![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2020\/06\/8.2-intro-image-flower-1-1024x614.jpg\" "\\"https:\/\/pixabay.com\/illustrations\/dispersion-flower-effect-flora-6184855\/\\"")
\nLearning Objectives<\/p>\n\n<\/header>\n
\n\nBy the end of this section the student should be able to:\n
- \n \t
- Determine the range for a data set.<\/li>\n \t
- Determine the standard deviation for a data set.<\/li>\n \t
- Determine the standard deviation from a histogram<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
Measures of Dispersion or Spread<\/h1>\nWe have seen that measures of central tendency,<\/strong> including the\u00a0[pb_glossary id=\"1845\"]mean[\/pb_glossary]\u00a0and [pb_glossary id=\"1844\"]median[\/pb_glossary], \u00a0are used to identify a central position within a data set. They indicate where the data clusters.\n\nConsider student A's scores on five tests:\u00a0 32%\u00a0 \u00a095%\u00a0 \u00a089%\u00a0 \u00a074%\u00a0 \u00a055%\u00a0\u00a0The mean,<\/strong> or average, is\u00a0 (32 + 95 + 89 + 74 + 55)\/5 = 69%\u00a0 and the median<\/strong> is 74%.\n\nConsider\u00a0 student B's scores on the same five tests:\u00a0 68%\u00a0 \u00a069%\u00a0 \u00a072%\u00a0 \u00a074%\u00a0 \u00a062%\u00a0 \u00a0The mean,<\/strong> or average, is\u00a0 (68 + 69 + 72 + 74 + 62)\/5 = 69%\u00a0 and the median<\/strong> is 69%.\n\nBoth student's have the same test average of 69% but there is a substantial difference in the spread or dispersion of their scores. Student A's test scores range\u00a0from a low score of 32% to a high score of\u00a0 95% so the spread in marks is 63 percentage points. Student B's test scores range from a low score of 62% to a high score of\u00a0 74% so the spread in marks is 12 percentage points.\n\nWhen we analyze data it is important to consider how dispersed<\/strong> or spread out the data values are. In this section we will consider two measures of dispersion.\n
Range<\/h1>\nRange is one measure of dispersion. <\/strong>A measure of dispersion is used to describe the spread of data.\n
\n Range<\/p>\n\n<\/header>\n
\n\nThe range<\/strong> indicates the total spread in data values. It is the difference between the highest and lowest data values.\n\nRange = highest data value - lowest data value\n\n<\/div>\n<\/div>\n\n EXAMPLE 1<\/p>\n\n<\/header>\n
\n\nThe table shows the daily high temperature (\u00b0C) over a three week period.\n\nDetermine the highest temperature, lowest temperature, and the range\u00a0 in daily high temperatures over the three weeks.\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Ex1-Range-1024x212.png\")
\n\nSolution<\/strong>\n\nThe highest temperature was 29\u00ba, the lowest temperature was 16\u00b0, and the range in temperatures was\u00a0 29 - 16 =\u00a0 13\u00ba.\n\n<\/div>\n<\/div>\n\n TRY IT 1<\/p>\n\n<\/header>\n
\n\nThe table shows the test score for a group of fourteen students.\n\nDetermine the highest test score, the lowest test score, and the range\u00a0 in test scores for the group of 14 students.\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Try-It-1-Range-1024x114.png\")
\n\nShow answer<\/summary>The highest score was 89%, the lowest score was 56%, and the range in scores was 33%.\n\n<\/details><\/div>\n<\/div>\nAn advantage of using the range as a measure of dispersion is that it involves a simple calculation. A disadvantage is that the range only provides a measure between the highest and lowest values so it disregards all other data values. If the highest or lowest data value is an [pb_glossary id=\"1846\"]outlier[\/pb_glossary] then the range will not provide a true measure of the spread in the typical values.\n
\n EXAMPLE 2<\/p>\n\n<\/header>\n
\n\nA student wrote five tests and earned the following five scores:\n92%\u00a0 \u00a095%\u00a0 \u00a089%\u00a0 \u00a094%\u00a0 \u00a035%<\/p>\n\n
- \n \t
- Determine the mean, median and the range for these five scores.<\/li>\n \t
- Which of the five scores is an outlier?<\/li>\n \t
- Remove the outlier and recalculate the mean, median and the range for the four remaining scores.<\/li>\n \t
- Comparing the results for the five scores versus four scores, which of the three measures was least impacted by the outlier?<\/li>\n \t
- Comparing the results for the five scores versus four scores, which of the three measures was most impacted by the outlier?<\/li>\n<\/ol>\nSolution<\/strong>\n\n1. The student's mean score: (92 + 95 + 89 + 94 + 35) \/ 5 = 81%\n\nThe median score is 92%.\n\nThe range in marks is 95% - 35% = 60%.\n\n2. The score of 35% is an outlier.\n\n3. The student's mean score: (92 + 95 + 89 + 94) \/ 4 = 92.5%\n\nThe median score is 93%.\n\nThe range in marks is 95% - 89% = 6%.\n\n4. The median was least impacted by the removal of the outlier.\n\n5. The range was most impacted by the outlier.\n\n<\/div>\n \n\n<\/div>\n
\n TRY IT 2<\/p>\n\n<\/header>\n
\n\nThe following seven values are salaries at a local computer company.\n$62,000\u00a0 \u00a0 \u00a0$95, 000\u00a0 \u00a0$120,000\u00a0 \u00a0$101, 000\u00a0 \u00a0 \u00a0$99,000\u00a0 \u00a0$98,000\u00a0 \u00a0 \u00a0$110,000<\/p>\n\n
- \n \t
- Determine the mean, median and the range for these seven salaries.<\/li>\n \t
- Which of the seven salaries is an outlier?<\/li>\n \t
- Remove the outlier and recalculate the mean, median and the range for the six remaining salaries.<\/li>\n \t
- Comparing the results for the seven versus six salaries, which of the three measures was least impacted by the outlier?<\/li>\n \t
- Comparing the results for the seven versus six salaries, which of the three measures was most impacted by the outlier?<\/li>\n<\/ol>\n
Show answer<\/summary>\n
- \n \t
- mean salary $97,857; median salary $99,000;\u00a0 range in salaries $58,000<\/li>\n \t
- \u00a0$62,000<\/li>\n \t
- mean salary $103, 833; median salary $100,000;\u00a0 range in salaries $25,000<\/li>\n \t
- the median<\/li>\n \t
- the range<\/li>\n<\/ol>\n<\/details><\/div>\n<\/div>\nRefer back to Example 2 and the measures that were calculated for five test scores.The student's mean score is 81% and the range in marks is 60%.\u00a0 The range of 60% does not capture the fact that if the outlier is removed then there is a spread of only 6% for the four remaining data values. The range depends on only the highest<\/strong> and lowest<\/strong> data values. The existence of an outlier can result in a misleading representation of the spread in data values.\n\nAn alternative measure of dispersion is called the standard deviation<\/strong>. It depends on all<\/strong> data values rather than on only the highest and lowest data values.\n
Standard Deviation<\/h1>\nStandard deviation\u00a0 measures the\u00a0 dispersion of the data values around the mean<\/strong>. Unlike the range, its value depends on every data value in the data set. The standard deviation is found by determining how much each data value differs from the mean.\n\nWhat does the standard deviation actually tell us? Consider two sets of test scores:\n\nSet A:\u00a0 \u00a0 76%\u00a0 \u00a074%\u00a0 \u00a086%\u00a0 \u00a084%\u00a0 \u00a085%\n\nSet\u00a0<\/span>B:\u00a0 \u00a0 53%\u00a0 \u00a095%\u00a0 \u00a062%\u00a0 \u00a099%\u00a0 \u00a096%<\/span>\n\nRefer to Figure 1<\/a>. For both sets the mean is 81%. If we plot the scores (indicated by the *<\/strong>) on a scale of 0% to 100% we see that the scores in Set A are much less spread out around\u00a0 the mean. The scores from\u00a0 Set B are much more dispersed.<\/a>\n\n[caption id=\"attachment_1760\" align=\"aligncenter\" width=\"1024\"]
![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Sets-AB-image-1024x417.png\" "\\"Created")
Fig. 1[\/caption]\n\nIf we compare the standard deviations for the two data sets we should find that although the mean is the same, the standard deviation for Set B will be greater since the data is more widely spread out from the mean.\nPopulation versus Sample Standard Deviation<\/h1>\nWhen working with standard deviation it is important to distinguish whether you are working with the entire population or a sample of the population.\u00a0Statisticians generally survey a [pb_glossary id=\"1847\"]sample[\/pb_glossary] of the population because it is often impossible to survey the entire population.\u00a0 As an example, assume your university wants to determine food preferences for its entire student body. The population would be all students enrolled in the university. Rather than attempt to survey every student, the university will survey only a [pb_glossary id=\"1847\"]sample[\/pb_glossary].\n\nThe symbolic representation of standard deviation is different for a population versus a sample. If\u00a0 you are working with an entire population,<\/strong> the symbol for standard deviation is the Greek letter sigma, \u03c3.<\/strong> The symbol for the standard deviation of a sample<\/strong> is\u00a0 s<\/em><\/strong>. Similarly, the calculation of standard deviation is different for a population versus a sample.\u00a0 Unless otherwise indicated in this chapter, we will assume that we are working with\u00a0 a sample, rather than an entire population.\n
\n Standard Deviation Formula<\/p>\n\n<\/header>\n
\n\nStandard Deviation<\/strong>\u00a0Formula for a Population<\/strong>\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-pop-std-dev-formula.png\")
\n\nThe Greek letter sigma\u00a0 \u2211 is the summation symbol. It indicates that all of the values (x \u2013 \u03bc)2<\/sup> must be added.\n\n \n\nStandard Deviation<\/strong>\u00a0Formula for a Sample<\/strong>\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-sample-std-dev-formula.png\")
\n\n<\/div>\n<\/div>\nCalculating the Standard Deviation<\/h1>\nWhen calculating the standard deviation with the aid of a scientific calculator it is helpful to record the steps using a table.\n
\n Calculating Standard Deviation<\/p>\n\n<\/header>\n
\n\nWe will use the formula for finding the standard deviation of a sample:\n\n![](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-sample-std-dev-formula-1.png\")
\n\nTo determine the standard deviation for a sample we will use the following steps :\n- \n \t
- Find the mean<\/li>\n \t
- Create a table with three columns:\u00a0 \u00a0 data value,\u00a0 \u00a0 data value - mean,\u00a0 \u00a0 \u00a0(data value - mean)2<\/sup><\/li>\n \t
- Fill in the data value<\/strong> column with all values from the sample<\/li>\n \t
- Subtract the mean from each data value:\u00a0 (data value - mean)<\/li>\n \t
- Square the results from step 4:\u00a0 \u00a0(data value - mean)2<\/sup><\/li>\n \t
- Sum the results in column 3 (from step 5)<\/li>\n \t
- Divide the sum (from step 6) by (n - 1)<\/li>\n \t
- Find the square root of the result in step 7<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
\n EXAMPLE 3<\/p>\n\n<\/header>\n
\n\na) Determine the standard deviation for the sample set A test scores:\u00a0 76%\u00a0 \u00a074%\u00a0 \u00a086%\u00a0 \u00a084%\u00a0 \u00a085%<\/span>\n\nb) Determine the standard deviation for the sample set B test scores: \u00a0 \u00a0 53%\u00a0 \u00a095%\u00a0 \u00a062%\u00a0 \u00a099%\u00a0 \u00a096%<\/span>\n\nc) Compare the means and standard deviations for Set A and Set B. Which set is more spread out (dispersed)?\n\nSolution<\/strong>\n\na)<\/span>\n\n<\/div>\n\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Example-3-1024x701.png\")
\n\nThe standard deviation for Set A is 5.57 (rounded to 2 decimal places)\n\nb)\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Example-4-1024x699.png\")
\n\nThe standard deviation for Set B is 21.74 (rounded to 2 decimal places).\n\nc)\n\nIn comparing the two data sets A and B:\n- \n \t
- The means are the same value of\u00a0 81.<\/li>\n \t
- The standard deviation for Set A is\u00a0 s = 5.57\u00a0 and for Set B it is\u00a0 s = 21.74.\u00a0 \u00a0The much larger standard deviation for Set B indicates that there is a much greater spread in the data values around the mean of 81%.<\/li>\n<\/ul>\n<\/div>\n<\/div>\nNote that the sum of the middle column (data value - mean) is 0. This will always be the case. That is why we must square the values before we add them, as is done in column 3.\n
\n TRY IT 3<\/p>\n\n<\/header>\n
\n\nThe average high temperatures (in \u00baC) for one week in April for two different cities in Canada were as follows:\n\nCity A:\u00a0 \u00a0 15\u00a0 \u00a0 \u00a0 19\u00a0 \u00a0 \u00a022\u00a0 \u00a0 26\u00a0 \u00a0 21\u00a0 \u00a0 19\u00a0 \u00a0 18\n\nCity B:\u00a0 \u00a0 \u00a06\u00a0 \u00a0 \u00a0 \u00a09\u00a0 \u00a0 \u00a0 15\u00a0 \u00a0 18\u00a0 \u00a0 \u00a020\u00a0 \u00a0 19\u00a0 \u00a0 21\n\na) Calculate the mean high temperature (if necessary round to 2 decimal places) for each city. Which of the two cities appears to have a wider spread in temperatures around their means?\n\nb) Calculate the standard deviation for each temperature set (if necessary round to 2 decimal places) to see if your observation is correct.\n\nShow answer<\/summary>The mean for City A is 20 \u02daC and the mean for City B was 15.43\u02daC.\n\nThe standard deviation for City A is 3.46 and for City B is 5.80. City B\u2019s temperatures are more widely spread out from the mean temperature.\n\n<\/details><\/div>\n<\/div>\nWe have seen that two measures of spread or dispersion are the [pb_glossary id=\"1848\"]range[\/pb_glossary]\u00a0 and the [pb_glossary id=\"1850\"]standard deviation[\/pb_glossary]. Although the range is a much simpler calculation, it only takes into consideration the highest and lowest data values. The existence of an outlier can result in a range that is not truly indicative of the spread in data values. The standard deviation is a more complex calculation but takes into consideration all data values. It is important to note that technology is often used to calculate the standard deviation which eliminates the need for tedious calculations.\n
\n EXAMPLE 4<\/p>\n\n<\/header>\n
\n\nIn Example 1 the range for the following set of temperature was determined to be\u00a0 13 \u00baC.\n\na) Determine the mean and the standard deviation.\n\nb) Explain why the values for the range and standard deviation are different.\n\n\n\nSolution<\/strong>\n\na) The mean (average) is 22.29 \u02daC and the standard deviation is 3.86.\nb) The reason for the difference in values is that the range only tells us the difference between the highest and lowest temperature whereas the standard deviation tells us how widespread the temperatures are in relation to the mean temperature of 22.29 \u02daC\n\n<\/div>\n<\/div>\n\n TRY IT 4<\/p>\n\n<\/header>\n
\n\nRefer back to TRY IT 1 and the test scores for a group of 14 students. The range was determined to be 33.\n\nDetermine the mean (rounded to the nearest whole number) and the standard deviation (rounded to the nearest 2 decimal places) for this set of test scores.\n\n\n\n \n\nShow answer<\/summary>The mean is 72 and the standard deviation is 8.19.\n\n<\/details><\/div>\n<\/div>\n
Histograms and the Dispersion of Data Values<\/strong><\/h6>\nWe have seen that for two data sets with the same mean, when the standard deviation is larger the data values are more spread out. A [pb_glossary id=\"1851\"]histogram[\/pb_glossary] can be used to illustrate the spread of data values.\n\nConsider a dance competition where teams comprised of seven dancers compete for the prize money in several different dance categories. There were four teams entered in the elite category. The dancers must be between the ages of\u00a0 18-24 years old. The age breakdown for the members of the four teams in the elite category is:\n\nTeam Unity:\u00a0 all seven dancers are age 21\n\nTeam Harmony:\u00a0 2 dancers are 20, 3 dancers are 21 and 2 dancers\u00a0 are 22\n\nTeam Mix:\u00a0 1 dancer is 19, 2 dancers are 20, 1 dancer is 21, 2 dancers are 22 and 1 dancer is 23\n\nTeam Extend: 3 dancers are 18, 1 dancer is 21, and 3 dancers are 24\n\nThe mean<\/strong> age for all four teams is 21 but the standard deviations for each of the four teams are different. Team Unity has a standard deviation of 0, Team Harmony has a standard deviation of 0.82, Team Mix has a standard deviation of 1.412 and Team Extend has a standard deviation of\u00a0 3.\n\nThe histograms for each of the teams appears in Figures 2a through Figures 2d below.\n\n \n\n[caption id=\"attachment_1760\" align=\"aligncenter\" width=\"505\"]
![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.3-Team-Unity-Histogram.png\")
Fig. 2a. Team Unity with\u00a0 s = 0<\/strong>[\/caption]\n\nIn Figure 2a\u00a0 Team Unity has a standard deviation of 0 since all ages are the same. None of the ages spread out from the mean of 21.\n\n \n\n[caption id=\"attachment_1760\" align=\"aligncenter\" width=\"486\"]![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.3-Team-Harmony-Histogram.png\")
Fig. 2b. Team Harmony with\u00a0 s = 0.82<\/strong>[\/caption]\n\nIn Figure 2b Team Harmony has a standard deviation of 0.82 years. The histogram illustrates that the ages are closely clustered around the mean of 21.\n\n \n\n[caption id=\"attachment_1760\" align=\"aligncenter\" width=\"490\"]![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.3-Team-Mix-Histogram.png\")
Fig. 2c Team Mix\u00a0 with\u00a0 s = 1.41<\/strong>[\/caption]\n\nIn Figure 2c Team Mix has a standard deviation of 1.41.\u00a0 The histogram illustrates that the data (age) spread is greater than for Team Unity and Team Harmony.\n\n \n\n[caption id=\"attachment_1760\" align=\"aligncenter\" width=\"422\"]![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.3-Team-Extend-Histogram.png\")
Fig. 2d. Team Extend\u00a0 \u00a0with\u00a0 s = 3<\/strong>[\/caption]\n\nIn Figure 2d Team Extend\u00a0 has a standard deviation of 3. The histogram clearly illustrates that the ages for this team are the most spread out from the mean of 21.\n\n EXAMPLE 5<\/p>\n\n<\/header>\n
\n\nTen participants in a group fitness class were asked to rank the class on a scale from 1 to 5. Determine the mean and standard deviation for the evaluation scores as depicted in the histogram below.\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Example5-4.png\")
<\/a>\n\nSolution<\/strong>\n\n<\/div>\n
- Fill in the data value<\/strong> column with all values from the sample<\/li>\n \t
![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-ex-5-solution-1024x1019.png\")
<\/a>\n\nThe mean score is 3 and the standard deviation is 1.49\n\n<\/div>\n<\/div>\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Try-It-5-Histogram-1.png\")
\n\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Ex-Set1-Table-1024x118.png\")
\n![\"\"](\"https:\/\/opentextbc.ca\/oerdiscipline\/wp-content\/uploads\/sites\/361\/2021\/08\/8.2-Ex-Set6-Histograms-1024x441.png\")
\n