{"id":258,"date":"2021-07-14T15:59:00","date_gmt":"2021-07-14T15:59:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-standard-normal-distribution\/"},"modified":"2023-12-05T09:18:56","modified_gmt":"2023-12-05T09:18:56","slug":"the-standard-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-standard-normal-distribution\/","title":{"raw":"What is the Standard Normal Distribution?","rendered":"What is 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>Calculate and interpret <em>z<\/em>-scores<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Distance<\/h3>\r\nDistance is how far one number on a number line is from another number on a number line. Distance is positive. In statistics we find the distance from a data point to the mean when we are finding <em>z<\/em>-scores. <em>Z<\/em>-scores are described below. If you have a negative <em>z<\/em>-score, the negative is directional, meaning below the mean.\r\n\r\n<\/div>\r\nThe <strong>standard normal distribution<\/strong> is a normal distribution of standardized values called\u00a0<strong><em data-redactor-tag=\"em\">z<\/em>-scores<\/strong>. A <em data-redactor-tag=\"em\">z<\/em>-score is measured in units of the standard deviation. 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\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7115018&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xgQhefFOXrM&amp;video_target=tpm-plugin-yqp2tlvm-xgQhefFOXrM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7115019&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=iiRiOlkLa6A&amp;video_target=tpm-plugin-wz1bvzwv-iiRiOlkLa6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n<h1><em data-redactor-tag=\"em\">Z<\/em>-Scores<\/h1>\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 (\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. Notice that:<\/strong> 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<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<em data-redactor-tag=\"em\">X<\/em> ~ <em data-redactor-tag=\"em\">N<\/em>(12,3)?\r\n\r\n[reveal-answer q=\"632427\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"632427\"]\r\n\r\n[latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67} [\/latex]\r\n\r\n[\/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 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 style=\"list-style-type: lower-alpha;\">\r\n \t<li>Suppose a person <strong>lost<\/strong> 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 <strong>gained<\/strong> 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 \t<li>Suppose the random variables\u00a0<em data-effect=\"italics\">X<\/em>\u00a0and\u00a0<em data-effect=\"italics\">Y<\/em>\u00a0have the following normal distributions:\u00a0<em data-effect=\"italics\">X<\/em>\u00a0~\u00a0<em data-effect=\"italics\">N<\/em>(5, 6) and\u00a0<em data-effect=\"italics\">Y<\/em>\u00a0~\u00a0<em data-effect=\"italics\">N<\/em>(2, 1). If\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 17, then\u00a0<em data-effect=\"italics\">z<\/em>\u00a0= 2. (This was previously shown.) If\u00a0<em data-effect=\"italics\">y<\/em>\u00a0= 4, what is\u00a0<em data-effect=\"italics\">z<\/em>?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"818954\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"818954\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\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\">four<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">left<\/strong> of the mean.<\/li>\r\n \t<li>[latex]z= \\frac{y- \\mu}{\\sigma} = \\frac{4-2}{1}[\/latex] where [latex]\\mu = 2[\/latex] and [latex]\\sigma = 1[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\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 <strong>relative to their means.<\/strong>\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=\"590652\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"590652\"]\r\n\r\n1.5, left, 16\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list4253\">\n<li>Calculate and interpret <em>z<\/em>-scores<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Distance<\/h3>\n<p>Distance is how far one number on a number line is from another number on a number line. Distance is positive. In statistics we find the distance from a data point to the mean when we are finding <em>z<\/em>-scores. <em>Z<\/em>-scores are described below. If you have a negative <em>z<\/em>-score, the negative is directional, meaning below the mean.<\/p>\n<\/div>\n<p>The <strong>standard normal distribution<\/strong> is a normal distribution of standardized values called\u00a0<strong><em data-redactor-tag=\"em\">z<\/em>-scores<\/strong>. A <em data-redactor-tag=\"em\">z<\/em>-score is measured in units of the standard deviation. 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\" src=\"\/\/plugin.3playmedia.com\/show?mf=7115018&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xgQhefFOXrM&amp;video_target=tpm-plugin-yqp2tlvm-xgQhefFOXrM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7115019&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=iiRiOlkLa6A&amp;video_target=tpm-plugin-wz1bvzwv-iiRiOlkLa6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<h1><em data-redactor-tag=\"em\">Z<\/em>-Scores<\/h1>\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 (\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. Notice that:<\/strong> 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<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<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=\"q632427\">Show Answer<\/span><\/p>\n<div id=\"q632427\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\displaystyle {z}=\\frac{{1-12}}{{3}} = -{3.67}[\/latex]<\/p>\n<\/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 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 style=\"list-style-type: lower-alpha;\">\n<li>Suppose a person <strong>lost<\/strong> 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 <strong>gained<\/strong> 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<li>Suppose the random variables\u00a0<em data-effect=\"italics\">X<\/em>\u00a0and\u00a0<em data-effect=\"italics\">Y<\/em>\u00a0have the following normal distributions:\u00a0<em data-effect=\"italics\">X<\/em>\u00a0~\u00a0<em data-effect=\"italics\">N<\/em>(5, 6) and\u00a0<em data-effect=\"italics\">Y<\/em>\u00a0~\u00a0<em data-effect=\"italics\">N<\/em>(2, 1). If\u00a0<em data-effect=\"italics\">x<\/em>\u00a0= 17, then\u00a0<em data-effect=\"italics\">z<\/em>\u00a0= 2. (This was previously shown.) If\u00a0<em data-effect=\"italics\">y<\/em>\u00a0= 4, what is\u00a0<em data-effect=\"italics\">z<\/em>?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q818954\">Show Answer<\/span><\/p>\n<div id=\"q818954\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\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\">four<\/strong> standard deviations to the <strong data-redactor-tag=\"strong\">left<\/strong> of the mean.<\/li>\n<li>[latex]z= \\frac{y- \\mu}{\\sigma} = \\frac{4-2}{1}[\/latex] where [latex]\\mu = 2[\/latex] and [latex]\\sigma = 1[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\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 <strong>relative to their means.<\/strong><\/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=\"q590652\">Show Answer<\/span><\/p>\n<div id=\"q590652\" class=\"hidden-answer\" style=\"display: none\">\n<p>1.5, left, 16<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-258\">\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>Statistics, The Standard Normal Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/statistics\/pages\/6-1-the-standard-normal-distribution\">https:\/\/openstax.org\/books\/statistics\/pages\/6-1-the-standard-normal-distribution<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction<\/li><li>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><\/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":169134,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Statistics, The Standard Normal Distribution\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/statistics\/pages\/6-1-the-standard-normal-distribution\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"Normal Distributionu2014Explained Simply (part 1)\",\"author\":\"how2stats\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/embed\/xgQhefFOXrM\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube license\"},{\"type\":\"copyrighted_video\",\"description\":\"Normal Distribution 2014Explained Simply (part 2)\",\"author\":\"how2stats\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/embed\/iiRiOlkLa6A\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube license\"},{\"type\":\"copyrighted_video\",\"description\":\"ck12.org normal distribution problems: z-score | Probability and Statistics | Khan Academy\",\"author\":\"Khan Academy\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=Wp2nVIzBsE8\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube license\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at 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