{"id":238,"date":"2016-04-21T22:43:42","date_gmt":"2016-04-21T22:43:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&#038;p=238"},"modified":"2021-09-27T16:14:45","modified_gmt":"2021-09-27T16:14:45","slug":"the-standard-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/the-standard-normal-distribution\/","title":{"raw":"The Standard Normal Distribution","rendered":"The Standard Normal Distribution"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul id=\"list4253\">\r\n \t<li>Recognize the standard normal probability distribution and apply it appropriately<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe <strong>standard normal distribution<\/strong> is a normal distribution of <strong>standardized values called <em data-redactor-tag=\"em\">z<\/em>-scores<\/strong>. <strong>A <em data-redactor-tag=\"em\">z<\/em>-score is measured in units of the standard deviation<\/strong>. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:\r\n\r\n<em>x<\/em> = <em>\u03bc<\/em> + (<em>z<\/em>)(<em>\u03c3<\/em>) = 5 + (3)(2) = 11\r\n\r\nThe <em>z<\/em>-score is three.\r\n\r\nThe mean for the standard normal distribution is zero, and the standard deviation is one. The transformation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean <em data-redactor-tag=\"em\">\u03bc<\/em> and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em>.\r\n\r\nThe following two videos give a description of what\u00a0it means to have a data set that is \"normally\" distributed.\r\n\r\nhttps:\/\/www.youtube.com\/embed\/xgQhefFOXrM\r\n\r\nhttps:\/\/www.youtube.com\/embed\/iiRiOlkLa6A\r\n<h2>Z-Scores<\/h2>\r\nIf <em data-redactor-tag=\"em\">X<\/em> is a normally distributed random variable and <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N(\u03bc, \u03c3)<\/em>, then the <em data-redactor-tag=\"em\">z<\/em>-score is:\r\n\r\n[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]\r\n\r\n<strong data-redactor-tag=\"strong\">The <em data-redactor-tag=\"em\">z<\/em>-score tells you how many standard deviations the value <em data-redactor-tag=\"em\">x<\/em> is above (to the right of) or below (to the left of) the mean, <em data-redactor-tag=\"em\">\u03bc<\/em>.<\/strong> Values of <em data-redactor-tag=\"em\">x<\/em> that are larger than the mean have positive <em data-redactor-tag=\"em\">z<\/em>-scores, and values of <em data-redactor-tag=\"em\">x<\/em> that are smaller than the mean have negative <em data-redactor-tag=\"em\">z<\/em>-scores. If <em data-redactor-tag=\"em\">x<\/em> equals the mean, then <em data-redactor-tag=\"em\">x<\/em> has a <em data-redactor-tag=\"em\">z<\/em>-score of zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N(5, 6)<\/em>. This says that <em data-redactor-tag=\"em\">x<\/em> is a normally distributed random variable with mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5 and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em> = 6. Suppose <em data-redactor-tag=\"em\">x<\/em> = 17. Then:\r\n\r\n[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]=\u00a0[latex]\\displaystyle{z}=\\frac{{17-5}}{{6}}={2}[\/latex]\r\n\r\nThis means that <em data-redactor-tag=\"em\">x<\/em> = 17 is<strong data-redactor-tag=\"strong\"> two standard deviations<\/strong> (2<em data-redactor-tag=\"em\">\u03c3<\/em>) above or to the right of the mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5. The standard deviation is <em data-redactor-tag=\"em\">\u03c3<\/em> = 6.\r\n\r\nNotice that: 5 + (2)(6) = 17 (The pattern is <em data-redactor-tag=\"em\">\u03bc<\/em> + <em data-redactor-tag=\"em\">z\u03c3<\/em> = <em data-redactor-tag=\"em\">x<\/em>)\r\n\r\nNow suppose <em data-redactor-tag=\"em\">x<\/em> = 1. Then:\u00a0[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] = [latex]\\displaystyle {z}=\\frac{{1-5}}{{6}} = -{0.67}[\/latex]\r\n\r\n(rounded to two decimal places)\r\n\r\n<strong data-redactor-tag=\"strong\">This means that <em data-redactor-tag=\"em\">x<\/em> = 1 is 0.67 standard deviations <\/strong>(\u20130.67<em data-redactor-tag=\"em\">\u03c3<\/em>) below or to the left of the mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5.\r\n\r\nNotice that: 5 + (\u20130.67)(6) is approximately equal to one (This has the pattern <em data-redactor-tag=\"em\">\u03bc<\/em> + (\u20130.67)\u03c3 = 1)\r\n\r\nSummarizing, when <em data-redactor-tag=\"em\">z<\/em> is positive, <em data-redactor-tag=\"em\">x<\/em> is above or to the right of <em data-redactor-tag=\"em\">\u03bc<\/em> and when <em data-redactor-tag=\"em\">z<\/em>is negative, <em data-redactor-tag=\"em\">x<\/em> is to the left of or below <em data-redactor-tag=\"em\">\u03bc<\/em>. Or, when <em data-redactor-tag=\"em\">z<\/em> is positive, <em data-redactor-tag=\"em\">x<\/em> is greater than <em data-redactor-tag=\"em\">\u03bc<\/em>, and when <em data-redactor-tag=\"em\">z<\/em> is negative <em data-redactor-tag=\"em\">x<\/em> is less than <em data-redactor-tag=\"em\">\u03bc<\/em>.\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWhat is the <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">x<\/em>, when <em data-redactor-tag=\"em\">x<\/em> = 1 and\u00a0<em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(12,3)?\r\n\r\n[reveal-answer q=\"280221\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"280221\"]\r\n\r\n[latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67} [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=Wp2nVIzBsE8\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSome doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 2). Fill in the blanks.\r\n<ol>\r\n \t<li>Suppose a person lost ten pounds in a month. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 10 pounds is <em data-redactor-tag=\"em\">z<\/em> = 2.5 (verify). This <em data-redactor-tag=\"em\">z<\/em>-score tells you that\u00a0<em data-redactor-tag=\"em\">x<\/em> = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/li>\r\n \t<li>Suppose a person gained three pounds (a negative weight loss). Then <em data-redactor-tag=\"em\">z<\/em> = __________. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = \u20133 is ________ standard deviations to the __________ (right or left) of the mean.<\/li>\r\n<\/ol>\r\nSolution:\r\n<ol>\r\n \t<li>This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 10 is <strong data-redactor-tag=\"strong\">2.5<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">right<\/strong> of the mean <strong data-redactor-tag=\"strong\">five<\/strong>.<\/li>\r\n \t<li><em data-redactor-tag=\"em\">z<\/em>= \u20134. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = \u20133 is <strong data-redactor-tag=\"strong\">4<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">left<\/strong> of the mean.<\/li>\r\n<\/ol>\r\nNow...suppose the random variables <em data-redactor-tag=\"em\">X<\/em> and <em data-redactor-tag=\"em\">Y<\/em> have the following normal distributions: <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 6) and <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(2, 1). If <em data-redactor-tag=\"em\">x<\/em> = 17, then <em data-redactor-tag=\"em\">z<\/em> = 2. (This was previously shown.) If <em data-redactor-tag=\"em\">y<\/em> = 4, what is <em data-redactor-tag=\"em\">z<\/em>? [latex]\\displaystyle {z}=\\frac{{y - \\mu}}{{\\sigma}} = \\frac{{4-2}}{{1}}[\/latex].\r\n\r\nThe <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">y<\/em> = 4 is <em data-redactor-tag=\"em\">z<\/em> = 2. This means that four is <em data-redactor-tag=\"em\">z<\/em> = 2 standard deviations to the right of the mean. Therefore, <em data-redactor-tag=\"em\">x<\/em> = 17 and <em data-redactor-tag=\"em\">y<\/em> = 4 are both two (of <strong data-redactor-tag=\"strong\">their own<\/strong>) standard deviations to the right of their respective means.\r\n\r\n<strong data-redactor-tag=\"strong\">The <em data-redactor-tag=\"em\">z<\/em>-score allows us to compare data that are scaled differently.<\/strong> To understand the concept, suppose <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since <em data-redactor-tag=\"em\">x<\/em> = 17 and <em data-redactor-tag=\"em\">y<\/em>= 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFill in the blanks.\r\n\r\nJerome averages 16 points a game with a standard deviation of four points. <em data-redactor-tag=\"em\">X<\/em> ~<em data-redactor-tag=\"em\">N<\/em>(16,4). Suppose Jerome scores ten points in a game. The <em data-redactor-tag=\"em\">z<\/em>\u2013score when <em data-redactor-tag=\"em\">x<\/em> = 10 is \u20131.5. This score tells you that <em data-redactor-tag=\"em\">x<\/em> = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).\r\n\r\n[reveal-answer q=\"19131\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"19131\"]1.5, left, 16[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(170, 6.28).\r\n\r\na. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 168 cm is <em data-redactor-tag=\"em\">z<\/em> = _______. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).\r\n\r\nb. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">z<\/em> = 1.27. What is the male's height? The <em data-redactor-tag=\"em\">z<\/em>-score (<em data-redactor-tag=\"em\">z<\/em> = 1.27) tells you that the male's height is ________ standard deviations to the __________ (right or left) of the mean.\r\n\r\nSolution:\r\n\r\na. \u20130.32, 0.32, left, 170\r\n\r\nb. 177.98 cm, 1.27, right\r\n\r\n<\/div>\r\n<h4><\/h4>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse the information in Example 3 to answer the following questions.\r\n<ol>\r\n \t<li>Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 176 cm is <em data-redactor-tag=\"em\">z<\/em> = _______. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/li>\r\n \t<li>Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">z<\/em> = \u20132. What is the male's height? The <em data-redactor-tag=\"em\">z<\/em>-score (<em data-redactor-tag=\"em\">z<\/em> = \u20132) tells you that the male's height is ________ standard deviations to the __________ (right or left) of the mean.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"152857\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"152857\"]\r\n\r\nSolve the equation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] for x. x = \u03bc + (z)(\u03c3) for <em data-redactor-tag=\"em\">x<\/em>. <em data-redactor-tag=\"em\">x<\/em> = <em data-redactor-tag=\"em\">\u03bc<\/em> + (<em data-redactor-tag=\"em\">z<\/em>)(<em data-redactor-tag=\"em\">\u03c3<\/em>)\r\n<ol>\r\n \t<li>z&lt;=[latex]\\displaystyle\\frac{{176-170}}{{0.96}}[\/latex], This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm.<\/li>\r\n \t<li><em data-redactor-tag=\"em\">X<\/em> = 157.44 cm, The <em data-redactor-tag=\"em\">z<\/em>-score(<em data-redactor-tag=\"em\">z<\/em> = \u20132) tells you that the male's height is two standard deviations to the left of the mean.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFrom 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let <em data-redactor-tag=\"em\">Y<\/em> = the height of 15 to 18-year-old males from 1984 to 1985. Then <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(172.36, 6.34).\r\n\r\nThe mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(170, 6.28).\r\n\r\nFind the <em data-redactor-tag=\"em\">z<\/em>-scores for <em data-redactor-tag=\"em\">x<\/em> = 160.58 cm and <em data-redactor-tag=\"em\">y<\/em> = 162.85 cm. Interpret each <em data-redactor-tag=\"em\">z<\/em>-score. What can you say about <em data-redactor-tag=\"em\">x<\/em> = 160.58 cm and <em data-redactor-tag=\"em\">y<\/em> = 162.85 cm?\r\n\r\nSolution:\r\n\r\nThe <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em> = 160.58 is <em data-redactor-tag=\"em\">z<\/em> = \u20131.5.\r\n\r\nThe <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">y<\/em> = 162.85 is <em data-redactor-tag=\"em\">z<\/em> = \u20131.5.\r\n\r\nBoth <em data-redactor-tag=\"em\">x<\/em> = 160.58 and <em data-redactor-tag=\"em\">y<\/em> = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIn 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean\u00a0<em data-redactor-tag=\"em\">\u00b5<\/em> = 496 and a standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em> = 114. Let <em data-redactor-tag=\"em\">X<\/em> = a SAT exam verbal section score in 2012. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(496, 114).\r\n\r\nFind the <em data-redactor-tag=\"em\">z<\/em>-scores for <em data-redactor-tag=\"em\">x<\/em>1 = 325 and <em data-redactor-tag=\"em\">x<\/em>2 = 366.21. Interpret each <em data-redactor-tag=\"em\">z<\/em>-score. What can you say about <em data-redactor-tag=\"em\">x<\/em>1 = 325 and <em data-redactor-tag=\"em\">x<\/em>2 = 366.21?\r\n\r\n[reveal-answer q=\"961333\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"961333\"]\r\n\r\nThe <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em>1 = 325 is <em data-redactor-tag=\"em\">z<\/em>1 = \u20131.5.\r\n\r\nThe <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em>2 = 366.21 is <em data-redactor-tag=\"em\">z<\/em>2 = \u20131.14.\r\n\r\nStudent 2 scored closer to the mean than Student 1. Both students scored below the mean, since they both had negative <em data-redactor-tag=\"em\">z<\/em>-scores, Student 2 had the better score.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<h2>The Empirical Rule<\/h2>\r\nIf <em data-redactor-tag=\"em\">X<\/em> is a random variable and has a normal distribution with mean <em data-redactor-tag=\"em\">\u00b5<\/em> and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em>, then the <strong data-redactor-tag=\"strong\">Empirical Rule<\/strong> says the following:\r\n<ul>\r\n \t<li>About 68% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> and +1<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em> (within one standard deviation of the mean).<\/li>\r\n \t<li>About 95% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> and +2<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em> (within two standard deviations of the mean).<\/li>\r\n \t<li>About 99.7% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> and +3<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em>(within three standard deviations of the mean). Notice that almost all the<em data-redactor-tag=\"em\">x<\/em> values lie within three standard deviations of the mean.<\/li>\r\n \t<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +1<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> are +1 and \u20131, respectively.<\/li>\r\n \t<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +2<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> are +2 and \u20132, respectively.<\/li>\r\n \t<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +3<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> are +3 and \u20133 respectively.<\/li>\r\n<\/ul>\r\nThe empirical rule is also known as the 68-95-99.7 rule.<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-7.34.36-PM.png\"><img class=\"aligncenter size-full wp-image-509\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214545\/Screen-Shot-2015-06-07-at-7.34.36-PM.png\" alt=\"Graph of the empirical Rule\" width=\"682\" height=\"392\" \/><\/a>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose <em data-redactor-tag=\"em\">x<\/em> has a normal distribution with mean 50 and standard deviation 6.\r\n<ul>\r\n \t<li>About 68% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20131)(6) = \u20136 and 1<em data-redactor-tag=\"em\">\u03c3<\/em> = (1)(6) = 6 of the mean 50. The values 50 \u2013 6 = 44 and 50 + 6 = 56 are within one standard deviation of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and +1 for 44 and 56, respectively.<\/li>\r\n \t<li>About 95% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20132)(6) = \u201312 and 2<em data-redactor-tag=\"em\">\u03c3<\/em> = (2)(6) = 12. The values 50 \u2013 12 = 38 and 50 + 12 = 62 are within two standard deviations of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and +2 for 38 and 62,respectively.<\/li>\r\n \t<li>About 99.7% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20133)(6) = \u201318 and 3<em data-redactor-tag=\"em\">\u03c3<\/em>= (3)(6) = 18 of the mean 50. The values 50 \u2013 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and +3 for 32 and 68, respectively<\/li>\r\n<\/ul>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSuppose <em data-redactor-tag=\"em\">X<\/em> has a normal distribution with mean 25 and standard deviation five. Between what values of <em data-redactor-tag=\"em\">x<\/em> do 68% of the values lie?\r\n\r\n[reveal-answer q=\"590629\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"590629\"]Between 20 and 30.[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFrom 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let <em data-redactor-tag=\"em\">Y<\/em> = the height of 15 to 18-year-old males in 1984 to 1985. Then <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(172.36, 6.34).\r\n<ol>\r\n \t<li>About 68% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 95% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________ respectively.<\/li>\r\n \t<li>About 99.7% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two\u00a0 values?\u00a0These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________ respectively<\/li>\r\n<\/ol>\r\n<strong>Solution:<\/strong>\r\n<ol>\r\n \t<li>About 68% of the values lie between the values 166.02 and 178.7 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and 1, respectively.<\/li>\r\n \t<li>About 95% of the values lie between the values 159.68 and 185.04 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and 2, respectively.<\/li>\r\n \t<li>About 99.7% of the values lie between the values 153.34 and 191.38 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3, respectively.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nThe scores on a college entrance exam have an approximate normal distribution with mean, <em data-redactor-tag=\"em\">\u00b5<\/em> = 52 points and a standard deviation, <em data-redactor-tag=\"em\">\u03c3<\/em> = 11 points.\r\n\r\n.7% of the values lie between 153.34 and 191.38. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3.\r\n<ol>\r\n \t<li>About 68% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 95% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 99.7% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"226606\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"226606\"]\r\n<ol>\r\n \t<li>About 68% of the values lie between the values 41 and 63 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and 1, respectively.<\/li>\r\n \t<li>About 95% of the values lie between the values 30 and 74 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and 2, respectively.<\/li>\r\n \t<li>About 99.7% of the values lie between the values 19 and 85 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3, respectively.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn general, If we know the z-score(s), we can find the amount of area that lies to the left, to the right, or between two z-scores using the Excel function\u00a0<strong>NORM.S.DIST<\/strong>.\u00a0 If we know the area we are interested in, we can also find what z-score(s) are boundaries using the Excel function\u00a0<strong>NORM.S.INV<\/strong>.\r\n<div class=\"textbox shaded\">\r\n\r\nExcel function\u00a0<strong>NORM.S.DIST\u00a0<\/strong>has the following syntax:\r\n\r\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Syntax \u2013 Standard Normal Distribution\">\r\n<p class=\"\" style=\"padding-left: 30px;\">NORM.S.DIST(z,cumulative)<\/p>\r\n<p style=\"padding-left: 30px;\">The NORM.S.DIST function syntax has the following arguments:<\/p>\r\n\r\n<ul>\r\n \t<li><b class=\"ocpRunInHead\">Z<\/b>\u00a0\u00a0\u00a0\u00a0 Required. The value for which you want the distribution.<\/li>\r\n \t<li><b class=\"ocpRunInHead\">Cumulative<\/b>\u00a0\u00a0\u00a0\u00a0 Required. Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, NORMS.DIST returns the cumulative distribution function; if FALSE, it returns the probability mass function.<\/li>\r\n<\/ul>\r\n<\/section><section class=\"ocpSection\" role=\"region\" aria-label=\"Remarks\">\r\n<h2 style=\"padding-left: 30px;\">Remarks<\/h2>\r\n<ul>\r\n \t<li>If z is nonnumeric, NORM.S.DIST returns the #VALUE! error value.<\/li>\r\n \t<li>The equation for the standard normal density function is:<img src=\"https:\/\/support.content.office.net\/en-us\/media\/bd99a6ad-418a-444f-aa09-533f65ec63b5.gif\" alt=\"Equation\" \/><\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 30px;\"><a href=\"https:\/\/support.microsoft.com\/en-us\/office\/norm-s-dist-function-1e787282-3832-4520-a9ae-bd2a8d99ba88\">Excel Support - Norm.s.dist<\/a><\/p>\r\n&nbsp;\r\n\r\nExcel function\u00a0<strong>NORM.S.INV\u00a0<\/strong>has the following syntax:\r\n\r\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Syntax\">\r\n<p class=\"\" style=\"padding-left: 30px;\">NORM.S.INV(probability)<\/p>\r\n<p style=\"padding-left: 30px;\">The NORM.S.INV function syntax has the following arguments:<\/p>\r\n\r\n<ul>\r\n \t<li>\r\n<p class=\"x-hidden-focus\"><b class=\"ocpRunInHead\">Probability<\/b>\u00a0\u00a0\u00a0\u00a0 Required. A probability corresponding to the normal distribution.<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/section><section class=\"ocpSection\" role=\"region\" aria-label=\"Remarks\">\r\n<h2 style=\"padding-left: 30px;\">Remarks<\/h2>\r\n<ul>\r\n \t<li>If probability is nonnumeric, NORMS.INV returns the #VALUE! error value.<\/li>\r\n \t<li>If probability &lt;= 0 or if probability &gt;= 1, NORMS.INV returns the #NUM! error value.<\/li>\r\n<\/ul>\r\n<p class=\"\" style=\"padding-left: 30px;\">Given a value for probability, NORM.S.INV seeks that value z such that NORM.S.DIST(z,TRUE) = probability. Thus, precision of NORM.S.INV depends on precision of NORM.S.DIST. NORM.S.INV uses an iterative search technique.<\/p>\r\n<p style=\"padding-left: 30px;\"><a href=\"https:\/\/support.microsoft.com\/en-us\/office\/norm-s-inv-function-d6d556b4-ab7f-49cd-b526-5a20918452b1\">Excel Suport - Norm.s.inv<\/a><\/p>\r\n\r\n<\/section><\/section><\/div>\r\n<div><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<ol>\r\n \t<li>Find the area to the left of [latex] z=1.26 [\/latex].<\/li>\r\n \t<li>Find the area to the right of [latex] z=-0.95 [\/latex].<\/li>\r\n \t<li>Find the area between [latex] z=-1.74 [\/latex] and [latex] z=2.12 [\/latex]<\/li>\r\n<\/ol>\r\n<strong>Solution:<\/strong>\r\n\r\n1. Start by drawing a sketch.\r\n\r\n<img class=\"wp-image-2744 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12231959\/norm.s.dist-1.jpg\" alt=\"\" width=\"317\" height=\"175\" \/>\r\n\r\n<strong>In Excel:<\/strong> Use =NORM.S.DIST(1.26,1)= 0.896.\u00a0 89.6% of values fall to the left of [latex] z=1.26 [\/latex].\r\n\r\n2. Start by drawing a sketch.\r\n\r\n<img class=\"wp-image-2745 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12232434\/norm.s.dist-2.jpg\" alt=\"\" width=\"314\" height=\"188\" \/>\r\n\r\n<strong>In Excel:<\/strong> Since the NORM.S.DIST function returns the area to the left of a z-score, we will need to use the fact that the total area under the standard normal curve is 1 to get to our answer.\u00a0 Use =1-NORM.S.DIST(-0.95, 1) = 1- 0.171 = 0.829.\u00a0 82.9% of values fall to the right of\u00a0[latex] z=-0.95 [\/latex].\r\n\r\n3. Start by drawing a sketch.\r\n\r\n<img class=\"wp-image-2746 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12233134\/norm.s.dist-3.jpg\" alt=\"\" width=\"314\" height=\"188\" \/>\r\n\r\n<strong>In Excel:\u00a0<\/strong>Use =NORM.S.DIST(2.12,1) - NORM.S.DIST(-1.74,1)= 0.983 - 0.041 = 0.943.\u00a0 94.3% of values fall between\u00a0 [latex] z=-1.74 [\/latex] and [latex] z=2.12 [\/latex].\u00a0 *Note, in this calculation we find the area to the left of the larger z-score then subtract the area to the left of the smaller z-score.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\"><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it<\/h3>\r\n<ol>\r\n \t<li>Find the z-score that has an area of 0.378 to its left.<\/li>\r\n \t<li>Find the z-score that has an area of 0.092 to its right.<\/li>\r\n \t<li>Find the symmetric z-scores that correspond to having 0.10 as the combined area in two tails.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"713595\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"713595\"]\r\n\r\n<strong>Solution:<\/strong>\r\n\r\n1. Start by drawing a sketch.\r\n\r\n<img class=\" wp-image-2751 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12235639\/norm.s.inv-1.jpg\" alt=\"\" width=\"346\" height=\"208\" \/>\r\n\r\n<strong>In Excel:<\/strong> Use =NORM.S.INV(0.378) = -0.31.\u00a0 The z-score -0.31 is the boundary that cuts off an area to the left of 0.378.\r\n\r\n2. Start by drawing a sketch.\r\n\r\n<img class=\"wp-image-2752 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/13000105\/norm.s.inv-2.jpg\" alt=\"\" width=\"386\" height=\"228\" \/>\r\n\r\n<strong>In Excel:<\/strong> Since the NORM.S.INV function takes as an input area to the left of a z-score, we will need to use the fact that the total area under the standard normal curve is 1 to get to our answer.\u00a0 So, 0.092 to the right means there is 1-0.092= 0.908 to the left of the z-score we are interested in.\r\n\r\nUse =NORM.S.INV(0.908) = 1.33.\u00a0 The z-score 1.33 is the boundary that cuts off an area of 0.092 to the right.\r\n\r\n3. Start by drawing a sketch.\r\n\r\n<img class=\"wp-image-2753 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/13000506\/norm.s.inv-3.jpg\" alt=\"\" width=\"344\" height=\"205\" \/>\r\n\r\n&nbsp;\r\n\r\n<strong>In Excel:\u00a0<\/strong>To have a combined area of 0.10 in the two tails cut off by symmetric z-scores, we have 0.05 in each tail.\u00a0 If we find the z-score that corresponds to the area in the left tail, the symmetric z-score that cuts off the right tail will just be the positive.\r\n\r\nUse =NORM.S.INV(0.05) =-1.65.\u00a0 So, the two symmetric z-scores are -1.65 and 1.65.\r\n\r\n&nbsp;\r\n\r\nPut Answer Here[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<h2>References<\/h2>\r\n\"Blood Pressure of Males and Females.\" StatCruch, 2013. Available online at http:\/\/www.statcrunch.com\/5.0\/viewreport.php?reportid=11960 (accessed May 14, 2013).\r\n\r\n\"The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.\" London School of Hygiene and Tropical Medicine, 2009. Available online at http:\/\/conflict.lshtm.ac.uk\/page_125.htm (accessed May 14, 2013).\r\n\r\n\"2012 College-Bound Seniors Total Group Profile Report.\" CollegeBoard, 2012. Available online at http:\/\/media.collegeboard.com\/digitalServices\/pdf\/research\/TotalGroup-2012.pdf (accessed May 14, 2013).\r\n\r\n\"Digest of Education Statistics: ACT score average and standard deviations by sex and race\/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.\" National Center for Education Statistics. Available online at http:\/\/nces.ed.gov\/programs\/digest\/d09\/tables\/dt09_147.asp (accessed May 14, 2013).\r\n\r\nData from the <em data-redactor-tag=\"em\">San Jose Mercury News<\/em>.\r\n\r\nData from <em data-redactor-tag=\"em\">The World Almanac and Book of Facts<\/em>.\r\n\r\n\"List of stadiums by capacity.\" Wikipedia. Available online at https:\/\/en.wikipedia.org\/wiki\/List_of_stadiums_by_capacity (accessed May 14, 2013).\r\n\r\nData from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).\r\n<h1>Concept Review<\/h1>\r\nA <em data-redactor-tag=\"em\">z<\/em>-score is a standardized value. Its distribution is the standard normal, <em data-redactor-tag=\"em\">Z<\/em> ~<em data-redactor-tag=\"em\">N<\/em>(0, 1). The mean of the <em data-redactor-tag=\"em\">z<\/em>-scores is zero and the standard deviation is one. If <em data-redactor-tag=\"em\">z\u00a0<\/em>is the <em data-redactor-tag=\"em\">z<\/em>-score for a value <em data-redactor-tag=\"em\">x<\/em> from the normal distribution <em data-redactor-tag=\"em\">N<\/em>(<em data-redactor-tag=\"em\">\u00b5<\/em>, <em data-redactor-tag=\"em\">\u03c3<\/em>) then <em data-redactor-tag=\"em\">z<\/em> tells you how many standard deviations <em data-redactor-tag=\"em\">x<\/em> is above (greater than) or below (less than) <em data-redactor-tag=\"em\">\u00b5<\/em>. The Empirical Rule provides approximation for the percentage of values that lie within key intervals in a normal distribution.\r\n<h1>Formula Review<\/h1>\r\n<em data-redactor-tag=\"em\">Z<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(0, 1)\r\n\r\n<em data-redactor-tag=\"em\">z<\/em> = a standardized value (<em data-redactor-tag=\"em\">z<\/em>-score)\r\n\r\nmean = 0; standard deviation = 1\r\n\r\nTo find an x value\u00a0when the <em data-redactor-tag=\"em\">z<\/em>-score is known: [latex]\\displaystyle{x}={\\mu}+{z}{\\sigma}[\/latex]\r\n\r\n<em data-redactor-tag=\"em\">z<\/em>-score:[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]\r\n\r\n<em data-redactor-tag=\"em\">Z<\/em> = the random variable for <em data-redactor-tag=\"em\">z<\/em>-scores\r\n\r\n<em data-redactor-tag=\"em\">Z<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(0, 1)","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list4253\">\n<li>Recognize the standard normal probability distribution and apply it appropriately<\/li>\n<\/ul>\n<\/div>\n<p>The <strong>standard normal distribution<\/strong> is a normal distribution of <strong>standardized values called <em data-redactor-tag=\"em\">z<\/em>-scores<\/strong>. <strong>A <em data-redactor-tag=\"em\">z<\/em>-score is measured in units of the standard deviation<\/strong>. For example, if the mean of a normal distribution is five and the standard deviation is two, the value 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:<\/p>\n<p><em>x<\/em> = <em>\u03bc<\/em> + (<em>z<\/em>)(<em>\u03c3<\/em>) = 5 + (3)(2) = 11<\/p>\n<p>The <em>z<\/em>-score is three.<\/p>\n<p>The mean for the standard normal distribution is zero, and the standard deviation is one. The transformation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] produces the distribution Z ~ N(0, 1). The value x comes from a normal distribution with mean <em data-redactor-tag=\"em\">\u03bc<\/em> and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em>.<\/p>\n<p>The following two videos give a description of what\u00a0it means to have a data set that is &#8220;normally&#8221; distributed.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Normal Distribution - Explained Simply (part 1)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xgQhefFOXrM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Normal Distribution - Explained Simply (part 2)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/iiRiOlkLa6A?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Z-Scores<\/h2>\n<p>If <em data-redactor-tag=\"em\">X<\/em> is a normally distributed random variable and <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N(\u03bc, \u03c3)<\/em>, then the <em data-redactor-tag=\"em\">z<\/em>-score is:<\/p>\n<p>[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]<\/p>\n<p><strong data-redactor-tag=\"strong\">The <em data-redactor-tag=\"em\">z<\/em>-score tells you how many standard deviations the value <em data-redactor-tag=\"em\">x<\/em> is above (to the right of) or below (to the left of) the mean, <em data-redactor-tag=\"em\">\u03bc<\/em>.<\/strong> Values of <em data-redactor-tag=\"em\">x<\/em> that are larger than the mean have positive <em data-redactor-tag=\"em\">z<\/em>-scores, and values of <em data-redactor-tag=\"em\">x<\/em> that are smaller than the mean have negative <em data-redactor-tag=\"em\">z<\/em>-scores. If <em data-redactor-tag=\"em\">x<\/em> equals the mean, then <em data-redactor-tag=\"em\">x<\/em> has a <em data-redactor-tag=\"em\">z<\/em>-score of zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N(5, 6)<\/em>. This says that <em data-redactor-tag=\"em\">x<\/em> is a normally distributed random variable with mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5 and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em> = 6. Suppose <em data-redactor-tag=\"em\">x<\/em> = 17. Then:<\/p>\n<p>[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]=\u00a0[latex]\\displaystyle{z}=\\frac{{17-5}}{{6}}={2}[\/latex]<\/p>\n<p>This means that <em data-redactor-tag=\"em\">x<\/em> = 17 is<strong data-redactor-tag=\"strong\"> two standard deviations<\/strong> (2<em data-redactor-tag=\"em\">\u03c3<\/em>) above or to the right of the mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5. The standard deviation is <em data-redactor-tag=\"em\">\u03c3<\/em> = 6.<\/p>\n<p>Notice that: 5 + (2)(6) = 17 (The pattern is <em data-redactor-tag=\"em\">\u03bc<\/em> + <em data-redactor-tag=\"em\">z\u03c3<\/em> = <em data-redactor-tag=\"em\">x<\/em>)<\/p>\n<p>Now suppose <em data-redactor-tag=\"em\">x<\/em> = 1. Then:\u00a0[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] = [latex]\\displaystyle {z}=\\frac{{1-5}}{{6}} = -{0.67}[\/latex]<\/p>\n<p>(rounded to two decimal places)<\/p>\n<p><strong data-redactor-tag=\"strong\">This means that <em data-redactor-tag=\"em\">x<\/em> = 1 is 0.67 standard deviations <\/strong>(\u20130.67<em data-redactor-tag=\"em\">\u03c3<\/em>) below or to the left of the mean <em data-redactor-tag=\"em\">\u03bc<\/em> = 5.<\/p>\n<p>Notice that: 5 + (\u20130.67)(6) is approximately equal to one (This has the pattern <em data-redactor-tag=\"em\">\u03bc<\/em> + (\u20130.67)\u03c3 = 1)<\/p>\n<p>Summarizing, when <em data-redactor-tag=\"em\">z<\/em> is positive, <em data-redactor-tag=\"em\">x<\/em> is above or to the right of <em data-redactor-tag=\"em\">\u03bc<\/em> and when <em data-redactor-tag=\"em\">z<\/em>is negative, <em data-redactor-tag=\"em\">x<\/em> is to the left of or below <em data-redactor-tag=\"em\">\u03bc<\/em>. Or, when <em data-redactor-tag=\"em\">z<\/em> is positive, <em data-redactor-tag=\"em\">x<\/em> is greater than <em data-redactor-tag=\"em\">\u03bc<\/em>, and when <em data-redactor-tag=\"em\">z<\/em> is negative <em data-redactor-tag=\"em\">x<\/em> is less than <em data-redactor-tag=\"em\">\u03bc<\/em>.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>What is the <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">x<\/em>, when <em data-redactor-tag=\"em\">x<\/em> = 1 and\u00a0<em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(12,3)?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q280221\">Show Answer<\/span><\/p>\n<div id=\"q280221\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"ck12.org normal distribution problems: z-score | Probability and Statistics | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Wp2nVIzBsE8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Some doctors believe that a person can lose five pounds, on the average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 2). Fill in the blanks.<\/p>\n<ol>\n<li>Suppose a person lost ten pounds in a month. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 10 pounds is <em data-redactor-tag=\"em\">z<\/em> = 2.5 (verify). This <em data-redactor-tag=\"em\">z<\/em>-score tells you that\u00a0<em data-redactor-tag=\"em\">x<\/em> = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/li>\n<li>Suppose a person gained three pounds (a negative weight loss). Then <em data-redactor-tag=\"em\">z<\/em> = __________. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = \u20133 is ________ standard deviations to the __________ (right or left) of the mean.<\/li>\n<\/ol>\n<p>Solution:<\/p>\n<ol>\n<li>This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 10 is <strong data-redactor-tag=\"strong\">2.5<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">right<\/strong> of the mean <strong data-redactor-tag=\"strong\">five<\/strong>.<\/li>\n<li><em data-redactor-tag=\"em\">z<\/em>= \u20134. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = \u20133 is <strong data-redactor-tag=\"strong\">4<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">left<\/strong> of the mean.<\/li>\n<\/ol>\n<p>Now&#8230;suppose the random variables <em data-redactor-tag=\"em\">X<\/em> and <em data-redactor-tag=\"em\">Y<\/em> have the following normal distributions: <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 6) and <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(2, 1). If <em data-redactor-tag=\"em\">x<\/em> = 17, then <em data-redactor-tag=\"em\">z<\/em> = 2. (This was previously shown.) If <em data-redactor-tag=\"em\">y<\/em> = 4, what is <em data-redactor-tag=\"em\">z<\/em>? [latex]\\displaystyle {z}=\\frac{{y - \\mu}}{{\\sigma}} = \\frac{{4-2}}{{1}}[\/latex].<\/p>\n<p>The <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">y<\/em> = 4 is <em data-redactor-tag=\"em\">z<\/em> = 2. This means that four is <em data-redactor-tag=\"em\">z<\/em> = 2 standard deviations to the right of the mean. Therefore, <em data-redactor-tag=\"em\">x<\/em> = 17 and <em data-redactor-tag=\"em\">y<\/em> = 4 are both two (of <strong data-redactor-tag=\"strong\">their own<\/strong>) standard deviations to the right of their respective means.<\/p>\n<p><strong data-redactor-tag=\"strong\">The <em data-redactor-tag=\"em\">z<\/em>-score allows us to compare data that are scaled differently.<\/strong> To understand the concept, suppose <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since <em data-redactor-tag=\"em\">x<\/em> = 17 and <em data-redactor-tag=\"em\">y<\/em>= 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means.<\/p>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Fill in the blanks.<\/p>\n<p>Jerome averages 16 points a game with a standard deviation of four points. <em data-redactor-tag=\"em\">X<\/em> ~<em data-redactor-tag=\"em\">N<\/em>(16,4). Suppose Jerome scores ten points in a game. The <em data-redactor-tag=\"em\">z<\/em>\u2013score when <em data-redactor-tag=\"em\">x<\/em> = 10 is \u20131.5. This score tells you that <em data-redactor-tag=\"em\">x<\/em> = 10 is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q19131\">Show Answer<\/span><\/p>\n<div id=\"q19131\" class=\"hidden-answer\" style=\"display: none\">1.5, left, 16<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(170, 6.28).<\/p>\n<p>a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 168 cm is <em data-redactor-tag=\"em\">z<\/em> = _______. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/p>\n<p>b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">z<\/em> = 1.27. What is the male&#8217;s height? The <em data-redactor-tag=\"em\">z<\/em>-score (<em data-redactor-tag=\"em\">z<\/em> = 1.27) tells you that the male&#8217;s height is ________ standard deviations to the __________ (right or left) of the mean.<\/p>\n<p>Solution:<\/p>\n<p>a. \u20130.32, 0.32, left, 170<\/p>\n<p>b. 177.98 cm, 1.27, right<\/p>\n<\/div>\n<h4><\/h4>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use the information in Example 3 to answer the following questions.<\/p>\n<ol>\n<li>Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The <em data-redactor-tag=\"em\">z<\/em>-score when <em data-redactor-tag=\"em\">x<\/em> = 176 cm is <em data-redactor-tag=\"em\">z<\/em> = _______. This <em data-redactor-tag=\"em\">z<\/em>-score tells you that <em data-redactor-tag=\"em\">x<\/em> = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).<\/li>\n<li>Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a <em data-redactor-tag=\"em\">z<\/em>-score of <em data-redactor-tag=\"em\">z<\/em> = \u20132. What is the male&#8217;s height? The <em data-redactor-tag=\"em\">z<\/em>-score (<em data-redactor-tag=\"em\">z<\/em> = \u20132) tells you that the male&#8217;s height is ________ standard deviations to the __________ (right or left) of the mean.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q152857\">Show Answer<\/span><\/p>\n<div id=\"q152857\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solve the equation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] for x. x = \u03bc + (z)(\u03c3) for <em data-redactor-tag=\"em\">x<\/em>. <em data-redactor-tag=\"em\">x<\/em> = <em data-redactor-tag=\"em\">\u03bc<\/em> + (<em data-redactor-tag=\"em\">z<\/em>)(<em data-redactor-tag=\"em\">\u03c3<\/em>)<\/p>\n<ol>\n<li>z&lt;=[latex]\\displaystyle\\frac{{176-170}}{{0.96}}[\/latex], This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm.<\/li>\n<li><em data-redactor-tag=\"em\">X<\/em> = 157.44 cm, The <em data-redactor-tag=\"em\">z<\/em>-score(<em data-redactor-tag=\"em\">z<\/em> = \u20132) tells you that the male&#8217;s height is two standard deviations to the left of the mean.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let <em data-redactor-tag=\"em\">Y<\/em> = the height of 15 to 18-year-old males from 1984 to 1985. Then <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(172.36, 6.34).<\/p>\n<p>The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let <em data-redactor-tag=\"em\">X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(170, 6.28).<\/p>\n<p>Find the <em data-redactor-tag=\"em\">z<\/em>-scores for <em data-redactor-tag=\"em\">x<\/em> = 160.58 cm and <em data-redactor-tag=\"em\">y<\/em> = 162.85 cm. Interpret each <em data-redactor-tag=\"em\">z<\/em>-score. What can you say about <em data-redactor-tag=\"em\">x<\/em> = 160.58 cm and <em data-redactor-tag=\"em\">y<\/em> = 162.85 cm?<\/p>\n<p>Solution:<\/p>\n<p>The <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em> = 160.58 is <em data-redactor-tag=\"em\">z<\/em> = \u20131.5.<\/p>\n<p>The <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">y<\/em> = 162.85 is <em data-redactor-tag=\"em\">z<\/em> = \u20131.5.<\/p>\n<p>Both <em data-redactor-tag=\"em\">x<\/em> = 160.58 and <em data-redactor-tag=\"em\">y<\/em> = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a mean\u00a0<em data-redactor-tag=\"em\">\u00b5<\/em> = 496 and a standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em> = 114. Let <em data-redactor-tag=\"em\">X<\/em> = a SAT exam verbal section score in 2012. Then <em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(496, 114).<\/p>\n<p>Find the <em data-redactor-tag=\"em\">z<\/em>-scores for <em data-redactor-tag=\"em\">x<\/em>1 = 325 and <em data-redactor-tag=\"em\">x<\/em>2 = 366.21. Interpret each <em data-redactor-tag=\"em\">z<\/em>-score. What can you say about <em data-redactor-tag=\"em\">x<\/em>1 = 325 and <em data-redactor-tag=\"em\">x<\/em>2 = 366.21?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q961333\">Show Answer<\/span><\/p>\n<div id=\"q961333\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em>1 = 325 is <em data-redactor-tag=\"em\">z<\/em>1 = \u20131.5.<\/p>\n<p>The <em data-redactor-tag=\"em\">z<\/em>-score for <em data-redactor-tag=\"em\">x<\/em>2 = 366.21 is <em data-redactor-tag=\"em\">z<\/em>2 = \u20131.14.<\/p>\n<p>Student 2 scored closer to the mean than Student 1. Both students scored below the mean, since they both had negative <em data-redactor-tag=\"em\">z<\/em>-scores, Student 2 had the better score.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<h2>The Empirical Rule<\/h2>\n<p>If <em data-redactor-tag=\"em\">X<\/em> is a random variable and has a normal distribution with mean <em data-redactor-tag=\"em\">\u00b5<\/em> and standard deviation <em data-redactor-tag=\"em\">\u03c3<\/em>, then the <strong data-redactor-tag=\"strong\">Empirical Rule<\/strong> says the following:<\/p>\n<ul>\n<li>About 68% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> and +1<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em> (within one standard deviation of the mean).<\/li>\n<li>About 95% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> and +2<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em> (within two standard deviations of the mean).<\/li>\n<li>About 99.7% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> and +3<em data-redactor-tag=\"em\">\u03c3<\/em> of the mean <em data-redactor-tag=\"em\">\u00b5<\/em>(within three standard deviations of the mean). Notice that almost all the<em data-redactor-tag=\"em\">x<\/em> values lie within three standard deviations of the mean.<\/li>\n<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +1<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> are +1 and \u20131, respectively.<\/li>\n<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +2<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> are +2 and \u20132, respectively.<\/li>\n<li>The <em data-redactor-tag=\"em\">z<\/em>-scores for +3<em data-redactor-tag=\"em\">\u03c3<\/em> and \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> are +3 and \u20133 respectively.<\/li>\n<\/ul>\n<p>The empirical rule is also known as the 68-95-99.7 rule.<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-07-at-7.34.36-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-509\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214545\/Screen-Shot-2015-06-07-at-7.34.36-PM.png\" alt=\"Graph of the empirical Rule\" width=\"682\" height=\"392\" \/><\/a><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose <em data-redactor-tag=\"em\">x<\/em> has a normal distribution with mean 50 and standard deviation 6.<\/p>\n<ul>\n<li>About 68% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20131<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20131)(6) = \u20136 and 1<em data-redactor-tag=\"em\">\u03c3<\/em> = (1)(6) = 6 of the mean 50. The values 50 \u2013 6 = 44 and 50 + 6 = 56 are within one standard deviation of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and +1 for 44 and 56, respectively.<\/li>\n<li>About 95% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20132<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20132)(6) = \u201312 and 2<em data-redactor-tag=\"em\">\u03c3<\/em> = (2)(6) = 12. The values 50 \u2013 12 = 38 and 50 + 12 = 62 are within two standard deviations of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and +2 for 38 and 62,respectively.<\/li>\n<li>About 99.7% of the <em data-redactor-tag=\"em\">x<\/em> values lie between \u20133<em data-redactor-tag=\"em\">\u03c3<\/em> = (\u20133)(6) = \u201318 and 3<em data-redactor-tag=\"em\">\u03c3<\/em>= (3)(6) = 18 of the mean 50. The values 50 \u2013 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and +3 for 32 and 68, respectively<\/li>\n<\/ul>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Suppose <em data-redactor-tag=\"em\">X<\/em> has a normal distribution with mean 25 and standard deviation five. Between what values of <em data-redactor-tag=\"em\">x<\/em> do 68% of the values lie?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q590629\">Show Answer<\/span><\/p>\n<div id=\"q590629\" class=\"hidden-answer\" style=\"display: none\">Between 20 and 30.<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let <em data-redactor-tag=\"em\">Y<\/em> = the height of 15 to 18-year-old males in 1984 to 1985. Then <em data-redactor-tag=\"em\">Y<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(172.36, 6.34).<\/p>\n<ol>\n<li>About 68% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 95% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________ respectively.<\/li>\n<li>About 99.7% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two\u00a0 values?\u00a0These values are ________________. The <em data-redactor-tag=\"em\">z<\/em>-scores are ________________ respectively<\/li>\n<\/ol>\n<p><strong>Solution:<\/strong><\/p>\n<ol>\n<li>About 68% of the values lie between the values 166.02 and 178.7 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and 1, respectively.<\/li>\n<li>About 95% of the values lie between the values 159.68 and 185.04 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and 2, respectively.<\/li>\n<li>About 99.7% of the values lie between the values 153.34 and 191.38 cm. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3, respectively.<\/li>\n<\/ol>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>The scores on a college entrance exam have an approximate normal distribution with mean, <em data-redactor-tag=\"em\">\u00b5<\/em> = 52 points and a standard deviation, <em data-redactor-tag=\"em\">\u03c3<\/em> = 11 points.<\/p>\n<p>.7% of the values lie between 153.34 and 191.38. The <em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3.<\/p>\n<ol>\n<li>About 68% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 95% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 99.7% of the <em data-redactor-tag=\"em\">y<\/em> values lie between what two values? These values are ________________. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are ________________, respectively.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226606\">Show Answer<\/span><\/p>\n<div id=\"q226606\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>About 68% of the values lie between the values 41 and 63 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20131 and 1, respectively.<\/li>\n<li>About 95% of the values lie between the values 30 and 74 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20132 and 2, respectively.<\/li>\n<li>About 99.7% of the values lie between the values 19 and 85 points. The\u00a0<em data-redactor-tag=\"em\">z<\/em>-scores are \u20133 and 3, respectively.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>In general, If we know the z-score(s), we can find the amount of area that lies to the left, to the right, or between two z-scores using the Excel function\u00a0<strong>NORM.S.DIST<\/strong>.\u00a0 If we know the area we are interested in, we can also find what z-score(s) are boundaries using the Excel function\u00a0<strong>NORM.S.INV<\/strong>.<\/p>\n<div class=\"textbox shaded\">\n<p>Excel function\u00a0<strong>NORM.S.DIST\u00a0<\/strong>has the following syntax:<\/p>\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Syntax \u2013 Standard Normal Distribution\">\n<p class=\"\" style=\"padding-left: 30px;\">NORM.S.DIST(z,cumulative)<\/p>\n<p style=\"padding-left: 30px;\">The NORM.S.DIST function syntax has the following arguments:<\/p>\n<ul>\n<li><b class=\"ocpRunInHead\">Z<\/b>\u00a0\u00a0\u00a0\u00a0 Required. The value for which you want the distribution.<\/li>\n<li><b class=\"ocpRunInHead\">Cumulative<\/b>\u00a0\u00a0\u00a0\u00a0 Required. Cumulative is a logical value that determines the form of the function. If cumulative is TRUE, NORMS.DIST returns the cumulative distribution function; if FALSE, it returns the probability mass function.<\/li>\n<\/ul>\n<\/section>\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Remarks\">\n<h2 style=\"padding-left: 30px;\">Remarks<\/h2>\n<ul>\n<li>If z is nonnumeric, NORM.S.DIST returns the #VALUE! error value.<\/li>\n<li>The equation for the standard normal density function is:<img decoding=\"async\" src=\"https:\/\/support.content.office.net\/en-us\/media\/bd99a6ad-418a-444f-aa09-533f65ec63b5.gif\" alt=\"Equation\" \/><\/li>\n<\/ul>\n<p style=\"padding-left: 30px;\"><a href=\"https:\/\/support.microsoft.com\/en-us\/office\/norm-s-dist-function-1e787282-3832-4520-a9ae-bd2a8d99ba88\">Excel Support &#8211; Norm.s.dist<\/a><\/p>\n<p>&nbsp;<\/p>\n<p>Excel function\u00a0<strong>NORM.S.INV\u00a0<\/strong>has the following syntax:<\/p>\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Syntax\">\n<p class=\"\" style=\"padding-left: 30px;\">NORM.S.INV(probability)<\/p>\n<p style=\"padding-left: 30px;\">The NORM.S.INV function syntax has the following arguments:<\/p>\n<ul>\n<li>\n<p class=\"x-hidden-focus\"><b class=\"ocpRunInHead\">Probability<\/b>\u00a0\u00a0\u00a0\u00a0 Required. A probability corresponding to the normal distribution.<\/p>\n<\/li>\n<\/ul>\n<\/section>\n<section class=\"ocpSection\" role=\"region\" aria-label=\"Remarks\">\n<h2 style=\"padding-left: 30px;\">Remarks<\/h2>\n<ul>\n<li>If probability is nonnumeric, NORMS.INV returns the #VALUE! error value.<\/li>\n<li>If probability &lt;= 0 or if probability &gt;= 1, NORMS.INV returns the #NUM! error value.<\/li>\n<\/ul>\n<p class=\"\" style=\"padding-left: 30px;\">Given a value for probability, NORM.S.INV seeks that value z such that NORM.S.DIST(z,TRUE) = probability. Thus, precision of NORM.S.INV depends on precision of NORM.S.DIST. NORM.S.INV uses an iterative search technique.<\/p>\n<p style=\"padding-left: 30px;\"><a href=\"https:\/\/support.microsoft.com\/en-us\/office\/norm-s-inv-function-d6d556b4-ab7f-49cd-b526-5a20918452b1\">Excel Suport &#8211; Norm.s.inv<\/a><\/p>\n<\/section>\n<\/section>\n<\/div>\n<div><\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<ol>\n<li>Find the area to the left of [latex]z=1.26[\/latex].<\/li>\n<li>Find the area to the right of [latex]z=-0.95[\/latex].<\/li>\n<li>Find the area between [latex]z=-1.74[\/latex] and [latex]z=2.12[\/latex]<\/li>\n<\/ol>\n<p><strong>Solution:<\/strong><\/p>\n<p>1. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2744 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12231959\/norm.s.dist-1.jpg\" alt=\"\" width=\"317\" height=\"175\" \/><\/p>\n<p><strong>In Excel:<\/strong> Use =NORM.S.DIST(1.26,1)= 0.896.\u00a0 89.6% of values fall to the left of [latex]z=1.26[\/latex].<\/p>\n<p>2. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2745 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12232434\/norm.s.dist-2.jpg\" alt=\"\" width=\"314\" height=\"188\" \/><\/p>\n<p><strong>In Excel:<\/strong> Since the NORM.S.DIST function returns the area to the left of a z-score, we will need to use the fact that the total area under the standard normal curve is 1 to get to our answer.\u00a0 Use =1-NORM.S.DIST(-0.95, 1) = 1- 0.171 = 0.829.\u00a0 82.9% of values fall to the right of\u00a0[latex]z=-0.95[\/latex].<\/p>\n<p>3. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2746 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12233134\/norm.s.dist-3.jpg\" alt=\"\" width=\"314\" height=\"188\" \/><\/p>\n<p><strong>In Excel:\u00a0<\/strong>Use =NORM.S.DIST(2.12,1) &#8211; NORM.S.DIST(-1.74,1)= 0.983 &#8211; 0.041 = 0.943.\u00a0 94.3% of values fall between\u00a0 [latex]z=-1.74[\/latex] and [latex]z=2.12[\/latex].\u00a0 *Note, in this calculation we find the area to the left of the larger z-score then subtract the area to the left of the smaller z-score.<\/p>\n<\/div>\n<div class=\"textbox exercises\"><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<ol>\n<li>Find the z-score that has an area of 0.378 to its left.<\/li>\n<li>Find the z-score that has an area of 0.092 to its right.<\/li>\n<li>Find the symmetric z-scores that correspond to having 0.10 as the combined area in two tails.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q713595\">Show Solution<\/span><\/p>\n<div id=\"q713595\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Solution:<\/strong><\/p>\n<p>1. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2751 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/12235639\/norm.s.inv-1.jpg\" alt=\"\" width=\"346\" height=\"208\" \/><\/p>\n<p><strong>In Excel:<\/strong> Use =NORM.S.INV(0.378) = -0.31.\u00a0 The z-score -0.31 is the boundary that cuts off an area to the left of 0.378.<\/p>\n<p>2. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2752 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/13000105\/norm.s.inv-2.jpg\" alt=\"\" width=\"386\" height=\"228\" \/><\/p>\n<p><strong>In Excel:<\/strong> Since the NORM.S.INV function takes as an input area to the left of a z-score, we will need to use the fact that the total area under the standard normal curve is 1 to get to our answer.\u00a0 So, 0.092 to the right means there is 1-0.092= 0.908 to the left of the z-score we are interested in.<\/p>\n<p>Use =NORM.S.INV(0.908) = 1.33.\u00a0 The z-score 1.33 is the boundary that cuts off an area of 0.092 to the right.<\/p>\n<p>3. Start by drawing a sketch.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2753 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/13000506\/norm.s.inv-3.jpg\" alt=\"\" width=\"344\" height=\"205\" \/><\/p>\n<p>&nbsp;<\/p>\n<p><strong>In Excel:\u00a0<\/strong>To have a combined area of 0.10 in the two tails cut off by symmetric z-scores, we have 0.05 in each tail.\u00a0 If we find the z-score that corresponds to the area in the left tail, the symmetric z-score that cuts off the right tail will just be the positive.<\/p>\n<p>Use =NORM.S.INV(0.05) =-1.65.\u00a0 So, the two symmetric z-scores are -1.65 and 1.65.<\/p>\n<p>&nbsp;<\/p>\n<p>Put Answer Here<\/p><\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<h2>References<\/h2>\n<p>&#8220;Blood Pressure of Males and Females.&#8221; StatCruch, 2013. Available online at http:\/\/www.statcrunch.com\/5.0\/viewreport.php?reportid=11960 (accessed May 14, 2013).<\/p>\n<p>&#8220;The Use of Epidemiological Tools in Conflict-affected populations: Open-access educational resources for policy-makers: Calculation of z-scores.&#8221; London School of Hygiene and Tropical Medicine, 2009. Available online at http:\/\/conflict.lshtm.ac.uk\/page_125.htm (accessed May 14, 2013).<\/p>\n<p>&#8220;2012 College-Bound Seniors Total Group Profile Report.&#8221; CollegeBoard, 2012. Available online at http:\/\/media.collegeboard.com\/digitalServices\/pdf\/research\/TotalGroup-2012.pdf (accessed May 14, 2013).<\/p>\n<p>&#8220;Digest of Education Statistics: ACT score average and standard deviations by sex and race\/ethnicity and percentage of ACT test takers, by selected composite score ranges and planned fields of study: Selected years, 1995 through 2009.&#8221; National Center for Education Statistics. Available online at http:\/\/nces.ed.gov\/programs\/digest\/d09\/tables\/dt09_147.asp (accessed May 14, 2013).<\/p>\n<p>Data from the <em data-redactor-tag=\"em\">San Jose Mercury News<\/em>.<\/p>\n<p>Data from <em data-redactor-tag=\"em\">The World Almanac and Book of Facts<\/em>.<\/p>\n<p>&#8220;List of stadiums by capacity.&#8221; Wikipedia. Available online at https:\/\/en.wikipedia.org\/wiki\/List_of_stadiums_by_capacity (accessed May 14, 2013).<\/p>\n<p>Data from the National Basketball Association. Available online at www.nba.com (accessed May 14, 2013).<\/p>\n<h1>Concept Review<\/h1>\n<p>A <em data-redactor-tag=\"em\">z<\/em>-score is a standardized value. Its distribution is the standard normal, <em data-redactor-tag=\"em\">Z<\/em> ~<em data-redactor-tag=\"em\">N<\/em>(0, 1). The mean of the <em data-redactor-tag=\"em\">z<\/em>-scores is zero and the standard deviation is one. If <em data-redactor-tag=\"em\">z\u00a0<\/em>is the <em data-redactor-tag=\"em\">z<\/em>-score for a value <em data-redactor-tag=\"em\">x<\/em> from the normal distribution <em data-redactor-tag=\"em\">N<\/em>(<em data-redactor-tag=\"em\">\u00b5<\/em>, <em data-redactor-tag=\"em\">\u03c3<\/em>) then <em data-redactor-tag=\"em\">z<\/em> tells you how many standard deviations <em data-redactor-tag=\"em\">x<\/em> is above (greater than) or below (less than) <em data-redactor-tag=\"em\">\u00b5<\/em>. The Empirical Rule provides approximation for the percentage of values that lie within key intervals in a normal distribution.<\/p>\n<h1>Formula Review<\/h1>\n<p><em data-redactor-tag=\"em\">Z<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(0, 1)<\/p>\n<p><em data-redactor-tag=\"em\">z<\/em> = a standardized value (<em data-redactor-tag=\"em\">z<\/em>-score)<\/p>\n<p>mean = 0; standard deviation = 1<\/p>\n<p>To find an x value\u00a0when the <em data-redactor-tag=\"em\">z<\/em>-score is known: [latex]\\displaystyle{x}={\\mu}+{z}{\\sigma}[\/latex]<\/p>\n<p><em data-redactor-tag=\"em\">z<\/em>-score:[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]<\/p>\n<p><em data-redactor-tag=\"em\">Z<\/em> = the random variable for <em data-redactor-tag=\"em\">z<\/em>-scores<\/p>\n<p><em data-redactor-tag=\"em\">Z<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(0, 1)<\/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-238\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>OpenStax, Statistics, The Standard Normal Distribution. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"\"><\/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>Introductory Statistics . <strong>Authored by<\/strong>: Barbara Illowski, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/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>: Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Normal Distributionu2014Explained Simply (part 1). <strong>Authored by<\/strong>: how2stats. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/embed\/xgQhefFOXrM\">https:\/\/www.youtube.com\/embed\/xgQhefFOXrM<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube license<\/li><li>Normal Distribution 2014Explained Simply (part 2). <strong>Authored by<\/strong>: how2stats. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/embed\/iiRiOlkLa6A\">https:\/\/www.youtube.com\/embed\/iiRiOlkLa6A<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube license<\/li><li>ck12.org normal distribution problems: z-score | Probability and Statistics | Khan Academy. <strong>Authored by<\/strong>: Khan Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=Wp2nVIzBsE8\">https:\/\/www.youtube.com\/watch?v=Wp2nVIzBsE8<\/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":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax, Statistics, The Standard Normal Distribution\",\"author\":\"\",\"organization\":\"\",\"url\":\"Download for free at http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44a\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"copyrighted_video\",\"description\":\"Normal Distributionu2014Explained 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