{"id":259,"date":"2021-07-14T15:59:01","date_gmt":"2021-07-14T15:59:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/using-the-normal-distribution\/"},"modified":"2023-12-05T09:20:02","modified_gmt":"2023-12-05T09:20:02","slug":"using-the-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/using-the-normal-distribution\/","title":{"raw":"Probabilities for Normal Distributions","rendered":"Probabilities for Normal Distributions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul id=\"list4253\">\r\n \t<li>Calculate normal distribution probabilities using technology<\/li>\r\n \t<li>Calculate percentiles for a normal distribution using technology<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Inequalities<\/h3>\r\nHere are some common inequalities seen in statistics:\r\n<ul>\r\n \t<li>&lt; indicates less than, for example x &lt; 5 indicates x is less than 5<\/li>\r\n \t<li>\u2264 indicates less than or equal to, for example x \u2264 5 indicates x is less than or equal to 5 (5 is included)<\/li>\r\n \t<li>&gt; indicates greater than, for example x &gt; 5 indicates x is greater than 5<\/li>\r\n \t<li>\u2265 indicates greater than or equal to, for example x \u2265 5 indicates x is greater than or equal to 5 (5 is included)<\/li>\r\n<\/ul>\r\nWhile trying to find the probability you may need to read the situation you are working within and determine which inequality above represents that situation. Below are some common phrases you may see in statistics and the inequality that represents the situation.\r\n<table style=\"border-collapse: collapse; width: 100%;\" border=\"1\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">[latex]&lt;[\/latex]<\/td>\r\n<td style=\"width: 50%; text-align: center;\">Less than\r\nTo the left of<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">[latex]\\leq[\/latex]<\/td>\r\n<td style=\"width: 50%; text-align: center;\">Less than or equal to\r\nNo more than<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">[latex]&gt;[\/latex]<\/td>\r\n<td style=\"width: 50%; text-align: center;\">Greater than\r\nTo the right of\r\nMore than<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 50%; text-align: center;\">[latex]\\geq[\/latex]<\/td>\r\n<td style=\"width: 50%; text-align: center;\">\u00a0Greater than or equal to\r\nAt least<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThe shaded area in the following graph indicates the area to the left of\u00a0<em>x<\/em>. This area is represented by the probability <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>). Normal tables, computers, and calculators provide or calculate the probability <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>).\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/5o4v-t4czc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. A value, x, is labeled on the horizontal axis, X. A vertical line extends from point x to the curve, and the area under the curve to the left of x is shaded. The area of this shaded section represents the probability that a value of the variable is less than x.\" \/>\r\n\r\nThe area to the right is then\u00a0<em>P<\/em>(<em>X<\/em> &gt; <em>x<\/em>) = 1 \u2013 <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>). Remember, <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = <strong>Area to the left <\/strong>of the vertical line through <em>x<\/em>. <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = 1 \u2013 <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = <strong>Area to the right<\/strong> of the vertical line through <em>x<\/em>. <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) is the same as <em>P<\/em>(<em>X<\/em> \u2264 <em>x<\/em>) and <em>P<\/em>(<em>X<\/em> &gt; <em>x<\/em>) is the same as <em>P<\/em>(<em>X<\/em> \u2265 <em>x<\/em>) for continuous distributions.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Complement<\/h3>\r\nThe word complement means what is needed to make your situation \u201cwhole.\u201d\u00a0 Since the probability of an event occurring is 1, then all of the outcomes in that event must sum to 1. We can use this and the complement rule to find the probability of some events. For example, what is the probability of not rolling doubles when you roll two 6-sided fair dice. There are 6 sets of doubles out of 36 outcomes. We can take [latex]\\frac{6}{36}[\/latex] away from 1 to get the probability of not rolling doubles:\r\n<p style=\"text-align: center;\">[latex]1-\\frac{6}{36}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]= \\frac{36}{36} - \\frac{6}{36}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]=\\frac{30}{36}[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2><strong>Calculations of Probabilities<\/strong><\/h2>\r\nProbabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. Additionally, <a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oervm\/stats\/probs.html\" target=\"_blank\" rel=\"noopener\">this link<\/a> houses a tool that allows you to explore the <strong>normal distribution<\/strong> with varying means and standard deviations as well as associated probabilities. The following video explains how to use the tool.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7115057&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rOs-jlJvcyM&amp;video_target=tpm-plugin-7y2thhjy-rOs-jlJvcyM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\n<hr \/>\r\n\r\n<h3>Note<\/h3>\r\nTo calculate the probability without the use of technology, use the probability tables provided\u00a0<a href=\"http:\/\/www.itl.nist.gov\/div898\/handbook\/eda\/section3\/eda367.htm\" target=\"_blank\" rel=\"noopener\">here<\/a>. The tables include instructions for how to use them.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf the area to the left is 0.0228, then the area to the right is 1 \u2013 0.0228 = 0.9772.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIf the area to the left of\u00a0<em>x<\/em> is 0.012, then what is the area to the right?\r\n\r\n[reveal-answer q=\"945562\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"945562\"]1 \u2212 0.012 = 0.988[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Inverse Function<\/h3>\r\nAn <strong>inverse function<\/strong> is a function where the input of the original function becomes the output of the inverse function. This naturally leads to the output of the original function becoming the input of the inverse function. We have been using the normal distribution to calculate the probability of a random variable. Given a probability, we can use a technique of working backwards to find the random variable. This is informally called the inverse of the normal distribution. TI-83+ and TI-84 calculators have a command\u00a0<code style=\"line-height: 1.6em; text-align: justify;\">invnormal<\/code> that will find the <em>x<\/em>-value given a probability similar to the\u00a0<code style=\"line-height: 1.6em; text-align: justify;\">normalcdf<\/code> command that will find the probability given the <em>x<\/em>-value.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.\r\n\r\na. Find the probability that a randomly selected student scored more than 65 on the exam.\r\n<p style=\"padding-left: 30px;\">[reveal-answer q=\"730417\"]Show Answer[\/reveal-answer][hidden-answer a=\"730417\"]<\/p>\r\nLet <em>X<\/em> = a score on the final exam. <em>X<\/em> ~ <em>N<\/em>(63, 5), where <em>\u03bc<\/em> = 63 and <em>\u03c3<\/em> = 5\r\n\r\nDraw a graph.\r\n\r\nThen, find <em>P<\/em>(<em>x<\/em> &gt; 65).\r\n\r\n<em>P<\/em>(<em>x<\/em> &gt; 65) = 0.3446\r\n\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/qsr8-lbczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. The point 65 is also labeled. A vertical line extends from point 65 to the curve. The probability area to the right of 65 is shaded; it is equal to 0.3446.\" \/>\r\n\r\nThe probability that any student selected at random scores more than 65 is 0.3446.\r\n\r\n&nbsp;\r\n\r\n<header>\r\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header><section>\r\n<div class=\"os-note-body\"><\/div>\r\n<\/section>Go into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>.\r\n\r\nAfter pressing <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>, press<code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code>.\r\n\r\nThe syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10<sup>99<\/sup>) by pressing <code style=\"line-height: 1.6em; text-align: justify;\">1<\/code>, the <code style=\"line-height: 1.6em; text-align: justify;\">EE<\/code> key (a 2nd key) and then <code style=\"line-height: 1.6em; text-align: justify;\">99<\/code>. Or, you can enter<code style=\"line-height: 1.6em; text-align: justify;\">10^99<\/code> instead. The number 10<sup>99<\/sup> is way out in the right tail of the normal curve. We are calculating the area between 65 and 10<sup>99<\/sup>. In some instances, the lower number of the area might be \u20131E99 (= \u201310<sup>99<\/sup>). The number \u201310<sup>99<\/sup> is way out in the left tail of the normal curve.\r\n\r\n&nbsp;\r\n\r\n<header>\r\n<h3 class=\"os-title\" data-type=\"title\"><span id=\"12\" class=\"os-title-label\" data-type=\"\">HISTORICAL NOTE<\/span><\/h3>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idm194168736\" class=\" \">The TI probability program calculates a\u00a0<em data-effect=\"italics\">z<\/em>-score and then the probability from the\u00a0<em data-effect=\"italics\">z<\/em>-score. Before technology, the\u00a0<em data-effect=\"italics\">z<\/em>-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the\u00a0<em data-effect=\"italics\">z<\/em>-score was used. You calculate the\u00a0<em data-effect=\"italics\">z<\/em>-score and look up the area to the left. The probability is the area to the right.<\/p>\r\n\r\n<\/div>\r\n<\/section>[latex]\\displaystyle{z}=\\frac{{{65}-{63}}}{{5}}={0.4}[\/latex]\r\n\r\nArea to the left is 0.6554.\r\n<em>P<\/em>(<em>x<\/em> &gt; 65) = <em>P<\/em>(<em>z<\/em> &gt; 0.4) = 1 \u2013 0.6554 = 0.3446\u00a0[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<header>\r\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-idm197852304\" class=\" \">Find the percentile for a student scoring 65:<\/p>\r\n\r\n<\/div>\r\n<\/section>Calculate the <em>z<\/em>-score:\r\n*Press <code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\">2nd Distr\r\n<\/code>*Press <code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\">\r\n<\/code>*Enter lower bound, upper bound, mean, standard deviation followed by )\r\n\r\n*Press <code style=\"line-height: 1.6em; text-align: justify;\">ENTER<\/code><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\">\r\n<\/code>For this Example, the steps are\r\n\r\n<code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr<\/code>\r\n\r\n<code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code>(65,1,2nd EE,99,63,5) <code style=\"line-height: 1.6em; text-align: justify;\">ENTER<\/code>\r\n\r\nThe probability that a selected student scored more than 65 is 0.3446.<span data-type=\"newline\">\r\n<\/span>To find the probability that a selected student scored\u00a0<em data-effect=\"italics\">more than<\/em>\u00a065, subtract the percentile from 1.\r\n\r\n&nbsp;\r\n\r\nb. Find the probability that a randomly selected student scored less than 85.\r\n\r\n[reveal-answer q=\"478792\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"478792\"]\r\n\r\nDraw a graph.\r\n\r\nThen find <em>P<\/em>(<em>x<\/em> &lt; 85), and shade the graph.\r\n\r\nUsing a computer or calculator, find <em>P<\/em>(<em>x<\/em> &lt; 85) = 1.\r\n\r\nnormalcdf(0,85,63,5) = 1 (rounds to one).\r\n\r\nThe probability that one student scores less than 85 is approximately one (or 100%).\r\n\r\n[\/hidden-answer]\r\n\r\nc. Find the 90th percentile (that is, find the score <em>k<\/em> that has 90% of the scores below <em>k<\/em> and 10% of the scores above <em>k<\/em>).\r\n\r\n[reveal-answer q=\"101756\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"101756\"]\r\n\r\nFind the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the <em>x<\/em>-axis. Shade the area that corresponds to the 90th percentile.\r\n\r\n<strong>Let <em data-redactor-tag=\"em\">k<\/em><\/strong><strong> = the 90th percentile.<\/strong> The variable <em>k<\/em> is located on the <em>x<\/em>-axis. <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) is the area to the left of <em>k<\/em>. The 90th percentile <em>k <\/em>separates the exam scores into those that are the same or lower than <em>k<\/em> and those that are the same or higher. Ninety percent of the test scores are the same or lower than <em>k<\/em>, and ten percent are the same or higher. The variable <em>k<\/em> is often called a <strong>critical value<\/strong>.\r\n\r\n<em>k<\/em> = 69.4\r\n\r\n<img style=\"font-size: 1rem; orphans: 1; text-align: initial;\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/82dc-7iczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. A point, k, is labeled to the right of 63. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. This represents the probability that x is less than k: P(x &lt; k) = 0.90\" \/>\r\n\r\nThe 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<header>\r\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header><section>\r\n<div class=\"os-note-body\"><\/div>\r\n<\/section><code style=\"line-height: 1.6em; text-align: justify;\">invNorm<\/code><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> in <\/span><code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">. invNorm(area to the left, mean, standard deviation)<\/span>\r\n\r\n<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">For this problem, invNorm(0.90,63,5) = 69.4<\/span>\r\n\r\nd. Find the 70th percentile (that is, find the score <em>k<\/em> such that 70% of scores are below <em>k<\/em> and 30% of the scores are above <em>k<\/em>).\r\n\r\n[reveal-answer q=\"726273\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"726273\"]\r\n\r\nFind the 70th percentile.\r\n\r\nDraw a new graph and label it appropriately. <em>k<\/em> = 65.6\r\n\r\nThe 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.\r\n\r\ninvNorm(0.70,63,5) = 65.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 2<\/h3>\r\nThe golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.\r\n\r\nFind the probability that a randomly selected golfer scored less than 65.\r\n\r\n[reveal-answer q=\"336580\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"336580\"]\r\n\r\nnormalcdf(10<sup>99<\/sup>,65,68,3) = 0.1587\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 3<\/h3>\r\nA personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.\r\n\r\na. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.[reveal-answer q=\"350551\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"350551\"]\r\n\r\na. Let <em>X\u00a0<\/em>= the amount of time (in hours) a household personal computer is used for entertainment. <em>X<\/em> ~ <em>N<\/em>(2, 0.5) where <em>\u03bc<\/em> = 2 and <em>\u03c3<\/em> = 0.5.\r\n\r\nFind <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75).\r\n\r\nThe probability for which you are looking is the area\u00a0<strong>between\u00a0<\/strong><em>x<\/em> = 1.8 and <em>x<\/em> = 2.75. <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75) = 0.5886\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/n694-bnczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.\" \/>\r\n\r\nnormalcdf(1.8,2.75,2,0.5) = 0.5886\r\n\r\nThe probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.\r\n\r\n[\/hidden-answer]\r\n\r\nb. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.\r\n\r\n[reveal-answer q=\"16307\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"16307\"]\r\n\r\nb. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, <strong>find the 25th percentile<\/strong>, <em>k<\/em>, where <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = 0.25.\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/kznj-7tczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The area under the left tail of the curve is shaded. The shaded area shows that the probability that x is less than k is 0.25. It follows that k = 1.67.\" \/>\r\n\r\ninvNorm(0.25,2,0.5) = 1.66\r\n\r\nThe maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 3<\/h3>\r\nThe golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.\r\n\r\n[reveal-answer q=\"692702\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"692702\"]\r\n\r\nnormalcdf(66,70,68,3) = 0.4950\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 4<\/h3>\r\nThere are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.\r\n\r\na. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.[reveal-answer q=\"568362\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"568362\"]\r\n\r\nnormalcdf(23,64.7,36.9,13.9) = 0.8186\r\n\r\n[\/hidden-answer]\r\n\r\nb. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.\r\n\r\n[reveal-answer q=\"493361\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"493361\"]\r\n\r\nnormalcdf(\u20131099,50.8,36.9,13.9) = 0.8413\r\n\r\n[\/hidden-answer]\r\n\r\nc. Find the 80th percentile of this distribution, and interpret it in a complete sentence.\r\n\r\n[reveal-answer q=\"766768\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"766768\"]\r\n<ul>\r\n \t<li>invNorm(0.80,36.9,13.9) = 48.6<\/li>\r\n \t<li>The 80th percentile is 48.6 years.<\/li>\r\n \t<li>80% of the smartphone users in the age range 13 \u2013 55+ are 48.6 years old or less.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 4<\/h3>\r\nUse the information in the previous Example to answer the following questions.\r\n\r\n1. Find the 30th percentile, and interpret it in a complete sentence.[reveal-answer q=\"816644\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"816644\"]\r\n\r\nLet\u00a0<em>X<\/em> = a smart phone user whose age is 13 to 55+. <em>X<\/em> ~ <em>N<\/em>(36.9, 13.9).\r\n\r\nTo find the 30th percentile, findk such that P(x &lt; k) = 0.30.invNorm(0.30, 36.9, 13.9) = 29.6 years. Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.\r\n\r\n[\/hidden-answer]\r\n\r\n2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old?[reveal-answer q=\"405257\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"405257\"]\r\n\r\nLet\u00a0<em>X<\/em> = a smart phone user whose age is 13 to 55+. <em>X<\/em> ~ <em>N<\/em>(36.9, 13.9).\r\n\r\nFind <em>P<\/em>(<em>x<\/em> &lt; 27)\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/3jtg-lxczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 36.9 on the horizontal axis. The point 27 is also labeled. A vertical line extends from 27 to the curve. The area under the curve to the left of 27 is shaded. The shaded area shows that P(x &lt; 27) = 0.2342.\" \/>\r\n\r\nnormalcdf(0,27,36.9,13.9) = 0.2342\r\n\r\n(Note that normalcdf(\u20131099,27,36.9,13.9) = 0.2382. The two answers differ only by 0.0040.)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 5<\/h3>\r\nThere are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).\r\n\r\na.\u00a0Calculate the interquartile range (IQR).\r\n\r\n[reveal-answer q=\"339354\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"339354\"]\r\n\r\na.\r\n<ul>\r\n \t<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub>Calculate<\/li>\r\n \t<li><em>Q<\/em><sub>3<\/sub> = 75th percentile and <em>Q<\/em><sub>1<\/sub> = 25th percentile.<\/li>\r\n \t<li>invNorm(0.75,36.9,13.9) = <em>Q<\/em><sub>3<\/sub> = 46.2754<\/li>\r\n \t<li>invNorm(0.25,36.9,13.9) = <em>Q<\/em><sub>1<\/sub> = 27.5246<\/li>\r\n \t<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub> = 18.7508<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n2. Forty percent of the ages that range from 13 to 55+ are at least what age?\r\n\r\n[reveal-answer q=\"141324\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"141324\"]\r\n\r\nb.\r\n<ul>\r\n \t<li>Find <em>k<\/em> where <em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = 0.40 (\"At least\" translates to \"greater than or equal to.\")<\/li>\r\n \t<li>0.40 = the area to the right.<\/li>\r\n \t<li>Area to the left = 1 \u2013 0.40 = 0.60.<\/li>\r\n \t<li>The area to the left of <em>k<\/em> = 0.60.<\/li>\r\n \t<li>invNorm(0.60,36.9,13.9) = 40.4215.<\/li>\r\n \t<li><em>k<\/em> = 40.42.<\/li>\r\n \t<li>Forty percent of the ages that range from 13 to 55+ are at least 40.4 years.<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 5<\/h3>\r\nTwo thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean\r\n<em>\u03bc<\/em> = 81 points and standard deviation <em>\u03c3<\/em> = 15 points.\r\n\r\na. Calculate the first- and third-quartile scores for this exam.\r\n\r\n[reveal-answer q=\"595065\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"595065\"]\r\n\r\n<em>Q<\/em><sub>1<\/sub> = 25th percentile = invNorm(0.25,81,15) = 70.9\r\n<em>Q<\/em><sub>3<\/sub> = 75th percentile = invNorm(0.75,81,15) = 91.9\r\n\r\n[\/hidden-answer]\r\n\r\nb. The middle 50% of the exam scores are between what two values?\r\n\r\n[reveal-answer q=\"936148\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"936148\"]\r\n\r\nThe middle 50% of the scores are between 70.9 and 91.1.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 6<\/h3>\r\nA citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.\r\n\r\na. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.\r\n\r\n[reveal-answer q=\"912943\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"912943\"]\r\n\r\nnormalcdf(6,10^99,5.85,0.24) = 0.2660\r\n<img src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/nj1i-s1dzc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nb. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.\r\n\r\n[reveal-answer q=\"710322\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"710322\"]\r\n\r\nb.\r\n<ul>\r\n \t<li>1 \u2013 0.20 = 0.80<\/li>\r\n \t<li>The tails of the graph of the normal distribution each have an area of 0.40.<\/li>\r\n \t<li>Find <em>k1<\/em>, the 40th percentile, and <em>k2<\/em>, the 60th percentile (0.40 + 0.20 = 0.60).<\/li>\r\n \t<li><em>k1<\/em> = invNorm(0.40,5.85,0.24) = 5.79 cm<\/li>\r\n \t<li><em>k2<\/em> = invNorm(0.60,5.85,0.24) = 5.91 cm<\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\nc. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.[reveal-answer q=\"993139\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"993139\"]\r\n\r\nc. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it 6<\/h3>\r\nUsing the information from the previous Example, answer the following:\r\n\r\na. The middle 45% of mandarin oranges from this farm are between ______ and ______.\r\n\r\n[reveal-answer q=\"262254\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"262254\"]\r\n\r\nThe middle area = 0.40, so each tail has an area of 0.30.1 \u2013 0.40 = 0.60. The tails of the graph of the normal distribution each have an area of 0.30. Find <em>k1<\/em>, the 30th percentile and <em>k2<\/em>, the 70th percentile (0.40 + 0.30 = 0.70).<em>k1<\/em> = invNorm(0.30,5.85,0.24) = 5.72 cm<em>k2<\/em> = invNorm(0.70,5.85,0.24) = 5.98 cm\r\n\r\n[\/hidden-answer]\r\n\r\nb. Find the 16th percentile and interpret it in a complete sentence.\r\n\r\n[reveal-answer q=\"378342\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"378342\"]\r\n\r\nnormalcdf(5,1099,5.85,0.24) = 0.9998\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list4253\">\n<li>Calculate normal distribution probabilities using technology<\/li>\n<li>Calculate percentiles for a normal distribution using technology<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Inequalities<\/h3>\n<p>Here are some common inequalities seen in statistics:<\/p>\n<ul>\n<li>&lt; indicates less than, for example x &lt; 5 indicates x is less than 5<\/li>\n<li>\u2264 indicates less than or equal to, for example x \u2264 5 indicates x is less than or equal to 5 (5 is included)<\/li>\n<li>&gt; indicates greater than, for example x &gt; 5 indicates x is greater than 5<\/li>\n<li>\u2265 indicates greater than or equal to, for example x \u2265 5 indicates x is greater than or equal to 5 (5 is included)<\/li>\n<\/ul>\n<p>While trying to find the probability you may need to read the situation you are working within and determine which inequality above represents that situation. Below are some common phrases you may see in statistics and the inequality that represents the situation.<\/p>\n<table style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 50%; text-align: center;\">[latex]<[\/latex]<\/td>\n<td style=\"width: 50%; text-align: center;\">Less than<br \/>\nTo the left of<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">[latex]\\leq[\/latex]<\/td>\n<td style=\"width: 50%; text-align: center;\">Less than or equal to<br \/>\nNo more than<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">[latex]>[\/latex]<\/td>\n<td style=\"width: 50%; text-align: center;\">Greater than<br \/>\nTo the right of<br \/>\nMore than<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50%; text-align: center;\">[latex]\\geq[\/latex]<\/td>\n<td style=\"width: 50%; text-align: center;\">\u00a0Greater than or equal to<br \/>\nAt least<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>The shaded area in the following graph indicates the area to the left of\u00a0<em>x<\/em>. This area is represented by the probability <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>). Normal tables, computers, and calculators provide or calculate the probability <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>).<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/5o4v-t4czc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. A value, x, is labeled on the horizontal axis, X. A vertical line extends from point x to the curve, and the area under the curve to the left of x is shaded. The area of this shaded section represents the probability that a value of the variable is less than x.\" \/><\/p>\n<p>The area to the right is then\u00a0<em>P<\/em>(<em>X<\/em> &gt; <em>x<\/em>) = 1 \u2013 <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>). Remember, <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = <strong>Area to the left <\/strong>of the vertical line through <em>x<\/em>. <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = 1 \u2013 <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = <strong>Area to the right<\/strong> of the vertical line through <em>x<\/em>. <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) is the same as <em>P<\/em>(<em>X<\/em> \u2264 <em>x<\/em>) and <em>P<\/em>(<em>X<\/em> &gt; <em>x<\/em>) is the same as <em>P<\/em>(<em>X<\/em> \u2265 <em>x<\/em>) for continuous distributions.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Complement<\/h3>\n<p>The word complement means what is needed to make your situation \u201cwhole.\u201d\u00a0 Since the probability of an event occurring is 1, then all of the outcomes in that event must sum to 1. We can use this and the complement rule to find the probability of some events. For example, what is the probability of not rolling doubles when you roll two 6-sided fair dice. There are 6 sets of doubles out of 36 outcomes. We can take [latex]\\frac{6}{36}[\/latex] away from 1 to get the probability of not rolling doubles:<\/p>\n<p style=\"text-align: center;\">[latex]1-\\frac{6}{36}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]= \\frac{36}{36} - \\frac{6}{36}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]=\\frac{30}{36}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2><strong>Calculations of Probabilities<\/strong><\/h2>\n<p>Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators. Additionally, <a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oervm\/stats\/probs.html\" target=\"_blank\" rel=\"noopener\">this link<\/a> houses a tool that allows you to explore the <strong>normal distribution<\/strong> with varying means and standard deviations as well as associated probabilities. The following video explains how to use the tool.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7115057&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rOs-jlJvcyM&amp;video_target=tpm-plugin-7y2thhjy-rOs-jlJvcyM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<hr \/>\n<h3>Note<\/h3>\n<p>To calculate the probability without the use of technology, use the probability tables provided\u00a0<a href=\"http:\/\/www.itl.nist.gov\/div898\/handbook\/eda\/section3\/eda367.htm\" target=\"_blank\" rel=\"noopener\">here<\/a>. The tables include instructions for how to use them.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If the area to the left is 0.0228, then the area to the right is 1 \u2013 0.0228 = 0.9772.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>If the area to the left of\u00a0<em>x<\/em> is 0.012, then what is the area to the right?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q945562\">Show Answer<\/span><\/p>\n<div id=\"q945562\" class=\"hidden-answer\" style=\"display: none\">1 \u2212 0.012 = 0.988<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Inverse Function<\/h3>\n<p>An <strong>inverse function<\/strong> is a function where the input of the original function becomes the output of the inverse function. This naturally leads to the output of the original function becoming the input of the inverse function. We have been using the normal distribution to calculate the probability of a random variable. Given a probability, we can use a technique of working backwards to find the random variable. This is informally called the inverse of the normal distribution. TI-83+ and TI-84 calculators have a command\u00a0<code style=\"line-height: 1.6em; text-align: justify;\">invnormal<\/code> that will find the <em>x<\/em>-value given a probability similar to the\u00a0<code style=\"line-height: 1.6em; text-align: justify;\">normalcdf<\/code> command that will find the probability given the <em>x<\/em>-value.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.<\/p>\n<p>a. Find the probability that a randomly selected student scored more than 65 on the exam.<\/p>\n<p style=\"padding-left: 30px;\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q730417\">Show Answer<\/span><\/p>\n<div id=\"q730417\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em>X<\/em> = a score on the final exam. <em>X<\/em> ~ <em>N<\/em>(63, 5), where <em>\u03bc<\/em> = 63 and <em>\u03c3<\/em> = 5<\/p>\n<p>Draw a graph.<\/p>\n<p>Then, find <em>P<\/em>(<em>x<\/em> &gt; 65).<\/p>\n<p><em>P<\/em>(<em>x<\/em> &gt; 65) = 0.3446<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/qsr8-lbczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. The point 65 is also labeled. A vertical line extends from point 65 to the curve. The probability area to the right of 65 is shaded; it is equal to 0.3446.\" \/><\/p>\n<p>The probability that any student selected at random scores more than 65 is 0.3446.<\/p>\n<p>&nbsp;<\/p>\n<header>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<\/header>\n<section>\n<div class=\"os-note-body\"><\/div>\n<\/section>\n<p>Go into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>.<\/p>\n<p>After pressing <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>, press<code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code>.<\/p>\n<p>The syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10<sup>99<\/sup>) by pressing <code style=\"line-height: 1.6em; text-align: justify;\">1<\/code>, the <code style=\"line-height: 1.6em; text-align: justify;\">EE<\/code> key (a 2nd key) and then <code style=\"line-height: 1.6em; text-align: justify;\">99<\/code>. Or, you can enter<code style=\"line-height: 1.6em; text-align: justify;\">10^99<\/code> instead. The number 10<sup>99<\/sup> is way out in the right tail of the normal curve. We are calculating the area between 65 and 10<sup>99<\/sup>. In some instances, the lower number of the area might be \u20131E99 (= \u201310<sup>99<\/sup>). The number \u201310<sup>99<\/sup> is way out in the left tail of the normal curve.<\/p>\n<p>&nbsp;<\/p>\n<header>\n<h3 class=\"os-title\" data-type=\"title\"><span id=\"12\" class=\"os-title-label\" data-type=\"\">HISTORICAL NOTE<\/span><\/h3>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idm194168736\" class=\"\">The TI probability program calculates a\u00a0<em data-effect=\"italics\">z<\/em>-score and then the probability from the\u00a0<em data-effect=\"italics\">z<\/em>-score. Before technology, the\u00a0<em data-effect=\"italics\">z<\/em>-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the\u00a0<em data-effect=\"italics\">z<\/em>-score was used. You calculate the\u00a0<em data-effect=\"italics\">z<\/em>-score and look up the area to the left. The probability is the area to the right.<\/p>\n<\/div>\n<\/section>\n<p>[latex]\\displaystyle{z}=\\frac{{{65}-{63}}}{{5}}={0.4}[\/latex]<\/p>\n<p>Area to the left is 0.6554.<br \/>\n<em>P<\/em>(<em>x<\/em> &gt; 65) = <em>P<\/em>(<em>z<\/em> &gt; 0.4) = 1 \u2013 0.6554 = 0.3446\u00a0<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<header>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-idm197852304\" class=\"\">Find the percentile for a student scoring 65:<\/p>\n<\/div>\n<\/section>\n<p>Calculate the <em>z<\/em>-score:<br \/>\n*Press <code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\">2nd Distr<br \/>\n<\/code>*Press <code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><br \/>\n<\/code>*Enter lower bound, upper bound, mean, standard deviation followed by )<\/p>\n<p>*Press <code style=\"line-height: 1.6em; text-align: justify;\">ENTER<\/code><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><br \/>\n<\/code>For this Example, the steps are<\/p>\n<p><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr<\/code><\/p>\n<p><code style=\"line-height: 1.6em; text-align: justify;\">2:normalcdf<\/code>(65,1,2nd EE,99,63,5) <code style=\"line-height: 1.6em; text-align: justify;\">ENTER<\/code><\/p>\n<p>The probability that a selected student scored more than 65 is 0.3446.<span data-type=\"newline\"><br \/>\n<\/span>To find the probability that a selected student scored\u00a0<em data-effect=\"italics\">more than<\/em>\u00a065, subtract the percentile from 1.<\/p>\n<p>&nbsp;<\/p>\n<p>b. Find the probability that a randomly selected student scored less than 85.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q478792\">Show Answer<\/span><\/p>\n<div id=\"q478792\" class=\"hidden-answer\" style=\"display: none\">\n<p>Draw a graph.<\/p>\n<p>Then find <em>P<\/em>(<em>x<\/em> &lt; 85), and shade the graph.<\/p>\n<p>Using a computer or calculator, find <em>P<\/em>(<em>x<\/em> &lt; 85) = 1.<\/p>\n<p>normalcdf(0,85,63,5) = 1 (rounds to one).<\/p>\n<p>The probability that one student scores less than 85 is approximately one (or 100%).<\/p>\n<\/div>\n<\/div>\n<p>c. Find the 90th percentile (that is, find the score <em>k<\/em> that has 90% of the scores below <em>k<\/em> and 10% of the scores above <em>k<\/em>).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q101756\">Show Answer<\/span><\/p>\n<div id=\"q101756\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the <em>x<\/em>-axis. Shade the area that corresponds to the 90th percentile.<\/p>\n<p><strong>Let <em data-redactor-tag=\"em\">k<\/em><\/strong><strong> = the 90th percentile.<\/strong> The variable <em>k<\/em> is located on the <em>x<\/em>-axis. <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) is the area to the left of <em>k<\/em>. The 90th percentile <em>k <\/em>separates the exam scores into those that are the same or lower than <em>k<\/em> and those that are the same or higher. Ninety percent of the test scores are the same or lower than <em>k<\/em>, and ten percent are the same or higher. The variable <em>k<\/em> is often called a <strong>critical value<\/strong>.<\/p>\n<p><em>k<\/em> = 69.4<\/p>\n<p><img decoding=\"async\" style=\"font-size: 1rem; orphans: 1; text-align: initial;\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/82dc-7iczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. A point, k, is labeled to the right of 63. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. This represents the probability that x is less than k: P(x &lt; k) = 0.90\" \/><\/p>\n<p>The 90th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<header>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<\/header>\n<section>\n<div class=\"os-note-body\"><\/div>\n<\/section>\n<p><code style=\"line-height: 1.6em; text-align: justify;\">invNorm<\/code><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> in <\/span><code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">. invNorm(area to the left, mean, standard deviation)<\/span><\/p>\n<p><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">For this problem, invNorm(0.90,63,5) = 69.4<\/span><\/p>\n<p>d. Find the 70th percentile (that is, find the score <em>k<\/em> such that 70% of scores are below <em>k<\/em> and 30% of the scores are above <em>k<\/em>).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726273\">Show Answer<\/span><\/p>\n<div id=\"q726273\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find the 70th percentile.<\/p>\n<p>Draw a new graph and label it appropriately. <em>k<\/em> = 65.6<\/p>\n<p>The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.<\/p>\n<p>invNorm(0.70,63,5) = 65.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it 2<\/h3>\n<p>The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.<\/p>\n<p>Find the probability that a randomly selected golfer scored less than 65.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q336580\">Show Answer<\/span><\/p>\n<div id=\"q336580\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(10<sup>99<\/sup>,65,68,3) = 0.1587<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 3<\/h3>\n<p>A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.<\/p>\n<p>a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q350551\">Show Answer<\/span><\/p>\n<div id=\"q350551\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Let <em>X\u00a0<\/em>= the amount of time (in hours) a household personal computer is used for entertainment. <em>X<\/em> ~ <em>N<\/em>(2, 0.5) where <em>\u03bc<\/em> = 2 and <em>\u03c3<\/em> = 0.5.<\/p>\n<p>Find <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75).<\/p>\n<p>The probability for which you are looking is the area\u00a0<strong>between\u00a0<\/strong><em>x<\/em> = 1.8 and <em>x<\/em> = 2.75. <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75) = 0.5886<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/n694-bnczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.\" \/><\/p>\n<p>normalcdf(1.8,2.75,2,0.5) = 0.5886<\/p>\n<p>The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.<\/p>\n<\/div>\n<\/div>\n<p>b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16307\">Show Answer<\/span><\/p>\n<div id=\"q16307\" class=\"hidden-answer\" style=\"display: none\">\n<p>b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, <strong>find the 25th percentile<\/strong>, <em>k<\/em>, where <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = 0.25.<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/kznj-7tczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The area under the left tail of the curve is shaded. The shaded area shows that the probability that x is less than k is 0.25. It follows that k = 1.67.\" \/><\/p>\n<p>invNorm(0.25,2,0.5) = 1.66<\/p>\n<p>The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it 3<\/h3>\n<p>The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q692702\">Show Answer<\/span><\/p>\n<div id=\"q692702\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(66,70,68,3) = 0.4950<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 4<\/h3>\n<p>There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.<\/p>\n<p>a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568362\">Show Answer<\/span><\/p>\n<div id=\"q568362\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(23,64.7,36.9,13.9) = 0.8186<\/p>\n<\/div>\n<\/div>\n<p>b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q493361\">Show Answer<\/span><\/p>\n<div id=\"q493361\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(\u20131099,50.8,36.9,13.9) = 0.8413<\/p>\n<\/div>\n<\/div>\n<p>c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q766768\">Show Answer<\/span><\/p>\n<div id=\"q766768\" class=\"hidden-answer\" style=\"display: none\">\n<ul>\n<li>invNorm(0.80,36.9,13.9) = 48.6<\/li>\n<li>The 80th percentile is 48.6 years.<\/li>\n<li>80% of the smartphone users in the age range 13 \u2013 55+ are 48.6 years old or less.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it 4<\/h3>\n<p>Use the information in the previous Example to answer the following questions.<\/p>\n<p>1. Find the 30th percentile, and interpret it in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q816644\">Show Answer<\/span><\/p>\n<div id=\"q816644\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u00a0<em>X<\/em> = a smart phone user whose age is 13 to 55+. <em>X<\/em> ~ <em>N<\/em>(36.9, 13.9).<\/p>\n<p>To find the 30th percentile, findk such that P(x &lt; k) = 0.30.invNorm(0.30, 36.9, 13.9) = 29.6 years. Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.<\/p>\n<\/div>\n<\/div>\n<p>2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q405257\">Show Answer<\/span><\/p>\n<div id=\"q405257\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u00a0<em>X<\/em> = a smart phone user whose age is 13 to 55+. <em>X<\/em> ~ <em>N<\/em>(36.9, 13.9).<\/p>\n<p>Find <em>P<\/em>(<em>x<\/em> &lt; 27)<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/3jtg-lxczc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 36.9 on the horizontal axis. The point 27 is also labeled. A vertical line extends from 27 to the curve. The area under the curve to the left of 27 is shaded. The shaded area shows that P(x &lt; 27) = 0.2342.\" \/><\/p>\n<p>normalcdf(0,27,36.9,13.9) = 0.2342<\/p>\n<p>(Note that normalcdf(\u20131099,27,36.9,13.9) = 0.2382. The two answers differ only by 0.0040.)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 5<\/h3>\n<p>There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).<\/p>\n<p>a.\u00a0Calculate the interquartile range (IQR).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q339354\">Show Answer<\/span><\/p>\n<div id=\"q339354\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.<\/p>\n<ul>\n<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub>Calculate<\/li>\n<li><em>Q<\/em><sub>3<\/sub> = 75th percentile and <em>Q<\/em><sub>1<\/sub> = 25th percentile.<\/li>\n<li>invNorm(0.75,36.9,13.9) = <em>Q<\/em><sub>3<\/sub> = 46.2754<\/li>\n<li>invNorm(0.25,36.9,13.9) = <em>Q<\/em><sub>1<\/sub> = 27.5246<\/li>\n<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub> = 18.7508<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>2. Forty percent of the ages that range from 13 to 55+ are at least what age?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141324\">Show Answer<\/span><\/p>\n<div id=\"q141324\" class=\"hidden-answer\" style=\"display: none\">\n<p>b.<\/p>\n<ul>\n<li>Find <em>k<\/em> where <em>P<\/em>(<em>x<\/em> &gt; <em>k<\/em>) = 0.40 (&#8220;At least&#8221; translates to &#8220;greater than or equal to.&#8221;)<\/li>\n<li>0.40 = the area to the right.<\/li>\n<li>Area to the left = 1 \u2013 0.40 = 0.60.<\/li>\n<li>The area to the left of <em>k<\/em> = 0.60.<\/li>\n<li>invNorm(0.60,36.9,13.9) = 40.4215.<\/li>\n<li><em>k<\/em> = 40.42.<\/li>\n<li>Forty percent of the ages that range from 13 to 55+ are at least 40.4 years.<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it 5<\/h3>\n<p>Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean<br \/>\n<em>\u03bc<\/em> = 81 points and standard deviation <em>\u03c3<\/em> = 15 points.<\/p>\n<p>a. Calculate the first- and third-quartile scores for this exam.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q595065\">Show Answer<\/span><\/p>\n<div id=\"q595065\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>Q<\/em><sub>1<\/sub> = 25th percentile = invNorm(0.25,81,15) = 70.9<br \/>\n<em>Q<\/em><sub>3<\/sub> = 75th percentile = invNorm(0.75,81,15) = 91.9<\/p>\n<\/div>\n<\/div>\n<p>b. The middle 50% of the exam scores are between what two values?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q936148\">Show Answer<\/span><\/p>\n<div id=\"q936148\" class=\"hidden-answer\" style=\"display: none\">\n<p>The middle 50% of the scores are between 70.9 and 91.1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 6<\/h3>\n<p>A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.<\/p>\n<p>a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q912943\">Show Answer<\/span><\/p>\n<div id=\"q912943\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(6,10^99,5.85,0.24) = 0.2660<br \/>\n<img decoding=\"async\" src=\"https:\/\/textimgs.s3.amazonaws.com\/DE\/stats\/nj1i-s1dzc27i#fixme#fixme#fixme\" alt=\"This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.\" \/><\/p>\n<\/div>\n<\/div>\n<p>b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q710322\">Show Answer<\/span><\/p>\n<div id=\"q710322\" class=\"hidden-answer\" style=\"display: none\">\n<p>b.<\/p>\n<ul>\n<li>1 \u2013 0.20 = 0.80<\/li>\n<li>The tails of the graph of the normal distribution each have an area of 0.40.<\/li>\n<li>Find <em>k1<\/em>, the 40th percentile, and <em>k2<\/em>, the 60th percentile (0.40 + 0.20 = 0.60).<\/li>\n<li><em>k1<\/em> = invNorm(0.40,5.85,0.24) = 5.79 cm<\/li>\n<li><em>k2<\/em> = invNorm(0.60,5.85,0.24) = 5.91 cm<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p>c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q993139\">Show Answer<\/span><\/p>\n<div id=\"q993139\" class=\"hidden-answer\" style=\"display: none\">\n<p>c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it 6<\/h3>\n<p>Using the information from the previous Example, answer the following:<\/p>\n<p>a. The middle 45% of mandarin oranges from this farm are between ______ and ______.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q262254\">Show Answer<\/span><\/p>\n<div id=\"q262254\" class=\"hidden-answer\" style=\"display: none\">\n<p>The middle area = 0.40, so each tail has an area of 0.30.1 \u2013 0.40 = 0.60. The tails of the graph of the normal distribution each have an area of 0.30. Find <em>k1<\/em>, the 30th percentile and <em>k2<\/em>, the 70th percentile (0.40 + 0.30 = 0.70).<em>k1<\/em> = invNorm(0.30,5.85,0.24) = 5.72 cm<em>k2<\/em> = invNorm(0.70,5.85,0.24) = 5.98 cm<\/p>\n<\/div>\n<\/div>\n<p>b. Find the 16th percentile and interpret it in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q378342\">Show Answer<\/span><\/p>\n<div id=\"q378342\" class=\"hidden-answer\" style=\"display: none\">\n<p>normalcdf(5,1099,5.85,0.24) = 0.9998<\/p>\n<\/div>\n<\/div>\n<\/div>\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-259\">\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, Using the Normal Distribution. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/statistics\/pages\/6-2-using-the-normal-distribution\">https:\/\/openstax.org\/books\/statistics\/pages\/6-2-using-the-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>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions\">https:\/\/openstax.org\/books\/precalculus\/pages\/1-introduction-to-functions<\/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\/precalculus\/pages\/1-introduction-to-functions<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>How to use the online Normal Distribution Calculator. <strong>Authored by<\/strong>: gatorpj. <strong>Provided by<\/strong>: https:\/\/www.youtube.com\/watch?v=rOs-jlJvcyM. <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":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Statistics, Using the Normal Distribution\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/statistics\/pages\/6-2-using-the-normal-distribution\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/statistics\/pages\/1-introduction\"},{\"type\":\"copyrighted_video\",\"description\":\"How to use the online Normal Distribution Calculator\",\"author\":\"gatorpj\",\"organization\":\"https:\/\/www.youtube.com\/watch?v=rOs-jlJvcyM\",\"url\":\"\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube 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