{"id":5372,"date":"2021-03-17T23:34:40","date_gmt":"2021-03-17T23:34:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/nwfsc-mathforliberalartscorequisite\/chapter\/the-standard-normal-distribution\/"},"modified":"2024-03-12T17:36:57","modified_gmt":"2024-03-12T17:36:57","slug":"the-standard-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/nwfsc-mathforliberalartscorequisite\/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>z<\/em>-scores<\/strong>. <strong>A <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>\u03bc<\/em> and standard deviation <em>\u03c3<\/em>.\r\n<h1><em>Z<\/em>-Scores<\/h1>\r\nIf <em>X<\/em> is a normally distributed random variable and <em>X<\/em> ~ <em>N(\u03bc, \u03c3)<\/em>, then the <em>z<\/em>-score is:\r\n\r\n[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]\r\n\r\n<strong>The <em>z<\/em>-score tells you how many standard deviations the value <em>x<\/em> is above (to the right of) or below (to the left of) the mean, <em>\u03bc<\/em>.<\/strong> Values of <em>x<\/em> that are larger than the mean have positive <em>z<\/em>-scores, and values of <em>x<\/em> that are smaller than the mean have negative <em>z<\/em>-scores. If <em>x<\/em> equals the mean, then <em>x<\/em> has a <em>z<\/em>-score of zero.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose <em>X<\/em> ~ <em>N(5, 6)<\/em>. This says that <em>x<\/em> is a normally distributed random variable with mean <em>\u03bc<\/em> = 5 and standard deviation <em>\u03c3<\/em> = 6. Suppose <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>x<\/em> = 17 is<strong> two standard deviations<\/strong> (2<em>\u03c3<\/em>) above or to the right of the mean <em>\u03bc<\/em> = 5. The standard deviation is <em>\u03c3<\/em> = 6.\r\n\r\nNotice that: 5 + (2)(6) = 17 (The pattern is <em>\u03bc<\/em> + <em>z\u03c3<\/em> = <em>x<\/em>)\r\n\r\nNow suppose <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>This means that <em>x<\/em> = 1 is 0.67 standard deviations (\u20130.67<em>\u03c3<\/em>) below or to the left of the mean <em>\u03bc<\/em> = 5. Notice that:<\/strong> 5 + (\u20130.67)(6) is approximately equal to one (This has the pattern <em>\u03bc<\/em> + (\u20130.67)\u03c3 = 1)\r\n\r\nSummarizing, when <em>z<\/em> is positive, <em>x<\/em> is above or to the right of <em>\u03bc<\/em> and when <em>z<\/em>is negative, <em>x<\/em> is to the left of or below <em>\u03bc<\/em>. Or, when <em>z<\/em> is positive, <em>x<\/em> is greater than <em>\u03bc<\/em>, and when <em>z<\/em> is negative <em>x<\/em> is less than <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>z<\/em>-score of <em>x<\/em>, when <em>x<\/em> = 1 and<em>X<\/em> ~ <em>N<\/em>(12,3)?\r\n\r\n[reveal-answer q=\"600161\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"600161\"][latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67} [\/latex][\/hidden-answer]\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>X<\/em> = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. <em>X<\/em> ~ <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>z<\/em>-score when <em>x<\/em> = 10 pounds is <em>z<\/em> = 2.5 (verify). This <em>z<\/em>-score tells you that<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>z<\/em> = __________. This <em>z<\/em>-score tells you that <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>z<\/em>-score tells you that <em>x<\/em> = 10 is <strong>2.5<\/strong> standard deviations to the <strong>right<\/strong> of the mean <strong>five<\/strong>.<\/li>\r\n \t<li><em>z<\/em>= \u20134. This <em>z<\/em>-score tells you that <em>x<\/em> = \u20133 is <strong>4<\/strong> standard deviations to the <strong>left<\/strong> of the mean.<\/li>\r\n<\/ol>\r\nSuppose the random variables <em>X<\/em> and <em>Y<\/em> have the following normal distributions: <em>X<\/em> ~ <em>N<\/em>(5, 6) and <em>Y<\/em> ~ <em>N<\/em>(2, 1). If <em>x<\/em> = 17, then <em>z<\/em> = 2. (This was previously shown.) If <em>y<\/em> = 4, what is <em>z<\/em>? [latex]\\displaystyle {z}=\\frac{{y - \\mu}}{{\\sigma}} = \\frac{{4-2}}{{1}}[\/latex].\r\n\r\nThe <em>z<\/em>-score for <em>y<\/em> = 4 is <em>z<\/em> = 2. This means that four is <em>z<\/em> = 2 standard deviations to the right of the mean. Therefore, <em>x<\/em> = 17 and <em>y<\/em> = 4 are both two (of <strong>their own<\/strong>) standard deviations to the right of their respective means.\r\n\r\n<strong>The <em>z<\/em>-score allows us to compare data that are scaled differently.<\/strong> To understand the concept, suppose <em>X<\/em> ~ <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>Y<\/em> ~ <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>x<\/em> = 17 and <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>X<\/em> ~<em>N<\/em>(16,4). Suppose Jerome scores ten points in a game. The <em>z<\/em>\u2013score when <em>x<\/em> = 10 is \u20131.5. This score tells you that <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=\"263084\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"263084\"]1.5, left, 16[\/hidden-answer]\r\n\r\n<\/div>\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>X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em>X<\/em> ~ <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>z<\/em>-score when <em>x<\/em> = 168 cm is <em>z<\/em> = _______. This <em>z<\/em>-score tells you that <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>z<\/em>-score of <em>z<\/em> = 1.27. What is the male's height? The <em>z<\/em>-score (<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, 1.27, right\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>try it<\/h3>\r\nUse the information from the last example 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>z<\/em>-score when <em>x<\/em> = 176 cm is <em>z<\/em> = _______. This <em>z<\/em>-score tells you that <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>z<\/em>-score of <em>z<\/em> = \u20132. What is the male's height? The <em>z<\/em>-score (<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\nSolve the equation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] for x. x = \u03bc + (z)(\u03c3) for <em>x<\/em>. <em>x<\/em> = <em>\u03bc<\/em> + (<em>z<\/em>)(<em>\u03c3<\/em>)\r\n<ol>\r\n \t<li>z=0.96. 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>X<\/em> = 157.44 cm, The <em>z<\/em>-score(<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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3><span style=\"font-size: 1.2em;\">Example<\/span><\/h3>\r\n<div class=\"textbox exercises\">\r\n\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>Y<\/em> = the height of 15 to 18-year-old males from 1984 to 1985. Then <em>Y<\/em> ~ <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>X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em>X<\/em> ~ <em>N<\/em>(170, 6.28).\r\n\r\nFind the <em>z<\/em>-scores for <em>x<\/em> = 160.58 cm and <em>y<\/em> = 162.85 cm. Interpret each <em>z<\/em>-score. What can you say about <em>x<\/em> = 160.58 cm and <em>y<\/em> = 162.85 cm?\r\n\r\nSolution:\r\n\r\nThe <em>z<\/em>-score for <em>x<\/em> = 160.58 is <em>z<\/em> = \u20131.5.\r\n\r\nThe <em>z<\/em>-score for <em>y<\/em> = 162.85 is <em>z<\/em> = \u20131.5.Both <em>x<\/em> = 160.58 and <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<\/div>\r\n<div class=\"textbox exercises\">\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<em>\u00b5<\/em> = 496 and a standard deviation <em>\u03c3<\/em> = 114. Let <em>X<\/em> = a SAT exam verbal section score in 2012. Then <em>X<\/em> ~ <em>N<\/em>(496, 114).\r\n\r\nFind the <em>z<\/em>-scores for <em>x<\/em>1 = 325 and <em>x<\/em>2 = 366.21. Interpret each <em>z<\/em>-score. What can you say about <em>x<\/em>1 = 325 and <em>x<\/em>2 = 366.21?\r\n\r\nThe <em>z<\/em>-score for <em>x<\/em>1 = 325 is <em>z<\/em>1 = \u20131.5.\r\n\r\nThe <em>z<\/em>-score for <em>x<\/em>2 = 366.21 is <em>z<\/em>2 = \u20131.14.\r\n\r\nStudent 2 scored closer to the mean than Student 1 and, since they both had negative <em>z<\/em>-scores, Student 2 had the better score.\r\n\r\n<\/div>\r\n<h2>The Empirical Rule<\/h2>\r\nIf <em>X<\/em> is a random variable and has a normal distribution with mean <em>\u00b5<\/em> and standard deviation <em>\u03c3<\/em>, then the <strong>Empirical Rule<\/strong> says the following:\r\n<ul>\r\n \t<li>About 68% of the <em>x<\/em> values lie between \u20131<em>\u03c3<\/em> and +1<em>\u03c3<\/em> of the mean <em>\u00b5<\/em> (within one standard deviation of the mean).<\/li>\r\n \t<li>About 95% of the <em>x<\/em> values lie between \u20132<em>\u03c3<\/em> and +2<em>\u03c3<\/em> of the mean <em>\u00b5<\/em> (within two standard deviations of the mean).<\/li>\r\n \t<li>About 99.7% of the <em>x<\/em> values lie between \u20133<em>\u03c3<\/em> and +3<em>\u03c3<\/em> of the mean <em>\u00b5<\/em>(within three standard deviations of the mean). Notice that almost all the<em>x<\/em> values lie within three standard deviations of the mean.<\/li>\r\n \t<li>The <em>z<\/em>-scores for +1<em>\u03c3<\/em> and \u20131<em>\u03c3<\/em> are +1 and \u20131, respectively.<\/li>\r\n \t<li>The <em>z<\/em>-scores for +2<em>\u03c3<\/em> and \u20132<em>\u03c3<\/em> are +2 and \u20132, respectively.<\/li>\r\n \t<li>The <em>z<\/em>-scores for +3<em>\u03c3<\/em> and \u20133<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.\r\n\r\n<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\" \/>\r\n<h3 style=\"text-align: start;\"><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose <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>x<\/em> values lie between \u20131<em>\u03c3<\/em> = (\u20131)(6) = \u20136 and 1<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>z<\/em>-scores are \u20131 and +1 for 44 and 56, respectively.<\/li>\r\n \t<li>About 95% of the <em>x<\/em> values lie between \u20132<em>\u03c3<\/em> = (\u20132)(6) = \u201312 and 2<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>z<\/em>-scores are \u20132 and +2 for 38 and 62, respectively.<\/li>\r\n \t<li>About 99.7% of the <em>x<\/em> values lie between \u20133<em>\u03c3<\/em> = (\u20133)(6) = \u201318 and 3<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>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>X<\/em> has a normal distribution with mean 25 and standard deviation five. Between what values of <em>x<\/em> do 68% of the values lie?\r\n\r\nSolution:\r\n\r\nBetween 20 and 30.\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>Y<\/em> = the height of 15 to 18-year-old males in 1984 to 1985. Then <em>Y<\/em> ~ <em>N<\/em>(172.36, 6.34).\r\n<ol>\r\n \t<li>About 68% of the <em>y<\/em> values lie between what two values? These values are ________________. The <em>z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 95% of the <em>y<\/em> values lie between what two values? These values are ________________. The <em>z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 99.7% of the <em>y<\/em> values lie between what two values? These values are ________________.\u00a0 The z-scores are ________________, respectively.<\/li>\r\n<\/ol>\r\nSOLUTION: 1. 166.02 and 178.7, -1 and 1\r\n\r\n2. 159.68 and 185.04, -2 and 2\r\n\r\n3. 153.34 and 191.38, -3 and 3\r\n\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>\u00b5<\/em> = 52 points and a standard deviation, <em>\u03c3<\/em> = 11 points.\r\n<ol>\r\n \t<li>About 68% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 95% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\r\n \t<li>About 99.7% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\r\n<\/ol>\r\nSolution:\r\n<ol>\r\n \t<li>About 68% of the values lie between the values 41 and 63. The\u00a0<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. The\u00a0<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. The\u00a0<em>z<\/em>-scores are \u20133 and 3, respectively.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\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>San Jose Mercury News<\/em>.\r\n\r\nData from <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).","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>z<\/em>-scores<\/strong>. <strong>A <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>\u03bc<\/em> and standard deviation <em>\u03c3<\/em>.<\/p>\n<h1><em>Z<\/em>-Scores<\/h1>\n<p>If <em>X<\/em> is a normally distributed random variable and <em>X<\/em> ~ <em>N(\u03bc, \u03c3)<\/em>, then the <em>z<\/em>-score is:<\/p>\n<p>[latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex]<\/p>\n<p><strong>The <em>z<\/em>-score tells you how many standard deviations the value <em>x<\/em> is above (to the right of) or below (to the left of) the mean, <em>\u03bc<\/em>.<\/strong> Values of <em>x<\/em> that are larger than the mean have positive <em>z<\/em>-scores, and values of <em>x<\/em> that are smaller than the mean have negative <em>z<\/em>-scores. If <em>x<\/em> equals the mean, then <em>x<\/em> has a <em>z<\/em>-score of zero.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose <em>X<\/em> ~ <em>N(5, 6)<\/em>. This says that <em>x<\/em> is a normally distributed random variable with mean <em>\u03bc<\/em> = 5 and standard deviation <em>\u03c3<\/em> = 6. Suppose <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>x<\/em> = 17 is<strong> two standard deviations<\/strong> (2<em>\u03c3<\/em>) above or to the right of the mean <em>\u03bc<\/em> = 5. The standard deviation is <em>\u03c3<\/em> = 6.<\/p>\n<p>Notice that: 5 + (2)(6) = 17 (The pattern is <em>\u03bc<\/em> + <em>z\u03c3<\/em> = <em>x<\/em>)<\/p>\n<p>Now suppose <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>This means that <em>x<\/em> = 1 is 0.67 standard deviations (\u20130.67<em>\u03c3<\/em>) below or to the left of the mean <em>\u03bc<\/em> = 5. Notice that:<\/strong> 5 + (\u20130.67)(6) is approximately equal to one (This has the pattern <em>\u03bc<\/em> + (\u20130.67)\u03c3 = 1)<\/p>\n<p>Summarizing, when <em>z<\/em> is positive, <em>x<\/em> is above or to the right of <em>\u03bc<\/em> and when <em>z<\/em>is negative, <em>x<\/em> is to the left of or below <em>\u03bc<\/em>. Or, when <em>z<\/em> is positive, <em>x<\/em> is greater than <em>\u03bc<\/em>, and when <em>z<\/em> is negative <em>x<\/em> is less than <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>z<\/em>-score of <em>x<\/em>, when <em>x<\/em> = 1 and<em>X<\/em> ~ <em>N<\/em>(12,3)?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q600161\">Show Answer<\/span><\/p>\n<div id=\"q600161\" class=\"hidden-answer\" style=\"display: none\">[latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67}[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" 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>X<\/em> = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. <em>X<\/em> ~ <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>z<\/em>-score when <em>x<\/em> = 10 pounds is <em>z<\/em> = 2.5 (verify). This <em>z<\/em>-score tells you that<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>z<\/em> = __________. This <em>z<\/em>-score tells you that <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>z<\/em>-score tells you that <em>x<\/em> = 10 is <strong>2.5<\/strong> standard deviations to the <strong>right<\/strong> of the mean <strong>five<\/strong>.<\/li>\n<li><em>z<\/em>= \u20134. This <em>z<\/em>-score tells you that <em>x<\/em> = \u20133 is <strong>4<\/strong> standard deviations to the <strong>left<\/strong> of the mean.<\/li>\n<\/ol>\n<p>Suppose the random variables <em>X<\/em> and <em>Y<\/em> have the following normal distributions: <em>X<\/em> ~ <em>N<\/em>(5, 6) and <em>Y<\/em> ~ <em>N<\/em>(2, 1). If <em>x<\/em> = 17, then <em>z<\/em> = 2. (This was previously shown.) If <em>y<\/em> = 4, what is <em>z<\/em>? [latex]\\displaystyle {z}=\\frac{{y - \\mu}}{{\\sigma}} = \\frac{{4-2}}{{1}}[\/latex].<\/p>\n<p>The <em>z<\/em>-score for <em>y<\/em> = 4 is <em>z<\/em> = 2. This means that four is <em>z<\/em> = 2 standard deviations to the right of the mean. Therefore, <em>x<\/em> = 17 and <em>y<\/em> = 4 are both two (of <strong>their own<\/strong>) standard deviations to the right of their respective means.<\/p>\n<p><strong>The <em>z<\/em>-score allows us to compare data that are scaled differently.<\/strong> To understand the concept, suppose <em>X<\/em> ~ <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>Y<\/em> ~ <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>x<\/em> = 17 and <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>X<\/em> ~<em>N<\/em>(16,4). Suppose Jerome scores ten points in a game. The <em>z<\/em>\u2013score when <em>x<\/em> = 10 is \u20131.5. This score tells you that <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=\"q263084\">Show Answer<\/span><\/p>\n<div id=\"q263084\" class=\"hidden-answer\" style=\"display: none\">1.5, left, 16<\/div>\n<\/div>\n<\/div>\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>X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em>X<\/em> ~ <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>z<\/em>-score when <em>x<\/em> = 168 cm is <em>z<\/em> = _______. This <em>z<\/em>-score tells you that <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>z<\/em>-score of <em>z<\/em> = 1.27. What is the male&#8217;s height? The <em>z<\/em>-score (<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, 1.27, right<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>try it<\/h3>\n<p>Use the information from the last example 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>z<\/em>-score when <em>x<\/em> = 176 cm is <em>z<\/em> = _______. This <em>z<\/em>-score tells you that <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>z<\/em>-score of <em>z<\/em> = \u20132. What is the male&#8217;s height? The <em>z<\/em>-score (<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<p>Solve the equation [latex]\\displaystyle{z}=\\frac{{x - \\mu}}{{\\sigma}}[\/latex] for x. x = \u03bc + (z)(\u03c3) for <em>x<\/em>. <em>x<\/em> = <em>\u03bc<\/em> + (<em>z<\/em>)(<em>\u03c3<\/em>)<\/p>\n<ol>\n<li>z=0.96. 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>X<\/em> = 157.44 cm, The <em>z<\/em>-score(<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 class=\"textbox exercises\">\n<h3><span style=\"font-size: 1.2em;\">Example<\/span><\/h3>\n<div class=\"textbox exercises\">\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>Y<\/em> = the height of 15 to 18-year-old males from 1984 to 1985. Then <em>Y<\/em> ~ <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>X<\/em> = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then <em>X<\/em> ~ <em>N<\/em>(170, 6.28).<\/p>\n<p>Find the <em>z<\/em>-scores for <em>x<\/em> = 160.58 cm and <em>y<\/em> = 162.85 cm. Interpret each <em>z<\/em>-score. What can you say about <em>x<\/em> = 160.58 cm and <em>y<\/em> = 162.85 cm?<\/p>\n<p>Solution:<\/p>\n<p>The <em>z<\/em>-score for <em>x<\/em> = 160.58 is <em>z<\/em> = \u20131.5.<\/p>\n<p>The <em>z<\/em>-score for <em>y<\/em> = 162.85 is <em>z<\/em> = \u20131.5.Both <em>x<\/em> = 160.58 and <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<\/div>\n<div class=\"textbox exercises\">\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<em>\u00b5<\/em> = 496 and a standard deviation <em>\u03c3<\/em> = 114. Let <em>X<\/em> = a SAT exam verbal section score in 2012. Then <em>X<\/em> ~ <em>N<\/em>(496, 114).<\/p>\n<p>Find the <em>z<\/em>-scores for <em>x<\/em>1 = 325 and <em>x<\/em>2 = 366.21. Interpret each <em>z<\/em>-score. What can you say about <em>x<\/em>1 = 325 and <em>x<\/em>2 = 366.21?<\/p>\n<p>The <em>z<\/em>-score for <em>x<\/em>1 = 325 is <em>z<\/em>1 = \u20131.5.<\/p>\n<p>The <em>z<\/em>-score for <em>x<\/em>2 = 366.21 is <em>z<\/em>2 = \u20131.14.<\/p>\n<p>Student 2 scored closer to the mean than Student 1 and, since they both had negative <em>z<\/em>-scores, Student 2 had the better score.<\/p>\n<\/div>\n<h2>The Empirical Rule<\/h2>\n<p>If <em>X<\/em> is a random variable and has a normal distribution with mean <em>\u00b5<\/em> and standard deviation <em>\u03c3<\/em>, then the <strong>Empirical Rule<\/strong> says the following:<\/p>\n<ul>\n<li>About 68% of the <em>x<\/em> values lie between \u20131<em>\u03c3<\/em> and +1<em>\u03c3<\/em> of the mean <em>\u00b5<\/em> (within one standard deviation of the mean).<\/li>\n<li>About 95% of the <em>x<\/em> values lie between \u20132<em>\u03c3<\/em> and +2<em>\u03c3<\/em> of the mean <em>\u00b5<\/em> (within two standard deviations of the mean).<\/li>\n<li>About 99.7% of the <em>x<\/em> values lie between \u20133<em>\u03c3<\/em> and +3<em>\u03c3<\/em> of the mean <em>\u00b5<\/em>(within three standard deviations of the mean). Notice that almost all the<em>x<\/em> values lie within three standard deviations of the mean.<\/li>\n<li>The <em>z<\/em>-scores for +1<em>\u03c3<\/em> and \u20131<em>\u03c3<\/em> are +1 and \u20131, respectively.<\/li>\n<li>The <em>z<\/em>-scores for +2<em>\u03c3<\/em> and \u20132<em>\u03c3<\/em> are +2 and \u20132, respectively.<\/li>\n<li>The <em>z<\/em>-scores for +3<em>\u03c3<\/em> and \u20133<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.<\/p>\n<p><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\" \/><\/p>\n<h3 style=\"text-align: start;\"><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose <em>x<\/em> has a normal distribution with mean 50 and standard deviation 6.<\/p>\n<ul>\n<li>About 68% of the <em>x<\/em> values lie between \u20131<em>\u03c3<\/em> = (\u20131)(6) = \u20136 and 1<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>z<\/em>-scores are \u20131 and +1 for 44 and 56, respectively.<\/li>\n<li>About 95% of the <em>x<\/em> values lie between \u20132<em>\u03c3<\/em> = (\u20132)(6) = \u201312 and 2<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>z<\/em>-scores are \u20132 and +2 for 38 and 62, respectively.<\/li>\n<li>About 99.7% of the <em>x<\/em> values lie between \u20133<em>\u03c3<\/em> = (\u20133)(6) = \u201318 and 3<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>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>X<\/em> has a normal distribution with mean 25 and standard deviation five. Between what values of <em>x<\/em> do 68% of the values lie?<\/p>\n<p>Solution:<\/p>\n<p>Between 20 and 30.<\/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>Y<\/em> = the height of 15 to 18-year-old males in 1984 to 1985. Then <em>Y<\/em> ~ <em>N<\/em>(172.36, 6.34).<\/p>\n<ol>\n<li>About 68% of the <em>y<\/em> values lie between what two values? These values are ________________. The <em>z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 95% of the <em>y<\/em> values lie between what two values? These values are ________________. The <em>z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 99.7% of the <em>y<\/em> values lie between what two values? These values are ________________.\u00a0 The z-scores are ________________, respectively.<\/li>\n<\/ol>\n<p>SOLUTION: 1. 166.02 and 178.7, -1 and 1<\/p>\n<p>2. 159.68 and 185.04, -2 and 2<\/p>\n<p>3. 153.34 and 191.38, -3 and 3<\/p>\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>\u00b5<\/em> = 52 points and a standard deviation, <em>\u03c3<\/em> = 11 points.<\/p>\n<ol>\n<li>About 68% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 95% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\n<li>About 99.7% of the <em>y<\/em> values lie between what two values? These values are ________________. The\u00a0<em>z<\/em>-scores are ________________, respectively.<\/li>\n<\/ol>\n<p>Solution:<\/p>\n<ol>\n<li>About 68% of the values lie between the values 41 and 63. The\u00a0<em>z<\/em>-scores are \u20131 and 1, respectively.<\/li>\n<li>About 95% of the values lie between the values 30 and 74. The\u00a0<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. The\u00a0<em>z<\/em>-scores are \u20133 and 3, respectively.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\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>San Jose Mercury News<\/em>.<\/p>\n<p>Data from <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\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-5372\">\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":167848,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax, Statistics, The Standard Normal 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