{"id":164,"date":"2016-04-21T22:43:43","date_gmt":"2016-04-21T22:43:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&#038;p=164"},"modified":"2022-03-06T01:46:39","modified_gmt":"2022-03-06T01:46:39","slug":"the-uniform-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/the-uniform-distribution\/","title":{"raw":"The Uniform Distribution","rendered":"The Uniform Distribution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Recognize the uniform probability distribution and apply it appropriately<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. It is often referred as the <strong>rectangular distribution<\/strong> because the graph of the pdf has the form of a rectangle. Notation: [latex]X{\\sim}U(a,b)[\/latex]. The mean is [latex]\\mu=\\frac{{a+b}}{{2}}[\/latex] and the standard deviation is [latex]\\sigma=\\sqrt{\\frac{{b-a}}{{12}}}[\/latex]. The probability density function is [latex]f(x)=\\frac{{1}}{{b-a}}[\/latex] for [latex]a&lt;x&lt;b[\/latex] or [latex]a{\\leq}x{\\leq}b[\/latex]. The cumulative distribution is [latex]P(X{\\leq}x)=\\frac{{x-a}}{{b-a}}[\/latex].\r\n\r\nWhen working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe data in the table below are 55 smiling times, in seconds, of an eight-week-old baby.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td scope=\"row\">10.4<\/td>\r\n<td>19.6<\/td>\r\n<td>18.8<\/td>\r\n<td>13.9<\/td>\r\n<td>17.8<\/td>\r\n<td>16.8<\/td>\r\n<td>21.6<\/td>\r\n<td>17.9<\/td>\r\n<td>12.5<\/td>\r\n<td>11.1<\/td>\r\n<td>4.9<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">12.8<\/td>\r\n<td>14.8<\/td>\r\n<td>22.8<\/td>\r\n<td>20.0<\/td>\r\n<td>15.9<\/td>\r\n<td>16.3<\/td>\r\n<td>13.4<\/td>\r\n<td>17.1<\/td>\r\n<td>14.5<\/td>\r\n<td>19.0<\/td>\r\n<td>22.8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">1.3<\/td>\r\n<td>0.7<\/td>\r\n<td>8.9<\/td>\r\n<td>11.9<\/td>\r\n<td>10.9<\/td>\r\n<td>7.3<\/td>\r\n<td>5.9<\/td>\r\n<td>3.7<\/td>\r\n<td>17.9<\/td>\r\n<td>19.2<\/td>\r\n<td>9.8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">5.8<\/td>\r\n<td>6.9<\/td>\r\n<td>2.6<\/td>\r\n<td>5.8<\/td>\r\n<td>21.7<\/td>\r\n<td>11.8<\/td>\r\n<td>3.4<\/td>\r\n<td>2.1<\/td>\r\n<td>4.5<\/td>\r\n<td>6.3<\/td>\r\n<td>10.7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">8.9<\/td>\r\n<td>9.4<\/td>\r\n<td>9.4<\/td>\r\n<td>7.6<\/td>\r\n<td>10.0<\/td>\r\n<td>3.3<\/td>\r\n<td>6.7<\/td>\r\n<td>7.8<\/td>\r\n<td>11.6<\/td>\r\n<td>13.8<\/td>\r\n<td>18.6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe sample mean[latex]=11.49[\/latex] and the sample standard deviation[latex]=6.23[\/latex].\r\n\r\nWe will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.\r\n\r\nLet [latex]X=[\/latex] length, in seconds, of an eight-week-old baby's smile.\r\n\r\nThe notation for the uniform distribution is [latex]X{\\sim}U(a,b)[\/latex] where [latex]a=[\/latex] the lowest value of <em>x<\/em> and\u00a0[latex]b=[\/latex] the highest value of <em>x<\/em>.\r\n\r\nThe probability density function is [latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}}[\/latex] for [latex]a{\\leq}x{\\leq}b[\/latex].\r\n\r\nFor this example, [latex]X{\\sim}U(0,23)[\/latex] and [latex]{f{{({x})}}}=\\frac{{1}}{{{23}-{0}}}[\/latex] for [latex]0{\\leq}X{\\leq}23[\/latex].\r\n\r\nFormulas for the theoretical mean and standard deviation are [latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}{\\quad\\text{and}\\quad}{\\sigma}=\\sqrt{{\\frac{{{({b}-{a})}^{{2}}}}{{12}}}}[\/latex]\r\n\r\nFor this problem, the theoretical mean and standard deviation are [latex]{\\mu}=\\frac{{{0}+{23}}}{{2}}={11.50} \\text{ seconds}{\\quad\\text{and}\\quad}{\\sigma}=\\sqrt{{\\frac{{{({23}-{0})}^{{2}}}}{{12}}}}={6.64} \\text{ seconds}[\/latex]\r\n\r\nNotice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe data that follow are the number of passengers on 35 different charter fishing boats. The sample mean [latex]=7.9[\/latex] and the sample standard deviation [latex]=4.33[\/latex]. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of\u00a0<em>a<\/em> and <em>b<\/em>. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td scope=\"row\">1<\/td>\r\n<td>12<\/td>\r\n<td>4<\/td>\r\n<td>10<\/td>\r\n<td>4<\/td>\r\n<td>14<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">7<\/td>\r\n<td>11<\/td>\r\n<td>4<\/td>\r\n<td>13<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">3<\/td>\r\n<td>10<\/td>\r\n<td>0<\/td>\r\n<td>12<\/td>\r\n<td>6<\/td>\r\n<td>9<\/td>\r\n<td>10<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">5<\/td>\r\n<td>13<\/td>\r\n<td>4<\/td>\r\n<td>10<\/td>\r\n<td>14<\/td>\r\n<td>12<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<tr>\r\n<td scope=\"row\">6<\/td>\r\n<td>10<\/td>\r\n<td>11<\/td>\r\n<td>0<\/td>\r\n<td>11<\/td>\r\n<td>13<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"810841\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"810841\"]\r\n\r\na is zero; b is 14; [latex]X{\\sim}U(0,14)[\/latex]; [latex]\\mu=7[\/latex] passengers; [latex]\\sigma=4.04[\/latex] passengers\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<ol>\r\n \t<li>Refer to the previous example. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?<\/li>\r\n \t<li>Find the 90th percentile for an eight-week-old baby's smiling time.<\/li>\r\n \t<li>Find the probability that a random eight-week-old baby smiles more than 12 seconds <strong>knowing<\/strong> that the baby smiles <strong>more than eight seconds<\/strong>.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"96316\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"96316\"]\r\n<ol>\r\n \t<li>Find <em>P<\/em>(2 &lt; <em>x<\/em> &lt; 18).\r\n[latex]{P}{({2}{&lt;}{x}{&lt;}{18})}={(\\text{base})}{(\\text{height})}={({18}-{2})}{(\\frac{{1}}{{23}})}={(\\frac{{16}}{{23}})}\\approx0.6957[\/latex]<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/80tq-575e627i#fixme#fixme#fixme\" alt=\"This graph shows a uniform distribution. The horizontal axis ranges from 0 to 15. The distribution is modeled by a rectangle extending from x = 0 to x = 15. A region from x = 2 to x = 18 is shaded inside the rectangle.\" \/><\/li>\r\n \t<li>Ninety percent of the smiling times fall below the 90th percentile, <em>k<\/em>, so <em>P<\/em>(<em>x<\/em> &lt;<em>k<\/em>) = 0.90\r\n[latex]P(x&lt;k)=0.90[\/latex]\r\n[latex](\\text{base})(\\text{height})=0.90[\/latex]\r\n[latex](k-0)(\\frac{1}{23})=0.90[\/latex]\r\n[latex]k=(23)(0.90)=20.7[\/latex]\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/5ivo-nc5e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 1\/15. A horiztonal line ranges from the point (0, 1\/15) to the point (15, 1\/15). A vertical line extends from the x-axis to the end of the line at point (15, 1\/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = k. The shaded area represents P(x &lt; k) = 0.90.\" \/><\/li>\r\n<\/ol>\r\n90% of the smile times are at 20.7 or less seconds.\r\n\r\n3. This probability question is a <strong>conditional<\/strong>. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you <strong>already know<\/strong> the baby has smiled for more than eight seconds.Find\u00a0<em>P<\/em>(<em>x<\/em> &gt; 12|<em>x<\/em> &gt; 8) There are two ways to do the problem.\r\n\r\nUse the fact that this is a <strong>conditional<\/strong> and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. Write a new\u00a0<em>f<\/em>(<em>x<\/em>):[latex]{f{{({x})}}}=\\frac{{1}}{{{23}-{8}}}=\\frac{{1}}{{15}}[\/latex]for 8 &lt; <em>x<\/em> &lt; 23\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/odic-jg5e627i#fixme#fixme#fixme\" alt=\"f(X)=1\/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1\/15 on the y-axis, a vertical upward line from points 8 and 23 on the x-axis, and the x-axis. A shaded region from points 12-23 occurs within this area.\" \/>\r\n\r\n[latex]{P}{({x}\\geq{12} given x &gt; 8 seconds)}={(\\text{base})}{(\\text{height})}={({23}-{12})}{(\\frac{{1}}{{15}})}={(\\frac{{11}}{{15}})}\\approx0.7333[\/latex]\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\nA distribution is given as [latex]X{\\sim}U(0,20)[\/latex]. What is [latex]P(2&lt;x&lt;18)[\/latex]? Find the 90th percentile.\r\n\r\n[reveal-answer q=\"306492\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"306492\"]\r\n\r\n[latex]P(2&lt;x&lt;18)=0.8[\/latex];\r\n\r\n90th percentile[latex]=18[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.\r\n<ol>\r\n \t<li>What is the probability that a person waits fewer than 12.5 minutes?<\/li>\r\n \t<li>On the average, how long must a person wait? Find the mean, <em>\u03bc<\/em>, and the standard deviation, <em>\u03c3<\/em>.<\/li>\r\n \t<li>Ninety percent of the time, the time a person must wait falls below what value? This asks for the <strong>90th percentile<\/strong>.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"112796\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"112796\"]\r\n<ol>\r\n \t<li>Let <em>X<\/em> = the number of minutes a person must wait for a bus. <em>a<\/em> = 0 and <em>b<\/em> = 15. <em>X<\/em>~ <em>U<\/em>(0, 15). Write the probability density function. [latex]{f{{({x})}}}=\\frac{{1}}{{{15}-{0}}}=\\frac{{1}}{{15}}[\/latex] for 0 \u2264<em>x<\/em> \u2264 15.Find\u00a0<em>P<\/em> (<em>x<\/em> &lt; 12.5). Draw a graph.\r\n[latex]{P}{({x}{&lt;}{k})}={(\\text{base})}{(\\text{height})}={({12.5}-{0})}{(\\frac{{1}}{{15}})}\\approx{0.8333}[\/latex]\r\nThe probability a person waits less than 12.5 minutes is approximately 0.8333.<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/jt07-1p5e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 1\/15. A horiztonal line ranges from the point (0, 1\/15) to the point (15, 1\/15). A vertical line extends from the x-axis to the end of the line at point (15, 1\/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.\" \/><\/li>\r\n \t<li>[latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}=\\frac{{{15}+{0}}}{{2}}={7.5}[\/latex]. On the average, a person must wait 7.5 minutes. [latex]{\\sigma}=\\sqrt{{\\frac{{{({b}-{a})}^{{2}}}}{{12}}}}=\\sqrt{{\\frac{{{({15}-{0})}^{{2}}}}{{12}}}}={4.3}[\/latex]The standard deviation is 4.3 minutes.<\/li>\r\n \t<li>Find the 90th percentile. Draw a graph. Let <em>k<\/em> = the 90th percentile.\r\n[latex]{P}(x&lt;k)=(\\text{base})(\\text{height})=(k-0)(\\frac{1}{15})[\/latex]\r\n[latex]{0.90}=(k)(\\frac{1}{15})[\/latex]\r\n[latex]k=(0.90)(15)=13.5[\/latex]\r\n<em>k<\/em> is sometimes called a critical value. The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/zovz-0t5e627i#fixme#fixme#fixme\" alt=\"f(X)=1\/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1\/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.\" \/><\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<h4><\/h4>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.\r\n<ol>\r\n \t<li>Find <em>a<\/em> and <em>b<\/em> and describe what they represent.<\/li>\r\n \t<li>Write the distribution.<\/li>\r\n \t<li>Find the mean and the standard deviation.<\/li>\r\n \t<li>What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?<\/li>\r\n \t<li>What is the 65th percentile for the duration of games for a team for the 2011 season?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"597587\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"597587\"]\r\n<ol>\r\n \t<li><em>a<\/em> is 447 hours, and <em>b<\/em> is 521 hours. <em>a<\/em> is the minimum duration of games for a team for the 2011 season, and <em>b<\/em> is the maximum duration of games for a team for the 2011 season.<\/li>\r\n \t<li><em>X<\/em> ~ <em>U<\/em> (447, 521).<\/li>\r\n \t<li><em>\u03bc<\/em> = 484 hours, and <em>\u03c3<\/em> = 21.36 hours\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/ld91-516e627i#fixme#fixme#fixme\" alt=\"\" \/><\/li>\r\n \t<li><em>P<\/em>(480 &lt; <em>x<\/em> &lt; 500) is approximately 0.2703<\/li>\r\n \t<li>The 65th percentile is 495.1 hours.<\/li>\r\n<\/ol>\r\n&nbsp;\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 the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let\u00a0<em>X<\/em> = the time, in minutes, it takes a student to finish a quiz. Then <em>X<\/em> ~ <em>U<\/em> (6, 15).\r\n\r\nFind the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.\r\n\r\n[reveal-answer q=\"341172\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"341172\"]\r\n\r\n<em>P<\/em> (<em>x<\/em> &gt; 8) is approximately 0.7778\r\n\r\n<em>P<\/em> (<em>x<\/em> &gt; 8 | x &gt; 7) = 0.875\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n<hr \/>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nAce Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let\u00a0<em>x<\/em> = the time needed to fix a furnace. Then <em>x<\/em> ~ <em>U<\/em> (1.5, 4).\r\n<ol>\r\n \t<li>Find the probability that a randomly selected furnace repair requires more than two hours.<\/li>\r\n \t<li>Find the probability that a randomly selected furnace repair requires less than three hours.<\/li>\r\n \t<li>Find the 30th percentile of furnace repair times.<\/li>\r\n \t<li>The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?<\/li>\r\n \t<li>Find the mean and standard deviation<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"311856\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"311856\"]\r\n<ol>\r\n \t<li>To find\u00a0<em>f<\/em>(<em>x<\/em>): [latex]{f{{({x})}}}=\\frac{{1}}{{{4}-{1.5}}}={12.5} \\text{ so } {f{{({x})}}}={0.4}[\/latex]\r\n<em>P<\/em>(<em>x<\/em> &gt; 2) = (base)(height) = (4 \u2013 2)(0.4) = 0.8\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/42bq-596e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.\" \/>\r\nUniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time\r\n<em>x<\/em> is greater than two<\/li>\r\n \t<li><em>P<\/em>(<em>x<\/em> &lt; 3) = (base)(height) = (3 \u2013 1.5)(0.4) = 0.6The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between\r\n<em>x<\/em> = 1.5 and <em>x<\/em> = 3. Note that the shaded area starts at <em>x<\/em> = 1.5 rather than at <em>x<\/em> = 0; since <em>X<\/em> ~ <em>U<\/em> (1.5, 4), <em>x <\/em>can not be less than 1.5.\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/3x2o-ad6e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.\" \/>Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time\u00a0<em>x<\/em> is less than three<\/li>\r\n \t<li><img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/1yjh-4h6e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = k. The shaded area represents P(x &lt; k) = 0.3.\" \/>\r\nUniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.\r\n<em>P<\/em> (<em>x<\/em> &lt; <em>k<\/em>) = 0.30\r\n<em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = (base)(height) = (<em>k<\/em> \u2013 1.5)(0.4)\u00a0 \u00a00.3 = (<em>k<\/em> \u2013 1.5) (0.4); Solve to find <em>k<\/em>: 0.75 =\u00a0<em>k<\/em> \u2013 1.5, obtained by dividing both sides by 0.4\r\n<em>k<\/em> = 2.25 , obtained by adding 1.5 to both sides. The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.<\/li>\r\n \t<li><img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/h8bp-3l6e627i#fixme#fixme#fixme\" alt=\"\" \/>\r\nUniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.\u00a0<em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = 0.25\r\n<em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = (base)(height) = (4 \u2013 <em>k<\/em>)(0.4)\u00a0 \u00a0 0.25 = (4 \u2013\u00a0<em>k<\/em>)(0.4); Solve for <em>k<\/em>:\u00a0 0.625 = 4 \u2212\u00a0<em>k<\/em>, obtained by dividing both sides by 0.4<\/li>\r\n<\/ol>\r\n\u22123.375 = \u2212\u00a0<em>k<\/em>, obtained by subtracting four from both sides: <em>k<\/em> = 3.375\r\n\r\nThe longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).\r\n\r\n<strong>Note:<\/strong> Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the <strong>75th percentile<\/strong> of furnace repair times.\r\n<ul>\r\n \t<li>[latex]{\\mu}={\\frac{a+b}{2}}\\text{ and }{\\sigma}=\\sqrt{\\frac{(b-a)^2}{12}}[\/latex]\r\n[latex]{\\mu}=\\frac{1.5+4}{2}=2.75\\text{ hours and }{\\sigma}=\\sqrt{\\frac{(4-1.5)^2}{12}}= 0.7217 \\text{ hours}[\/latex]<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let\u00a0<em>X<\/em> = the time needed to change the oil on a car.\r\n<ol>\r\n \t<li>Write the random variable <em>X<\/em> in words. <em>X<\/em> = __________________.<\/li>\r\n \t<li>Write the distribution.<\/li>\r\n \t<li>Graph the distribution.<\/li>\r\n \t<li>Find <em>P<\/em> (<em>x<\/em> &gt; 19).<\/li>\r\n \t<li>Find the 50th percentile.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"54558\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"54558\"]\r\n<ol>\r\n \t<li>Let <em>X<\/em> = the time needed to change the oil in a car.<\/li>\r\n \t<li><em>X<\/em> ~ <em>U<\/em> (11, 21).<\/li>\r\n \t<li><\/li>\r\n \t<li><em>P<\/em> (<em>x<\/em> &gt; 19) = 0.2<\/li>\r\n \t<li>the 50th percentile is 16 minutes.<\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2><\/h2>\r\n<h2>References<\/h2>\r\nMcDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.\r\n<h2>Concept Review<\/h2>\r\nIf\u00a0<em>X<\/em> has a uniform distribution where <em>a<\/em> &lt; <em>x<\/em> &lt; <em>b<\/em> or <em>a<\/em> \u2264 <em>x<\/em> \u2264 <em>b<\/em>, then <em>X<\/em> takes on values between <em>a <\/em>and <em>b<\/em> (may include <em>a<\/em> and <em>b<\/em>). All values <em>x<\/em> are equally likely. We write <em>X<\/em> \u223c <em>U<\/em>(<em>a<\/em>, <em>b<\/em>). The mean of <em>X <\/em>is [latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}[\/latex]. <em>X<\/em> is continuous.\r\n\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/38jr-7p6e627i#fixme#fixme#fixme\" alt=\"The graph shows a rectangle with total area equal to 1. The rectangle extends from x = a to x = b on the x-axis and has a height of 1\/(b-a).\" \/>\r\n\r\nThe probability\u00a0<em>P<\/em>(<em>c<\/em> &lt; <em>X<\/em> &lt; <em>d<\/em>) may be found by computing the area under <em>f<\/em>(<em>x<\/em>), between <em>c<\/em> and <em>d<\/em>. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.\r\n<h2>Formula Review<\/h2>\r\n<em>X<\/em> = a real number between <em>a<\/em> and <em>b<\/em> (in some instances, <em>X<\/em> can take on the values <em>a<\/em> and <em>b<\/em>). <em>a<\/em> = smallest <em>X<\/em>; <em>b<\/em> = largest <em>X<\/em>\r\n\r\n<em>X<\/em> ~ <em>U<\/em> (a, b)\r\n\r\nThe mean is\u00a0[latex]\\mu=\\frac{{{a} + {b}}}{{2}}[\/latex]\r\n\r\nThe standard deviation is\u00a0[latex]\\sigma=\\sqrt{{\\frac{{({b}-{a})}^{{2}}}{{12}}}}[\/latex]\r\n\r\nProbability density function:\u00a0[latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}} \\text{ for } {a}\\leq{X}\\leq{b}[\/latex]\r\n\r\nArea to the Left of\u00a0<em>x<\/em>: [latex]{P}{({X}{&lt;}{x})}={({x}-{a})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]\r\n\r\nArea to the Right of\u00a0<em>x<\/em>: [latex]{P}{({X}{&gt;}{x})}={({b}-{x})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]\r\n\r\nArea Between\u00a0<em>c<\/em> and <em>d<\/em>: [latex]{P}{({c}{&lt;}{x}{&lt;}{d})}={(\\text{base})}{(\\text{height})}={({d}-{c})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]\r\n\r\nUniform:\u00a0<em>X<\/em> ~ <em>U<\/em>(<em>a<\/em>, <em>b<\/em>) where <em>a<\/em> &lt; <em>x<\/em> &lt; <em>b<\/em>\r\n<ul>\r\n \t<li>pdf: [latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}}[\/latex] for <em>a \u2264 x \u2264 b<\/em><\/li>\r\n \t<li>cdf: <em>P<\/em>(<em>X<\/em> \u2264 <em>x<\/em>) = [latex]\\frac{{{x}-{a}}}{{{b}-{a}}}[\/latex]<\/li>\r\n \t<li>mean: [latex]\\mu=\\frac{{{a} + {b}}}{{2}}[\/latex]<\/li>\r\n \t<li>standard deviation: [latex]\\sigma=\\sqrt{{\\frac{{({b}-{a})}^{{2}}}{{12}}}}[\/latex]<\/li>\r\n \t<li><em>P<\/em>(<em>c<\/em> &lt; <em>X<\/em> &lt; <em>d<\/em>) = (<em>d<\/em> \u2013 <em>c<\/em>)<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Recognize the uniform probability distribution and apply it appropriately<\/li>\n<\/ul>\n<\/div>\n<p>The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. It is often referred as the <strong>rectangular distribution<\/strong> because the graph of the pdf has the form of a rectangle. Notation: [latex]X{\\sim}U(a,b)[\/latex]. The mean is [latex]\\mu=\\frac{{a+b}}{{2}}[\/latex] and the standard deviation is [latex]\\sigma=\\sqrt{\\frac{{b-a}}{{12}}}[\/latex]. The probability density function is [latex]f(x)=\\frac{{1}}{{b-a}}[\/latex] for [latex]a<x<b[\/latex] or [latex]a{\\leq}x{\\leq}b[\/latex]. The cumulative distribution is [latex]P(X{\\leq}x)=\\frac{{x-a}}{{b-a}}[\/latex].\n\nWhen working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.\n\n\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The data in the table below are 55 smiling times, in seconds, of an eight-week-old baby.<\/p>\n<table>\n<tbody>\n<tr>\n<td scope=\"row\">10.4<\/td>\n<td>19.6<\/td>\n<td>18.8<\/td>\n<td>13.9<\/td>\n<td>17.8<\/td>\n<td>16.8<\/td>\n<td>21.6<\/td>\n<td>17.9<\/td>\n<td>12.5<\/td>\n<td>11.1<\/td>\n<td>4.9<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">12.8<\/td>\n<td>14.8<\/td>\n<td>22.8<\/td>\n<td>20.0<\/td>\n<td>15.9<\/td>\n<td>16.3<\/td>\n<td>13.4<\/td>\n<td>17.1<\/td>\n<td>14.5<\/td>\n<td>19.0<\/td>\n<td>22.8<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">1.3<\/td>\n<td>0.7<\/td>\n<td>8.9<\/td>\n<td>11.9<\/td>\n<td>10.9<\/td>\n<td>7.3<\/td>\n<td>5.9<\/td>\n<td>3.7<\/td>\n<td>17.9<\/td>\n<td>19.2<\/td>\n<td>9.8<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">5.8<\/td>\n<td>6.9<\/td>\n<td>2.6<\/td>\n<td>5.8<\/td>\n<td>21.7<\/td>\n<td>11.8<\/td>\n<td>3.4<\/td>\n<td>2.1<\/td>\n<td>4.5<\/td>\n<td>6.3<\/td>\n<td>10.7<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">8.9<\/td>\n<td>9.4<\/td>\n<td>9.4<\/td>\n<td>7.6<\/td>\n<td>10.0<\/td>\n<td>3.3<\/td>\n<td>6.7<\/td>\n<td>7.8<\/td>\n<td>11.6<\/td>\n<td>13.8<\/td>\n<td>18.6<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The sample mean[latex]=11.49[\/latex] and the sample standard deviation[latex]=6.23[\/latex].<\/p>\n<p>We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.<\/p>\n<p>Let [latex]X=[\/latex] length, in seconds, of an eight-week-old baby&#8217;s smile.<\/p>\n<p>The notation for the uniform distribution is [latex]X{\\sim}U(a,b)[\/latex] where [latex]a=[\/latex] the lowest value of <em>x<\/em> and\u00a0[latex]b=[\/latex] the highest value of <em>x<\/em>.<\/p>\n<p>The probability density function is [latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}}[\/latex] for [latex]a{\\leq}x{\\leq}b[\/latex].<\/p>\n<p>For this example, [latex]X{\\sim}U(0,23)[\/latex] and [latex]{f{{({x})}}}=\\frac{{1}}{{{23}-{0}}}[\/latex] for [latex]0{\\leq}X{\\leq}23[\/latex].<\/p>\n<p>Formulas for the theoretical mean and standard deviation are [latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}{\\quad\\text{and}\\quad}{\\sigma}=\\sqrt{{\\frac{{{({b}-{a})}^{{2}}}}{{12}}}}[\/latex]<\/p>\n<p>For this problem, the theoretical mean and standard deviation are [latex]{\\mu}=\\frac{{{0}+{23}}}{{2}}={11.50} \\text{ seconds}{\\quad\\text{and}\\quad}{\\sigma}=\\sqrt{{\\frac{{{({23}-{0})}^{{2}}}}{{12}}}}={6.64} \\text{ seconds}[\/latex]<\/p>\n<p>Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.<\/p>\n<\/div>\n<hr \/>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean [latex]=7.9[\/latex] and the sample standard deviation [latex]=4.33[\/latex]. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of\u00a0<em>a<\/em> and <em>b<\/em>. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.<\/p>\n<table>\n<tbody>\n<tr>\n<td scope=\"row\">1<\/td>\n<td>12<\/td>\n<td>4<\/td>\n<td>10<\/td>\n<td>4<\/td>\n<td>14<\/td>\n<td>11<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">7<\/td>\n<td>11<\/td>\n<td>4<\/td>\n<td>13<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">3<\/td>\n<td>10<\/td>\n<td>0<\/td>\n<td>12<\/td>\n<td>6<\/td>\n<td>9<\/td>\n<td>10<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">5<\/td>\n<td>13<\/td>\n<td>4<\/td>\n<td>10<\/td>\n<td>14<\/td>\n<td>12<\/td>\n<td>11<\/td>\n<\/tr>\n<tr>\n<td scope=\"row\">6<\/td>\n<td>10<\/td>\n<td>11<\/td>\n<td>0<\/td>\n<td>11<\/td>\n<td>13<\/td>\n<td>2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q810841\">Show Solution<\/span><\/p>\n<div id=\"q810841\" class=\"hidden-answer\" style=\"display: none\">\n<p>a is zero; b is 14; [latex]X{\\sim}U(0,14)[\/latex]; [latex]\\mu=7[\/latex] passengers; [latex]\\sigma=4.04[\/latex] passengers<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<ol>\n<li>Refer to the previous example. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?<\/li>\n<li>Find the 90th percentile for an eight-week-old baby&#8217;s smiling time.<\/li>\n<li>Find the probability that a random eight-week-old baby smiles more than 12 seconds <strong>knowing<\/strong> that the baby smiles <strong>more than eight seconds<\/strong>.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96316\">Show Solution<\/span><\/p>\n<div id=\"q96316\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Find <em>P<\/em>(2 &lt; <em>x<\/em> &lt; 18).<br \/>\n[latex]{P}{({2}{<}{x}{<}{18})}={(\\text{base})}{(\\text{height})}={({18}-{2})}{(\\frac{{1}}{{23}})}={(\\frac{{16}}{{23}})}\\approx0.6957[\/latex]<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/80tq-575e627i#fixme#fixme#fixme\" alt=\"This graph shows a uniform distribution. The horizontal axis ranges from 0 to 15. The distribution is modeled by a rectangle extending from x = 0 to x = 15. A region from x = 2 to x = 18 is shaded inside the rectangle.\" \/><\/li>\n<li>Ninety percent of the smiling times fall below the 90th percentile, <em>k<\/em>, so <em>P<\/em>(<em>x<\/em> &lt;<em>k<\/em>) = 0.90<br \/>\n[latex]P(x<k)=0.90[\/latex]\n[latex](\\text{base})(\\text{height})=0.90[\/latex]\n[latex](k-0)(\\frac{1}{23})=0.90[\/latex]\n[latex]k=(23)(0.90)=20.7[\/latex]\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/5ivo-nc5e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 1\/15. A horiztonal line ranges from the point (0, 1\/15) to the point (15, 1\/15). A vertical line extends from the x-axis to the end of the line at point (15, 1\/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = k. The shaded area represents P(x &lt; k) = 0.90.\" \/><\/li>\n<\/ol>\n<p>90% of the smile times are at 20.7 or less seconds.<\/p>\n<p>3. This probability question is a <strong>conditional<\/strong>. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you <strong>already know<\/strong> the baby has smiled for more than eight seconds.Find\u00a0<em>P<\/em>(<em>x<\/em> &gt; 12|<em>x<\/em> &gt; 8) There are two ways to do the problem.<\/p>\n<p>Use the fact that this is a <strong>conditional<\/strong> and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. Write a new\u00a0<em>f<\/em>(<em>x<\/em>):[latex]{f{{({x})}}}=\\frac{{1}}{{{23}-{8}}}=\\frac{{1}}{{15}}[\/latex]for 8 &lt; <em>x<\/em> &lt; 23<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/odic-jg5e627i#fixme#fixme#fixme\" alt=\"f(X)=1\/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1\/15 on the y-axis, a vertical upward line from points 8 and 23 on the x-axis, and the x-axis. A shaded region from points 12-23 occurs within this area.\" \/><\/p>\n<p>[latex]{P}{({x}\\geq{12} given x > 8 seconds)}={(\\text{base})}{(\\text{height})}={({23}-{12})}{(\\frac{{1}}{{15}})}={(\\frac{{11}}{{15}})}\\approx0.7333[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A distribution is given as [latex]X{\\sim}U(0,20)[\/latex]. What is [latex]P(2<x<18)[\/latex]? Find the 90th percentile.\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q306492\">Show Solution<\/span><\/p>\n<div id=\"q306492\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]P(2<x<18)=0.8[\/latex];\n\n90th percentile[latex]=18[\/latex]\n\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.<\/p>\n<ol>\n<li>What is the probability that a person waits fewer than 12.5 minutes?<\/li>\n<li>On the average, how long must a person wait? Find the mean, <em>\u03bc<\/em>, and the standard deviation, <em>\u03c3<\/em>.<\/li>\n<li>Ninety percent of the time, the time a person must wait falls below what value? This asks for the <strong>90th percentile<\/strong>.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q112796\">Show Solution<\/span><\/p>\n<div id=\"q112796\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Let <em>X<\/em> = the number of minutes a person must wait for a bus. <em>a<\/em> = 0 and <em>b<\/em> = 15. <em>X<\/em>~ <em>U<\/em>(0, 15). Write the probability density function. [latex]{f{{({x})}}}=\\frac{{1}}{{{15}-{0}}}=\\frac{{1}}{{15}}[\/latex] for 0 \u2264<em>x<\/em> \u2264 15.Find\u00a0<em>P<\/em> (<em>x<\/em> &lt; 12.5). Draw a graph.<br \/>\n[latex]{P}{({x}{<}{k})}={(\\text{base})}{(\\text{height})}={({12.5}-{0})}{(\\frac{{1}}{{15}})}\\approx{0.8333}[\/latex]\nThe probability a person waits less than 12.5 minutes is approximately 0.8333.<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/jt07-1p5e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 1\/15. A horiztonal line ranges from the point (0, 1\/15) to the point (15, 1\/15). A vertical line extends from the x-axis to the end of the line at point (15, 1\/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.\" \/><\/li>\n<li>[latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}=\\frac{{{15}+{0}}}{{2}}={7.5}[\/latex]. On the average, a person must wait 7.5 minutes. [latex]{\\sigma}=\\sqrt{{\\frac{{{({b}-{a})}^{{2}}}}{{12}}}}=\\sqrt{{\\frac{{{({15}-{0})}^{{2}}}}{{12}}}}={4.3}[\/latex]The standard deviation is 4.3 minutes.<\/li>\n<li>Find the 90th percentile. Draw a graph. Let <em>k<\/em> = the 90th percentile.<br \/>\n[latex]{P}(x<k)=(\\text{base})(\\text{height})=(k-0)(\\frac{1}{15})[\/latex]\n[latex]{0.90}=(k)(\\frac{1}{15})[\/latex]\n[latex]k=(0.90)(15)=13.5[\/latex]\n<em>k<\/em> is sometimes called a critical value. The 90th percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/zovz-0t5e627i#fixme#fixme#fixme\" alt=\"f(X)=1\/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1\/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.\" \/><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<hr \/>\n<h4><\/h4>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.<\/p>\n<ol>\n<li>Find <em>a<\/em> and <em>b<\/em> and describe what they represent.<\/li>\n<li>Write the distribution.<\/li>\n<li>Find the mean and the standard deviation.<\/li>\n<li>What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?<\/li>\n<li>What is the 65th percentile for the duration of games for a team for the 2011 season?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q597587\">Show Answer<\/span><\/p>\n<div id=\"q597587\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><em>a<\/em> is 447 hours, and <em>b<\/em> is 521 hours. <em>a<\/em> is the minimum duration of games for a team for the 2011 season, and <em>b<\/em> is the maximum duration of games for a team for the 2011 season.<\/li>\n<li><em>X<\/em> ~ <em>U<\/em> (447, 521).<\/li>\n<li><em>\u03bc<\/em> = 484 hours, and <em>\u03c3<\/em> = 21.36 hours<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/ld91-516e627i#fixme#fixme#fixme\" alt=\"\" \/><\/li>\n<li><em>P<\/em>(480 &lt; <em>x<\/em> &lt; 500) is approximately 0.2703<\/li>\n<li>The 65th percentile is 495.1 hours.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let\u00a0<em>X<\/em> = the time, in minutes, it takes a student to finish a quiz. Then <em>X<\/em> ~ <em>U<\/em> (6, 15).<\/p>\n<p>Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q341172\">Show Answer<\/span><\/p>\n<div id=\"q341172\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>P<\/em> (<em>x<\/em> &gt; 8) is approximately 0.7778<\/p>\n<p><em>P<\/em> (<em>x<\/em> &gt; 8 | x &gt; 7) = 0.875<\/p>\n<\/div>\n<\/div>\n<\/div>\n<hr \/>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let\u00a0<em>x<\/em> = the time needed to fix a furnace. Then <em>x<\/em> ~ <em>U<\/em> (1.5, 4).<\/p>\n<ol>\n<li>Find the probability that a randomly selected furnace repair requires more than two hours.<\/li>\n<li>Find the probability that a randomly selected furnace repair requires less than three hours.<\/li>\n<li>Find the 30th percentile of furnace repair times.<\/li>\n<li>The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?<\/li>\n<li>Find the mean and standard deviation<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q311856\">Show Solution<\/span><\/p>\n<div id=\"q311856\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>To find\u00a0<em>f<\/em>(<em>x<\/em>): [latex]{f{{({x})}}}=\\frac{{1}}{{{4}-{1.5}}}={12.5} \\text{ so } {f{{({x})}}}={0.4}[\/latex]<br \/>\n<em>P<\/em>(<em>x<\/em> &gt; 2) = (base)(height) = (4 \u2013 2)(0.4) = 0.8<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/42bq-596e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.\" \/><br \/>\nUniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time<br \/>\n<em>x<\/em> is greater than two<\/li>\n<li><em>P<\/em>(<em>x<\/em> &lt; 3) = (base)(height) = (3 \u2013 1.5)(0.4) = 0.6The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between<br \/>\n<em>x<\/em> = 1.5 and <em>x<\/em> = 3. Note that the shaded area starts at <em>x<\/em> = 1.5 rather than at <em>x<\/em> = 0; since <em>X<\/em> ~ <em>U<\/em> (1.5, 4), <em>x <\/em>can not be less than 1.5.<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/3x2o-ad6e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.\" \/>Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time\u00a0<em>x<\/em> is less than three<\/li>\n<li><img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/1yjh-4h6e627i#fixme#fixme#fixme\" alt=\"This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = k. The shaded area represents P(x &lt; k) = 0.3.\" \/><br \/>\nUniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.<br \/>\n<em>P<\/em> (<em>x<\/em> &lt; <em>k<\/em>) = 0.30<br \/>\n<em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = (base)(height) = (<em>k<\/em> \u2013 1.5)(0.4)\u00a0 \u00a00.3 = (<em>k<\/em> \u2013 1.5) (0.4); Solve to find <em>k<\/em>: 0.75 =\u00a0<em>k<\/em> \u2013 1.5, obtained by dividing both sides by 0.4<br \/>\n<em>k<\/em> = 2.25 , obtained by adding 1.5 to both sides. The 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.<\/li>\n<li><img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/h8bp-3l6e627i#fixme#fixme#fixme\" alt=\"\" \/><br \/>\nUniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.\u00a0<em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = 0.25<br \/>\n<em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = (base)(height) = (4 \u2013 <em>k<\/em>)(0.4)\u00a0 \u00a0 0.25 = (4 \u2013\u00a0<em>k<\/em>)(0.4); Solve for <em>k<\/em>:\u00a0 0.625 = 4 \u2212\u00a0<em>k<\/em>, obtained by dividing both sides by 0.4<\/li>\n<\/ol>\n<p>\u22123.375 = \u2212\u00a0<em>k<\/em>, obtained by subtracting four from both sides: <em>k<\/em> = 3.375<\/p>\n<p>The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).<\/p>\n<p><strong>Note:<\/strong> Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the <strong>75th percentile<\/strong> of furnace repair times.<\/p>\n<ul>\n<li>[latex]{\\mu}={\\frac{a+b}{2}}\\text{ and }{\\sigma}=\\sqrt{\\frac{(b-a)^2}{12}}[\/latex]<br \/>\n[latex]{\\mu}=\\frac{1.5+4}{2}=2.75\\text{ hours and }{\\sigma}=\\sqrt{\\frac{(4-1.5)^2}{12}}= 0.7217 \\text{ hours}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let\u00a0<em>X<\/em> = the time needed to change the oil on a car.<\/p>\n<ol>\n<li>Write the random variable <em>X<\/em> in words. <em>X<\/em> = __________________.<\/li>\n<li>Write the distribution.<\/li>\n<li>Graph the distribution.<\/li>\n<li>Find <em>P<\/em> (<em>x<\/em> &gt; 19).<\/li>\n<li>Find the 50th percentile.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q54558\">Show Answer<\/span><\/p>\n<div id=\"q54558\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Let <em>X<\/em> = the time needed to change the oil in a car.<\/li>\n<li><em>X<\/em> ~ <em>U<\/em> (11, 21).<\/li>\n<li><\/li>\n<li><em>P<\/em> (<em>x<\/em> &gt; 19) = 0.2<\/li>\n<li>the 50th percentile is 16 minutes.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\n<h2>References<\/h2>\n<p>McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.<\/p>\n<h2>Concept Review<\/h2>\n<p>If\u00a0<em>X<\/em> has a uniform distribution where <em>a<\/em> &lt; <em>x<\/em> &lt; <em>b<\/em> or <em>a<\/em> \u2264 <em>x<\/em> \u2264 <em>b<\/em>, then <em>X<\/em> takes on values between <em>a <\/em>and <em>b<\/em> (may include <em>a<\/em> and <em>b<\/em>). All values <em>x<\/em> are equally likely. We write <em>X<\/em> \u223c <em>U<\/em>(<em>a<\/em>, <em>b<\/em>). The mean of <em>X <\/em>is [latex]{\\mu}=\\frac{{{a}+{b}}}{{2}}[\/latex]. <em>X<\/em> is continuous.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/38jr-7p6e627i#fixme#fixme#fixme\" alt=\"The graph shows a rectangle with total area equal to 1. The rectangle extends from x = a to x = b on the x-axis and has a height of 1\/(b-a).\" \/><\/p>\n<p>The probability\u00a0<em>P<\/em>(<em>c<\/em> &lt; <em>X<\/em> &lt; <em>d<\/em>) may be found by computing the area under <em>f<\/em>(<em>x<\/em>), between <em>c<\/em> and <em>d<\/em>. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.<\/p>\n<h2>Formula Review<\/h2>\n<p><em>X<\/em> = a real number between <em>a<\/em> and <em>b<\/em> (in some instances, <em>X<\/em> can take on the values <em>a<\/em> and <em>b<\/em>). <em>a<\/em> = smallest <em>X<\/em>; <em>b<\/em> = largest <em>X<\/em><\/p>\n<p><em>X<\/em> ~ <em>U<\/em> (a, b)<\/p>\n<p>The mean is\u00a0[latex]\\mu=\\frac{{{a} + {b}}}{{2}}[\/latex]<\/p>\n<p>The standard deviation is\u00a0[latex]\\sigma=\\sqrt{{\\frac{{({b}-{a})}^{{2}}}{{12}}}}[\/latex]<\/p>\n<p>Probability density function:\u00a0[latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}} \\text{ for } {a}\\leq{X}\\leq{b}[\/latex]<\/p>\n<p>Area to the Left of\u00a0<em>x<\/em>: [latex]{P}{({X}{<}{x})}={({x}-{a})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]\n\nArea to the Right of\u00a0<em>x<\/em>: [latex]{P}{({X}{>}{x})}={({b}-{x})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]<\/p>\n<p>Area Between\u00a0<em>c<\/em> and <em>d<\/em>: [latex]{P}{({c}{<}{x}{<}{d})}={(\\text{base})}{(\\text{height})}={({d}-{c})}{(\\frac{{1}}{{{b}-{a}}})}[\/latex]\n\nUniform:\u00a0<em>X<\/em> ~ <em>U<\/em>(<em>a<\/em>, <em>b<\/em>) where <em>a<\/em> &lt; <em>x<\/em> &lt; <em>b<\/em><\/p>\n<ul>\n<li>pdf: [latex]{f{{({x})}}}=\\frac{{1}}{{{b}-{a}}}[\/latex] for <em>a \u2264 x \u2264 b<\/em><\/li>\n<li>cdf: <em>P<\/em>(<em>X<\/em> \u2264 <em>x<\/em>) = [latex]\\frac{{{x}-{a}}}{{{b}-{a}}}[\/latex]<\/li>\n<li>mean: [latex]\\mu=\\frac{{{a} + {b}}}{{2}}[\/latex]<\/li>\n<li>standard deviation: [latex]\\sigma=\\sqrt{{\\frac{{({b}-{a})}^{{2}}}{{12}}}}[\/latex]<\/li>\n<li><em>P<\/em>(<em>c<\/em> &lt; <em>X<\/em> &lt; <em>d<\/em>) = (<em>d<\/em> \u2013 <em>c<\/em>)<\/li>\n<\/ul>\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-164\">\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, Statistics, The Uniform Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:36\/Introductory_Statistics\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:36\/Introductory_Statistics<\/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>Introductory Statistics . <strong>Authored by<\/strong>: Barbara Illowski, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/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>: Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/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":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax, Statistics, The Uniform Distribution\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:36\/Introductory_Statistics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics \",\"author\":\"Barbara Illowski, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-164","chapter","type-chapter","status-publish","hentry"],"part":157,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/164","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":34,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/164\/revisions"}],"predecessor-version":[{"id":2852,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/164\/revisions\/2852"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/parts\/157"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapters\/164\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/media?parent=164"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/pressbooks\/v2\/chapter-type?post=164"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/contributor?post=164"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/wp-json\/wp\/v2\/license?post=164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}