{"id":39,"date":"2021-06-22T15:30:11","date_gmt":"2021-06-22T15:30:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/measures-of-the-spread-of-data\/"},"modified":"2023-12-05T08:57:52","modified_gmt":"2023-12-05T08:57:52","slug":"measures-of-the-spread-of-data","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/measures-of-the-spread-of-data\/","title":{"raw":"Measures of Spread: Standard Deviation","rendered":"Measures of Spread: Standard Deviation"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul id=\"list123523\">\r\n \t<li>Calculate standard deviation for a set of data using technology<\/li>\r\n<\/ul>\r\n<\/div>\r\nAn important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The <strong>standard deviation<\/strong> is a number that measures how far data values are from their mean.\r\n\r\nThe standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean.\r\n<h3>The standard deviation<\/h3>\r\n<ul>\r\n \t<li>provides a measure of the overall variation in a data set, and<\/li>\r\n \t<li>can be used to determine whether a particular data value is close to or far from the mean.<\/li>\r\n<\/ul>\r\n<h4 data-type=\"title\">The standard deviation provides a measure of the overall variation in a data set<\/h4>\r\nThe standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.\r\n\r\nSuppose that we are studying the amount of time customers wait in line at the checkout at supermarket [latex]A[\/latex] and supermarket [latex]B[\/latex]. The average wait time at both supermarkets is five minutes. At supermarket [latex]A[\/latex], the standard deviation for the wait time is two minutes; at supermarket [latex]B[\/latex] the standard deviation for the wait time is four minutes.\r\n\r\nBecause supermarket [latex]B[\/latex] has a higher standard deviation, we know that there is more variation in the wait times at supermarket [latex]B[\/latex]. Overall, wait times at supermarket [latex]B[\/latex] are more spread out from the average; wait times at supermarket [latex]A[\/latex] are more concentrated near the average.\r\n<h4>The standard deviation can be used to determine whether a data value is close to or far from the mean<\/h4>\r\nSuppose that Rosa and Binh both shop at supermarket [latex]A[\/latex]. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket [latex]A[\/latex], the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.\r\n\r\n<strong>Rosa waits for seven minutes:<\/strong>\r\n<ul>\r\n \t<li>Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.<\/li>\r\n \t<li>Rosa's wait time of seven minutes is <strong>two minutes longer than the average<\/strong> of five minutes.<\/li>\r\n \t<li>Rosa's wait time of seven minutes is <strong>one standard deviation above the average <\/strong>of five minutes.<\/li>\r\n<\/ul>\r\n<strong>Binh waits for one minute.<\/strong>\r\n<ul>\r\n \t<li>One is four minutes less than the average of five; four minutes is equal to two standard deviations.<\/li>\r\n \t<li>Binh's wait time of one minute is <strong>four minutes less than the average<\/strong> of five minutes.<\/li>\r\n \t<li>Binh's wait time of one minute is <strong>two standard deviations below the average<\/strong> of five minutes.<\/li>\r\n \t<li>A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate \"rule of thumb\" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)<\/li>\r\n<\/ul>\r\nThe number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is\u00a0<strong>one<\/strong> standard deviation to the <strong>right<\/strong> of five because [latex]5 + (1)(2) = 7[\/latex].\r\n\r\nIf one were also part of the data set, then one is <strong>two<\/strong> standard deviations to the <strong>left<\/strong> of five because [latex]5 + (\u20132)(2) = 1[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/1ofl-icac027i#fixme#fixme#fixme\" alt=\"This shows a number line in intervals of 1 from 0 to 7.\" \/>\r\n<ul>\r\n \t<li>In general, a <strong>va<\/strong><strong>lue = mean + (#ofSTDEV)(standard deviation)<\/strong><\/li>\r\n \t<li>Where #ofSTDEVs = the number of standard deviations<\/li>\r\n \t<li>#ofSTDEV does not need to be an integer<\/li>\r\n \t<li>One is <strong>two s<\/strong><strong>tandard deviations less than the mean<\/strong> of five because: [latex]1 = 5 + (\u20132)(2)[\/latex]<\/li>\r\n<\/ul>\r\nThe equation <strong>value = mean + (#ofSTDEVs)(standard deviation)<\/strong> can be expressed for a sample and for a population.\r\n<ul>\r\n \t<li>Sample: [latex]\\displaystyle{x}=\\overline{{x}}+[\/latex](# of STDEV)[latex]{({s})}[\/latex]<\/li>\r\n \t<li>Population: [latex]\\displaystyle{x}=\\mu+[\/latex](# of STDEV)[latex]{(\\sigma)}[\/latex]<\/li>\r\n<\/ul>\r\nThe lower case letter [latex]s[\/latex] represents the sample standard deviation and the Greek letter [latex]\u03c3[\/latex] (sigma, lower case) represents the population standard deviation.\r\n\r\nThe symbol [latex]\\displaystyle\\overline{{x}}[\/latex] is the sample mean and the Greek symbol [latex]\u03bc[\/latex] is the population mean.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Finding the Distance Between TWo Points<\/h3>\r\nTo calculate the distance between two points on a number line, take the larger number and subtract the smaller number. When you think about numbers on a number line, zero is in the middle and the numbers to the left are negative and the numbers to the right are positive. The negative numbers are below zero and the positive numbers are above zero. Distance measures how far apart two numbers are from each other, therefore it is always positive. If you calculate a negative distance, that only means the direction of the distance.\r\n\r\n<\/div>\r\n<h2>Calculating the Standard Deviation<\/h2>\r\nIf [latex]x[\/latex] is a number, then the difference \"[latex]x[\/latex] \u2013 mean\" is called its <strong>deviation<\/strong>. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is [latex]x \u2013 \u03bc[\/latex]. For sample data, in symbols a deviation is [latex]\\displaystyle{x}-\\overline{{x}}[\/latex].\r\n\r\nThe procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore, the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter [latex]s[\/latex] represents the sample standard deviation and the Greek letter [latex]\u03c3[\/latex] (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then [latex]s[\/latex] should be a good estimate of [latex]\u03c3[\/latex].\r\n\r\nTo calculate the standard deviation, we need to calculate the variance first. The\u00a0<strong>variance<\/strong> is the <strong>average of the squares of the deviations<\/strong> (the [latex]x[\/latex] \u2013 [latex]\\displaystyle\\overline{{x}}[\/latex] values for a sample, or the [latex]x \u2013 \u03bc[\/latex] values for a population). The symbol [latex]\u03c3^2[\/latex] represents the population variance; the population standard deviation [latex]\u03c3[\/latex] is the square root of the population variance. The symbol [latex]s^2[\/latex] represents the sample variance; the sample standard deviation [latex]s[\/latex] is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.\r\n\r\nIf the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by [latex]N[\/latex], the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by <strong>[latex]n \u2013 1[\/latex]<\/strong>, one less than the number of items in the sample.\r\n\r\nIn the following video an example of calculating the variance and standard deviation of a set of data is presented.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7114972&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qqOyy_NjflU&amp;video_target=tpm-plugin-1x1mrptn-qqOyy_NjflU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Exponential Notation<\/h3>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224353\/CNX_BMath_Figure_10_02_013_img.png\" alt=\"On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.\" \/>\r\nThis is read [latex]a[\/latex] to the [latex]{m}^{\\mathrm{th}}[\/latex] power.\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Exponential Notation<\/h3>\r\nThe symbol for the square root is called a <strong>radical symbol<\/strong> and looks like this: [latex]\\sqrt{\\,\\,\\,}[\/latex]. The expression [latex] \\sqrt{25}[\/latex] is read \u201cthe square root of twenty-five\u201d or \u201cradical twenty-five.\u201d The number that is written under the radical symbol is called the <strong>radicand<\/strong>.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200220\/CNX_CAT_Figure_01_03_002.jpg\" alt=\"The expression: square root of twenty-five is enclosed in a circle. The circle has an arrow pointing to it labeled: Radical expression. The square root symbol has an arrow pointing to it labeled: Radical. The number twenty-five has an arrow pointing to it labeled: Radicand.\" \/>\r\n<h4>\u00a0The square root of a square<\/h4>\r\nFor a nonnegative real number, a, [latex]\\sqrt{a^2}=a[\/latex]\r\n\r\nThe radicand represents the same number being multiplied to itself. For example, for [latex]\\sqrt{25} = \\sqrt{5 \\cdot 5} = 5[\/latex].\r\n\r\n<\/div>\r\n<h2>Formulas for the Sample Standard Deviation<\/h2>\r\n[latex]\\displaystyle{s}=\\sqrt{{\\frac{{\\sum{({x}-\\overline{{x}})}^{{2}}}}{{{n}-{1}}}}}{\\quad\\text{or}\\quad}{s}=\\sqrt{{\\frac{{\\sum{f{{({x}-\\overline{{x}})}}}^{{2}}}}{{{n}-{1}}}}}[\/latex]\r\n<p class=\"p1\">For the sample standard deviation, the denominator is [latex]n \u2013 1[\/latex], that is the sample size MINUS [latex]1[\/latex].<\/p>\r\n\r\n<h2>Formulas for the Population Standard Deviation<\/h2>\r\n[latex]\\displaystyle\\sigma=\\sqrt{{\\frac{{\\sum{({x}-\\mu)}^{{2}}}}{{{N}}}}}{\\quad\\text{or}\\quad}\\sigma=\\sqrt{{\\frac{{\\sum{f{{({x}-\\mu)}}}^{{2}}}}{{{N}}}}}[\/latex]\r\n\r\nFor the population standard deviation, the denominator is [latex]N[\/latex], the number of items in the population.\r\n\r\nIn these formulas, [latex]f[\/latex] represents the frequency with which a value appears. For example, if a value appears once, [latex]f[\/latex] is one. If a value appears three times in the data set or population, [latex]f[\/latex] is three.\r\n<h2>Sampling Variability of a Statistic<\/h2>\r\nThe statistic of a sampling distribution was discussed in\u00a0Descriptive Statistics: Measuring the Center of the Data. How much the statistic varies from one sample to another is known as the <strong>sampling variability of a statistic<\/strong>. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean when you learn about The Central Limit Theorem (not now). The notation for the standard error of the mean is [latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex] where [latex]\u03c3[\/latex] is the standard deviation of the population and [latex]n[\/latex] is the size of the sample.\r\n<div class=\"textbox shaded\">\r\n<h3>Note<\/h3>\r\n<strong>In practice, use a calculator or computer software to calculate the standard deviation. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation [latex]\u03c3_x[\/latex] or [latex]s_x[\/latex] from the summary statistics.<\/strong> We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIn a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a sample of [latex]n = 20[\/latex] fifth grade students. The ages are rounded to the nearest half year:\r\n\r\n[latex]\\displaystyle {9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5;}[\/latex]\r\n\r\nSolution:\r\n\r\n[latex]\\displaystyle\\overline{x} = \\frac{9+9.5(2)+10(4)+10.5(4)+11(6)+11.5(3)}{20}={10.525}[\/latex]\r\nThe average age is [latex]10.53[\/latex] years, rounded to two places.\r\n\r\nThe variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating [latex]s[\/latex].\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Data<\/th>\r\n<th>Freq.<\/th>\r\n<th>Deviations<\/th>\r\n<th>[latex]Deviations^2[\/latex]<\/th>\r\n<th>(Freq.)( [latex]Deviations^2[\/latex])<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]x[\/latex]<\/td>\r\n<td>[latex]f[\/latex]<\/td>\r\n<td>( [latex]x[\/latex] \u2013 [latex]\\displaystyle\\overline{x}[\/latex])<\/td>\r\n<td>( [latex]x[\/latex] \u2013[latex]\\displaystyle\\overline{x}[\/latex])<sup data-redactor-tag=\"sup\">2<\/sup><\/td>\r\n<td>( [latex]f[\/latex])([latex]x[\/latex] \u2013[latex]\\displaystyle\\overline{x}[\/latex])<sup data-redactor-tag=\"sup\">2<\/sup><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]9[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]9 \u2013 10.525 = \u20131.525[\/latex]<\/td>\r\n<td>[latex](\u20131.525)^2 = 2.325625[\/latex]<\/td>\r\n<td>[latex]1 \u00d7 2.325625 = 2.325625[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]9.5[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]9.5 \u2013 10.525 = \u20131.025[\/latex]<\/td>\r\n<td>[latex](\u20131.025)^2 = 1.050625[\/latex]<\/td>\r\n<td>[latex]2 \u00d7 1.050625 = 2.101250[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]10 \u2013 10.525 = \u20130.525[\/latex]<\/td>\r\n<td>[latex](\u20130.525)^2 = 0.275625[\/latex]<\/td>\r\n<td>[latex]4 \u00d7 0.275625 = 1.1025[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]10.5[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]10.5 \u2013 10.525 = \u20130.025[\/latex]<\/td>\r\n<td>[latex](\u20130.025)^2 = 0.000625[\/latex]<\/td>\r\n<td>[latex]4 \u00d7 0.000625 = 0.0025[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]11[\/latex]<\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]11 \u2013 10.525 = 0.475[\/latex]<\/td>\r\n<td>[latex](0.475)^2 = 0.225625[\/latex]<\/td>\r\n<td>[latex]6 \u00d7 0.225625 = 1.35375[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]11.5[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]11.5 \u2013 10.525 = 0.975[\/latex]<\/td>\r\n<td>[latex](0.975)^2 = 0.950625[\/latex]<\/td>\r\n<td>[latex]3 \u00d7 0.950625 = 2.851875[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td><\/td>\r\n<td>The total is [latex]9.7375[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe sample variance, [latex]\\displaystyle{s}^{2}[\/latex], is equal to the sum of the last column [latex](9.7375)[\/latex] divided by the total number of data values minus one [latex](20 \u2013 1)[\/latex]:\r\n[latex]s^2 =\\frac{9.7375}{20-1} =0.5125[\/latex]\r\n\r\nThe <strong>sample standard deviation<\/strong> [latex]s[\/latex] is equal to the square root of the sample variance: [latex]s = \\sqrt{0.5125} = 0.715891[\/latex] which is rounded to two decimal places, [latex]s[\/latex] = 0.72.\r\n\r\n<strong data-redactor-tag=\"strong\">Typically, you do the calculation for the standard deviation on your calculator or computer.<\/strong> The intermediate results are not rounded. This is done for accuracy.\r\n<ul>\r\n \t<li>For the following problems, recall that <strong data-redactor-tag=\"strong\">value = mean + (#ofSTDEVs)(standard deviation)<\/strong>. Verify the mean and standard deviation or a calculator or computer.<\/li>\r\n \t<li>For a sample: [latex]x[\/latex] =[latex]\\displaystyle\\overline{x}[\/latex]\u00a0+ (#ofSTDEVs)([latex]s[\/latex])<\/li>\r\n \t<li>For a population: [latex]x[\/latex] = [latex]\u03bc[\/latex] + (#ofSTDEVs)([latex]\u03c3[\/latex])<\/li>\r\n \t<li>For this example, use [latex]x[\/latex] =[latex]\\displaystyle\\overline{x}[\/latex]\u00a0+ (#ofSTDEVs)([latex]s[\/latex]) because the data is from a sample<\/li>\r\n<\/ul>\r\n<ol>\r\n \t<li>Verify the mean and standard deviation on your calculator or computer.<\/li>\r\n \t<li>Find the value that is one standard deviation above the mean. Find ([latex]\\displaystyle\\overline{x}[\/latex]+ [latex]1s[\/latex]).<\/li>\r\n \t<li>Find the value that is two standard deviations below the mean. Find ([latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u2013 [latex]2s[\/latex]).<\/li>\r\n \t<li>Find the values that are [latex]1.5[\/latex] standard deviations <strong>from<\/strong> (below and above) the mean.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124075\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>USING THE TI-83, 83+, 84, 84+ CALCULATOR\r\n<ul>\r\n \t<li>Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.<\/li>\r\n \t<li>Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.<\/li>\r\n \t<li>Put the data values ([latex]9[\/latex], [latex]9.5[\/latex], [latex]10[\/latex], [latex]10.5[\/latex], [latex]11[\/latex], [latex]11.5[\/latex]) into list L1 and the frequencies ([latex]1[\/latex], [latex]2[\/latex], [latex]4[\/latex], [latex]4[\/latex], [latex]6[\/latex], [latex]3[\/latex]) into list L2. Use the arrow keys to move around.<\/li>\r\n \t<li>Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.<\/li>\r\n \t<li>[latex]\\displaystyle\\overline{x}[\/latex]\u00a0= [latex]10.525[\/latex]<\/li>\r\n \t<li>Use Sx because this is sample data (not a population): Sx=[latex]0.715891[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>([latex]\\displaystyle\\overline{x}+ 1s) = 10.53 + (1)(0.72) = 11.25[\/latex]<\/li>\r\n \t<li>([latex]\\displaystyle\\overline{x}\u2013 2s) = 10.53 \u2013 (2)(0.72) = 9.09[\/latex]\r\n<ul>\r\n \t<li>([latex]\\displaystyle\\overline{x}\u2013 1.5s) = 10.53 \u2013 (1.5)(0.72) = 9.45[\/latex]<\/li>\r\n \t<li>([latex]\\displaystyle\\overline{x}+ 1.5s) = 10.53 + (1.5)(0.72) = 11.61[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nOn a baseball team, the ages of each of the players are as follows:\r\n\r\n[latex]\\displaystyle {21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40}[\/latex]\r\n\r\nUse your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.\r\n\r\n[reveal-answer q=\"124076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124076\"]\r\n[latex]\u03bc = 30.68[\/latex]\r\n\r\n[latex]s = 6.09[\/latex]\r\n\r\n([latex]\\displaystyle\\overline{x}+ 2s) = 30.68 + (2)(6.09) = 42.86[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Explanation of the Standard Deviation Calculation Shown in the Table<\/h2>\r\nThe deviations show how spread out the data are about the mean. The data value [latex]11.5[\/latex] is farther from the mean than is the data value [latex]11[\/latex] which is indicated by the deviations [latex]0.97[\/latex] and [latex]0.47[\/latex]. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is [latex]\u20131.525[\/latex] for the data value nine. <strong data-redactor-tag=\"strong\">If you add the deviations, the sum is always zero. <\/strong>(For Example 1, there are [latex]n = 20[\/latex] deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.\r\n\r\nThe variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.\r\n\r\nNotice that instead of dividing by [latex]n= 20[\/latex], the calculation divided by [latex]n \u2013 1 = 20 \u2013 1 = 19[\/latex] because the data is a sample. For the sample variance, we divide by the sample size minus one ([latex]n \u2013 1[\/latex]). Why not divide by [latex]n[\/latex]? The answer has to do with the population variance. <strong data-redactor-tag=\"strong\">The sample variance is an estimate of the population variance.<\/strong> Based on the theoretical mathematics that lies behind these calculations, dividing by ([latex]n \u2013 1[\/latex]) gives a better estimate of the population variance.\r\n<div class=\"textbox shaded\">\r\n<h3>Note<\/h3>\r\nYour concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.\r\n\r\n<\/div>\r\nThe standard deviation, [latex]s[\/latex] or [latex]\u03c3[\/latex], is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make [latex]s[\/latex] or [latex]\u03c3[\/latex] very large.\r\n\r\nThe standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better \"feel\" for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, <strong data-redactor-tag=\"strong\">always graph your data<\/strong>. Display your data in a histogram or a box plot.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nUse the following data (first exam scores) from Susan Dean's spring pre-calculus class:\r\n\r\n[latex]\\displaystyle {33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100}[\/latex]\r\n<ol>\r\n \t<li>Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.<\/li>\r\n \t<li>Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:\r\n<ol>\r\n \t<li>The sample mean<\/li>\r\n \t<li>The sample standard deviation<\/li>\r\n \t<li>The median<\/li>\r\n \t<li>The first quartile<\/li>\r\n \t<li>The third quartile<\/li>\r\n \t<li>[latex]IQR[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"124077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124077\"]\r\n<ol>\r\n \t<li>\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Data<\/th>\r\n<th>Frequency<\/th>\r\n<th>Relative Frequency<\/th>\r\n<th>Cumulative Relative Frequency<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]33[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]42[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.064[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]49[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]0.065[\/latex]<\/td>\r\n<td>[latex]0.129[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]53[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.161[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]55[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]0.065[\/latex]<\/td>\r\n<td>[latex]0.226[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]61[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.258[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]63[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.29[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]67[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.322[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]68[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]0.065[\/latex]<\/td>\r\n<td>[latex]0.387[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]69[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]0.065[\/latex]<\/td>\r\n<td>[latex]0.452[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]72[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.484[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]73[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.516[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]74[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.548[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]78[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.580[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]80[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.612[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]83[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.644[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]88[\/latex]<\/td>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]0.097[\/latex]<\/td>\r\n<td>[latex]0.741[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]90[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.773[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]92[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.805[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]94[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]0.129[\/latex]<\/td>\r\n<td>[latex]0.934[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]96[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.966[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]100[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]0.032[\/latex]<\/td>\r\n<td>[latex]0.998[\/latex] (Why isn't this value [latex]1[\/latex]?)<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<ol>\r\n \t<li>The sample mean = [latex]73.5[\/latex]<\/li>\r\n \t<li>The sample standard deviation = [latex]17.9[\/latex]<\/li>\r\n \t<li>The median = [latex]73[\/latex]<\/li>\r\n \t<li>The first quartile = [latex]61[\/latex]<\/li>\r\n \t<li>The third quartile = [latex]90[\/latex]<\/li>\r\n \t<li>[latex]IQR = 90 \u2013 61 = 29[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The [latex]x[\/latex]-axis goes from [latex]32.5[\/latex] to [latex]100.5[\/latex]; [latex]y[\/latex]-axis goes from [latex]\u20132.4[\/latex] to [latex]15[\/latex] for the histogram. The number of intervals is five, so the width of an interval is [latex](100.5 \u2013 32.5)[\/latex] divided by five, is equal to [latex]13.6[\/latex]. Endpoints of the intervals are as follows: the starting point is [latex]32.5, 32.5 + 13.6 = 46.1[\/latex], [latex]46.1 + 13.6 = 59.7[\/latex], [latex]59.7 + 13.6 = 73.3[\/latex], [latex]73.3 + 13.6 = 86.9[\/latex], [latex]86.9 + 13.6 = 100.5[\/latex] = the ending value; No data values fall on an interval boundary.<\/li>\r\n \t<li><img class=\"aligncenter wp-image-3394 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/06\/19165655\/f52363d0d4686738d58f1174609c14cf3971f2a3.jpeg\" alt=\"A hybrid image displaying both a histogram and box plot described in detail in the answer solution above.\" width=\"487\" height=\"318\" \/><\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\nThe long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower [latex]50[\/latex]% is greater ([latex]73 \u2013 33 = 40[\/latex]) than the spread in the upper [latex]50[\/latex]% ([latex]100 \u2013 73 = 27[\/latex]). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades ([latex]80[\/latex]s, [latex]90[\/latex]s, and [latex]100[\/latex]). The histogram clearly shows this. The box plot shows us that the middle [latex]50[\/latex]% of the exam scores ([latex]IQR[\/latex] = [latex]29[\/latex]) are Ds, Cs, and Bs. The box plot also shows us that the lower [latex]25[\/latex]% of the exam scores are Ds and Fs.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe following data show the different types of pet food stores in the area carry.\r\n\r\n[latex]\\displaystyle {6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;}[\/latex]\r\nCalculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.\r\n\r\n[reveal-answer q=\"124078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124078\"]\r\n[latex]\u03bc = 9.3[\/latex]\r\n\r\n[latex]s = 2.2[\/latex]\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Standard Deviation of Grouped Frequency Tables<\/h2>\r\nRecall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:\r\n\r\n<em>Mean of Frequency Table<\/em> =[latex]\\displaystyle\\frac{{\\sum(fm)}}{{\\sum(f)}}[\/latex]\r\n\r\nwhere [latex]f[\/latex] = interval frequencies and [latex]m[\/latex] = interval midpoints.\r\n\r\nJust as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how \"unusual\" individual data is compared to the mean.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the standard deviation for the data in the table below.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Class<\/th>\r\n<th>Frequency, [latex]f[\/latex]<\/th>\r\n<th>Midpoint, [latex]m[\/latex]<\/th>\r\n<th>[latex]m^2[\/latex]<\/th>\r\n<th>[latex]\\displaystyle\\overline{x}^2[\/latex]<\/th>\r\n<th>[latex]fm^2[\/latex]<\/th>\r\n<th>Standard Deviation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]0\u20132[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\u20135[\/latex]<\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<td>[latex]4[\/latex]<\/td>\r\n<td>[latex]16[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]96[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6\u20138[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<td>[latex]49[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]490[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]9\u201311[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<td>[latex]100[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]700[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12\u201314[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]13[\/latex]<\/td>\r\n<td>[latex]169[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]15\u201317[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<td>[latex]16[\/latex]<\/td>\r\n<td>[latex]256[\/latex]<\/td>\r\n<td>[latex]7.58[\/latex]<\/td>\r\n<td>[latex]512[\/latex]<\/td>\r\n<td>[latex]3.5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFor this data set, we have the mean, [latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u00a0= [latex]7.58[\/latex] and the standard deviation, [latex]\\displaystyle{s}_{x} = 3.5[\/latex]. This means that a randomly selected data value would be expected to be [latex]3.5[\/latex] units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since [latex]7.58 \u2013 3.5 \u2013 3.5 = 0.58[\/latex]. While the formula for calculating the standard deviation is not complicated, [latex]\\displaystyle{s}_{x}=\\sqrt{{\\frac{{f{(m-\\overline{x})}^{2}}}{{n-1}}}}[\/latex] where [latex]\\displaystyle{s}_{x} = [\/latex]sample standard deviation, [latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u00a0= sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the standard deviation for the data from the previous example\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Class<\/th>\r\n<th>Frequency, [latex]f[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]0\u20132[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\u20135[\/latex]<\/td>\r\n<td>[latex]6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6\u20138[\/latex]<\/td>\r\n<td>[latex]10[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]9\u201311[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12\u201314[\/latex]<\/td>\r\n<td>[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]15\u201317[\/latex]<\/td>\r\n<td>[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFirst, press the <strong>STAT <\/strong>key and select <strong>1:Edit<\/strong>\r\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.08.28-PM.png\"><img class=\"size-full wp-image-499 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214250\/Screen-Shot-2015-06-07-at-6.08.28-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.08.28 PM\" width=\"330\" height=\"217\" \/><\/a>\r\n\r\nInput the midpoint values into <strong data-redactor-tag=\"strong\">L1<\/strong> and the frequencies into <strong data-redactor-tag=\"strong\">L2<\/strong>\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.09.59-PM.png\"><img class=\"aligncenter size-full wp-image-500\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214253\/Screen-Shot-2015-06-07-at-6.09.59-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.09.59 PM\" width=\"328\" height=\"221\" \/><\/a>\r\n\r\nSelect <strong data-redactor-tag=\"strong\">STAT<\/strong>, <strong data-redactor-tag=\"strong\">CALC<\/strong>, and <strong data-redactor-tag=\"strong\">1: 1-Var Stats<\/strong>\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.10.46-PM.png\"><img class=\"aligncenter size-full wp-image-501\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214255\/Screen-Shot-2015-06-07-at-6.10.46-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.10.46 PM\" width=\"327\" height=\"222\" \/><\/a>\r\n\r\nSelect <strong data-redactor-tag=\"strong\">2nd<\/strong> then <strong data-redactor-tag=\"strong\">1<\/strong> then , <strong data-redactor-tag=\"strong\">2nd<\/strong> then <strong data-redactor-tag=\"strong\">2<\/strong><strong data-redactor-tag=\"strong\">Enter<\/strong>\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.11.19-PM.png\"><img class=\"aligncenter size-full wp-image-502\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214257\/Screen-Shot-2015-06-07-at-6.11.19-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.11.19 PM\" width=\"330\" height=\"222\" \/><\/a>\r\n\r\nYou will see displayed both a population standard deviation, <em data-redactor-tag=\"em\">\u03c3_x<\/em>, and the sample standard deviation, [latex]s_x[\/latex].\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list123523\">\n<li>Calculate standard deviation for a set of data using technology<\/li>\n<\/ul>\n<\/div>\n<p>An important characteristic of any set of data is the variation in the data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The <strong>standard deviation<\/strong> is a number that measures how far data values are from their mean.<\/p>\n<p>The standard deviation provides a numerical measure of the overall amount of variation in a data set, and can be used to determine whether a particular data value is close to or far from the mean.<\/p>\n<h3>The standard deviation<\/h3>\n<ul>\n<li>provides a measure of the overall variation in a data set, and<\/li>\n<li>can be used to determine whether a particular data value is close to or far from the mean.<\/li>\n<\/ul>\n<h4 data-type=\"title\">The standard deviation provides a measure of the overall variation in a data set<\/h4>\n<p>The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation.<\/p>\n<p>Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket [latex]A[\/latex] and supermarket [latex]B[\/latex]. The average wait time at both supermarkets is five minutes. At supermarket [latex]A[\/latex], the standard deviation for the wait time is two minutes; at supermarket [latex]B[\/latex] the standard deviation for the wait time is four minutes.<\/p>\n<p>Because supermarket [latex]B[\/latex] has a higher standard deviation, we know that there is more variation in the wait times at supermarket [latex]B[\/latex]. Overall, wait times at supermarket [latex]B[\/latex] are more spread out from the average; wait times at supermarket [latex]A[\/latex] are more concentrated near the average.<\/p>\n<h4>The standard deviation can be used to determine whether a data value is close to or far from the mean<\/h4>\n<p>Suppose that Rosa and Binh both shop at supermarket [latex]A[\/latex]. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket [latex]A[\/latex], the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.<\/p>\n<p><strong>Rosa waits for seven minutes:<\/strong><\/p>\n<ul>\n<li>Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.<\/li>\n<li>Rosa&#8217;s wait time of seven minutes is <strong>two minutes longer than the average<\/strong> of five minutes.<\/li>\n<li>Rosa&#8217;s wait time of seven minutes is <strong>one standard deviation above the average <\/strong>of five minutes.<\/li>\n<\/ul>\n<p><strong>Binh waits for one minute.<\/strong><\/p>\n<ul>\n<li>One is four minutes less than the average of five; four minutes is equal to two standard deviations.<\/li>\n<li>Binh&#8217;s wait time of one minute is <strong>four minutes less than the average<\/strong> of five minutes.<\/li>\n<li>Binh&#8217;s wait time of one minute is <strong>two standard deviations below the average<\/strong> of five minutes.<\/li>\n<li>A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate &#8220;rule of thumb&#8221; than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)<\/li>\n<\/ul>\n<p>The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is\u00a0<strong>one<\/strong> standard deviation to the <strong>right<\/strong> of five because [latex]5 + (1)(2) = 7[\/latex].<\/p>\n<p>If one were also part of the data set, then one is <strong>two<\/strong> standard deviations to the <strong>left<\/strong> of five because [latex]5 + (\u20132)(2) = 1[\/latex].<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/1ofl-icac027i#fixme#fixme#fixme\" alt=\"This shows a number line in intervals of 1 from 0 to 7.\" \/><\/p>\n<ul>\n<li>In general, a <strong>va<\/strong><strong>lue = mean + (#ofSTDEV)(standard deviation)<\/strong><\/li>\n<li>Where #ofSTDEVs = the number of standard deviations<\/li>\n<li>#ofSTDEV does not need to be an integer<\/li>\n<li>One is <strong>two s<\/strong><strong>tandard deviations less than the mean<\/strong> of five because: [latex]1 = 5 + (\u20132)(2)[\/latex]<\/li>\n<\/ul>\n<p>The equation <strong>value = mean + (#ofSTDEVs)(standard deviation)<\/strong> can be expressed for a sample and for a population.<\/p>\n<ul>\n<li>Sample: [latex]\\displaystyle{x}=\\overline{{x}}+[\/latex](# of STDEV)[latex]{({s})}[\/latex]<\/li>\n<li>Population: [latex]\\displaystyle{x}=\\mu+[\/latex](# of STDEV)[latex]{(\\sigma)}[\/latex]<\/li>\n<\/ul>\n<p>The lower case letter [latex]s[\/latex] represents the sample standard deviation and the Greek letter [latex]\u03c3[\/latex] (sigma, lower case) represents the population standard deviation.<\/p>\n<p>The symbol [latex]\\displaystyle\\overline{{x}}[\/latex] is the sample mean and the Greek symbol [latex]\u03bc[\/latex] is the population mean.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Finding the Distance Between TWo Points<\/h3>\n<p>To calculate the distance between two points on a number line, take the larger number and subtract the smaller number. When you think about numbers on a number line, zero is in the middle and the numbers to the left are negative and the numbers to the right are positive. The negative numbers are below zero and the positive numbers are above zero. Distance measures how far apart two numbers are from each other, therefore it is always positive. If you calculate a negative distance, that only means the direction of the distance.<\/p>\n<\/div>\n<h2>Calculating the Standard Deviation<\/h2>\n<p>If [latex]x[\/latex] is a number, then the difference &#8220;[latex]x[\/latex] \u2013 mean&#8221; is called its <strong>deviation<\/strong>. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is [latex]x \u2013 \u03bc[\/latex]. For sample data, in symbols a deviation is [latex]\\displaystyle{x}-\\overline{{x}}[\/latex].<\/p>\n<p>The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore, the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter [latex]s[\/latex] represents the sample standard deviation and the Greek letter [latex]\u03c3[\/latex] (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then [latex]s[\/latex] should be a good estimate of [latex]\u03c3[\/latex].<\/p>\n<p>To calculate the standard deviation, we need to calculate the variance first. The\u00a0<strong>variance<\/strong> is the <strong>average of the squares of the deviations<\/strong> (the [latex]x[\/latex] \u2013 [latex]\\displaystyle\\overline{{x}}[\/latex] values for a sample, or the [latex]x \u2013 \u03bc[\/latex] values for a population). The symbol [latex]\u03c3^2[\/latex] represents the population variance; the population standard deviation [latex]\u03c3[\/latex] is the square root of the population variance. The symbol [latex]s^2[\/latex] represents the sample variance; the sample standard deviation [latex]s[\/latex] is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.<\/p>\n<p>If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by [latex]N[\/latex], the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by <strong>[latex]n \u2013 1[\/latex]<\/strong>, one less than the number of items in the sample.<\/p>\n<p>In the following video an example of calculating the variance and standard deviation of a set of data is presented.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7114972&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=qqOyy_NjflU&amp;video_target=tpm-plugin-1x1mrptn-qqOyy_NjflU\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Exponential Notation<\/h3>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224353\/CNX_BMath_Figure_10_02_013_img.png\" alt=\"On the left side, a raised to the m is shown. The m is labeled in blue as an exponent. The a is labeled in red as the base. On the right, it says a to the m means multiply m factors of a. Below this, it says a to the m equals a times a times a times a, with m factors written below in blue.\" \/><br \/>\nThis is read [latex]a[\/latex] to the [latex]{m}^{\\mathrm{th}}[\/latex] power.<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Exponential Notation<\/h3>\n<p>The symbol for the square root is called a <strong>radical symbol<\/strong> and looks like this: [latex]\\sqrt{\\,\\,\\,}[\/latex]. The expression [latex]\\sqrt{25}[\/latex] is read \u201cthe square root of twenty-five\u201d or \u201cradical twenty-five.\u201d The number that is written under the radical symbol is called the <strong>radicand<\/strong>.<br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200220\/CNX_CAT_Figure_01_03_002.jpg\" alt=\"The expression: square root of twenty-five is enclosed in a circle. The circle has an arrow pointing to it labeled: Radical expression. The square root symbol has an arrow pointing to it labeled: Radical. The number twenty-five has an arrow pointing to it labeled: Radicand.\" \/><\/p>\n<h4>\u00a0The square root of a square<\/h4>\n<p>For a nonnegative real number, a, [latex]\\sqrt{a^2}=a[\/latex]<\/p>\n<p>The radicand represents the same number being multiplied to itself. For example, for [latex]\\sqrt{25} = \\sqrt{5 \\cdot 5} = 5[\/latex].<\/p>\n<\/div>\n<h2>Formulas for the Sample Standard Deviation<\/h2>\n<p>[latex]\\displaystyle{s}=\\sqrt{{\\frac{{\\sum{({x}-\\overline{{x}})}^{{2}}}}{{{n}-{1}}}}}{\\quad\\text{or}\\quad}{s}=\\sqrt{{\\frac{{\\sum{f{{({x}-\\overline{{x}})}}}^{{2}}}}{{{n}-{1}}}}}[\/latex]<\/p>\n<p class=\"p1\">For the sample standard deviation, the denominator is [latex]n \u2013 1[\/latex], that is the sample size MINUS [latex]1[\/latex].<\/p>\n<h2>Formulas for the Population Standard Deviation<\/h2>\n<p>[latex]\\displaystyle\\sigma=\\sqrt{{\\frac{{\\sum{({x}-\\mu)}^{{2}}}}{{{N}}}}}{\\quad\\text{or}\\quad}\\sigma=\\sqrt{{\\frac{{\\sum{f{{({x}-\\mu)}}}^{{2}}}}{{{N}}}}}[\/latex]<\/p>\n<p>For the population standard deviation, the denominator is [latex]N[\/latex], the number of items in the population.<\/p>\n<p>In these formulas, [latex]f[\/latex] represents the frequency with which a value appears. For example, if a value appears once, [latex]f[\/latex] is one. If a value appears three times in the data set or population, [latex]f[\/latex] is three.<\/p>\n<h2>Sampling Variability of a Statistic<\/h2>\n<p>The statistic of a sampling distribution was discussed in\u00a0Descriptive Statistics: Measuring the Center of the Data. How much the statistic varies from one sample to another is known as the <strong>sampling variability of a statistic<\/strong>. You typically measure the sampling variability of a statistic by its standard error. The standard error of the mean is an example of a standard error. It is a special standard deviation and is known as the standard deviation of the sampling distribution of the mean. You will cover the standard error of the mean when you learn about The Central Limit Theorem (not now). The notation for the standard error of the mean is [latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex] where [latex]\u03c3[\/latex] is the standard deviation of the population and [latex]n[\/latex] is the size of the sample.<\/p>\n<div class=\"textbox shaded\">\n<h3>Note<\/h3>\n<p><strong>In practice, use a calculator or computer software to calculate the standard deviation. If you are using a TI-83, 83+, 84+ calculator, you need to select the appropriate standard deviation [latex]\u03c3_x[\/latex] or [latex]s_x[\/latex] from the summary statistics.<\/strong> We will concentrate on using and interpreting the information that the standard deviation gives us. However you should study the following step-by-step example to help you understand how the standard deviation measures variation from the mean. (The calculator instructions appear at the end of this example.)<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>In a fifth grade class, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a sample of [latex]n = 20[\/latex] fifth grade students. The ages are rounded to the nearest half year:<\/p>\n<p>[latex]\\displaystyle {9; 9.5; 9.5; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5;}[\/latex]<\/p>\n<p>Solution:<\/p>\n<p>[latex]\\displaystyle\\overline{x} = \\frac{9+9.5(2)+10(4)+10.5(4)+11(6)+11.5(3)}{20}={10.525}[\/latex]<br \/>\nThe average age is [latex]10.53[\/latex] years, rounded to two places.<\/p>\n<p>The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating [latex]s[\/latex].<\/p>\n<table>\n<thead>\n<tr>\n<th>Data<\/th>\n<th>Freq.<\/th>\n<th>Deviations<\/th>\n<th>[latex]Deviations^2[\/latex]<\/th>\n<th>(Freq.)( [latex]Deviations^2[\/latex])<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]x[\/latex]<\/td>\n<td>[latex]f[\/latex]<\/td>\n<td>( [latex]x[\/latex] \u2013 [latex]\\displaystyle\\overline{x}[\/latex])<\/td>\n<td>( [latex]x[\/latex] \u2013[latex]\\displaystyle\\overline{x}[\/latex])<sup data-redactor-tag=\"sup\">2<\/sup><\/td>\n<td>( [latex]f[\/latex])([latex]x[\/latex] \u2013[latex]\\displaystyle\\overline{x}[\/latex])<sup data-redactor-tag=\"sup\">2<\/sup><\/td>\n<\/tr>\n<tr>\n<td>[latex]9[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]9 \u2013 10.525 = \u20131.525[\/latex]<\/td>\n<td>[latex](\u20131.525)^2 = 2.325625[\/latex]<\/td>\n<td>[latex]1 \u00d7 2.325625 = 2.325625[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]9.5[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]9.5 \u2013 10.525 = \u20131.025[\/latex]<\/td>\n<td>[latex](\u20131.025)^2 = 1.050625[\/latex]<\/td>\n<td>[latex]2 \u00d7 1.050625 = 2.101250[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]10 \u2013 10.525 = \u20130.525[\/latex]<\/td>\n<td>[latex](\u20130.525)^2 = 0.275625[\/latex]<\/td>\n<td>[latex]4 \u00d7 0.275625 = 1.1025[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]10.5[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]10.5 \u2013 10.525 = \u20130.025[\/latex]<\/td>\n<td>[latex](\u20130.025)^2 = 0.000625[\/latex]<\/td>\n<td>[latex]4 \u00d7 0.000625 = 0.0025[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]11[\/latex]<\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]11 \u2013 10.525 = 0.475[\/latex]<\/td>\n<td>[latex](0.475)^2 = 0.225625[\/latex]<\/td>\n<td>[latex]6 \u00d7 0.225625 = 1.35375[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]11.5[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]11.5 \u2013 10.525 = 0.975[\/latex]<\/td>\n<td>[latex](0.975)^2 = 0.950625[\/latex]<\/td>\n<td>[latex]3 \u00d7 0.950625 = 2.851875[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td><\/td>\n<td>The total is [latex]9.7375[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The sample variance, [latex]\\displaystyle{s}^{2}[\/latex], is equal to the sum of the last column [latex](9.7375)[\/latex] divided by the total number of data values minus one [latex](20 \u2013 1)[\/latex]:<br \/>\n[latex]s^2 =\\frac{9.7375}{20-1} =0.5125[\/latex]<\/p>\n<p>The <strong>sample standard deviation<\/strong> [latex]s[\/latex] is equal to the square root of the sample variance: [latex]s = \\sqrt{0.5125} = 0.715891[\/latex] which is rounded to two decimal places, [latex]s[\/latex] = 0.72.<\/p>\n<p><strong data-redactor-tag=\"strong\">Typically, you do the calculation for the standard deviation on your calculator or computer.<\/strong> The intermediate results are not rounded. This is done for accuracy.<\/p>\n<ul>\n<li>For the following problems, recall that <strong data-redactor-tag=\"strong\">value = mean + (#ofSTDEVs)(standard deviation)<\/strong>. Verify the mean and standard deviation or a calculator or computer.<\/li>\n<li>For a sample: [latex]x[\/latex] =[latex]\\displaystyle\\overline{x}[\/latex]\u00a0+ (#ofSTDEVs)([latex]s[\/latex])<\/li>\n<li>For a population: [latex]x[\/latex] = [latex]\u03bc[\/latex] + (#ofSTDEVs)([latex]\u03c3[\/latex])<\/li>\n<li>For this example, use [latex]x[\/latex] =[latex]\\displaystyle\\overline{x}[\/latex]\u00a0+ (#ofSTDEVs)([latex]s[\/latex]) because the data is from a sample<\/li>\n<\/ul>\n<ol>\n<li>Verify the mean and standard deviation on your calculator or computer.<\/li>\n<li>Find the value that is one standard deviation above the mean. Find ([latex]\\displaystyle\\overline{x}[\/latex]+ [latex]1s[\/latex]).<\/li>\n<li>Find the value that is two standard deviations below the mean. Find ([latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u2013 [latex]2s[\/latex]).<\/li>\n<li>Find the values that are [latex]1.5[\/latex] standard deviations <strong>from<\/strong> (below and above) the mean.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124075\">Show Solution<\/span><\/p>\n<div id=\"q124075\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>USING THE TI-83, 83+, 84, 84+ CALCULATOR\n<ul>\n<li>Clear lists L1 and L2. Press STAT 4:ClrList. Enter 2nd 1 for L1, the comma (,), and 2nd 2 for L2.<\/li>\n<li>Enter data into the list editor. Press STAT 1:EDIT. If necessary, clear the lists by arrowing up into the name. Press CLEAR and arrow down.<\/li>\n<li>Put the data values ([latex]9[\/latex], [latex]9.5[\/latex], [latex]10[\/latex], [latex]10.5[\/latex], [latex]11[\/latex], [latex]11.5[\/latex]) into list L1 and the frequencies ([latex]1[\/latex], [latex]2[\/latex], [latex]4[\/latex], [latex]4[\/latex], [latex]6[\/latex], [latex]3[\/latex]) into list L2. Use the arrow keys to move around.<\/li>\n<li>Press STAT and arrow to CALC. Press 1:1-VarStats and enter L1 (2nd 1), L2 (2nd 2). Do not forget the comma. Press ENTER.<\/li>\n<li>[latex]\\displaystyle\\overline{x}[\/latex]\u00a0= [latex]10.525[\/latex]<\/li>\n<li>Use Sx because this is sample data (not a population): Sx=[latex]0.715891[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li>([latex]\\displaystyle\\overline{x}+ 1s) = 10.53 + (1)(0.72) = 11.25[\/latex]<\/li>\n<li>([latex]\\displaystyle\\overline{x}\u2013 2s) = 10.53 \u2013 (2)(0.72) = 9.09[\/latex]\n<ul>\n<li>([latex]\\displaystyle\\overline{x}\u2013 1.5s) = 10.53 \u2013 (1.5)(0.72) = 9.45[\/latex]<\/li>\n<li>([latex]\\displaystyle\\overline{x}+ 1.5s) = 10.53 + (1.5)(0.72) = 11.61[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>On a baseball team, the ages of each of the players are as follows:<\/p>\n<p>[latex]\\displaystyle {21; 21; 22; 23; 24; 24; 25; 25; 28; 29; 29; 31; 32; 33; 33; 34; 35; 36; 36; 36; 36; 38; 38; 38; 40}[\/latex]<\/p>\n<p>Use your calculator or computer to find the mean and standard deviation. Then find the value that is two standard deviations above the mean.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124076\">Show Solution<\/span><\/p>\n<div id=\"q124076\" class=\"hidden-answer\" style=\"display: none\">\n[latex]\u03bc = 30.68[\/latex]<\/p>\n<p>[latex]s = 6.09[\/latex]<\/p>\n<p>([latex]\\displaystyle\\overline{x}+ 2s) = 30.68 + (2)(6.09) = 42.86[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Explanation of the Standard Deviation Calculation Shown in the Table<\/h2>\n<p>The deviations show how spread out the data are about the mean. The data value [latex]11.5[\/latex] is farther from the mean than is the data value [latex]11[\/latex] which is indicated by the deviations [latex]0.97[\/latex] and [latex]0.47[\/latex]. A positive deviation occurs when the data value is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean. The deviation is [latex]\u20131.525[\/latex] for the data value nine. <strong data-redactor-tag=\"strong\">If you add the deviations, the sum is always zero. <\/strong>(For Example 1, there are [latex]n = 20[\/latex] deviations.) So you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation.<\/p>\n<p>The variance is a squared measure and does not have the same units as the data. Taking the square root solves the problem. The standard deviation measures the spread in the same units as the data.<\/p>\n<p>Notice that instead of dividing by [latex]n= 20[\/latex], the calculation divided by [latex]n \u2013 1 = 20 \u2013 1 = 19[\/latex] because the data is a sample. For the sample variance, we divide by the sample size minus one ([latex]n \u2013 1[\/latex]). Why not divide by [latex]n[\/latex]? The answer has to do with the population variance. <strong data-redactor-tag=\"strong\">The sample variance is an estimate of the population variance.<\/strong> Based on the theoretical mathematics that lies behind these calculations, dividing by ([latex]n \u2013 1[\/latex]) gives a better estimate of the population variance.<\/p>\n<div class=\"textbox shaded\">\n<h3>Note<\/h3>\n<p>Your concentration should be on what the standard deviation tells us about the data. The standard deviation is a number which measures how far the data are spread from the mean. Let a calculator or computer do the arithmetic.<\/p>\n<\/div>\n<p>The standard deviation, [latex]s[\/latex] or [latex]\u03c3[\/latex], is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, the all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make [latex]s[\/latex] or [latex]\u03c3[\/latex] very large.<\/p>\n<p>The standard deviation, when first presented, can seem unclear. By graphing your data, you can get a better &#8220;feel&#8221; for the deviations and the standard deviation. You will find that in symmetrical distributions, the standard deviation can be very helpful but in skewed distributions, the standard deviation may not be much help. The reason is that the two sides of a skewed distribution have different spreads. In a skewed distribution, it is better to look at the first quartile, the median, the third quartile, the smallest value, and the largest value. Because numbers can be confusing, <strong data-redactor-tag=\"strong\">always graph your data<\/strong>. Display your data in a histogram or a box plot.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Use the following data (first exam scores) from Susan Dean&#8217;s spring pre-calculus class:<\/p>\n<p>[latex]\\displaystyle {33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100}[\/latex]<\/p>\n<ol>\n<li>Create a chart containing the data, frequencies, relative frequencies, and cumulative relative frequencies to three decimal places.<\/li>\n<li>Calculate the following to one decimal place using a TI-83+ or TI-84 calculator:\n<ol>\n<li>The sample mean<\/li>\n<li>The sample standard deviation<\/li>\n<li>The median<\/li>\n<li>The first quartile<\/li>\n<li>The third quartile<\/li>\n<li>[latex]IQR[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Construct a box plot and a histogram on the same set of axes. Make comments about the box plot, the histogram, and the chart.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124077\">Show Solution<\/span><\/p>\n<div id=\"q124077\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>\n<table>\n<thead>\n<tr>\n<th>Data<\/th>\n<th>Frequency<\/th>\n<th>Relative Frequency<\/th>\n<th>Cumulative Relative Frequency<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]33[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]42[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.064[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]49[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]0.065[\/latex]<\/td>\n<td>[latex]0.129[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]53[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.161[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]55[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]0.065[\/latex]<\/td>\n<td>[latex]0.226[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]61[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.258[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]63[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.29[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]67[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.322[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]68[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]0.065[\/latex]<\/td>\n<td>[latex]0.387[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]69[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]0.065[\/latex]<\/td>\n<td>[latex]0.452[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]72[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.484[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]73[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.516[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]74[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.548[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]78[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.580[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]80[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.612[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]83[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.644[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]88[\/latex]<\/td>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]0.097[\/latex]<\/td>\n<td>[latex]0.741[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]90[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.773[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]92[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.805[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]94[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]0.129[\/latex]<\/td>\n<td>[latex]0.934[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]96[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.966[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]100[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]0.032[\/latex]<\/td>\n<td>[latex]0.998[\/latex] (Why isn&#8217;t this value [latex]1[\/latex]?)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<ol>\n<li>The sample mean = [latex]73.5[\/latex]<\/li>\n<li>The sample standard deviation = [latex]17.9[\/latex]<\/li>\n<li>The median = [latex]73[\/latex]<\/li>\n<li>The first quartile = [latex]61[\/latex]<\/li>\n<li>The third quartile = [latex]90[\/latex]<\/li>\n<li>[latex]IQR = 90 \u2013 61 = 29[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>The [latex]x[\/latex]-axis goes from [latex]32.5[\/latex] to [latex]100.5[\/latex]; [latex]y[\/latex]-axis goes from [latex]\u20132.4[\/latex] to [latex]15[\/latex] for the histogram. The number of intervals is five, so the width of an interval is [latex](100.5 \u2013 32.5)[\/latex] divided by five, is equal to [latex]13.6[\/latex]. Endpoints of the intervals are as follows: the starting point is [latex]32.5, 32.5 + 13.6 = 46.1[\/latex], [latex]46.1 + 13.6 = 59.7[\/latex], [latex]59.7 + 13.6 = 73.3[\/latex], [latex]73.3 + 13.6 = 86.9[\/latex], [latex]86.9 + 13.6 = 100.5[\/latex] = the ending value; No data values fall on an interval boundary.<\/li>\n<li><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-3394 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5668\/2021\/06\/19165655\/f52363d0d4686738d58f1174609c14cf3971f2a3.jpeg\" alt=\"A hybrid image displaying both a histogram and box plot described in detail in the answer solution above.\" width=\"487\" height=\"318\" \/><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p>The long left whisker in the box plot is reflected in the left side of the histogram. The spread of the exam scores in the lower [latex]50[\/latex]% is greater ([latex]73 \u2013 33 = 40[\/latex]) than the spread in the upper [latex]50[\/latex]% ([latex]100 \u2013 73 = 27[\/latex]). The histogram, box plot, and chart all reflect this. There are a substantial number of A and B grades ([latex]80[\/latex]s, [latex]90[\/latex]s, and [latex]100[\/latex]). The histogram clearly shows this. The box plot shows us that the middle [latex]50[\/latex]% of the exam scores ([latex]IQR[\/latex] = [latex]29[\/latex]) are Ds, Cs, and Bs. The box plot also shows us that the lower [latex]25[\/latex]% of the exam scores are Ds and Fs.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The following data show the different types of pet food stores in the area carry.<\/p>\n<p>[latex]\\displaystyle {6; 6; 6; 6; 7; 7; 7; 7; 7; 8; 9; 9; 9; 9; 10; 10; 10; 10; 10; 11; 11; 11; 11; 12; 12; 12; 12; 12; 12;}[\/latex]<br \/>\nCalculate the sample mean and the sample standard deviation to one decimal place using a TI-83+ or TI-84 calculator.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q124078\">Show Solution<\/span><\/p>\n<div id=\"q124078\" class=\"hidden-answer\" style=\"display: none\">\n[latex]\u03bc = 9.3[\/latex]<\/p>\n<p>[latex]s = 2.2[\/latex]\n<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Standard Deviation of Grouped Frequency Tables<\/h2>\n<p>Recall that for grouped data we do not know individual data values, so we cannot describe the typical value of the data with precision. In other words, we cannot find the exact mean, median, or mode. We can, however, determine the best estimate of the measures of center by finding the mean of the grouped data with the formula:<\/p>\n<p><em>Mean of Frequency Table<\/em> =[latex]\\displaystyle\\frac{{\\sum(fm)}}{{\\sum(f)}}[\/latex]<\/p>\n<p>where [latex]f[\/latex] = interval frequencies and [latex]m[\/latex] = interval midpoints.<\/p>\n<p>Just as we could not find the exact mean, neither can we find the exact standard deviation. Remember that standard deviation describes numerically the expected deviation a data value has from the mean. In simple English, the standard deviation allows us to compare how &#8220;unusual&#8221; individual data is compared to the mean.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the standard deviation for the data in the table below.<\/p>\n<table>\n<thead>\n<tr>\n<th>Class<\/th>\n<th>Frequency, [latex]f[\/latex]<\/th>\n<th>Midpoint, [latex]m[\/latex]<\/th>\n<th>[latex]m^2[\/latex]<\/th>\n<th>[latex]\\displaystyle\\overline{x}^2[\/latex]<\/th>\n<th>[latex]fm^2[\/latex]<\/th>\n<th>Standard Deviation<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]0\u20132[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\u20135[\/latex]<\/td>\n<td>[latex]6[\/latex]<\/td>\n<td>[latex]4[\/latex]<\/td>\n<td>[latex]16[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]96[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6\u20138[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<td>[latex]49[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]490[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]9\u201311[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<td>[latex]100[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]700[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12\u201314[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]13[\/latex]<\/td>\n<td>[latex]169[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]15\u201317[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<td>[latex]16[\/latex]<\/td>\n<td>[latex]256[\/latex]<\/td>\n<td>[latex]7.58[\/latex]<\/td>\n<td>[latex]512[\/latex]<\/td>\n<td>[latex]3.5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>For this data set, we have the mean, [latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u00a0= [latex]7.58[\/latex] and the standard deviation, [latex]\\displaystyle{s}_{x} = 3.5[\/latex]. This means that a randomly selected data value would be expected to be [latex]3.5[\/latex] units from the mean. If we look at the first class, we see that the class midpoint is equal to one. This is almost two full standard deviations from the mean since [latex]7.58 \u2013 3.5 \u2013 3.5 = 0.58[\/latex]. While the formula for calculating the standard deviation is not complicated, [latex]\\displaystyle{s}_{x}=\\sqrt{{\\frac{{f{(m-\\overline{x})}^{2}}}{{n-1}}}}[\/latex] where [latex]\\displaystyle{s}_{x} =[\/latex]sample standard deviation, [latex]\\displaystyle\\overline{x}[\/latex]\u00a0\u00a0= sample mean, the calculations are tedious. It is usually best to use technology when performing the calculations.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the standard deviation for the data from the previous example<\/p>\n<table>\n<thead>\n<tr>\n<th>Class<\/th>\n<th>Frequency, [latex]f[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]0\u20132[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\u20135[\/latex]<\/td>\n<td>[latex]6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6\u20138[\/latex]<\/td>\n<td>[latex]10[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]9\u201311[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12\u201314[\/latex]<\/td>\n<td>[latex]0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]15\u201317[\/latex]<\/td>\n<td>[latex]2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>First, press the <strong>STAT <\/strong>key and select <strong>1:Edit<\/strong><br \/>\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.08.28-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-499 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214250\/Screen-Shot-2015-06-07-at-6.08.28-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.08.28 PM\" width=\"330\" height=\"217\" \/><\/a><\/p>\n<p>Input the midpoint values into <strong data-redactor-tag=\"strong\">L1<\/strong> and the frequencies into <strong data-redactor-tag=\"strong\">L2<\/strong><\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.09.59-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-500\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214253\/Screen-Shot-2015-06-07-at-6.09.59-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.09.59 PM\" width=\"328\" height=\"221\" \/><\/a><\/p>\n<p>Select <strong data-redactor-tag=\"strong\">STAT<\/strong>, <strong data-redactor-tag=\"strong\">CALC<\/strong>, and <strong data-redactor-tag=\"strong\">1: 1-Var Stats<\/strong><\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.10.46-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-501\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214255\/Screen-Shot-2015-06-07-at-6.10.46-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.10.46 PM\" width=\"327\" height=\"222\" \/><\/a><\/p>\n<p>Select <strong data-redactor-tag=\"strong\">2nd<\/strong> then <strong data-redactor-tag=\"strong\">1<\/strong> then , <strong data-redactor-tag=\"strong\">2nd<\/strong> then <strong data-redactor-tag=\"strong\">2<\/strong><strong data-redactor-tag=\"strong\">Enter<\/strong><\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-6.11.19-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-502\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214257\/Screen-Shot-2015-06-07-at-6.11.19-PM.png\" alt=\"Screen Shot 2015-06-07 at 6.11.19 PM\" width=\"330\" height=\"222\" \/><\/a><\/p>\n<p>You will see displayed both a population standard deviation, <em data-redactor-tag=\"em\">\u03c3_x<\/em>, and the sample standard deviation, [latex]s_x[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\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-39\">\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>OpenStax Statistics. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction<\/li><li>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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>Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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 class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>How to calculate Standard Deviation and Variance. <strong>Authored by<\/strong>: statisticsfun. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qqOyy_NjflU\">https:\/\/youtu.be\/qqOyy_NjflU<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/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":17533,"menu_order":34,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax Statistics\",\"author\":\"\",\"organization\":\"\",\"url\":\"https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"How to calculate Standard Deviation and Variance\",\"author\":\"statisticsfun\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/qqOyy_NjflU\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\"},{\"type\":\"cc\",\"description\":\"Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\" http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\" http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-39","chapter","type-chapter","status-publish","hentry"],"part":31,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/39","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":24,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/39\/revisions"}],"predecessor-version":[{"id":3486,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/39\/revisions\/3486"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/parts\/31"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapters\/39\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/media?parent=39"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=39"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/contributor?post=39"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/wp-json\/wp\/v2\/license?post=39"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}