{"id":2033,"date":"2021-05-04T20:47:01","date_gmt":"2021-05-04T20:47:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/?post_type=chapter&p=2033"},"modified":"2023-01-10T19:20:35","modified_gmt":"2023-01-10T19:20:35","slug":"placeholder","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/placeholder\/","title":{"raw":"Frequency Distributions","rendered":"Frequency Distributions"},"content":{"raw":"

Ungrouped Frequency Distributions<\/strong><\/span><\/h2>\r\n

Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows:\r\n5, 6, 3, 3, 2, 4, 7, 5, 2, 3, 5, 6, 5, 4, 4, 3, 5, 2, 5, 3<\/span>.<\/p>\r\n

The following table lists the different data values in ascending order and their frequencies.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Frequency Table of Student Work Hours<\/caption>\r\n
DATA VALUE<\/th>\r\nFREQUENCY<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
2<\/span><\/td>\r\n3<\/td>\r\n<\/tr>\r\n
3<\/span><\/td>\r\n5<\/td>\r\n<\/tr>\r\n
4<\/span><\/td>\r\n3<\/td>\r\n<\/tr>\r\n
5<\/span><\/td>\r\n6<\/td>\r\n<\/tr>\r\n
6<\/span><\/td>\r\n2<\/td>\r\n<\/tr>\r\n
7<\/span><\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

In this research, 3 students studied for 2 hours. 5 students studies for 3 hours.<\/p>\r\nA frequency is the number of times a value of the data occurs. According to the table, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.\r\n

A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample\u2013in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.<\/p>\r\n

Relative frequency = [latex]\\frac{\\text{frequency of the class}}{\\text{total}}[\/latex]<\/strong><\/p>\r\n

Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in the table below.<\/p>\r\n

Cumulative relative frequency = sum of previous relative frequencies + current class frequency\u00a0<\/strong><\/p>\r\n\r\n\r\n

\r\n\r\n\r\n\r\n
\r\n\r\n

Example 1<\/strong><\/span><\/h2>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies<\/caption>\r\n
DATA VALUE<\/th>\r\nFREQUENCY<\/th>\r\nRELATIVE\r\n
<\/div>\r\nFREQUENCY<\/th>\r\n
CUMULATIVE RELATIVE\r\n
<\/div>\r\nFREQUENCY<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
2<\/span><\/td>\r\n3<\/td>\r\n[latex]\\frac{3}{20}[\/latex]\u00a0or 0.15<\/td>\r\n0.15<\/td>\r\n<\/tr>\r\n
3<\/span><\/td>\r\n5<\/td>\r\n[latex]\\frac{5}{20}[\/latex]\u00a0or 0.25<\/td>\r\n0.15 + 0.25 = 0.40<\/td>\r\n<\/tr>\r\n
4<\/span><\/td>\r\n3<\/td>\r\n[latex]\\frac{3}{20}[\/latex]\u00a0or 0.15<\/td>\r\n0.40 + 0.15 = 0.55<\/td>\r\n<\/tr>\r\n
5<\/span><\/td>\r\n6<\/td>\r\n[latex]\\frac{6}{20}[\/latex]\u00a0or 0.30<\/td>\r\n0.55 + 0.30 = 0.85<\/td>\r\n<\/tr>\r\n
6<\/span><\/td>\r\n2<\/td>\r\n[latex]\\frac{2}{20}[\/latex]\u00a0or 0.10<\/td>\r\n0.85 + 0.10 = 0.95<\/td>\r\n<\/tr>\r\n
7<\/span><\/td>\r\n1<\/td>\r\n[latex]\\frac{1}{20}[\/latex]\u00a0or 0.05<\/td>\r\n0.95 + 0.05 = 1.00<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.\r\n\r\n
\r\n\r\n

Grouped Frequency Distributions<\/strong><\/span><\/h2>\r\n

Example 2<\/span><\/h2>\r\nWe sample the height of 100 soccer players. The result is shown 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\r\n
Height (inches)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\n<\/tr>\r\n
59.95 - 61.95<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
61.95 - 63.95<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
63.95 - 65.95<\/td>\r\n15<\/td>\r\n<\/tr>\r\n
65.95 - 67.95<\/td>\r\n40<\/td>\r\n<\/tr>\r\n
67.95 - 69.95<\/td>\r\n17<\/td>\r\n<\/tr>\r\n
69.95 - 71.95<\/td>\r\n12<\/td>\r\n<\/tr>\r\n
71.95 - 73.95<\/td>\r\n7<\/td>\r\n<\/tr>\r\n
73.95 - 75.95<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
<\/td>\r\nTotal = 100<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind:\r\n\r\na. the relative frequency for each class.\r\n[reveal-answer q=\"816210\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"816210\"]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Height (Inches)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\nRelative Frequency<\/strong><\/td>\r\nCumulative Relative Frequency<\/strong><\/td>\r\n<\/tr>\r\n
59.95 - 61.95<\/td>\r\n5<\/td>\r\n[latex]\\frac{5}{100}[\/latex] or 0.05<\/td>\r\n0.05<\/td>\r\n<\/tr>\r\n
61.95 - 63.95<\/td>\r\n3<\/td>\r\n[latex]\\frac{3}{100}[\/latex] or 0.03<\/td>\r\n0.05 + 0.03 = 0.08<\/td>\r\n<\/tr>\r\n
63.95 - 65.95<\/td>\r\n15<\/td>\r\n[latex]\\frac{15}{100}[\/latex] or 0.15<\/td>\r\n0.08 + 0.15 = 0.23<\/td>\r\n<\/tr>\r\n
65.95 - 67.95<\/td>\r\n40<\/td>\r\n[latex]\\frac{4}{100}[\/latex] or 0.04<\/td>\r\n0.23 + 0.40 = 0.63<\/td>\r\n<\/tr>\r\n
67.95 - 69.95<\/td>\r\n17<\/td>\r\n[latex]\\frac{17}{100}[\/latex] or 0.17<\/td>\r\n0.63 + 0.17 = 0.80<\/td>\r\n<\/tr>\r\n
69.95 - 71.95<\/td>\r\n12<\/td>\r\n[latex]\\frac{12}{100}[\/latex] or 0.12<\/td>\r\n0.80 + 0.12 = 0.92<\/td>\r\n<\/tr>\r\n
71.95 - 73.95<\/td>\r\n7<\/td>\r\n[latex]\\frac{7}{100}[\/latex] or 0.07<\/td>\r\n0.92 + 0.07 = 0.99<\/td>\r\n<\/tr>\r\n
73.95 - 75.95<\/td>\r\n1<\/td>\r\n[latex]\\frac{1}{100}[\/latex] or 0.01<\/td>\r\n0.99 + 0.01 = 1.00<\/td>\r\n<\/tr>\r\n
<\/td>\r\nTotal = 100<\/strong><\/td>\r\nTotal = 1<\/strong><\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\nb.\u00a0the percentage for height that is less than 63.95 inches.\r\n[reveal-answer q=\"26068\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"26068\"][latex]\\frac{5+3}{100}[\/latex] = 0.08 = 8%[\/hidden-answer]\r\n\r\nc. the percentage for height that is between 69.95 inches and 73.95 inches.\r\n[reveal-answer q=\"839825\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"839825\"][latex]\\frac{12}{100}[\/latex] + [latex]\\frac{7}{100}[\/latex] = 0.12 + 0.07 = 0.19[\/hidden-answer]\r\n\r\nIn this sample, there are five<\/strong> players whose heights fall within the interval 59.95\u201361.95 inches, three<\/strong> players whose heights fall within the interval 61.95\u201363.95 inches, 15<\/strong> players whose heights fall within the interval 63.95\u201365.95 inches, 40<\/strong> players whose heights fall within the interval 65.95\u201367.95 inches, 17<\/strong> players whose heights fall within the interval 67.95\u201369.95 inches, 12<\/strong> players whose heights fall within the interval 69.95\u201371.95, seven<\/strong> players whose heights fall within the interval 71.95\u201373.95, and one<\/strong> player whose heights fall within the interval 73.95\u201375.95. All heights fall between the endpoints of an interval and not at the endpoints.\r\n\r\n
\r\n\r\n\r\n\r\n
\r\n\r\n

Example 3<\/span><\/h2>\r\nThe table shows the amount, in inches, of annual rainfall in a sample of towns.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Rainfall (inches)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\n<\/tr>\r\n
2.95 - 4.97<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
4.97 - 6.99<\/td>\r\n7<\/td>\r\n<\/tr>\r\n
6.99 - 9.01<\/td>\r\n15<\/td>\r\n<\/tr>\r\n
9.01 - 11.03<\/td>\r\n8<\/td>\r\n<\/tr>\r\n
11.03 - 13.05<\/td>\r\n9<\/td>\r\n<\/tr>\r\n
13.05 - 15.07<\/td>\r\n5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind\r\n
    \r\n \t
  1. the relative frequency and cumulative relative frequency for each class.\r\n[reveal-answer q=\"118523\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"118523\"]\r\nTotal = sum of all frequencies = 6 + 7 + 15 + 8 + 9 + 5 = 50\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Rainfall (inches)<\/strong><\/td>\r\nFrequency<\/strong><\/td>\r\nRelative frequency<\/strong><\/td>\r\nCumulative relative frequency<\/strong><\/td>\r\n<\/tr>\r\n
    2.95 - 4.97<\/td>\r\n6<\/td>\r\n[latex]\\frac{6}{50}[\/latex] = 0.12<\/td>\r\n0.12<\/td>\r\n<\/tr>\r\n
    4.97 - 6.99<\/td>\r\n7<\/td>\r\n[latex]\\frac{7}{50}[\/latex] = 0.14<\/td>\r\n0.12 + 0.14 = 0.26<\/td>\r\n<\/tr>\r\n
    6.99 - 9.01<\/td>\r\n15<\/td>\r\n[latex]\\frac{15}{50}[\/latex] = 0.30<\/td>\r\n0.26 + 0.30 = 0.56<\/td>\r\n<\/tr>\r\n
    9.01 - 11.03<\/td>\r\n8<\/td>\r\n[latex]\\frac{8}{50}[\/latex] = 0.16<\/td>\r\n0.56 + 0.16 = 0.72<\/td>\r\n<\/tr>\r\n
    11.03 - 13.05<\/td>\r\n9<\/td>\r\n[latex]\\frac{9}{50}[\/latex] = 0.18<\/td>\r\n0.72 + 0.18 = 0.90<\/td>\r\n<\/tr>\r\n
    13.05 - 15.07<\/td>\r\n5<\/td>\r\n[latex]\\frac{5}{50}[\/latex] = 0.10<\/td>\r\n0.90 + 0.10 = 1.00<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]<\/li>\r\n \t
  2. the percentage of rainfall that is less than 9.01 inches.\r\n[reveal-answer q=\"355649\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"355649\"]The percentage of rainfall that is less than 9.01 inches = 0.12 + 0.14 + 0.30 = 0.56 = 56%[\/hidden-answer]<\/li>\r\n \t
  3. the percentage of rainfall amounts between 6.99 inches and 11.03 inches.\r\n[reveal-answer q=\"94434\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"94434\"]The percentage of rainfall values between 6.99 inches and 11.03 inches = [latex]\\frac{15}{50}[\/latex] +\u00a0[latex]\\frac{8}{50}[\/latex] = 0.46=46%[\/hidden-answer]<\/li>\r\n<\/ol>\r\n \r\n\r\n \r\n\r\n
    \r\n\r\n\r\n\r\n
    \r\n\r\n
    \r\n

    Try It<\/h3>\r\nThe table contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Year<\/td>\r\nTotal Number of Deaths<\/td>\r\n<\/tr>\r\n
    2000<\/td>\r\n231<\/td>\r\n<\/tr>\r\n
    2001<\/td>\r\n21,357<\/td>\r\n<\/tr>\r\n
    2002<\/td>\r\n11,685<\/td>\r\n<\/tr>\r\n
    2003<\/td>\r\n33,819<\/td>\r\n<\/tr>\r\n
    2004<\/td>\r\n228,802<\/td>\r\n<\/tr>\r\n
    2005<\/td>\r\n87,503<\/td>\r\n<\/tr>\r\n
    2006<\/td>\r\n6,605<\/td>\r\n<\/tr>\r\n
    2007<\/td>\r\n712<\/td>\r\n<\/tr>\r\n
    2008<\/td>\r\n88,011<\/td>\r\n<\/tr>\r\n
    2009<\/td>\r\n1,790<\/td>\r\n<\/tr>\r\n
    2010<\/td>\r\n320,120<\/td>\r\n<\/tr>\r\n
    2011<\/td>\r\n21,953<\/td>\r\n<\/tr>\r\n
    2012<\/td>\r\n768<\/td>\r\n<\/tr>\r\n
    Total<\/td>\r\n823,356<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
      \r\n \t
    1. \u00a0What is the frequency of deaths measured from 2006 through 2009?\r\n[reveal-answer q=\"337996\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"337996\"]97,118 [\/hidden-answer]<\/li>\r\n \t
    2. What percentage of deaths occurred after 2009?\r\n[reveal-answer q=\"175141\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"175141\"]41.6%[\/hidden-answer]<\/li>\r\n \t
    3. What is the relative frequency of deaths that occurred in 2003 or earlier?\r\n[reveal-answer q=\"294919\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"294919\"][latex]\\frac{67,092}{823,356}[\/latex] = 0.081[\/hidden-answer]<\/li>\r\n \t
    4. What is the percentage of deaths that occurred in 2004?\r\n[reveal-answer q=\"154890\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"154890\"]27.8%[\/hidden-answer]<\/li>\r\n \t
    5. What kind of data are the numbers of deaths?\r\n[reveal-answer q=\"292106\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"292106\"]Quantitative discrete[\/hidden-answer]<\/li>\r\n \t
    6. The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers?\r\n[reveal-answer q=\"359271\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"359271\"]Quantitative continuous[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n \r\n\r\n
      \r\n\r\n\r\n\r\n
      \r\n\r\n

      Example 4<\/span><\/h2>\r\n

      The table contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Year<\/th>\r\nTotal Number of Crashes<\/th>\r\nYear<\/th>\r\nTotal Number of Crashes<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      1994<\/td>\r\n36,254<\/td>\r\n2004<\/td>\r\n38,444<\/td>\r\n<\/tr>\r\n
      1995<\/td>\r\n37,241<\/td>\r\n2005<\/td>\r\n39,252<\/td>\r\n<\/tr>\r\n
      1996<\/td>\r\n37,494<\/td>\r\n2006<\/td>\r\n38,648<\/td>\r\n<\/tr>\r\n
      1997<\/td>\r\n37,324<\/td>\r\n2007<\/td>\r\n37,435<\/td>\r\n<\/tr>\r\n
      1998<\/td>\r\n37,107<\/td>\r\n2008<\/td>\r\n34,172<\/td>\r\n<\/tr>\r\n
      1999<\/td>\r\n37,140<\/td>\r\n2009<\/td>\r\n30,862<\/td>\r\n<\/tr>\r\n
      2000<\/td>\r\n37,526<\/td>\r\n2010<\/td>\r\n30,296<\/td>\r\n<\/tr>\r\n
      2001<\/td>\r\n37,862<\/td>\r\n2011<\/td>\r\n29,757<\/td>\r\n<\/tr>\r\n
      2002<\/td>\r\n38,491<\/td>\r\nTotal<\/td>\r\n653,782<\/td>\r\n<\/tr>\r\n
      2003<\/td>\r\n38,477<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \r\n \t
      1. What is the frequency of deaths measured from 2000 through 2004?\r\n[reveal-answer q=\"514149\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"514149\"]37,526 + 37,862 + 38,491 + 38,477 + 38,444 = 190,800 [\/hidden-answer]<\/li>\r\n \t
      2. What percentage of deaths occurred after 2006?\r\n[reveal-answer q=\"107848\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"107848\"][latex]\\frac{37,435 + 34,172 + 30,862 + 30,296 + 29,757}{653,782}[\/latex] or 24.9%[\/hidden-answer]<\/li>\r\n \t
      3. What is the relative frequency of deaths that occurred in 2000 or before?\r\n[reveal-answer q=\"897794\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"897794\"][latex]\\frac{260,086}{653,782}[\/latex] or 39.8%[\/hidden-answer]<\/li>\r\n \t
      4. What is the percentage of deaths that occurred in 2011?\r\n[reveal-answer q=\"43840\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"43840\"][latex]\\frac{29,757}{653,782}[\/latex] or 4.6%[\/hidden-answer]<\/li>\r\n \t
      5. What is the cumulative relative frequency for 2006? Explain what this number tells you about the data.\r\n[reveal-answer q=\"366185\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"366185\"]75.1% of all fatal traffic crashes for the period from 1994 to 2011 happened from 1994 to 2006.[\/hidden-answer]<\/li>\r\n<\/ol>\r\n\r\n
        \r\n\r\n ","rendered":"

        Ungrouped Frequency Distributions<\/strong><\/span><\/h2>\n

        Twenty students were asked how many hours they worked per day. Their responses, in hours, are as follows:
        \n5, 6, 3, 3, 2, 4, 7, 5, 2, 3, 5, 6, 5, 4, 4, 3, 5, 2, 5, 3<\/span>.<\/p>\n

        The following table lists the different data values in ascending order and their frequencies.<\/p>\n\n\n\n\n\n\n\n\n\n\n
        Frequency Table of Student Work Hours<\/caption>\n
        DATA VALUE<\/th>\nFREQUENCY<\/th>\n<\/tr>\n<\/thead>\n
        2<\/span><\/td>\n3<\/td>\n<\/tr>\n
        3<\/span><\/td>\n5<\/td>\n<\/tr>\n
        4<\/span><\/td>\n3<\/td>\n<\/tr>\n
        5<\/span><\/td>\n6<\/td>\n<\/tr>\n
        6<\/span><\/td>\n2<\/td>\n<\/tr>\n
        7<\/span><\/td>\n1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        In this research, 3 students studied for 2 hours. 5 students studies for 3 hours.<\/p>\n

        A frequency is the number of times a value of the data occurs. According to the table, there are three students who work two hours, five students who work three hours, and so on. The sum of the values in the frequency column, 20, represents the total number of students included in the sample.<\/p>\n

        A relative frequency is the ratio (fraction or proportion) of the number of times a value of the data occurs in the set of all outcomes to the total number of outcomes. To find the relative frequencies, divide each frequency by the total number of students in the sample\u2013in this case, 20. Relative frequencies can be written as fractions, percents, or decimals.<\/p>\n

        Relative frequency = [latex]\\frac{\\text{frequency of the class}}{\\text{total}}[\/latex]<\/strong><\/p>\n

        Cumulative relative frequency is the accumulation of the previous relative frequencies. To find the cumulative relative frequencies, add all the previous relative frequencies to the relative frequency for the current row, as shown in the table below.<\/p>\n

        Cumulative relative frequency = sum of previous relative frequencies + current class frequency\u00a0<\/strong><\/p>\n

        \n
        \n

        Example 1<\/strong><\/span><\/h2>\n\n\n\n\n\n\n\n\n\n\n
        Frequency Table of Student Work Hours with Relative and Cumulative Relative Frequencies<\/caption>\n
        DATA VALUE<\/th>\nFREQUENCY<\/th>\nRELATIVE<\/p>\n
        <\/div>\n

        FREQUENCY<\/th>\n

        		CUMULATIVE RELATIVE<\/p>\n
        <\/div>\n

        FREQUENCY<\/th>\n<\/tr>\n<\/thead>\n

        2<\/span><\/td>\n3<\/td>\n[latex]\\frac{3}{20}[\/latex]\u00a0or 0.15<\/td>\n0.15<\/td>\n<\/tr>\n
        3<\/span><\/td>\n5<\/td>\n[latex]\\frac{5}{20}[\/latex]\u00a0or 0.25<\/td>\n0.15 + 0.25 = 0.40<\/td>\n<\/tr>\n
        4<\/span><\/td>\n3<\/td>\n[latex]\\frac{3}{20}[\/latex]\u00a0or 0.15<\/td>\n0.40 + 0.15 = 0.55<\/td>\n<\/tr>\n
        5<\/span><\/td>\n6<\/td>\n[latex]\\frac{6}{20}[\/latex]\u00a0or 0.30<\/td>\n0.55 + 0.30 = 0.85<\/td>\n<\/tr>\n
        6<\/span><\/td>\n2<\/td>\n[latex]\\frac{2}{20}[\/latex]\u00a0or 0.10<\/td>\n0.85 + 0.10 = 0.95<\/td>\n<\/tr>\n
        7<\/span><\/td>\n1<\/td>\n[latex]\\frac{1}{20}[\/latex]\u00a0or 0.05<\/td>\n0.95 + 0.05 = 1.00<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        The last entry of the cumulative relative frequency column is one, indicating that one hundred percent of the data has been accumulated.<\/p>\n

        \n

        Grouped Frequency Distributions<\/strong><\/span><\/h2>\n

        Example 2<\/span><\/h2>\n

        We sample the height of 100 soccer players. The result is shown below.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
        Height (inches)<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
        59.95 – 61.95<\/td>\n5<\/td>\n<\/tr>\n
        61.95 – 63.95<\/td>\n3<\/td>\n<\/tr>\n
        63.95 – 65.95<\/td>\n15<\/td>\n<\/tr>\n
        65.95 – 67.95<\/td>\n40<\/td>\n<\/tr>\n
        67.95 – 69.95<\/td>\n17<\/td>\n<\/tr>\n
        69.95 – 71.95<\/td>\n12<\/td>\n<\/tr>\n
        71.95 – 73.95<\/td>\n7<\/td>\n<\/tr>\n
        73.95 – 75.95<\/td>\n1<\/td>\n<\/tr>\n
        <\/td>\nTotal = 100<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Find:<\/p>\n

        a. the relative frequency for each class.<\/p>\n

        Show Answer<\/span><\/p>\n
        \n\n\n\n\n\n\n\n\n\n\n\n\n
        Height (Inches)<\/strong><\/td>\nFrequency<\/strong><\/td>\nRelative Frequency<\/strong><\/td>\nCumulative Relative Frequency<\/strong><\/td>\n<\/tr>\n
        59.95 – 61.95<\/td>\n5<\/td>\n[latex]\\frac{5}{100}[\/latex] or 0.05<\/td>\n0.05<\/td>\n<\/tr>\n
        61.95 – 63.95<\/td>\n3<\/td>\n[latex]\\frac{3}{100}[\/latex] or 0.03<\/td>\n0.05 + 0.03 = 0.08<\/td>\n<\/tr>\n
        63.95 – 65.95<\/td>\n15<\/td>\n[latex]\\frac{15}{100}[\/latex] or 0.15<\/td>\n0.08 + 0.15 = 0.23<\/td>\n<\/tr>\n
        65.95 – 67.95<\/td>\n40<\/td>\n[latex]\\frac{4}{100}[\/latex] or 0.04<\/td>\n0.23 + 0.40 = 0.63<\/td>\n<\/tr>\n
        67.95 – 69.95<\/td>\n17<\/td>\n[latex]\\frac{17}{100}[\/latex] or 0.17<\/td>\n0.63 + 0.17 = 0.80<\/td>\n<\/tr>\n
        69.95 – 71.95<\/td>\n12<\/td>\n[latex]\\frac{12}{100}[\/latex] or 0.12<\/td>\n0.80 + 0.12 = 0.92<\/td>\n<\/tr>\n
        71.95 – 73.95<\/td>\n7<\/td>\n[latex]\\frac{7}{100}[\/latex] or 0.07<\/td>\n0.92 + 0.07 = 0.99<\/td>\n<\/tr>\n
        73.95 – 75.95<\/td>\n1<\/td>\n[latex]\\frac{1}{100}[\/latex] or 0.01<\/td>\n0.99 + 0.01 = 1.00<\/td>\n<\/tr>\n
        <\/td>\nTotal = 100<\/strong><\/td>\nTotal = 1<\/strong><\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n

        b.\u00a0the percentage for height that is less than 63.95 inches.<\/p>\n

        Show Answer<\/span><\/p>\n
        [latex]\\frac{5+3}{100}[\/latex] = 0.08 = 8%<\/div>\n<\/div>\n

        c. the percentage for height that is between 69.95 inches and 73.95 inches.<\/p>\n

        Show Answer<\/span><\/p>\n
        [latex]\\frac{12}{100}[\/latex] + [latex]\\frac{7}{100}[\/latex] = 0.12 + 0.07 = 0.19<\/div>\n<\/div>\n

        In this sample, there are five<\/strong> players whose heights fall within the interval 59.95\u201361.95 inches, three<\/strong> players whose heights fall within the interval 61.95\u201363.95 inches, 15<\/strong> players whose heights fall within the interval 63.95\u201365.95 inches, 40<\/strong> players whose heights fall within the interval 65.95\u201367.95 inches, 17<\/strong> players whose heights fall within the interval 67.95\u201369.95 inches, 12<\/strong> players whose heights fall within the interval 69.95\u201371.95, seven<\/strong> players whose heights fall within the interval 71.95\u201373.95, and one<\/strong> player whose heights fall within the interval 73.95\u201375.95. All heights fall between the endpoints of an interval and not at the endpoints.<\/p>\n

        \n
        \n

        Example 3<\/span><\/h2>\n

        The table shows the amount, in inches, of annual rainfall in a sample of towns.<\/p>\n\n\n\n\n\n\n\n\n\n
        Rainfall (inches)<\/strong><\/td>\nFrequency<\/strong><\/td>\n<\/tr>\n
        2.95 – 4.97<\/td>\n6<\/td>\n<\/tr>\n
        4.97 – 6.99<\/td>\n7<\/td>\n<\/tr>\n
        6.99 – 9.01<\/td>\n15<\/td>\n<\/tr>\n
        9.01 – 11.03<\/td>\n8<\/td>\n<\/tr>\n
        11.03 – 13.05<\/td>\n9<\/td>\n<\/tr>\n
        13.05 – 15.07<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Find<\/p>\n

          \n
        1. the relative frequency and cumulative relative frequency for each class.\n
          Show Answer<\/span><\/p>\n
          \nTotal = sum of all frequencies = 6 + 7 + 15 + 8 + 9 + 5 = 50<\/p>\n\n\n\n\n\n\n\n\n\n
          Rainfall (inches)<\/strong><\/td>\nFrequency<\/strong><\/td>\nRelative frequency<\/strong><\/td>\nCumulative relative frequency<\/strong><\/td>\n<\/tr>\n
          2.95 – 4.97<\/td>\n6<\/td>\n[latex]\\frac{6}{50}[\/latex] = 0.12<\/td>\n0.12<\/td>\n<\/tr>\n
          4.97 – 6.99<\/td>\n7<\/td>\n[latex]\\frac{7}{50}[\/latex] = 0.14<\/td>\n0.12 + 0.14 = 0.26<\/td>\n<\/tr>\n
          6.99 – 9.01<\/td>\n15<\/td>\n[latex]\\frac{15}{50}[\/latex] = 0.30<\/td>\n0.26 + 0.30 = 0.56<\/td>\n<\/tr>\n
          9.01 – 11.03<\/td>\n8<\/td>\n[latex]\\frac{8}{50}[\/latex] = 0.16<\/td>\n0.56 + 0.16 = 0.72<\/td>\n<\/tr>\n
          11.03 – 13.05<\/td>\n9<\/td>\n[latex]\\frac{9}{50}[\/latex] = 0.18<\/td>\n0.72 + 0.18 = 0.90<\/td>\n<\/tr>\n
          13.05 – 15.07<\/td>\n5<\/td>\n[latex]\\frac{5}{50}[\/latex] = 0.10<\/td>\n0.90 + 0.10 = 1.00<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/li>\n
        2. the percentage of rainfall that is less than 9.01 inches.\n
          Show Answer<\/span><\/p>\n
          The percentage of rainfall that is less than 9.01 inches = 0.12 + 0.14 + 0.30 = 0.56 = 56%<\/div>\n<\/div>\n<\/li>\n
        3. the percentage of rainfall amounts between 6.99 inches and 11.03 inches.\n
          Show Answer<\/span><\/p>\n
          The percentage of rainfall values between 6.99 inches and 11.03 inches = [latex]\\frac{15}{50}[\/latex] +\u00a0[latex]\\frac{8}{50}[\/latex] = 0.46=46%<\/div>\n<\/div>\n<\/li>\n<\/ol>\n

           <\/p>\n

           <\/p>\n

          \n
          \n
          \n

          Try It<\/h3>\n

          The table contains the total number of deaths worldwide as a result of earthquakes for the period from 2000 to 2012.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
          Year<\/td>\nTotal Number of Deaths<\/td>\n<\/tr>\n
          2000<\/td>\n231<\/td>\n<\/tr>\n
          2001<\/td>\n21,357<\/td>\n<\/tr>\n
          2002<\/td>\n11,685<\/td>\n<\/tr>\n
          2003<\/td>\n33,819<\/td>\n<\/tr>\n
          2004<\/td>\n228,802<\/td>\n<\/tr>\n
          2005<\/td>\n87,503<\/td>\n<\/tr>\n
          2006<\/td>\n6,605<\/td>\n<\/tr>\n
          2007<\/td>\n712<\/td>\n<\/tr>\n
          2008<\/td>\n88,011<\/td>\n<\/tr>\n
          2009<\/td>\n1,790<\/td>\n<\/tr>\n
          2010<\/td>\n320,120<\/td>\n<\/tr>\n
          2011<\/td>\n21,953<\/td>\n<\/tr>\n
          2012<\/td>\n768<\/td>\n<\/tr>\n
          Total<\/td>\n823,356<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
            \n
          1. \u00a0What is the frequency of deaths measured from 2006 through 2009?\n
            Show Answer<\/span><\/p>\n
            97,118 <\/div>\n<\/div>\n<\/li>\n
          2. What percentage of deaths occurred after 2009?\n
            Show Answer<\/span><\/p>\n
            41.6%<\/div>\n<\/div>\n<\/li>\n
          3. What is the relative frequency of deaths that occurred in 2003 or earlier?\n
            Show Answer<\/span><\/p>\n
            [latex]\\frac{67,092}{823,356}[\/latex] = 0.081<\/div>\n<\/div>\n<\/li>\n
          4. What is the percentage of deaths that occurred in 2004?\n
            Show Answer<\/span><\/p>\n
            27.8%<\/div>\n<\/div>\n<\/li>\n
          5. What kind of data are the numbers of deaths?\n
            Show Answer<\/span><\/p>\n
            Quantitative discrete<\/div>\n<\/div>\n<\/li>\n
          6. The Richter scale is used to quantify the energy produced by an earthquake. Examples of Richter scale numbers are 2.3, 4.0, 6.1, and 7.0. What kind of data are these numbers?\n
            Show Answer<\/span><\/p>\n
            Quantitative continuous<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

             <\/p>\n

             <\/p>\n

            \n
            \n

            Example 4<\/span><\/h2>\n

            The table contains the total number of fatal motor vehicle traffic crashes in the United States for the period from 1994 to 2011.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
            Year<\/th>\nTotal Number of Crashes<\/th>\nYear<\/th>\nTotal Number of Crashes<\/th>\n<\/tr>\n<\/thead>\n
            1994<\/td>\n36,254<\/td>\n2004<\/td>\n38,444<\/td>\n<\/tr>\n
            1995<\/td>\n37,241<\/td>\n2005<\/td>\n39,252<\/td>\n<\/tr>\n
            1996<\/td>\n37,494<\/td>\n2006<\/td>\n38,648<\/td>\n<\/tr>\n
            1997<\/td>\n37,324<\/td>\n2007<\/td>\n37,435<\/td>\n<\/tr>\n
            1998<\/td>\n37,107<\/td>\n2008<\/td>\n34,172<\/td>\n<\/tr>\n
            1999<\/td>\n37,140<\/td>\n2009<\/td>\n30,862<\/td>\n<\/tr>\n
            2000<\/td>\n37,526<\/td>\n2010<\/td>\n30,296<\/td>\n<\/tr>\n
            2001<\/td>\n37,862<\/td>\n2011<\/td>\n29,757<\/td>\n<\/tr>\n
            2002<\/td>\n38,491<\/td>\nTotal<\/td>\n653,782<\/td>\n<\/tr>\n
            2003<\/td>\n38,477<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
              \n
            1. What is the frequency of deaths measured from 2000 through 2004?\n
              Show Answer<\/span><\/p>\n
              37,526 + 37,862 + 38,491 + 38,477 + 38,444 = 190,800 <\/div>\n<\/div>\n<\/li>\n
            2. What percentage of deaths occurred after 2006?\n
              Show Answer<\/span><\/p>\n
              [latex]\\frac{37,435 + 34,172 + 30,862 + 30,296 + 29,757}{653,782}[\/latex] or 24.9%<\/div>\n<\/div>\n<\/li>\n
            3. What is the relative frequency of deaths that occurred in 2000 or before?\n
              Show Answer<\/span><\/p>\n
              [latex]\\frac{260,086}{653,782}[\/latex] or 39.8%<\/div>\n<\/div>\n<\/li>\n
            4. What is the percentage of deaths that occurred in 2011?\n
              Show Answer<\/span><\/p>\n
              [latex]\\frac{29,757}{653,782}[\/latex] or 4.6%<\/div>\n<\/div>\n<\/li>\n
            5. What is the cumulative relative frequency for 2006? Explain what this number tells you about the data.\n
              Show Answer<\/span><\/p>\n
              75.1% of all fatal traffic crashes for the period from 1994 to 2011 happened from 1994 to 2006.<\/div>\n<\/div>\n<\/li>\n<\/ol>\n
              \n

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