{"id":6072,"date":"2021-03-09T23:20:10","date_gmt":"2021-03-09T23:20:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/?post_type=chapter&p=6072"},"modified":"2022-01-07T00:39:41","modified_gmt":"2022-01-07T00:39:41","slug":"4-14-comparison-of-mean-and-median","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-mathforliberalartscorequisite1\/chapter\/4-14-comparison-of-mean-and-median\/","title":{"raw":"4.14 Comparison of Mean and Median","rendered":"4.14 Comparison of Mean and Median"},"content":{"raw":"
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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
\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\nThe following example illustrates one of the downfalls of the arithmetic mean.\u00a0 The mean is not resistant to extremely high or low data values (outliers).\r\n
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Example - Mean is not resistant to Outliers<\/h3>\r\nOne hundred families in a particular neighborhood are asked their annual household income, to the nearest $5 thousand dollars.\u00a0The results are summarized in a frequency table below.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Income (thousands of dollars)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\n<\/tr>\r\n
15<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
20<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
25<\/td>\r\n11<\/td>\r\n<\/tr>\r\n
30<\/td>\r\n17<\/td>\r\n<\/tr>\r\n
35<\/td>\r\n19<\/td>\r\n<\/tr>\r\n
40<\/td>\r\n20<\/td>\r\n<\/tr>\r\n
45<\/td>\r\n12<\/td>\r\n<\/tr>\r\n
50<\/td>\r\n7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe calculate the mean and find the\u00a0mean household income of our sample to be 33.9 thousand dollars ($33,900).\r\n\r\nNow, suppose a new family moves into the neighborhood example that has a household income of $5 million ($5000 thousand).\u00a0 Adding the new family to our data.\u00a0 Our new mean is now 83.1 thousand dollars ($83,069).\r\n\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\n<\/div>\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\"Drawing\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\"Drawing\r\n\r\nFor this reason, when working with data that have outliers<\/strong> \u2013 values far outside the primary grouping \u2013 it is common to use a different measure of center, the median<\/strong>.\r\n
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Median<\/h3>\r\nThe\u00a0median\u00a0of a set of data is the value in the middle when the data is in order.\r\n
    \r\n \t
  • To find the median, begin by listing the data in order from smallest to largest, or largest to smallest.<\/li>\r\n \t
  • If the number of data values, N<\/em>, is odd, then the median is the middle data value. This value can be found by rounding N<\/em>\/2 up to the next whole number.<\/li>\r\n \t
  • 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 N<\/em>\/2 and N<\/em>\/2 + 1)<\/li>\r\n<\/ul>\r\n<\/div>\r\n
    \r\n

    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
    \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
    \r\n

    Example<\/h3>\r\nLet us return now to our original household income data\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Income (thousands of dollars)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\n<\/tr>\r\n
    15<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
    20<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
    25<\/td>\r\n11<\/td>\r\n<\/tr>\r\n
    30<\/td>\r\n17<\/td>\r\n<\/tr>\r\n
    35<\/td>\r\n19<\/td>\r\n<\/tr>\r\n
    40<\/td>\r\n20<\/td>\r\n<\/tr>\r\n
    45<\/td>\r\n12<\/td>\r\n<\/tr>\r\n
    50<\/td>\r\n7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nHere we have 100 data values. 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
    \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 \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 mode<\/strong>.\r\n
    \r\n

    Mode<\/h3>\r\nThe 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
    \r\n

    Example<\/h3>\r\nIn our vehicle color survey earlier in this section, we collected the data\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Color<\/th>\r\nFrequency<\/th>\r\n<\/tr>\r\n
    Blue<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
    Green<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
    Red<\/td>\r\n4<\/td>\r\n<\/tr>\r\n
    White<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
    Black<\/td>\r\n2<\/td>\r\n<\/tr>\r\n
    Grey<\/td>\r\n3<\/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
    \r\n

    Try It<\/h3>\r\nReviewers were asked to rate a product on a scale of 1 to 5. Find\r\n
      \r\n \t
    1. The mean rating<\/li>\r\n \t
    2. The median rating<\/li>\r\n \t
    3. The mode rating<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Rating<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\n<\/tr>\r\n
      1<\/td>\r\n4<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
      3<\/td>\r\n7<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
      5<\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

      Click <Next> to read Section 4.15<\/span><\/strong><\/p>","rendered":"

      \n

      examples<\/h3>\n

      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

      Show Solution<\/span><\/p>\n
      [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
      \n

      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

      37 33 33 32 29 28 28 23 22 22 22 21 21 21 20<\/p>\n

      20 19 19 18 18 18 18 16 15 14 14 14 12 12 9 6<\/p>\n

      What is the mean number of TD passes?<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      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

      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

      Both examples are described further in the following video.<\/p>\n