule of Exponents<\/h4>\r\nFor any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that\r\n<div style=\"text-align: center;\">[latex]{a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet\u00a0<em>X<\/em> = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.\r\n\r\n<em>X<\/em> is a continuous random variable since time is measured. It is given that <em>\u03bc<\/em> = 4 minutes. To do any calculations, you must know <em>m<\/em>, the decay parameter.\r\n\r\n$latex {m}=\\frac{1}{\\mu}$. Therefore,\u00a0$latex {m}=\\frac{1}{4}={0.25}$\r\n\r\nThe standard deviation, <em data-effect=\"italics\">\u03c3<\/em>, is the same as the mean. <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">\u03c3<\/em>\r\n<p id=\"fs-idp111556368\">The distribution notation is <em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(<em data-effect=\"italics\">m<\/em>). Therefore, <em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.25).<\/p>\r\n<p id=\"fs-idm18148336\" class=\"finger\">The probability density function is <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = <em data-effect=\"italics\">me<\/em><sup>-<em data-effect=\"italics\">mx<\/em><\/sup>. The number <em data-effect=\"italics\">e<\/em> = 2.71828182846... It is a number that is used often in mathematics. Scientific calculators have the key \"<em data-effect=\"italics\">e<sup>x<\/sup><\/em>.\" If you enter one for <em data-effect=\"italics\">x<\/em>, the calculator will display the value <em data-effect=\"italics\">e<\/em>.<\/p>\r\n<p id=\"fs-idm74833712\">The curve is:<\/p>\r\n<p id=\"fs-idp44233136\"><em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.25<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">x<\/em><\/sup> where <em data-effect=\"italics\">x<\/em> is at least zero and <em data-effect=\"italics\">m<\/em> = 0.25.<\/p>\r\n<p id=\"fs-idp127275616\">For example, <em data-effect=\"italics\">f<\/em>(5) = 0.25<em data-effect=\"italics\">e<\/em><sup>\u2212(0.25)(5)<\/sup> = 0.072. The value 0.072 is the height of the curve when\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 5. Below, you will learn how to find probabilities using the decay parameter.<\/p>\r\nThe graph is as follows:\r\n<p id=\"fs-idm23774336\"><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214449\/fig-ch05_04_01N.jpg\" alt=\"Exponential graph with increments of 2 from 0-20 on the x-axis of \u03bc = 4 and increments of 0.05 from 0.05-0.25 on the y-axis of m = 0.25. The curved line begins at the top at point (0, 0.25) and curves down to point (20, 0). The x-axis is equal to a continuous random variable.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/p>\r\n\r\n<div id=\"fs-idp58220144\" class=\"example\" data-type=\"example\"><section>\r\n<p id=\"fs-idm725296\">Notice the graph is a declining curve. When <em data-effect=\"italics\">x<\/em> = 0,\u00a0<em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.25<em data-effect=\"italics\">e<\/em><sup>(\u22120.25)(0)<\/sup> = (0.25)(1) = 0.25 = <em data-effect=\"italics\">m<\/em>. The maximum value on the <em data-effect=\"italics\">y<\/em>-axis is <em data-effect=\"italics\">m<\/em>.<\/p>\r\n\r\n<\/section><\/div>\r\n<\/div>\r\n<div id=\"fs-idm39150960\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><header>\r\n<div class=\"textbox key-takeaways\"><header>\r\n<h3 class=\"title\" data-type=\"title\">Try It<\/h3>\r\n<\/header><section>\r\n<div id=\"fs-idm37492592\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\r\n<p id=\"fs-idm18125728\">The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.<\/p>\r\n[reveal-answer q=\"826945\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"826945\"]\r\n\r\n<em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.125); <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.125e<sup>\u20130.125<em data-effect=\"italics\">x<\/em><\/sup>;\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214451\/CNX_Stats_C05_M04_tryit001annoN.jpg\" alt=\"\" width=\"380\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/header><\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Definition of the Natural Logarithm<\/h3>\r\nA natural logarithm is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number x\u00a0satisfies the following definition.\r\n\r\nFor [latex]x&gt;0[\/latex],\r\n\r\n[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ can be written as }{e}^{y}=x[\/latex]\r\nWe read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, \"the logarithm with base e\u00a0of x\" or \"the natural logarithm of x.\"\r\n\r\nThe logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.\r\n\r\nSince the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all x\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for x\u00a0&gt;\u00a0[latex]0[\/latex].\r\n<h3>Example<\/h3>\r\n<p style=\"text-align: center;\">[latex]e^x=0.25[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]ln(e^x)=ln(0.25)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x \\cdot ln(e)= ln(0.25)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{ln(0.25)}{ln(e)}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.38629}{1}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x=-1.38629[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idm39150960\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\"><section>\r\n<div id=\"fs-idm37492592\" class=\"exercise\" data-type=\"exercise\"><section>\r\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\r\n<div class=\"textbox exercises\">\r\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\r\n<h3>\u00a0Example<\/h3>\r\n<ol>\r\n \t<li>Using the information in the previous Example, find the probability that a clerk spends four to five minutes with a randomly selected customer.<\/li>\r\n \t<li>Half of all customers are finished within how long? (Find the 50<sup>th<\/sup>\u00a0percentile.)<span data-type=\"newline\" data-count=\"2\">\r\n<\/span><\/li>\r\n \t<li>Which is larger, the mean or the median?<span data-type=\"newline\" data-count=\"2\">\r\n<\/span><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"126461\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"126461\"]\r\n<ol>\r\n \t<li>Find\u00a0<em data-effect=\"italics\">P<\/em>(4 &lt;\u00a0<em data-effect=\"italics\">x<\/em>\u00a0&lt; 5).<span data-type=\"newline\">\r\n<\/span>The\u00a0<span id=\"term116\" data-type=\"term\">cumulative distribution function (CDF)<\/span>\u00a0gives the area to the left.<span data-type=\"newline\">\r\n<\/span><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt;\u00a0<em data-effect=\"italics\">x<\/em>) = 1 \u2013\u00a0<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em><span data-type=\"newline\">\r\n<\/span><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 5) = 1 \u2013\u00a0<em data-effect=\"italics\">e<\/em><sup>(\u22120.25)(5)<\/sup>\u00a0= 0.7135 and\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 4) = 1 \u2013\u00a0<em data-effect=\"italics\">e<\/em><sup>(\u20130.25)(4)<\/sup>\u00a0= 0.6321\r\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.29.28-PM.png\"><img class=\"aligncenter wp-image-697 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214453\/Screen-Shot-2015-06-15-at-12.29.28-PM.png\" alt=\"Graph with shaded area representing the probability P(4 &lt; X &lt; 5)\" width=\"382\" height=\"185\" \/><\/a>NOTE: You can do these calculations easily on a calculator.\r\nThe probability that a postal clerk spends four to five minutes with a randomly selected customer is <em data-effect=\"italics\">P<\/em>(4 &lt; <em data-effect=\"italics\">x<\/em> &lt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 5) \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 4) = 0.7135 \u2212 0.6321 = 0.0814.\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\nOn the home screen, enter (1 \u2013 e^(\u20130.25*5))\u2013(1\u2013e^(\u20130.25*4)) or enter e^(\u20130.25*4) \u2013 e^(\u20130.25*5).<\/li>\r\n \t<li>Find the 50<sup>th<\/sup>\u00a0percentile.\r\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.32.43-PM.png\"><img class=\"aligncenter size-full wp-image-699\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214455\/Screen-Shot-2015-06-15-at-12.32.43-PM.png\" alt=\"Graph of shaded area that represents P(x &gt; K) = 0.50\" width=\"385\" height=\"181\" \/><\/a>\r\n<p id=\"fs-idp138263568\" class=\"finger\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 0.50, <em data-effect=\"italics\">k<\/em> = 2.8 minutes (calculator or computer)<\/p>\r\n<p id=\"fs-idp70510832\">Half of all customers are finished within 2.8 minutes.<\/p>\r\n<p id=\"fs-idp142531232\">You can also do the calculation as follows:<\/p>\r\n<p id=\"fs-idp56204048\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 0.50 and <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 1 \u2013<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">k<\/em><\/sup><\/p>\r\n<p id=\"fs-idm67235152\">Therefore, 0.50 = 1 \u2212 <em data-effect=\"italics\">e<\/em><sup>\u22120.25<em data-effect=\"italics\">k<\/em><\/sup> and <em data-effect=\"italics\">e<\/em><sup>\u22120.25<em data-effect=\"italics\">k<\/em><\/sup> = 1 \u2212 0.50 = 0.5<\/p>\r\n<p id=\"fs-idp105448400\">Take natural logs: <em data-effect=\"italics\">ln<\/em>(<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">k<\/em><\/sup>) = <em data-effect=\"italics\">ln<\/em>(0.50). So, \u20130.25<em data-effect=\"italics\">k<\/em> = <em data-effect=\"italics\">ln<\/em>(0.50)<\/p>\r\nSolve for <em data-effect=\"italics\">k<\/em>: \u00a0$latex {k}=\\frac{ln0.50}{-0.25}={0.25}=2.8$ minutes.\u00a0The calculator simplifies the calculation for percentile\u00a0<em data-effect=\"italics\">k<\/em>. See the following two notes.\r\n<h4 class=\"os-title\" data-type=\"title\"><span id=\"6\" class=\"os-title-label\" data-type=\"\">NOTE<\/span><\/h4>\r\nA formula for the percentile\u00a0<em>k<\/em> is [latex]k=\\frac{ln(1-\\mathrm{Area \\ to \\ the \\ left})}{-m}[\/latex]. where [latex]ln[\/latex] is the natural log.\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\nOn the home screen, enter ln(1 \u2013 0.50)\/\u20130.25. Press the (-) for the negative.<\/li>\r\n \t<li>From part b, the median or 50<sup>th<\/sup>\u00a0percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?\r\n\r\n[reveal-answer q=\"745855\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"745855\"]\r\n<p id=\"fs-idp307360\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 10) = 0.4866<\/p>\r\n<p id=\"fs-idp137414800\">50<sup>th<\/sup> percentile = 10.40<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox tryit\"><header>\r\n<h3 class=\"title\" data-type=\"title\">Activity<\/h3>\r\n<\/header>\r\n<p id=\"fs-idm38401216\" class=\" \">Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.<\/p>\r\n<p id=\"fs-idp24026224\" class=\" \">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the amount of money a student in your class has in his or her pocket or purse.<\/p>\r\n<p id=\"fs-idm20762208\" class=\" \">The distribution for\u00a0<em data-effect=\"italics\">X<\/em>\u00a0is approximately exponential with mean,\u00a0<em data-effect=\"italics\">\u03bc<\/em>\u00a0= _______ and\u00a0<em data-effect=\"italics\">m<\/em>\u00a0= _______. The standard deviation,\u00a0<em data-effect=\"italics\">\u03c3<\/em>\u00a0= ________.<\/p>\r\n<p id=\"fs-idp99610512\" class=\" \">Draw the appropriate exponential graph. You should label the <em>x<\/em>\u2013 and <em>y<\/em>\u2013axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $.40 in his or her pocket or purse. (Shade\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 0.40)).<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div id=\"fs-idp55383888\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-idp93704912\">On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>What is the probability that a computer part lasts more than 7 years?<span data-type=\"newline\" data-count=\"2\">\r\n<\/span><\/li>\r\n \t<li>On the average, how long would five computer parts last if they are used one after another?<span data-type=\"newline\" data-count=\"2\">\r\n<\/span><\/li>\r\n \t<li>Eighty percent of computer parts last at most how long?<\/li>\r\n \t<li>What is the probability that a computer part lasts between nine and 11 years?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"889353\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"889353\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Let <em data-effect=\"italics\">x<\/em> = the amount of time (in years) a computer part lasts.$latex \\mu = {10}$ so m = $latex \\frac{1}{\\mu} = \\frac{1}{10}={0.10}$\r\n<em>P(x &gt; 7)<\/em>. Draw the graph.\r\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &gt; 7) = 1 \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 7).<\/div>\r\n<div data-type=\"newline\">Since <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em> then <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013(1 \u2013<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em>) = <em data-effect=\"italics\">e<sup>-mx<\/sup><\/em><\/div>\r\n<div data-type=\"newline\">\r\n\r\n<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &gt; 7) = <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(7)<\/sup> = 0.4966. The probability that a computer part lasts more than seven years is 0.4966.\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<div data-type=\"newline\">On the home screen, enter <em>e<\/em>^(-.1*7).<\/div>\r\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.57.00-PM.png\"><img class=\"aligncenter size-full wp-image-711\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214458\/Screen-Shot-2015-06-15-at-12.57.00-PM.png\" alt=\"Graph of shaded area that represents probability P(x &gt;7)\" width=\"383\" height=\"205\" \/><\/a>\r\n\r\n<\/div><\/li>\r\n \t<li>On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.<\/li>\r\n \t<li>Find the 80<sup>th<\/sup> percentile. Draw the graph. Let <em data-effect=\"italics\">k<\/em> = the 80<sup>th<\/sup> percentile.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.58.59-PM.png\"><img class=\"aligncenter size-full wp-image-712\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214500\/Screen-Shot-2015-06-15-at-12.58.59-PM.png\" alt=\"Graph of shaded area that represents probability P(x &lt; k) = 0.80\" width=\"382\" height=\"182\" \/><\/a>\r\n<p id=\"fs-idm26221856\">Solve for <em data-effect=\"italics\">k<\/em>:\u00a0$latex {k}=\\frac{ln(1-0.80)}{-0.1}={16.1}$<\/p>\r\nEighty percent of the computer parts last at most 16.1 years.\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idm132774320\" class=\" \">On the home screen, enter [latex]\\frac{ln(1-0.80)}{-0.1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/section><\/li>\r\n \t<li>Find <em data-effect=\"italics\">P<\/em>(9 &lt; <em data-effect=\"italics\">x<\/em> &lt; 11). Draw the graph.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-1.00.58-PM.png\"><img class=\"aligncenter size-full wp-image-713\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214502\/Screen-Shot-2015-06-15-at-1.00.58-PM.png\" alt=\"Shaded area represents probability that P(9 &lt; x&lt; 11)\" width=\"383\" height=\"206\" \/><\/a><em data-effect=\"italics\">P<\/em>(9 &lt; <em data-effect=\"italics\">x<\/em> &lt; 11) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 11) \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 9) = (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(11)<\/sup>) \u2013 (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(9)<\/sup>) = 0.6671 \u2013 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737.\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp127070960\" class=\" \">On the home screen, enter\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.1*9) \u2013\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.1*11).<\/p>\r\n\r\n<\/div>\r\n<\/section><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nOn average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?\r\n\r\n<\/div>\r\n<div id=\"fs-idp64620208\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"mfrac\"><span class=\"mn\">1<\/span><span class=\"mrow\"><span class=\"mn\">12<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let <em data-effect=\"italics\">X<\/em> = the length of a phone call, in minutes.\r\n\r\nWhat is <em data-effect=\"italics\">m<\/em>, <em data-effect=\"italics\">\u03bc<\/em>, and <em data-effect=\"italics\">\u03c3<\/em>? The probability that you must wait more than five minutes is _______ .\r\n\r\n[reveal-answer q=\"32855\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"32855\"]\r\n\r\n<em>m<\/em> = \u00a0$latex \\frac{1}{12}$\r\n\r\n$latex \\mu $ = 12\r\n\r\n$latex \\sigma $ = 12\r\n\r\n<em>P(x &gt; 5)<\/em> = 0.6592\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter [latex]\\frac{1}{20}[\/latex].\u00a0Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the distance people are willing to commute in miles. What is\u00a0<em data-effect=\"italics\">m<\/em>,\u00a0<em data-effect=\"italics\">\u03bc<\/em>, and\u00a0<em data-effect=\"italics\">\u03c3<\/em>? What is the probability that a person is willing to commute more than 25 miles?\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>On average, how many minutes elapse between two successive arrivals?<\/li>\r\n \t<li>When the store first opens, how long on average does it take for three customers to arrive?<\/li>\r\n \t<li>After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.<\/li>\r\n \t<li>After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.<\/li>\r\n \t<li>Seventy percent of the customers arrive within how many minutes of the previous customer?<\/li>\r\n \t<li>Is an exponential distribution reasonable for this situation?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"186371\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"186371\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.<\/li>\r\n \t<li>Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.<\/li>\r\n \t<li>Let <em data-effect=\"italics\">X<\/em> = the time between arrivals, in minutes. By part a, <em data-effect=\"italics\">\u03bc<\/em> = 2, so <em data-effect=\"italics\">m<\/em> = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"mfrac\"><span class=\"mn\">1<\/span><span class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 0.5.\r\nTherefore, <em data-effect=\"italics\">X<\/em> \u223c <em data-effect=\"italics\">Exp<\/em>(0.5).\r\nThe cumulative distribution function is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013 <em data-effect=\"italics\">e<\/em>(\u20130.5<em data-effect=\"italics\">x<\/em>)<sup><em data-effect=\"italics\">e<\/em><\/sup>.\r\n<strong>Therefore<\/strong> <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 1) = 1 \u2013 e<sup>(\u20130.5)(1)<\/sup> \u2248 0.3935<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idm49583248\" class=\" \">1 -\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.5) \u2248 0.3935<\/p>\r\n\r\n<\/div>\r\n<\/section><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-1.56.22-PM.png\"><img class=\"aligncenter size-full wp-image-715\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214504\/Screen-Shot-2015-06-15-at-1.56.22-PM.png\" alt=\"Graph of 1-e^(-0.5x * e) with shaded region showing probability that x &lt; 0.3935\" width=\"385\" height=\"215\" \/><\/a><\/li>\r\n \t<li>\r\n<div class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\r\n\r\n<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 5) = 1 \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 5) = 1 \u2013 (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20135)(0.5)<\/sup>) = e<sup>\u20132.5<\/sup> \u2248 0.0821.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.02.27-PM.png\"><img class=\"aligncenter size-full wp-image-716\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214506\/Screen-Shot-2015-06-15-at-2.02.27-PM.png\" alt=\"Graph of 1-(1-e^(-0.5x)) with probability P(x &lt; 5) shaded\" width=\"385\" height=\"216\" \/><\/a>\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idm142138736\" class=\" \">1 \u2013 (1 \u2013 e^((-0.50)(5))) or e^( \u2013 5*0.5)<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div><\/li>\r\n \t<li>We want to solve 0.70 = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) for <em data-effect=\"italics\">x<\/em>.\r\nSubstituting in the cumulative distribution function gives 0.70 = 1 \u2013 <em data-effect=\"italics\">e<\/em><sup>\u20130.5<em data-effect=\"italics\">x<\/em><\/sup>, so that <em data-effect=\"italics\">e<\/em><sup>\u20130.5x<\/sup> = 0.30. Converting this to logarithmic form gives \u20130.5<em data-effect=\"italics\">x<\/em> = <em data-effect=\"italics\">ln<\/em>(0.30), or [latex]x=\\frac{ln(0.30)}{-0.5} \\approx 2.41[\/latex]\u00a0minutes. Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.\r\nYou are finding the 70<sup>th<\/sup>\u00a0percentile\u00a0<em data-effect=\"italics\">k<\/em>\u00a0so you can use the formula [latex]k=\\frac{ln(1- \\mathrm{Area \\ to \\ the \\ left \\ of} \\ k)}{-m}[\/latex].\r\n[latex]k=\\frac{ln(1-0.70)}{(-0.5)} \\approx 2.41[\/latex] minutes\r\n<img class=\"wp-image-591 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/09184725\/8be07b48b42e98a31cc2be82fc246ea9e66aaab2.jpeg\" alt=\"Exponential graph with the graph beginning at point (0, 0.5) and curving down towards the horizontal axis which is an asymptote. A vertical line segment extends from the horizontal axis to the curve at x = 2.41. The area under the curve between the y-axis and this segment is shaded. Text states \u201cShaded area represents probability 0.70.\u201d\" width=\"487\" height=\"271\" \/><\/li>\r\n \t<li>This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-idp88700464\" class=\" \">Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.<\/p>\r\n\r\n<ol id=\"fs-idm57552288\" type=\"a\">\r\n \t<li>On average, how many seconds elapse between two successive cars?<\/li>\r\n \t<li>After a car passes by, how long on average will it take for another seven cars to pass by?<\/li>\r\n \t<li>Find the probability that after a car passes by, the next car will pass within the next 20 seconds.<\/li>\r\n \t<li>Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h1 data-type=\"title\">Memorylessness of the Exponential Distribution<\/h1>\r\n<p id=\"fs-idm15452368\">From the first Example,\u00a0recall that the amount of time between customers is exponentially distributed with a mean of two minutes (<em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em> (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case\u2014the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the <strong>memoryless property<\/strong>. Specifically, the memoryless property says that<\/p>\r\n<p id=\"fs-idp48694608\"><em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">r<\/em> + <em data-effect=\"italics\">t<\/em> | <em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">r<\/em>) = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">t<\/em>) for all <em data-effect=\"italics\">r<\/em> \u2265 0 and <em data-effect=\"italics\">t<\/em> \u2265 0<\/p>\r\n<p id=\"fs-idp98282704\">For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using <em data-effect=\"italics\">r<\/em> = 5 and <em data-effect=\"italics\">t<\/em> = 1 in the foregoing equation.<\/p>\r\n<p id=\"fs-idm55172592\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 5 + 1 | <em data-effect=\"italics\">X<\/em> &gt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 1) = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"msup\"><span class=\"mi\">e<\/span><span class=\"mrow\"><span class=\"mrow\"><span class=\"mo\">(<\/span><span class=\"mrow\"><span class=\"mo\">\u2013<\/span><span class=\"mn\">0.5<\/span><\/span><span class=\"mo\">)<\/span><\/span><span class=\"mrow\"><span class=\"mo\">(<\/span><span class=\"mn\">1<\/span><span class=\"mo\">)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> \u2248 0.6065.<\/p>\r\n<p id=\"fs-idm9174224\"><em data-effect=\"italics\">This is the same<\/em> probability as that of waiting more than one minute for a customer to arrive after the previous arrival.<\/p>\r\n<p id=\"fs-idm25008304\">The exponential distribution is often used to model the longevity of an electrical or mechanical device. In an earlier Example, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (<em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 17|<em data-effect=\"italics\">X<\/em> &gt; 10) =<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 7) = 0.4966.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nRefer to the first Example, where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?\r\n\r\nThe decay parameter of X is m = 14 = 0.25, so X \u223c Exp(0.25).\r\nThe cumulative distribution function is P(X &lt; x) = 1 \u2013 e\u20130.25x. We want to find P(X &gt; 7|X &gt; 4). The memoryless property says that P(X &gt; 7|X &gt; 4) = P (X &gt; 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.\r\nThis is P(X &gt; 3) = 1 \u2013 P (X &lt; 3) = 1 \u2013 (1 \u2013 e\u20130.25\u22c53) = e\u20130.75 \u2248 0.4724.\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.08.39-PM.png\"><img class=\"aligncenter size-full wp-image-717\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214508\/Screen-Shot-2015-06-15-at-2.08.39-PM.png\" alt=\"graph with shaded region representing P(x &gt; 3)\" width=\"389\" height=\"212\" \/><\/a>\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp114743232\" class=\" \">1\u2013(1\u2013e^(\u20130.25*3)) = e^(\u20130.25*3)<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSuppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Factorial<\/h3>\r\nFactorial is the product of a whole number and the whole numbers that came before it (to the left) on a number line, excluding zero. For example:\r\n<p style=\"text-align: center;\">[latex]4!=4 \\cdot 3 \\cdot 2 \\cdot 1 = 24[\/latex]<\/p>\r\n\r\n<\/div>\r\n<h1 data-type=\"title\">Relationship between the Poisson and the Exponential Distribution<\/h1>\r\n<p id=\"fs-idp96158416\">There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of <em data-effect=\"italics\">\u03bc<\/em> units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean <em data-effect=\"italics\">\u03bb<\/em> = 1\/\u03bc. Recall\u00a0that if <em data-effect=\"italics\">X<\/em> has the Poisson distribution with mean <em data-effect=\"italics\">\u03bb<\/em>, then [latex]P(X=k)=\\frac{{\\lambda}^{k}{e}^{-\\lambda}}{k!}[\/latex]. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. (<em data-effect=\"italics\">k<\/em>! = <em data-effect=\"italics\">k<\/em>*(<em data-effect=\"italics\">k<\/em>-1*)(<em data-effect=\"italics\">k<\/em>\u20132)*(<em data-effect=\"italics\">k<\/em>-3)\u20263*2*1)<\/p>\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp35150608\" class=\" \">Suppose\u00a0<em data-effect=\"italics\">X<\/em>\u00a0has the Poisson distribution with mean\u00a0<em data-effect=\"italics\">\u03bb<\/em>. Compute\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em>\u00a0=\u00a0<em data-effect=\"italics\">k<\/em>) by entering 2<sup>nd<\/sup>, VARS(DISTR), C: poissonpdf(<em data-effect=\"italics\">\u03bb<\/em>,\u00a0<em data-effect=\"italics\">k<\/em>). To compute\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em>\u00a0\u2264\u00a0<em data-effect=\"italics\">k<\/em>), enter 2<sup>nd<\/sup>, VARS (DISTR), D:poissoncdf(<em data-effect=\"italics\">\u03bb<\/em>,\u00a0<em data-effect=\"italics\">k<\/em>).<\/p>\r\n\r\n<\/div>\r\n<\/section>\r\n<div class=\"textbox exercises\">\r\n<h3 id=\"fs-idm148932176\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\">Example<\/h3>\r\n<p class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\">At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find the average time between two successive calls.<\/li>\r\n \t<li>Find the probability that after a call is received, the next call occurs in less than ten seconds.<\/li>\r\n \t<li>Find the probability that exactly five calls occur within a minute.<\/li>\r\n \t<li>Find the probability that less than five calls occur within a minute.<\/li>\r\n \t<li>Find the probability that more than 40 calls occur in an eight-minute period.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"873276\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"873276\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>On average there are four calls occur per minute, so 15 seconds, or $latex \\frac{15}{60} $= 0.25 minutes\u00a0occur between successive calls on average.<\/li>\r\n \t<li>Let <em data-effect=\"italics\">T<\/em> = time elapsed between calls.\u00a0From part a, $latex \\mu = {0.25} $, so m = $latex \\frac{1}{0.25} $ = 4. Thus, T ~ Exp(4). \u00a0The cumulative distribution function is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">T<\/em> &lt; <em data-effect=\"italics\">t<\/em>) = 1 \u2013 <em data-effect=\"italics\">e<\/em><sup>\u20134<em data-effect=\"italics\">t<\/em><\/sup>.\u00a0The probability that the next call occurs in less than ten seconds (ten seconds = 1\/6 minute) is P(T &lt; $latex \\frac{1}{6}$) = 1 - \u00a0$latex {e}^{-4\\frac{1}{6}} \\approx{0.4866} $\u00a0<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.58.18-PM.png\"><img class=\"aligncenter size-full wp-image-718\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214509\/Screen-Shot-2015-06-15-at-2.58.18-PM.png\" alt=\"Graph with shaded area representing probability P(x &lt; 1\/6) = 0.4866\" width=\"385\" height=\"253\" \/><\/a><\/li>\r\n \t<li>Let <em data-effect=\"italics\">X<\/em> = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.\u00a0Therefore, <em data-effect=\"italics\">X<\/em> \u223c <em data-effect=\"italics\">Poisson<\/em>(4), and so <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 5) = [latex]\\frac{{4}^{5}{e}^{-4}}{5!}\\approx[\/latex] 0.1563. (5! = (5)(4)(3)(2)(1))\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp115054976\" class=\" \">poissonpdf(4, 5) = 0.1563.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/li>\r\n \t<li>Keep in mind that <em data-effect=\"italics\">X<\/em> must be a whole number, so <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> \u2264 4).\r\n<div data-type=\"newline\">To compute this, we could take <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 0) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 1) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 2) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 3) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 4).<\/div>\r\n<div data-type=\"newline\">\r\n\r\nUsing technology, we see that <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> \u2264 4) = 0.6288.\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp165812304\" class=\" \">poisssoncdf(4, 4) = 0.6288<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div><\/li>\r\n \t<li>Let <em data-effect=\"italics\">Y<\/em> = the number of calls that occur during an eight minute period.\r\n<div data-type=\"newline\">Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period.<\/div>\r\n<div data-type=\"newline\">\r\n\r\nHence, <em data-effect=\"italics\">Y<\/em> \u223c <em data-effect=\"italics\">Poisson<\/em>(32). Therefore, <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">Y<\/em> &gt; 40) = 1 \u2013 <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">Y<\/em> \u2264 40) = 1 \u2013 0.9294 = 0.0707.\r\n\r\n<header>\r\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idp137845728\" class=\" \">1 \u2013 poissoncdf(32, 40). = 0.0707<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-idm15154368\" class=\" \">In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.<\/p>\r\n\r\n<ol id=\"fs-idp53070848\" type=\"a\">\r\n \t<li>Calculate the probability that there are at most 2 accidents occur in any given week.<\/li>\r\n \t<li>What is the probability that there is at least two weeks between any 2 accidents?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/section><\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Given the mean of an exponential distribution, create a model for the distribution<\/li>\n<li>Calculate probabilities for an exponential distribution<\/li>\n<li>Calculate percentiles for an exponential distribution<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Exponents<\/h3>\n<p>Exponents are used to represent repeated multiplication of the same number. For example,<\/p>\n<p style=\"text-align: center;\">[latex]x^3 = x \\cdot x \\cdot x[\/latex]<\/p>\n<p>so, multiplying [latex]x[\/latex] by iteself [latex]3[\/latex] times.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224353\/CNX_BMath_Figure_10_02_013_img.png\" alt=\"On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.\" \/><br \/>\nThis is read [latex]a[\/latex] to the [latex]{m}^{\\mathrm{th}}[\/latex] power.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: [latex]e[\/latex]<\/h3>\n<p id=\"fs-id1165135511335\">The letter <em>e<\/em> represents the irrational number<\/p>\n<div id=\"eip-id1165135378658\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{\\left(1+\\frac{1}{n}\\right)}^{n}[\/latex]<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">as n increases without bound<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165135369344\">The letter <em>e <\/em>is used as a base for many real-world exponential models. To work with base <em>e<\/em>, we use the approximation, [latex]e\\approx 2.718282[\/latex]. The constant was named by the Swiss mathematician Leonhard Euler (1707\u20131783) who first investigated and discovered many of its properties.<\/p>\n<\/div>\n<p>The\u00a0<strong>exponential distribution<\/strong> is often concerned with the amount of time until some specific event occurs. For example, the amount of time (beginning now) until an earthquake occurs has an exponential distribution. Other examples include the length, in minutes, of long distance business telephone calls, and the amount of time, in months, a car battery lasts. It can be shown, too, that the value of the change that you have in your pocket or purse approximately follows an exponential distribution.<\/p>\n<p>Values for an exponential random variable occur in the following way. There are fewer large values and more small values. For example, the amount of money customers spend in one trip to the supermarket follows an exponential distribution. There are more people who spend small amounts of money and fewer people who spend large amounts of money.<\/p>\n<p>The exponential distribution is widely used in the field of reliability. Reliability deals with the amount of time a product lasts.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Negative Exponents<\/h3>\n<h4>The Negative Rule of Exponents<\/h4>\n<p>For any nonzero real number [latex]a[\/latex] and natural number [latex]n[\/latex], the negative rule of exponents states that<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{-n}=\\frac{1}{{a}^{n}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let\u00a0<em>X<\/em> = amount of time (in minutes) a postal clerk spends with his or her customer. The time is known to have an exponential distribution with the average amount of time equal to four minutes.<\/p>\n<p><em>X<\/em> is a continuous random variable since time is measured. It is given that <em>\u03bc<\/em> = 4 minutes. To do any calculations, you must know <em>m<\/em>, the decay parameter.<\/p>\n<p>[latex]{m}=\\frac{1}{\\mu}[\/latex]. Therefore,\u00a0[latex]{m}=\\frac{1}{4}={0.25}[\/latex]<\/p>\n<p>The standard deviation, <em data-effect=\"italics\">\u03c3<\/em>, is the same as the mean. <em data-effect=\"italics\">\u03bc<\/em> = <em data-effect=\"italics\">\u03c3<\/em><\/p>\n<p id=\"fs-idp111556368\">The distribution notation is <em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(<em data-effect=\"italics\">m<\/em>). Therefore, <em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.25).<\/p>\n<p id=\"fs-idm18148336\" class=\"finger\">The probability density function is <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = <em data-effect=\"italics\">me<\/em><sup>&#8211;<em data-effect=\"italics\">mx<\/em><\/sup>. The number <em data-effect=\"italics\">e<\/em> = 2.71828182846&#8230; It is a number that is used often in mathematics. Scientific calculators have the key &#8220;<em data-effect=\"italics\">e<sup>x<\/sup><\/em>.&#8221; If you enter one for <em data-effect=\"italics\">x<\/em>, the calculator will display the value <em data-effect=\"italics\">e<\/em>.<\/p>\n<p id=\"fs-idm74833712\">The curve is:<\/p>\n<p id=\"fs-idp44233136\"><em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.25<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">x<\/em><\/sup> where <em data-effect=\"italics\">x<\/em> is at least zero and <em data-effect=\"italics\">m<\/em> = 0.25.<\/p>\n<p id=\"fs-idp127275616\">For example, <em data-effect=\"italics\">f<\/em>(5) = 0.25<em data-effect=\"italics\">e<\/em><sup>\u2212(0.25)(5)<\/sup> = 0.072. The value 0.072 is the height of the curve when\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 5. Below, you will learn how to find probabilities using the decay parameter.<\/p>\n<p>The graph is as follows:<\/p>\n<p id=\"fs-idm23774336\"><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214449\/fig-ch05_04_01N.jpg\" alt=\"Exponential graph with increments of 2 from 0-20 on the x-axis of \u03bc = 4 and increments of 0.05 from 0.05-0.25 on the y-axis of m = 0.25. The curved line begins at the top at point (0, 0.25) and curves down to point (20, 0). The x-axis is equal to a continuous random variable.\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/p>\n<div id=\"fs-idp58220144\" class=\"example\" data-type=\"example\">\n<section>\n<p id=\"fs-idm725296\">Notice the graph is a declining curve. When <em data-effect=\"italics\">x<\/em> = 0,\u00a0<em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.25<em data-effect=\"italics\">e<\/em><sup>(\u22120.25)(0)<\/sup> = (0.25)(1) = 0.25 = <em data-effect=\"italics\">m<\/em>. The maximum value on the <em data-effect=\"italics\">y<\/em>-axis is <em data-effect=\"italics\">m<\/em>.<\/p>\n<\/section>\n<\/div>\n<\/div>\n<div id=\"fs-idm39150960\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<header>\n<div class=\"textbox key-takeaways\"><\/div>\n<\/header>\n<header>\n<h3 class=\"title\" data-type=\"title\">Try It<\/h3>\n<\/header>\n<section>\n<div id=\"fs-idm37492592\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\n<p id=\"fs-idm18125728\">The amount of time spouses shop for anniversary cards can be modeled by an exponential distribution with the average amount of time equal to eight minutes. Write the distribution, state the probability density function, and graph the distribution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q826945\">Show Answer<\/span><\/p>\n<div id=\"q826945\" class=\"hidden-answer\" style=\"display: none\">\n<p><em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.125); <em data-effect=\"italics\">f<\/em>(<em data-effect=\"italics\">x<\/em>) = 0.125e<sup>\u20130.125<em data-effect=\"italics\">x<\/em><\/sup>;<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214451\/CNX_Stats_C05_M04_tryit001annoN.jpg\" alt=\"\" width=\"380\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Definition of the Natural Logarithm<\/h3>\n<p>A natural logarithm is a logarithm with base <em>e<\/em>. We write [latex]{\\mathrm{log}}_{e}\\left(x\\right)[\/latex] simply as [latex]\\mathrm{ln}\\left(x\\right)[\/latex]. The natural logarithm of a positive number x\u00a0satisfies the following definition.<\/p>\n<p>For [latex]x>0[\/latex],<\/p>\n<p>[latex]y=\\mathrm{ln}\\left(x\\right)\\text{ can be written as }{e}^{y}=x[\/latex]<br \/>\nWe read [latex]\\mathrm{ln}\\left(x\\right)[\/latex] as, &#8220;the logarithm with base e\u00a0of x&#8221; or &#8220;the natural logarithm of x.&#8221;<\/p>\n<p>The logarithm <em>y<\/em>\u00a0is the exponent to which <em>e<\/em>\u00a0must be raised to get <em>x<\/em>.<\/p>\n<p>Since the functions [latex]y=e{}^{x}[\/latex] and [latex]y=\\mathrm{ln}\\left(x\\right)[\/latex] are inverse functions, [latex]\\mathrm{ln}\\left({e}^{x}\\right)=x[\/latex] for all x\u00a0and [latex]e{}^{\\mathrm{ln}\\left(x\\right)}=x[\/latex] for x\u00a0&gt;\u00a0[latex]0[\/latex].<\/p>\n<h3>Example<\/h3>\n<p style=\"text-align: center;\">[latex]e^x=0.25[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]ln(e^x)=ln(0.25)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x \\cdot ln(e)= ln(0.25)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{ln(0.25)}{ln(e)}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=\\frac{-1.38629}{1}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-1.38629[\/latex]<\/p>\n<\/div>\n<div id=\"fs-idm39150960\" class=\"note statistics try ui-has-child-title\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<section>\n<div id=\"fs-idm37492592\" class=\"exercise\" data-type=\"exercise\">\n<section>\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\n<div class=\"textbox exercises\">\n<div id=\"fs-idp79313392\" class=\"problem\" data-type=\"problem\">\n<h3>\u00a0Example<\/h3>\n<ol>\n<li>Using the information in the previous Example, find the probability that a clerk spends four to five minutes with a randomly selected customer.<\/li>\n<li>Half of all customers are finished within how long? (Find the 50<sup>th<\/sup>\u00a0percentile.)<span data-type=\"newline\" data-count=\"2\"><br \/>\n<\/span><\/li>\n<li>Which is larger, the mean or the median?<span data-type=\"newline\" data-count=\"2\"><br \/>\n<\/span><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q126461\">Show Answer<\/span><\/p>\n<div id=\"q126461\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Find\u00a0<em data-effect=\"italics\">P<\/em>(4 &lt;\u00a0<em data-effect=\"italics\">x<\/em>\u00a0&lt; 5).<span data-type=\"newline\"><br \/>\n<\/span>The\u00a0<span id=\"term116\" data-type=\"term\">cumulative distribution function (CDF)<\/span>\u00a0gives the area to the left.<span data-type=\"newline\"><br \/>\n<\/span><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt;\u00a0<em data-effect=\"italics\">x<\/em>) = 1 \u2013\u00a0<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em><span data-type=\"newline\"><br \/>\n<\/span><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 5) = 1 \u2013\u00a0<em data-effect=\"italics\">e<\/em><sup>(\u22120.25)(5)<\/sup>\u00a0= 0.7135 and\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 4) = 1 \u2013\u00a0<em data-effect=\"italics\">e<\/em><sup>(\u20130.25)(4)<\/sup>\u00a0= 0.6321<br \/>\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.29.28-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-697 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214453\/Screen-Shot-2015-06-15-at-12.29.28-PM.png\" alt=\"Graph with shaded area representing the probability P(4 &lt; X &lt; 5)\" width=\"382\" height=\"185\" \/><\/a>NOTE: You can do these calculations easily on a calculator.<br \/>\nThe probability that a postal clerk spends four to five minutes with a randomly selected customer is <em data-effect=\"italics\">P<\/em>(4 &lt; <em data-effect=\"italics\">x<\/em> &lt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 5) \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 4) = 0.7135 \u2212 0.6321 = 0.0814.<\/p>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<p>On the home screen, enter (1 \u2013 e^(\u20130.25*5))\u2013(1\u2013e^(\u20130.25*4)) or enter e^(\u20130.25*4) \u2013 e^(\u20130.25*5).<\/li>\n<li>Find the 50<sup>th<\/sup>\u00a0percentile.<br \/>\n<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.32.43-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-699\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214455\/Screen-Shot-2015-06-15-at-12.32.43-PM.png\" alt=\"Graph of shaded area that represents P(x &gt; K) = 0.50\" width=\"385\" height=\"181\" \/><\/a><\/p>\n<p id=\"fs-idp138263568\" class=\"finger\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 0.50, <em data-effect=\"italics\">k<\/em> = 2.8 minutes (calculator or computer)<\/p>\n<p id=\"fs-idp70510832\">Half of all customers are finished within 2.8 minutes.<\/p>\n<p id=\"fs-idp142531232\">You can also do the calculation as follows:<\/p>\n<p id=\"fs-idp56204048\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 0.50 and <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; <em data-effect=\"italics\">k<\/em>) = 1 \u2013<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">k<\/em><\/sup><\/p>\n<p id=\"fs-idm67235152\">Therefore, 0.50 = 1 \u2212 <em data-effect=\"italics\">e<\/em><sup>\u22120.25<em data-effect=\"italics\">k<\/em><\/sup> and <em data-effect=\"italics\">e<\/em><sup>\u22120.25<em data-effect=\"italics\">k<\/em><\/sup> = 1 \u2212 0.50 = 0.5<\/p>\n<p id=\"fs-idp105448400\">Take natural logs: <em data-effect=\"italics\">ln<\/em>(<em data-effect=\"italics\">e<\/em><sup>\u20130.25<em data-effect=\"italics\">k<\/em><\/sup>) = <em data-effect=\"italics\">ln<\/em>(0.50). So, \u20130.25<em data-effect=\"italics\">k<\/em> = <em data-effect=\"italics\">ln<\/em>(0.50)<\/p>\n<p>Solve for <em data-effect=\"italics\">k<\/em>: \u00a0[latex]{k}=\\frac{ln0.50}{-0.25}={0.25}=2.8[\/latex] minutes.\u00a0The calculator simplifies the calculation for percentile\u00a0<em data-effect=\"italics\">k<\/em>. See the following two notes.<\/p>\n<h4 class=\"os-title\" data-type=\"title\"><span id=\"6\" class=\"os-title-label\" data-type=\"\">NOTE<\/span><\/h4>\n<p>A formula for the percentile\u00a0<em>k<\/em> is [latex]k=\\frac{ln(1-\\mathrm{Area \\ to \\ the \\ left})}{-m}[\/latex]. where [latex]ln[\/latex] is the natural log.<\/p>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<p>On the home screen, enter ln(1 \u2013 0.50)\/\u20130.25. Press the (-) for the negative.<\/li>\n<li>From part b, the median or 50<sup>th<\/sup>\u00a0percentile is 2.8 minutes. The theoretical mean is four minutes. The mean is larger.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q745855\">Show Answer<\/span><\/p>\n<div id=\"q745855\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-idp307360\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 10) = 0.4866<\/p>\n<p id=\"fs-idp137414800\">50<sup>th<\/sup> percentile = 10.40<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<header>\n<h3 class=\"title\" data-type=\"title\">Activity<\/h3>\n<\/header>\n<p id=\"fs-idm38401216\" class=\"\">Have each class member count the change he or she has in his or her pocket or purse. Your instructor will record the amounts in dollars and cents. Construct a histogram of the data taken by the class. Use five intervals. Draw a smooth curve through the bars. The graph should look approximately exponential. Then calculate the mean.<\/p>\n<p id=\"fs-idp24026224\" class=\"\">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the amount of money a student in your class has in his or her pocket or purse.<\/p>\n<p id=\"fs-idm20762208\" class=\"\">The distribution for\u00a0<em data-effect=\"italics\">X<\/em>\u00a0is approximately exponential with mean,\u00a0<em data-effect=\"italics\">\u03bc<\/em>\u00a0= _______ and\u00a0<em data-effect=\"italics\">m<\/em>\u00a0= _______. The standard deviation,\u00a0<em data-effect=\"italics\">\u03c3<\/em>\u00a0= ________.<\/p>\n<p id=\"fs-idp99610512\" class=\"\">Draw the appropriate exponential graph. You should label the <em>x<\/em>\u2013 and <em>y<\/em>\u2013axes, the decay rate, and the mean. Shade the area that represents the probability that one student has less than $.40 in his or her pocket or purse. (Shade\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em>\u00a0&lt; 0.40)).<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"fs-idp55383888\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-idp93704912\">On the average, a certain computer part lasts ten years. The length of time the computer part lasts is exponentially distributed.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>What is the probability that a computer part lasts more than 7 years?<span data-type=\"newline\" data-count=\"2\"><br \/>\n<\/span><\/li>\n<li>On the average, how long would five computer parts last if they are used one after another?<span data-type=\"newline\" data-count=\"2\"><br \/>\n<\/span><\/li>\n<li>Eighty percent of computer parts last at most how long?<\/li>\n<li>What is the probability that a computer part lasts between nine and 11 years?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889353\">Show Answer<\/span><\/p>\n<div id=\"q889353\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Let <em data-effect=\"italics\">x<\/em> = the amount of time (in years) a computer part lasts.[latex]\\mu = {10}[\/latex] so m = [latex]\\frac{1}{\\mu} = \\frac{1}{10}={0.10}[\/latex]<br \/>\n<em>P(x &gt; 7)<\/em>. Draw the graph.<\/p>\n<div data-type=\"newline\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &gt; 7) = 1 \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 7).<\/div>\n<div data-type=\"newline\">Since <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em> then <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013(1 \u2013<em data-effect=\"italics\">e<sup>\u2013mx<\/sup><\/em>) = <em data-effect=\"italics\">e<sup>-mx<\/sup><\/em><\/div>\n<div data-type=\"newline\">\n<p><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &gt; 7) = <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(7)<\/sup> = 0.4966. The probability that a computer part lasts more than seven years is 0.4966.<\/p>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<div data-type=\"newline\">On the home screen, enter <em>e<\/em>^(-.1*7).<\/div>\n<p><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.57.00-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-711\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214458\/Screen-Shot-2015-06-15-at-12.57.00-PM.png\" alt=\"Graph of shaded area that represents probability P(x &gt;7)\" width=\"383\" height=\"205\" \/><\/a><\/p>\n<\/div>\n<\/li>\n<li>On the average, one computer part lasts ten years. Therefore, five computer parts, if they are used one right after the other would last, on the average, (5)(10) = 50 years.<\/li>\n<li>Find the 80<sup>th<\/sup> percentile. Draw the graph. Let <em data-effect=\"italics\">k<\/em> = the 80<sup>th<\/sup> percentile.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-12.58.59-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-712\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214500\/Screen-Shot-2015-06-15-at-12.58.59-PM.png\" alt=\"Graph of shaded area that represents probability P(x &lt; k) = 0.80\" width=\"382\" height=\"182\" \/><\/a>\n<p id=\"fs-idm26221856\">Solve for <em data-effect=\"italics\">k<\/em>:\u00a0[latex]{k}=\\frac{ln(1-0.80)}{-0.1}={16.1}[\/latex]<\/p>\n<p>Eighty percent of the computer parts last at most 16.1 years.<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idm132774320\" class=\"\">On the home screen, enter [latex]\\frac{ln(1-0.80)}{-0.1}[\/latex]<\/p>\n<\/div>\n<\/section>\n<\/li>\n<li>Find <em data-effect=\"italics\">P<\/em>(9 &lt; <em data-effect=\"italics\">x<\/em> &lt; 11). Draw the graph.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-1.00.58-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-713\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214502\/Screen-Shot-2015-06-15-at-1.00.58-PM.png\" alt=\"Shaded area represents probability that P(9 &lt; x&lt; 11)\" width=\"383\" height=\"206\" \/><\/a><em data-effect=\"italics\">P<\/em>(9 &lt; <em data-effect=\"italics\">x<\/em> &lt; 11) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 11) \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">x<\/em> &lt; 9) = (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(11)<\/sup>) \u2013 (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20130.1)(9)<\/sup>) = 0.6671 \u2013 0.5934 = 0.0737. The probability that a computer part lasts between nine and 11 years is 0.0737.<br \/>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp127070960\" class=\"\">On the home screen, enter\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.1*9) \u2013\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.1*11).<\/p>\n<\/div>\n<\/section>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>On average, a pair of running shoes can last 18 months if used every day. The length of time running shoes last is exponentially distributed. What is the probability that a pair of running shoes last more than 15 months? On average, how long would six pairs of running shoes last if they are used one after the other? Eighty percent of running shoes last at most how long if used every day?<\/p>\n<\/div>\n<div id=\"fs-idp64620208\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose that the length of a phone call, in minutes, is an exponential random variable with decay parameter = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"mfrac\"><span class=\"mn\">1<\/span><span class=\"mrow\"><span class=\"mn\">12<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>. If another person arrives at a public telephone just before you, find the probability that you will have to wait more than five minutes. Let <em data-effect=\"italics\">X<\/em> = the length of a phone call, in minutes.<\/p>\n<p>What is <em data-effect=\"italics\">m<\/em>, <em data-effect=\"italics\">\u03bc<\/em>, and <em data-effect=\"italics\">\u03c3<\/em>? The probability that you must wait more than five minutes is _______ .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q32855\">Show Answer<\/span><\/p>\n<div id=\"q32855\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>m<\/em> = \u00a0[latex]\\frac{1}{12}[\/latex]<\/p>\n<p>[latex]\\mu[\/latex] = 12<\/p>\n<p>[latex]\\sigma[\/latex] = 12<\/p>\n<p><em>P(x &gt; 5)<\/em> = 0.6592<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose that the distance, in miles, that people are willing to commute to work is an exponential random variable with a decay parameter [latex]\\frac{1}{20}[\/latex].\u00a0Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the distance people are willing to commute in miles. What is\u00a0<em data-effect=\"italics\">m<\/em>,\u00a0<em data-effect=\"italics\">\u03bc<\/em>, and\u00a0<em data-effect=\"italics\">\u03c3<\/em>? What is the probability that a person is willing to commute more than 25 miles?<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The time spent waiting between events is often modeled using the exponential distribution. For example, suppose that an average of 30 customers per hour arrive at a store and the time between arrivals is exponentially distributed.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>On average, how many minutes elapse between two successive arrivals?<\/li>\n<li>When the store first opens, how long on average does it take for three customers to arrive?<\/li>\n<li>After a customer arrives, find the probability that it takes less than one minute for the next customer to arrive.<\/li>\n<li>After a customer arrives, find the probability that it takes more than five minutes for the next customer to arrive.<\/li>\n<li>Seventy percent of the customers arrive within how many minutes of the previous customer?<\/li>\n<li>Is an exponential distribution reasonable for this situation?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q186371\">Show Answer<\/span><\/p>\n<div id=\"q186371\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Since we expect 30 customers to arrive per hour (60 minutes), we expect on average one customer to arrive every two minutes on average.<\/li>\n<li>Since one customer arrives every two minutes on average, it will take six minutes on average for three customers to arrive.<\/li>\n<li>Let <em data-effect=\"italics\">X<\/em> = the time between arrivals, in minutes. By part a, <em data-effect=\"italics\">\u03bc<\/em> = 2, so <em data-effect=\"italics\">m<\/em> = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"mfrac\"><span class=\"mn\">1<\/span><span class=\"mn\">2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> = 0.5.<br \/>\nTherefore, <em data-effect=\"italics\">X<\/em> \u223c <em data-effect=\"italics\">Exp<\/em>(0.5).<br \/>\nThe cumulative distribution function is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) = 1 \u2013 <em data-effect=\"italics\">e<\/em>(\u20130.5<em data-effect=\"italics\">x<\/em>)<sup><em data-effect=\"italics\">e<\/em><\/sup>.<br \/>\n<strong>Therefore<\/strong> <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 1) = 1 \u2013 e<sup>(\u20130.5)(1)<\/sup> \u2248 0.3935<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idm49583248\" class=\"\">1 &#8211;\u00a0<em data-effect=\"italics\">e<\/em>^(\u20130.5) \u2248 0.3935<\/p>\n<\/div>\n<\/section>\n<p><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-1.56.22-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-715\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214504\/Screen-Shot-2015-06-15-at-1.56.22-PM.png\" alt=\"Graph of 1-e^(-0.5x * e) with shaded region showing probability that x &lt; 0.3935\" width=\"385\" height=\"215\" \/><\/a><\/li>\n<li>\n<div class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\" data-label=\"\">\n<p><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 5) = 1 \u2013 <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 5) = 1 \u2013 (1 \u2013 <em data-effect=\"italics\">e<\/em><sup>(\u20135)(0.5)<\/sup>) = e<sup>\u20132.5<\/sup> \u2248 0.0821.<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.02.27-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-716\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214506\/Screen-Shot-2015-06-15-at-2.02.27-PM.png\" alt=\"Graph of 1-(1-e^(-0.5x)) with probability P(x &lt; 5) shaded\" width=\"385\" height=\"216\" \/><\/a><\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idm142138736\" class=\"\">1 \u2013 (1 \u2013 e^((-0.50)(5))) or e^( \u2013 5*0.5)<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/li>\n<li>We want to solve 0.70 = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; <em data-effect=\"italics\">x<\/em>) for <em data-effect=\"italics\">x<\/em>.<br \/>\nSubstituting in the cumulative distribution function gives 0.70 = 1 \u2013 <em data-effect=\"italics\">e<\/em><sup>\u20130.5<em data-effect=\"italics\">x<\/em><\/sup>, so that <em data-effect=\"italics\">e<\/em><sup>\u20130.5x<\/sup> = 0.30. Converting this to logarithmic form gives \u20130.5<em data-effect=\"italics\">x<\/em> = <em data-effect=\"italics\">ln<\/em>(0.30), or [latex]x=\\frac{ln(0.30)}{-0.5} \\approx 2.41[\/latex]\u00a0minutes. Thus, seventy percent of customers arrive within 2.41 minutes of the previous customer.<br \/>\nYou are finding the 70<sup>th<\/sup>\u00a0percentile\u00a0<em data-effect=\"italics\">k<\/em>\u00a0so you can use the formula [latex]k=\\frac{ln(1- \\mathrm{Area \\ to \\ the \\ left \\ of} \\ k)}{-m}[\/latex].<br \/>\n[latex]k=\\frac{ln(1-0.70)}{(-0.5)} \\approx 2.41[\/latex] minutes<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"wp-image-591 size-full aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/07\/09184725\/8be07b48b42e98a31cc2be82fc246ea9e66aaab2.jpeg\" alt=\"Exponential graph with the graph beginning at point (0, 0.5) and curving down towards the horizontal axis which is an asymptote. A vertical line segment extends from the horizontal axis to the curve at x = 2.41. The area under the curve between the y-axis and this segment is shaded. Text states \u201cShaded area represents probability 0.70.\u201d\" width=\"487\" height=\"271\" \/><\/li>\n<li>This model assumes that a single customer arrives at a time, which may not be reasonable since people might shop in groups, leading to several customers arriving at the same time. It also assumes that the flow of customers does not change throughout the day, which is not valid if some times of the day are busier than others.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-idp88700464\" class=\"\">Suppose that on a certain stretch of highway, cars pass at an average rate of five cars per minute. Assume that the duration of time between successive cars follows the exponential distribution.<\/p>\n<ol id=\"fs-idm57552288\" type=\"a\">\n<li>On average, how many seconds elapse between two successive cars?<\/li>\n<li>After a car passes by, how long on average will it take for another seven cars to pass by?<\/li>\n<li>Find the probability that after a car passes by, the next car will pass within the next 20 seconds.<\/li>\n<li>Find the probability that after a car passes by, the next car will not pass for at least another 15 seconds.<\/li>\n<\/ol>\n<\/div>\n<h1 data-type=\"title\">Memorylessness of the Exponential Distribution<\/h1>\n<p id=\"fs-idm15452368\">From the first Example,\u00a0recall that the amount of time between customers is exponentially distributed with a mean of two minutes (<em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em> (0.5)). Suppose that five minutes have elapsed since the last customer arrived. Since an unusually long amount of time has now elapsed, it would seem to be more likely for a customer to arrive within the next minute. With the exponential distribution, this is not the case\u2014the additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the <strong>memoryless property<\/strong>. Specifically, the memoryless property says that<\/p>\n<p id=\"fs-idp48694608\"><em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">r<\/em> + <em data-effect=\"italics\">t<\/em> | <em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">r<\/em>) = <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">X<\/em> &gt; <em data-effect=\"italics\">t<\/em>) for all <em data-effect=\"italics\">r<\/em> \u2265 0 and <em data-effect=\"italics\">t<\/em> \u2265 0<\/p>\n<p id=\"fs-idp98282704\">For example, if five minutes has elapsed since the last customer arrived, then the probability that more than one minute will elapse before the next customer arrives is computed by using <em data-effect=\"italics\">r<\/em> = 5 and <em data-effect=\"italics\">t<\/em> = 1 in the foregoing equation.<\/p>\n<p id=\"fs-idm55172592\"><em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 5 + 1 | <em data-effect=\"italics\">X<\/em> &gt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 1) = <span class=\"MathJax\"><span class=\"math\"><span class=\"mrow\"><span class=\"semantics\"><span class=\"mrow\"><span class=\"mrow\"><span class=\"msup\"><span class=\"mi\">e<\/span><span class=\"mrow\"><span class=\"mrow\"><span class=\"mo\">(<\/span><span class=\"mrow\"><span class=\"mo\">\u2013<\/span><span class=\"mn\">0.5<\/span><\/span><span class=\"mo\">)<\/span><\/span><span class=\"mrow\"><span class=\"mo\">(<\/span><span class=\"mn\">1<\/span><span class=\"mo\">)<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> \u2248 0.6065.<\/p>\n<p id=\"fs-idm9174224\"><em data-effect=\"italics\">This is the same<\/em> probability as that of waiting more than one minute for a customer to arrive after the previous arrival.<\/p>\n<p id=\"fs-idm25008304\">The exponential distribution is often used to model the longevity of an electrical or mechanical device. In an earlier Example, the lifetime of a certain computer part has the exponential distribution with a mean of ten years (<em data-effect=\"italics\">X<\/em> ~ <em data-effect=\"italics\">Exp<\/em>(0.1)). The memoryless property says that knowledge of what has occurred in the past has no effect on future probabilities. In this case it means that an old part is not any more likely to break down at any particular time than a brand new part. In other words, the part stays as good as new until it suddenly breaks. For example, if the part has already lasted ten years, then the probability that it lasts another seven years is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 17|<em data-effect=\"italics\">X<\/em> &gt; 10) =<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &gt; 7) = 0.4966.<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Refer to the first Example, where the time a postal clerk spends with his or her customer has an exponential distribution with a mean of four minutes. Suppose a customer has spent four minutes with a postal clerk. What is the probability that he or she will spend at least an additional three minutes with the postal clerk?<\/p>\n<p>The decay parameter of X is m = 14 = 0.25, so X \u223c Exp(0.25).<br \/>\nThe cumulative distribution function is P(X &lt; x) = 1 \u2013 e\u20130.25x. We want to find P(X &gt; 7|X &gt; 4). The memoryless property says that P(X &gt; 7|X &gt; 4) = P (X &gt; 3), so we just need to find the probability that a customer spends more than three minutes with a postal clerk.<br \/>\nThis is P(X &gt; 3) = 1 \u2013 P (X &lt; 3) = 1 \u2013 (1 \u2013 e\u20130.25\u22c53) = e\u20130.75 \u2248 0.4724.<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.08.39-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-717\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214508\/Screen-Shot-2015-06-15-at-2.08.39-PM.png\" alt=\"graph with shaded region representing P(x &gt; 3)\" width=\"389\" height=\"212\" \/><\/a><\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp114743232\" class=\"\">1\u2013(1\u2013e^(\u20130.25*3)) = e^(\u20130.25*3)<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. If a bulb has already lasted 12 years, find the probability that it will last a total of over 19 years.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Factorial<\/h3>\n<p>Factorial is the product of a whole number and the whole numbers that came before it (to the left) on a number line, excluding zero. For example:<\/p>\n<p style=\"text-align: center;\">[latex]4!=4 \\cdot 3 \\cdot 2 \\cdot 1 = 24[\/latex]<\/p>\n<\/div>\n<h1 data-type=\"title\">Relationship between the Poisson and the Exponential Distribution<\/h1>\n<p id=\"fs-idp96158416\">There is an interesting relationship between the exponential distribution and the Poisson distribution. Suppose that the time that elapses between two successive events follows the exponential distribution with a mean of <em data-effect=\"italics\">\u03bc<\/em> units of time. Also assume that these times are independent, meaning that the time between events is not affected by the times between previous events. If these assumptions hold, then the number of events per unit time follows a Poisson distribution with mean <em data-effect=\"italics\">\u03bb<\/em> = 1\/\u03bc. Recall\u00a0that if <em data-effect=\"italics\">X<\/em> has the Poisson distribution with mean <em data-effect=\"italics\">\u03bb<\/em>, then [latex]P(X=k)=\\frac{{\\lambda}^{k}{e}^{-\\lambda}}{k!}[\/latex]. Conversely, if the number of events per unit time follows a Poisson distribution, then the amount of time between events follows the exponential distribution. (<em data-effect=\"italics\">k<\/em>! = <em data-effect=\"italics\">k<\/em>*(<em data-effect=\"italics\">k<\/em>-1*)(<em data-effect=\"italics\">k<\/em>\u20132)*(<em data-effect=\"italics\">k<\/em>-3)\u20263*2*1)<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp35150608\" class=\"\">Suppose\u00a0<em data-effect=\"italics\">X<\/em>\u00a0has the Poisson distribution with mean\u00a0<em data-effect=\"italics\">\u03bb<\/em>. Compute\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em>\u00a0=\u00a0<em data-effect=\"italics\">k<\/em>) by entering 2<sup>nd<\/sup>, VARS(DISTR), C: poissonpdf(<em data-effect=\"italics\">\u03bb<\/em>,\u00a0<em data-effect=\"italics\">k<\/em>). To compute\u00a0<em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em>\u00a0\u2264\u00a0<em data-effect=\"italics\">k<\/em>), enter 2<sup>nd<\/sup>, VARS (DISTR), D:poissoncdf(<em data-effect=\"italics\">\u03bb<\/em>,\u00a0<em data-effect=\"italics\">k<\/em>).<\/p>\n<\/div>\n<\/section>\n<div class=\"textbox exercises\">\n<h3 id=\"fs-idm148932176\" class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\">Example<\/h3>\n<p class=\"note statistics calculator\" data-type=\"note\" data-has-label=\"true\">At a police station in a large city, calls come in at an average rate of four calls per minute. Assume that the time that elapses from one call to the next has the exponential distribution. Take note that we are concerned only with the rate at which calls come in, and we are ignoring the time spent on the phone. We must also assume that the times spent between calls are independent. This means that a particularly long delay between two calls does not mean that there will be a shorter waiting period for the next call. We may then deduce that the total number of calls received during a time period has the Poisson distribution.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Find the average time between two successive calls.<\/li>\n<li>Find the probability that after a call is received, the next call occurs in less than ten seconds.<\/li>\n<li>Find the probability that exactly five calls occur within a minute.<\/li>\n<li>Find the probability that less than five calls occur within a minute.<\/li>\n<li>Find the probability that more than 40 calls occur in an eight-minute period.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q873276\">Show Answer<\/span><\/p>\n<div id=\"q873276\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>On average there are four calls occur per minute, so 15 seconds, or [latex]\\frac{15}{60}[\/latex]= 0.25 minutes\u00a0occur between successive calls on average.<\/li>\n<li>Let <em data-effect=\"italics\">T<\/em> = time elapsed between calls.\u00a0From part a, [latex]\\mu = {0.25}[\/latex], so m = [latex]\\frac{1}{0.25}[\/latex] = 4. Thus, T ~ Exp(4). \u00a0The cumulative distribution function is <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">T<\/em> &lt; <em data-effect=\"italics\">t<\/em>) = 1 \u2013 <em data-effect=\"italics\">e<\/em><sup>\u20134<em data-effect=\"italics\">t<\/em><\/sup>.\u00a0The probability that the next call occurs in less than ten seconds (ten seconds = 1\/6 minute) is P(T &lt; [latex]\\frac{1}{6}[\/latex]) = 1 &#8211; \u00a0[latex]{e}^{-4\\frac{1}{6}} \\approx{0.4866}[\/latex]\u00a0<a href=\"https:\/\/courses.candelalearning.com\/masterystats1x6xmaster\/wp-content\/uploads\/sites\/419\/2015\/06\/Screen-Shot-2015-06-15-at-2.58.18-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-718\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214509\/Screen-Shot-2015-06-15-at-2.58.18-PM.png\" alt=\"Graph with shaded area representing probability P(x &lt; 1\/6) = 0.4866\" width=\"385\" height=\"253\" \/><\/a><\/li>\n<li>Let <em data-effect=\"italics\">X<\/em> = the number of calls per minute. As previously stated, the number of calls per minute has a Poisson distribution, with a mean of four calls per minute.\u00a0Therefore, <em data-effect=\"italics\">X<\/em> \u223c <em data-effect=\"italics\">Poisson<\/em>(4), and so <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 5) = [latex]\\frac{{4}^{5}{e}^{-4}}{5!}\\approx[\/latex] 0.1563. (5! = (5)(4)(3)(2)(1))<br \/>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp115054976\" class=\"\">poissonpdf(4, 5) = 0.1563.<\/p>\n<\/div>\n<\/section>\n<\/li>\n<li>Keep in mind that <em data-effect=\"italics\">X<\/em> must be a whole number, so <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> &lt; 5) = <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> \u2264 4).\n<div data-type=\"newline\">To compute this, we could take <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 0) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 1) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 2) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 3) + <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> = 4).<\/div>\n<div data-type=\"newline\">\n<p>Using technology, we see that <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">X<\/em> \u2264 4) = 0.6288.<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp165812304\" class=\"\">poisssoncdf(4, 4) = 0.6288<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/li>\n<li>Let <em data-effect=\"italics\">Y<\/em> = the number of calls that occur during an eight minute period.\n<div data-type=\"newline\">Since there is an average of four calls per minute, there is an average of (8)(4) = 32 calls during each eight minute period.<\/div>\n<div data-type=\"newline\">\n<p>Hence, <em data-effect=\"italics\">Y<\/em> \u223c <em data-effect=\"italics\">Poisson<\/em>(32). Therefore, <em data-effect=\"italics\">P<\/em>(<em data-effect=\"italics\">Y<\/em> &gt; 40) = 1 \u2013 <em data-effect=\"italics\">P<\/em> (<em data-effect=\"italics\">Y<\/em> \u2264 40) = 1 \u2013 0.9294 = 0.0707.<\/p>\n<header>\n<h4 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idp137845728\" class=\"\">1 \u2013 poissoncdf(32, 40). = 0.0707<\/p>\n<\/div>\n<\/section>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-idm15154368\" class=\"\">In a small city, the number of automobile accidents occur with a Poisson distribution at an average of three per week.<\/p>\n<ol id=\"fs-idp53070848\" type=\"a\">\n<li>Calculate the probability that there are at most 2 accidents occur in any given week.<\/li>\n<li>What is the probability that there is at least two weeks between any 2 accidents?<\/li>\n<\/ol>\n<\/div>\n<\/section>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-253\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>OpenStax, The Exponential Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/5-3-the-exponential-distribution\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/5-3-the-exponential-distribution<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions\">https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction\">https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169134,"menu_order":14,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax, The Exponential Distribution\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/5-3-the-exponential-distribution\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"},{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\" http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions\"},{\"type\":\"cc-attribution\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-253","chapter","type-chapter","status-publish","hentry"],"part":249,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/253","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/users\/169134"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/253\/revisions"}],"predecessor-version":[{"id":3625,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/253\/revisions\/3625"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/parts\/249"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/253\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/media?parent=253"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=253"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/contributor?post=253"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/license?post=253"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}