{"id":422,"date":"2016-10-12T18:31:40","date_gmt":"2016-10-12T18:31:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=422"},"modified":"2019-05-30T16:59:31","modified_gmt":"2019-05-30T16:59:31","slug":"measures-of-central-tendency","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/chapter\/measures-of-central-tendency\/","title":{"raw":"Measures of Central Tendency","rendered":"Measures of Central Tendency"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the mean, median, and mode of a set of data<\/li>\r\n \t<li>Calculate the range of a data set, and recognize it's limitations in fully describing the behavior of a data set<\/li>\r\n \t<li>Calculate the standard deviation for a data set, and determine it's units<\/li>\r\n \t<li>Identify the difference between population variance and sample variance<\/li>\r\n \t<li>Identify the quartiles for a data set, and the calculations used to define them<\/li>\r\n \t<li>Identify the parts of a five number summary for a set of data, and create a box plot using it<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Mean, Median, and Mode<\/h2>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/22190716\/8027247398_96f7013a1f_z.jpg\"><img class=\"aligncenter size-full wp-image-955\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/22190716\/8027247398_96f7013a1f_z.jpg\" alt=\"Silver sphere which has red smaller spheres clustered on its left side, with magnetic attraction\" width=\"640\" height=\"427\" \/><\/a>\r\n\r\nLet's begin by trying to find the most \"typical\" value of a data set.\r\n\r\nNote that we just used the word \"typical\" although in many cases you might think of using the word \"average.\" We need to be careful with the word \"average\" as it means different things to different people in different contexts.\u00a0 One of the most common uses of the word \"average\" is what mathematicians and statisticians call the <strong>arithmetic mean<\/strong>, or just plain old <strong>mean<\/strong> for short.\u00a0 \"Arithmetic mean\" sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word \"average.\"\r\n<div class=\"textbox\">\r\n<h3>Mean<\/h3>\r\nThe <strong>mean<\/strong> of a set of data is the sum of the data values divided by the number of values.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nMarci\u2019s exam scores for her last math class were 79, 86, 82, and 94. What would the mean of these values would be?\r\n[reveal-answer q=\"162631\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"162631\"][latex]\\frac{79+86+82+94}{4}=85.25[\/latex]. Typically we round means to one more decimal place than the original data had. In this case, we would round 85.25 to 85.3.[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nThe number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season are shown below.\r\n\r\n37 33 33 32 29 28 28 23 22 22 22 21 21 21 20\r\n\r\n20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6\r\n\r\nWhat is the mean number of TD passes?\r\n[reveal-answer q=\"244479\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"244479\"]\r\n\r\nAdding these values, we get 634 total TDs. Dividing by 31, the number of data values, we get 634\/31 = 20.4516. It would be appropriate to round this to 20.5.\r\n\r\nIt would be most correct for us to report that \u201cThe mean number of touchdown passes thrown in the NFL in the 2000 season was 20.5 passes,\u201d but it is not uncommon to see the more casual word \u201caverage\u201d used in place of \u201cmean.\u201d\r\n\r\n[\/hidden-answer]\r\n\r\nBoth examples are described further in the following video.\r\n\r\nhttps:\/\/youtu.be\/3if9Le2sO0c\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe price of a jar of peanut butter at 5 stores was $3.29, $3.59, $3.79, $3.75, and $3.99. Find the mean price.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>examples<\/h3>\r\nThe one hundred families in a particular neighborhood are asked their annual household income, to the nearest $5 thousand dollars. The results are summarized in a frequency table below.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Income (thousands of dollars)<\/strong><\/td>\r\n<td><strong>Frequency<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>15<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>20<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>25<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>30<\/td>\r\n<td>17<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>35<\/td>\r\n<td>19<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>40<\/td>\r\n<td>20<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>45<\/td>\r\n<td>12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>50<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhat is the mean average income in this neighborhood?\r\n[reveal-answer q=\"231491\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"231491\"]\r\n\r\nCalculating the mean by hand could get tricky if we try to type in all 100 values:\r\n\r\n[latex]\\frac{\\overbrace{15+\\cdots+15}^{\\text{6terms}}+\\overbrace{20+\\cdots+20}^{\\text{8terms}}+\\overbrace{25+\\cdots+25}^{\\text{11terms}}+\\cdots}{\\text{100}}[\/latex]\r\n\r\nWe could calculate this more easily by noticing that adding 15 to itself six times is the same as = 90. Using this simplification, we get\r\n\r\n[latex]\\frac{15\\cdot6+20\\cdot8+25\\cdot11+30\\cdot17+35\\cdot19+40\\cdot20+45\\cdot12+50\\cdot7}{\\text{100}}=\\frac{3390}{100}=33.9[\/latex]\r\n\r\nThe mean household income of our sample is 33.9 thousand dollars ($33,900).\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nExtending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand).\r\n\r\nWhat is the new mean of this neighborhood's income?\r\n[reveal-answer q=\"820039\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"820039\"]\r\n\r\nAdding this to our sample, our mean is now:\r\n\r\n[latex]\\frac{15\\cdot6+20\\cdot8+25\\cdot11+30\\cdot17+35\\cdot19+40\\cdot20+45\\cdot12+50\\cdot7+5000\\cdot1}{\\text{101}}=\\frac{8390}{101}=83.069[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nBoth situations are explained further in this video.\r\n\r\nhttps:\/\/youtu.be\/1_4Hxcq8DpQ\r\n\r\n<\/div>\r\nWhile 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a \u201ctypical\u201d value.\r\n\r\nImagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below.\r\n\r\n<img class=\"aligncenter wp-image-423\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12182802\/balance1.png\" alt=\"Drawing of a balance bar. A large blue block is on left end, and two smaller blue rectangles are on right end of balance point. One is close to the balance, one is further away. \" width=\"349\" height=\"61\" \/>\r\n\r\nIf we graph our household data, the $5 million data value is so far out to the right that the mean has to adjust up to keep things in balance.\r\n\r\n<img class=\"aligncenter wp-image-424\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12182831\/balance2.png\" alt=\"Drawing of a balance bar. A large blue block is on left end and two smaller blue rectangles are on also on left of balance point. On right, a small blue rectangle is significantly far away from the balance point.\" width=\"1256\" height=\"100\" \/>\r\n\r\nFor this reason, when working with data that have <strong>outliers<\/strong> \u2013 values far outside the primary grouping \u2013 it is common to use a different measure of center, the <strong>median<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Median<\/h3>\r\nThe <strong>median<\/strong> of a set of data is the value in the middle when the data is in order.\r\n<ul>\r\n \t<li>To find the median, begin by listing the data in order from smallest to largest, or largest to smallest.<\/li>\r\n \t<li>If the number of data values, <em>N<\/em>, is odd, then the median is the middle data value. This value can be found by rounding <em>N<\/em>\/2 up to the next whole number.<\/li>\r\n \t<li>If the number of data values is even, there is no one middle value, so we find the mean of the two middle values (values <em>N<\/em>\/2 and <em>N<\/em>\/2 + 1)<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nReturning to the football touchdown data, we would start by listing the data in order. Luckily, it was already in decreasing order, so we can work with it without needing to reorder it first.\r\n\r\n37 33 33 32 29 28 28 23 22 22 22 21 21 21 20\r\n\r\n20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6\r\n\r\nWhat is the median TD value?\r\n[reveal-answer q=\"866224\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"866224\"]Since there are 31 data values, an odd number, the median will be the middle number, the 16th data value (31\/2 = 15.5, round up to 16, leaving 15 values below and 15 above). The 16th data value is 20, so the median number of touchdown passes in the 2000 season was 20 passes. Notice that for this data, the median is fairly close to the mean we calculated earlier, 20.5.[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nFind the median of these quiz scores: 5 10 8 6 4 8 2 5 7 7\r\n[reveal-answer q=\"282623\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"282623\"]\r\n\r\nWe start by listing the data in order: 2 4 5 5 6 7 7 8 8 10\r\n\r\nSince there are 10 data values, an even number, there is no one middle number. So we find the mean of the two middle numbers, 6 and 7, and get (6+7)\/2 = 6.5.\r\n\r\nThe median quiz score was 6.5.\r\n\r\n[\/hidden-answer]\r\n\r\nLearn more about these median examples in this video.\r\n\r\nhttps:\/\/youtu.be\/WEdr_rSRObk\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe price of a jar of peanut butter at 5 stores was\u00a0$3.29, $3.59, $3.79, $3.75, and $3.99. Find the median price.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLet us return now to our original household income data\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Income (thousands of dollars)<\/strong><\/td>\r\n<td><strong>Frequency<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>15<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>20<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>25<\/td>\r\n<td>11<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>30<\/td>\r\n<td>17<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>35<\/td>\r\n<td>19<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>40<\/td>\r\n<td>20<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>45<\/td>\r\n<td>12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>50<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhat is the mean of this neighborhood's household income?\r\n[reveal-answer q=\"941761\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"941761\"]\r\n\r\nHere we have 100 data values. If we didn\u2019t already know that, we could find it by adding the frequencies. Since 100 is an even number, we need to find the mean of the middle two data values - the 50th and 51st data values. To find these, we start counting up from the bottom:\r\n\r\nThere are 6 data values of $15, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 1 to 6 are $15 thousand\r\n\r\nThe next 8 data values are $20, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 7 to (6+8)=14 are $20 thousand\r\n\r\nThe next 11 data values are $25, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 15 to (14+11)=25 are $25 thousand\r\n\r\nThe next 17 data values are $30, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 26 to (25+17)=42 are $30 thousand\r\n\r\nThe next 19 data values are $35, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 43 to (42+19)=61 are $35 thousand\r\n\r\nFrom this we can tell that values 50 and 51 will be $35 thousand, and the mean of these two values is $35 thousand. The median income in this neighborhood is $35 thousand.\r\n\r\n[\/hidden-answer]\r\n\r\n<hr \/>\r\n\r\nIf we add in the new neighbor with a $5 million household income, then there will be 101 data values, and the 51st value will be the median. As we discovered in the last example, the 51st value is $35 thousand. Notice that the new neighbor did not affect the median in this case. The median is not swayed as much by outliers as the mean is.\r\n\r\nView more about the median of this neighborhood's household incomes here.\r\n\r\nhttps:\/\/youtu.be\/kqEu9EDkmfU\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn addition to the mean and the median, there is one other common measurement of the \"typical\" value of a data set: the <strong>mode<\/strong>.\r\n<div class=\"textbox\">\r\n<h3>Mode<\/h3>\r\nThe <strong>mode<\/strong> is the element of the data set that occurs most frequently.\r\n\r\n<\/div>\r\nThe mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn our vehicle color survey earlier in this section, we collected the data\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th>Color<\/th>\r\n<th>Frequency<\/th>\r\n<\/tr>\r\n<tr>\r\n<td>Blue<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Green<\/td>\r\n<td>5<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Red<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>White<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Black<\/td>\r\n<td>2<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Grey<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhich color is the mode?\r\n[reveal-answer q=\"638793\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"638793\"]For this data, Green is the mode, since it is the data value that occurred the most frequently.[\/hidden-answer]\r\n\r\nMode in this example is explained by the video here.\r\n\r\nhttps:\/\/youtu.be\/pFpkWrib3Jk\r\n\r\n<\/div>\r\nIt is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nReviewers were asked to rate a product on a scale of 1 to 5. Find\r\n<ol>\r\n \t<li>The mean rating<\/li>\r\n \t<li>The median rating<\/li>\r\n \t<li>The mode rating<\/li>\r\n<\/ol>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td><strong>Rating<\/strong><\/td>\r\n<td><strong>Frequency<\/strong><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>1<\/td>\r\n<td>4<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>2<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>3<\/td>\r\n<td>7<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>4<\/td>\r\n<td>3<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>5<\/td>\r\n<td>1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the mean, median, and mode of a set of data<\/li>\n<li>Calculate the range of a data set, and recognize it&#8217;s limitations in fully describing the behavior of a data set<\/li>\n<li>Calculate the standard deviation for a data set, and determine it&#8217;s units<\/li>\n<li>Identify the difference between population variance and sample variance<\/li>\n<li>Identify the quartiles for a data set, and the calculations used to define them<\/li>\n<li>Identify the parts of a five number summary for a set of data, and create a box plot using it<\/li>\n<\/ul>\n<\/div>\n<h2>Mean, Median, and Mode<\/h2>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/22190716\/8027247398_96f7013a1f_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-955\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/22190716\/8027247398_96f7013a1f_z.jpg\" alt=\"Silver sphere which has red smaller spheres clustered on its left side, with magnetic attraction\" width=\"640\" height=\"427\" \/><\/a><\/p>\n<p>Let&#8217;s begin by trying to find the most &#8220;typical&#8221; value of a data set.<\/p>\n<p>Note that we just used the word &#8220;typical&#8221; although in many cases you might think of using the word &#8220;average.&#8221; We need to be careful with the word &#8220;average&#8221; as it means different things to different people in different contexts.\u00a0 One of the most common uses of the word &#8220;average&#8221; is what mathematicians and statisticians call the <strong>arithmetic mean<\/strong>, or just plain old <strong>mean<\/strong> for short.\u00a0 &#8220;Arithmetic mean&#8221; sounds rather fancy, but you have likely calculated a mean many times without realizing it; the mean is what most people think of when they use the word &#8220;average.&#8221;<\/p>\n<div class=\"textbox\">\n<h3>Mean<\/h3>\n<p>The <strong>mean<\/strong> of a set of data is the sum of the data values divided by the number of values.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>Marci\u2019s exam scores for her last math class were 79, 86, 82, and 94. What would the mean of these values would be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q162631\">Show Solution<\/span><\/p>\n<div id=\"q162631\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{79+86+82+94}{4}=85.25[\/latex]. Typically we round means to one more decimal place than the original data had. In this case, we would round 85.25 to 85.3.<\/div>\n<\/div>\n<hr \/>\n<p>The number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season are shown below.<\/p>\n<p>37 33 33 32 29 28 28 23 22 22 22 21 21 21 20<\/p>\n<p>20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6<\/p>\n<p>What is the mean number of TD passes?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q244479\">Show Solution<\/span><\/p>\n<div id=\"q244479\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding these values, we get 634 total TDs. Dividing by 31, the number of data values, we get 634\/31 = 20.4516. It would be appropriate to round this to 20.5.<\/p>\n<p>It would be most correct for us to report that \u201cThe mean number of touchdown passes thrown in the NFL in the 2000 season was 20.5 passes,\u201d but it is not uncommon to see the more casual word \u201caverage\u201d used in place of \u201cmean.\u201d<\/p>\n<\/div>\n<\/div>\n<p>Both examples are described further in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Finding the mean of a data set\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3if9Le2sO0c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The price of a jar of peanut butter at 5 stores was $3.29, $3.59, $3.79, $3.75, and $3.99. Find the mean price.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>examples<\/h3>\n<p>The one hundred families in a particular neighborhood are asked their annual household income, to the nearest $5 thousand dollars. The results are summarized in a frequency table below.<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>Income (thousands of dollars)<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td>15<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>25<\/td>\n<td>11<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>17<\/td>\n<\/tr>\n<tr>\n<td>35<\/td>\n<td>19<\/td>\n<\/tr>\n<tr>\n<td>40<\/td>\n<td>20<\/td>\n<\/tr>\n<tr>\n<td>45<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>50<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>What is the mean average income in this neighborhood?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q231491\">Show Solution<\/span><\/p>\n<div id=\"q231491\" class=\"hidden-answer\" style=\"display: none\">\n<p>Calculating the mean by hand could get tricky if we try to type in all 100 values:<\/p>\n<p>[latex]\\frac{\\overbrace{15+\\cdots+15}^{\\text{6terms}}+\\overbrace{20+\\cdots+20}^{\\text{8terms}}+\\overbrace{25+\\cdots+25}^{\\text{11terms}}+\\cdots}{\\text{100}}[\/latex]<\/p>\n<p>We could calculate this more easily by noticing that adding 15 to itself six times is the same as = 90. Using this simplification, we get<\/p>\n<p>[latex]\\frac{15\\cdot6+20\\cdot8+25\\cdot11+30\\cdot17+35\\cdot19+40\\cdot20+45\\cdot12+50\\cdot7}{\\text{100}}=\\frac{3390}{100}=33.9[\/latex]<\/p>\n<p>The mean household income of our sample is 33.9 thousand dollars ($33,900).<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>Extending off the last example, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand).<\/p>\n<p>What is the new mean of this neighborhood&#8217;s income?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q820039\">Show Solution<\/span><\/p>\n<div id=\"q820039\" class=\"hidden-answer\" style=\"display: none\">\n<p>Adding this to our sample, our mean is now:<\/p>\n<p>[latex]\\frac{15\\cdot6+20\\cdot8+25\\cdot11+30\\cdot17+35\\cdot19+40\\cdot20+45\\cdot12+50\\cdot7+5000\\cdot1}{\\text{101}}=\\frac{8390}{101}=83.069[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>Both situations are explained further in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Mean from a frequency table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/1_4Hxcq8DpQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>While 83.1 thousand dollars ($83,069) is the correct mean household income, it no longer represents a \u201ctypical\u201d value.<\/p>\n<p>Imagine the data values on a see-saw or balance scale. The mean is the value that keeps the data in balance, like in the picture below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-423\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12182802\/balance1.png\" alt=\"Drawing of a balance bar. A large blue block is on left end, and two smaller blue rectangles are on right end of balance point. One is close to the balance, one is further away.\" width=\"349\" height=\"61\" \/><\/p>\n<p>If we graph our household data, the $5 million data value is so far out to the right that the mean has to adjust up to keep things in balance.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-424\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/12182831\/balance2.png\" alt=\"Drawing of a balance bar. A large blue block is on left end and two smaller blue rectangles are on also on left of balance point. On right, a small blue rectangle is significantly far away from the balance point.\" width=\"1256\" height=\"100\" \/><\/p>\n<p>For this reason, when working with data that have <strong>outliers<\/strong> \u2013 values far outside the primary grouping \u2013 it is common to use a different measure of center, the <strong>median<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Median<\/h3>\n<p>The <strong>median<\/strong> of a set of data is the value in the middle when the data is in order.<\/p>\n<ul>\n<li>To find the median, begin by listing the data in order from smallest to largest, or largest to smallest.<\/li>\n<li>If the number of data values, <em>N<\/em>, is odd, then the median is the middle data value. This value can be found by rounding <em>N<\/em>\/2 up to the next whole number.<\/li>\n<li>If the number of data values is even, there is no one middle value, so we find the mean of the two middle values (values <em>N<\/em>\/2 and <em>N<\/em>\/2 + 1)<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Returning to the football touchdown data, we would start by listing the data in order. Luckily, it was already in decreasing order, so we can work with it without needing to reorder it first.<\/p>\n<p>37 33 33 32 29 28 28 23 22 22 22 21 21 21 20<\/p>\n<p>20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6<\/p>\n<p>What is the median TD value?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q866224\">Show Solution<\/span><\/p>\n<div id=\"q866224\" class=\"hidden-answer\" style=\"display: none\">Since there are 31 data values, an odd number, the median will be the middle number, the 16th data value (31\/2 = 15.5, round up to 16, leaving 15 values below and 15 above). The 16th data value is 20, so the median number of touchdown passes in the 2000 season was 20 passes. Notice that for this data, the median is fairly close to the mean we calculated earlier, 20.5.<\/div>\n<\/div>\n<hr \/>\n<p>Find the median of these quiz scores: 5 10 8 6 4 8 2 5 7 7<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q282623\">Show Solution<\/span><\/p>\n<div id=\"q282623\" class=\"hidden-answer\" style=\"display: none\">\n<p>We start by listing the data in order: 2 4 5 5 6 7 7 8 8 10<\/p>\n<p>Since there are 10 data values, an even number, there is no one middle number. So we find the mean of the two middle numbers, 6 and 7, and get (6+7)\/2 = 6.5.<\/p>\n<p>The median quiz score was 6.5.<\/p>\n<\/div>\n<\/div>\n<p>Learn more about these median examples in this video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Median from a data list\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/WEdr_rSRObk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The price of a jar of peanut butter at 5 stores was\u00a0$3.29, $3.59, $3.79, $3.75, and $3.99. Find the median price.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Let us return now to our original household income data<\/p>\n<table>\n<tbody>\n<tr>\n<td><strong>Income (thousands of dollars)<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td>15<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td>20<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>25<\/td>\n<td>11<\/td>\n<\/tr>\n<tr>\n<td>30<\/td>\n<td>17<\/td>\n<\/tr>\n<tr>\n<td>35<\/td>\n<td>19<\/td>\n<\/tr>\n<tr>\n<td>40<\/td>\n<td>20<\/td>\n<\/tr>\n<tr>\n<td>45<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td>50<\/td>\n<td>7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>What is the mean of this neighborhood&#8217;s household income?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q941761\">Show Solution<\/span><\/p>\n<div id=\"q941761\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here we have 100 data values. If we didn\u2019t already know that, we could find it by adding the frequencies. Since 100 is an even number, we need to find the mean of the middle two data values &#8211; the 50th and 51st data values. To find these, we start counting up from the bottom:<\/p>\n<p>There are 6 data values of $15, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 1 to 6 are $15 thousand<\/p>\n<p>The next 8 data values are $20, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 7 to (6+8)=14 are $20 thousand<\/p>\n<p>The next 11 data values are $25, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 15 to (14+11)=25 are $25 thousand<\/p>\n<p>The next 17 data values are $30, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 26 to (25+17)=42 are $30 thousand<\/p>\n<p>The next 19 data values are $35, so \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Values 43 to (42+19)=61 are $35 thousand<\/p>\n<p>From this we can tell that values 50 and 51 will be $35 thousand, and the mean of these two values is $35 thousand. The median income in this neighborhood is $35 thousand.<\/p>\n<\/div>\n<\/div>\n<hr \/>\n<p>If we add in the new neighbor with a $5 million household income, then there will be 101 data values, and the 51st value will be the median. As we discovered in the last example, the 51st value is $35 thousand. Notice that the new neighbor did not affect the median in this case. The median is not swayed as much by outliers as the mean is.<\/p>\n<p>View more about the median of this neighborhood&#8217;s household incomes here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Median from a frequency table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/kqEu9EDkmfU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In addition to the mean and the median, there is one other common measurement of the &#8220;typical&#8221; value of a data set: the <strong>mode<\/strong>.<\/p>\n<div class=\"textbox\">\n<h3>Mode<\/h3>\n<p>The <strong>mode<\/strong> is the element of the data set that occurs most frequently.<\/p>\n<\/div>\n<p>The mode is fairly useless with data like weights or heights where there are a large number of possible values. The mode is most commonly used for categorical data, for which median and mean cannot be computed.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In our vehicle color survey earlier in this section, we collected the data<\/p>\n<table>\n<tbody>\n<tr>\n<th>Color<\/th>\n<th>Frequency<\/th>\n<\/tr>\n<tr>\n<td>Blue<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>Green<\/td>\n<td>5<\/td>\n<\/tr>\n<tr>\n<td>Red<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>White<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>Black<\/td>\n<td>2<\/td>\n<\/tr>\n<tr>\n<td>Grey<\/td>\n<td>3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Which color is the mode?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q638793\">Show Solution<\/span><\/p>\n<div id=\"q638793\" class=\"hidden-answer\" style=\"display: none\">For this data, Green is the mode, since it is the data value that occurred the most frequently.<\/div>\n<\/div>\n<p>Mode in this example is explained by the video here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Mode for categorical data\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pFpkWrib3Jk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>It is possible for a data set to have more than one mode if several categories have the same frequency, or no modes if each every category occurs only once.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Reviewers were asked to rate a product on a scale of 1 to 5. Find<\/p>\n<ol>\n<li>The mean rating<\/li>\n<li>The median rating<\/li>\n<li>The mode rating<\/li>\n<\/ol>\n<table>\n<tbody>\n<tr>\n<td><strong>Rating<\/strong><\/td>\n<td><strong>Frequency<\/strong><\/td>\n<\/tr>\n<tr>\n<td>1<\/td>\n<td>4<\/td>\n<\/tr>\n<tr>\n<td>2<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td>3<\/td>\n<td>7<\/td>\n<\/tr>\n<tr>\n<td>4<\/td>\n<td>3<\/td>\n<\/tr>\n<tr>\n<td>5<\/td>\n<td>1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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-422\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Measures of Central Tendency. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>Project<\/strong>: Math in Society. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Magnetic. <strong>Authored by<\/strong>: Philippe Put. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/flic.kr\/p\/dekJZd\">https:\/\/flic.kr\/p\/dekJZd<\/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>Finding the mean of a data set. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3if9Le2sO0c\">https:\/\/youtu.be\/3if9Le2sO0c<\/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>Mean from a frequency table. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/1_4Hxcq8DpQ\">https:\/\/youtu.be\/1_4Hxcq8DpQ<\/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>Median from a data list. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/WEdr_rSRObk\">https:\/\/youtu.be\/WEdr_rSRObk<\/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>Median from a frequency table. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/kqEu9EDkmfU\">https:\/\/youtu.be\/kqEu9EDkmfU<\/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>Mode for categorical data. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/pFpkWrib3Jk\">https:\/\/youtu.be\/pFpkWrib3Jk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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":20,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Measures of Central Tendency\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Magnetic\",\"author\":\"Philippe Put\",\"organization\":\"\",\"url\":\"https:\/\/flic.kr\/p\/dekJZd\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Finding the mean of a data set\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/3if9Le2sO0c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Mean from a frequency table\",\"author\":\"OCLPhase2\\'s 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