{"id":239,"date":"2016-04-21T22:43:42","date_gmt":"2016-04-21T22:43:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstats1xmaster\/?post_type=chapter&#038;p=239"},"modified":"2021-07-08T23:08:48","modified_gmt":"2021-07-08T23:08:48","slug":"using-the-normal-distribution","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/frontrange-introstats1\/chapter\/using-the-normal-distribution\/","title":{"raw":"Using the Normal Distribution","rendered":"Using the 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 normal probability distribution and apply it appropriately.<\/li>\r\n \t<li>Compare normal probabilities by converting to the standard normal distribution and directly using technology.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\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 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\nRemember, <em>P<\/em>(<em>X<\/em> &lt; <em>x<\/em>) = <strong>Area to the left <\/strong>of the vertical line through <em>x<\/em>.\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>) = <strong>Area to the right<\/strong> of the vertical line through <em>x<\/em>.\r\n\r\n<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\r\n<\/div>\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<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=\"337707\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"337707\"] 1 \u2212 0.012 = 0.988 [\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Calculation of Probabilities and Finding Values from Probabilities<\/h2>\r\nProbabilities are calculated using technology. We can also find values in the distribution (such as percentiles) from given probabilities using technology.\r\n\r\n<strong>Regardless of the method used to find the probabilities or numbers, always draw a sketch of the normal distribution, shade the relevant area under the curve and label the known and unknown values.\u00a0<\/strong>You can use the fact that the mean is in the center of the distribution (50% of the area to its left and 50% of the area to its right) and the Empirical Rule percentages in the previous section to determine if your answers are reasonable.\r\n<div class=\"textbox shaded\" style=\"text-align: left;\">\r\n<h3>Finding Probabilities FOR THE NORMAL DISTRIBUTION Using EXCEL<\/h3>\r\nAccess the norm.dist function (under\u00a0[latex]f_{x}[\/latex]). There are four inputs required [latex]x[\/latex], the mean, the standard deviation, and cumulative. <span style=\"color: #000000;\"><strong>For all work in this class, we are using this function to find P([latex]X &lt; x[\/latex]), so we will only be using cumulative = 1 (or True).<\/strong>\u00a0<\/span>\r\n\r\n<img class=\"size-medium wp-image-2547 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/08203237\/norm.dist_-300x94.png\" alt=\"\" width=\"300\" height=\"94\" \/>\r\n<ul>\r\n \t<li><\/li>\r\n \t<li><\/li>\r\n \t<li><\/li>\r\n \t<li><\/li>\r\n \t<li><strong>Area\/Probability to the Left of a Number:<\/strong> To calculate <em>P<\/em>([latex]x &lt; [\/latex] number),\u00a0 NORM.DIST(number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\r\n \t<li><strong>Area\/Probability Between Two Numbers:<\/strong> To calculate <em>P<\/em>([latex]number1 &lt; x &lt; number2 [\/latex]), subtract the smaller area from the larger area. NORM.DIST(number2, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1) -\u00a0NORM.DIST(number1, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\r\n \t<li><strong>Area\/Probability to the Right of a Number:<\/strong> To calculate <em>P<\/em>([latex]x &gt; [\/latex] number), subtract the area from one. 1 -\u00a0NORM.DIST(number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\r\n<\/ul>\r\n<em>If you convert your x values to z-scores, your distribution will be the Standard normal distribution. You can enter a mean of zero and a standard deviation of one into this NORM.DIST function or run the NORM.S.DIST function which only requires two inputs [the z-score and cumulative = 1 (or True)].<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Finding Probabilities FOR THE NORMAL DISTRIBUTION Using TI-83, 83+, 84, 84+<\/h3>\r\nGo into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>. 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>. The syntax for the instructions are as follows:\u00a0 normalcdf(lower value, upper value, mean, standard deviation)\r\n\r\nYou 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.\u00a0The number \u201310<sup>99<\/sup> is way out in the left tail of the normal curve.\r\n<ul>\r\n \t<li><strong>Area\/Probability to the Left of a Number:<\/strong> To calculate <em>P<\/em>([latex]x &lt; [\/latex] number),\u00a0 normcdf(-1E99, number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex]).<\/li>\r\n \t<li><strong>Area\/Probability Between Two Numbers:<\/strong> To calculate <em>P<\/em>([latex]number1 &lt; x&lt; number2 [\/latex]). Number2 is greater than number1\u00a0 normalcdf(number1,number2,[latex]\\displaystyle{\\mu},{\\sigma}[\/latex])<\/li>\r\n \t<li><strong>Area\/Probability to the Right of a Number:<\/strong> To calculate <em>P<\/em>([latex]x &gt; [\/latex] number),\u00a0 normalcdf(number, 1E99, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex]).<\/li>\r\n<\/ul>\r\n<em>If the mean and standard deviation are left out of the function, the calculator will use the mean of zero and standard deviation of one for the Standard Normal Distribution.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" style=\"text-align: left;\">\r\n<h3>Finding VALUES FOR THE NORMAL DISTRIBUTION Using EXCEL<\/h3>\r\nAccess the norm.inv function (under\u00a0[latex]f_{x}[\/latex]). There are three inputs required (the area to the left, the mean and the standard deviation).\r\n\r\n<img class=\"size-medium wp-image-2558 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/08210800\/norm.inv_-300x76.png\" alt=\"\" width=\"300\" height=\"76\" \/>\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n<ul>\r\n \t<li><strong>Number with a given probability\/area to its left:<\/strong> NORM.INV(probability\/area to the left, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex].<\/li>\r\n<\/ul>\r\n<em>- If the problem gives you an area to the right of your unknown number, convert it to the area to the left by subtracting it from one. Example: the x value that has area 0.015 its right also has 1 - 0.015 or 0.985 to its left<\/em><em>.<\/em>\r\n\r\n<em>- The NORM.S.INV function provides z-scores from the\u00a0<\/em><em>Standard normal distribution without having to enter the mean of zero and a standard deviation of one.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Finding VALUES FOR THE NORMAL DISTRIBUTION Using TI-83, 83+, 84, 84+<code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><\/code><\/code><\/h3>\r\nGo into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>. After pressing <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>, press<code style=\"line-height: 1.6em; text-align: justify;\">3:invNorm<\/code>. The syntax for the instructions are as follows: invNorm(probability\/area to the left, mean, standard deviation)\r\n<ul>\r\n \t<li><strong style=\"font-size: 0.9em;\">Number with a given probability\/area to its left:<\/strong><span style=\"font-size: 0.9em;\">\u00a0invNorm(probability\/area to the left, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex].<\/span><\/li>\r\n<\/ul>\r\n<em>- If the problem gives you an area to the right of your unknown number, convert it to the area to the left by subtracting it from one. Example: the x value that has area 0.015 its right also has 1 - 0.015 or 0.985 to its left<\/em><em>.<\/em>\r\n\r\n<em>- If the mean and standard deviation are left out of the function, the calculator will use the mean of zero and standard deviation of one for the Standard Normal Distribution.<\/em>\r\n\r\n<\/div>\r\n<h4>Note<\/h4>\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\nThe final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.\r\n<ol>\r\n \t<li>Find the probability that a randomly selected student scored more than 65 on the exam.<\/li>\r\n \t<li>Find the probability that a randomly selected student scored less than 85.<\/li>\r\n \t<li>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>).<\/li>\r\n \t<li>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>).<\/li>\r\n<\/ol>\r\nSolution:\u00a0 Let\u00a0<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<ol>\r\n \t<li>Draw a graph. Then, find <em>P<\/em>(<em>x<\/em> &gt; 65).\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.\" \/><\/li>\r\n<\/ol>\r\nUsing <strong>Excel <\/strong>1 - NORM.DIST(65, 63, 5, 1 or True)\u00a0The result is\u00a0[latex]P(x &gt; 65)=0.3446[\/latex].\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(65,1E99,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">). The result is\u00a0[latex]P(x &gt; 65)=0.3446[\/latex].\u00a0<\/span>\r\n\r\n<span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The probability that any student selected at random scores more than 65 is 0.3446.<\/span>\r\n\r\n2. Draw a graph. Then find <em>P<\/em>(<em>x<\/em> &lt; 85), and shade the graph.\r\n\r\nUsing <strong>Excel <\/strong>NORM.DIST(85, 63, 5, 1 or True)\u00a0The result is\u00a0[latex]P(x &lt; 85)=1.0000[\/latex].\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(-1E99,85,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">). The result is\u00a0[latex]P(x &lt; 85)=1.0000[\/latex].\u00a0<\/span>\r\n\r\nThe probability that one student scores less than 85 is approximately one (or 100%).\r\n\r\n3. 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.\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<img 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\nUsing <strong>Excel <\/strong>NORM.INV(0.90, 63, 5) The result is [latex]\\approx69.4[\/latex]\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 3:invNorm(0.90,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">)\u00a0The result is [latex]\\approx69.4[\/latex]<\/span>\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.\r\n\r\n4. Find the 70th percentile. Draw a new graph and label it appropriately.\r\n\r\nUsing <strong>Excel <\/strong>NORM.INV(0.70, 63, 5) The result is [latex]\\approx65.6[\/latex]\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 3:invNorm(0.70,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">)\u00a0The result is [latex]\\approx65.6[\/latex]<\/span>\r\n\r\n<em>k<\/em> = 65.6 The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.6 and 30% fall at or above 65.6.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/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=\"315155\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"315155\"]\r\n\r\nDraw a graph. Then find <em>P<\/em>(<em>x<\/em> &lt; 65), and shade the graph.\r\n\r\nUsing <strong>Excel <\/strong>NORM.DIST(65, 68, 3, 1) = 0.1587\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(-1E99,65,68,3<\/code><span style=\"font-size: 1rem; text-align: initial;\">) = 0.1587\u00a0<\/span>\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<\/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<ol>\r\n \t<li>Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.<\/li>\r\n \t<li>Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"425730\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425730\"]\r\n<ol>\r\n \t<li>Let <em>X<\/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. Find <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75).The probability for which you are looking is the area\r\n<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 class=\"alignleft\" 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.\" \/><\/li>\r\n<\/ol>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\nUsing <strong>Excel <\/strong>NORM.DIST(2.75, 2, 0.5, 1) - NORM.DIST(1.8, 2, 0.5, 1)= 0.5886\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong>normalcdf(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\n2. 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\nUsing <strong>Excel <\/strong>NORM.INV(0.25,2,0.5) = 1.66\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong>invNorm(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<\/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=\"577099\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"577099\"]\r\n\r\nUsing <strong>Excel <\/strong>NORM.DIST(70, 68, 3, 1) - NORM.DIST(66, 68, 3, 1)= 0.4950\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong>normalcdf(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<\/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<ol>\r\n \t<li>Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.<\/li>\r\n \t<li>Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.<\/li>\r\n \t<li>Find the 80th percentile of this distribution, and interpret it in a complete sentence.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"588265\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"588265\"]\r\n<ol>\r\n \t<li>Using <strong>Excel <\/strong>NORM.DIST(64.7,36.9,13.9, 1) - NORM.DIST(23,36.9,13.9, 1)= 0.8186Using <strong>TI83\/84\u00a0<\/strong><span style=\"font-size: 0.9em;\">normalcdf(23,64.7,36.9,13.9) = 0.8186<\/span><\/li>\r\n \t<li>Using <strong>Excel <\/strong>NORM.DIST(50.8,36.9,13.9, 1) = 0.8413Using <strong>TI83\/84\u00a0<\/strong><span style=\"font-size: 1rem; text-align: initial;\">normalcdf(\u201310<\/span><sup style=\"text-align: initial;\">99<\/sup><span style=\"font-size: 1rem; text-align: initial;\">,50.8,36.9,13.9) = 0.8413<\/span><\/li>\r\n \t<li>Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Excel <\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">NORM.INV(0.8,36.9,13.9) = 48.6\u00a0 \u00a0<\/span>Using <strong>TI83\/84 <\/strong><span style=\"font-size: 1rem; text-align: initial;\">invNorm(0.80,36.9,13.9) = 48.6\u00a0<\/span>The 80th percentile is 48.6 years. 80% of the smartphone users in the age range 13 \u2013 55+ are 48.6 years old or less.<\/li>\r\n<\/ol>\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<\/h3>\r\nUse the information in previous example to answer the following questions.\r\n<ol>\r\n \t<li>Find the 30th percentile, and interpret it in a complete sentence.<\/li>\r\n \t<li>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.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"699940\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"699940\"]\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\n1.To find the 30th percentile, find\u00a0<em>k<\/em> such that <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = 0.30. Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Excel\u00a0<\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">NORM.INV(0.8,36.9,13.9) = 48.6\u00a0<\/span>Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">TI83\/84\u00a0<\/strong>invNorm(0.30, 36.9, 13.9) = 29.6\r\n\r\nThirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.\r\n\r\n2. Find <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\n2.Using <strong>Excel <\/strong>NORM.DIST(27,36.9,13.9, 1) = 0.2382 <em>(ignore the slight discrepancy from the graph above)<\/em>\r\n\r\nUsing <strong>TI83\/84\u00a0<\/strong>normalcdf(\u201310<sup>99<\/sup>,27,36.9,13.9) = 0.2382\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<\/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<ol>\r\n \t<li>Calculate the interquartile range (<em>IQR<\/em>).<\/li>\r\n \t<li>Forty percent of the ages that range from 13 to 55+ are at least what age?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"845498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"845498\"]\r\n<ol>\r\n \t<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1\u00a0<\/sub>Calculate\u00a0<em>Q<\/em><sub>3<\/sub> = 75th percentile and <em>Q<\/em><sub>1<\/sub> = 25th percentile. Using <strong>Excel<\/strong> NORM.INV(0.75,36.9, 13.9) = 46.2754 and NORM.INV(0.25, 36.9,13.9) = 27.5246. Using the <strong>TI 83\/84<\/strong>\u00a0invNorm(0.75,36.9,13.9) = <em>Q<\/em><sub>3<\/sub> = 46.2754 invNorm(0.25,36.9,13.9) = <em>Q<\/em><sub>1<\/sub> = 27.5246\r\n<em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub> = 18.7508 years<\/li>\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.\") 0.40 = the area to the right. Area to the left = 1 \u2013 0.40 = 0.60. The area to the left of <em>k<\/em> = 0.60.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.60, 36.9, 13.9) using <strong>TI 83\/84<\/strong> invNorm(0.60,36.9,13.9) = 40.4215. <em>k<\/em> = 40.42. Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.<\/li>\r\n<\/ol>\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<\/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<ol>\r\n \t<li>Calculate the first- and third-quartile scores for this exam.<\/li>\r\n \t<li>The middle 50% of the exam scores are between what two values?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"889848\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"889848\"]\r\n<ol>\r\n \t<li><em>Q<\/em><sub>1<\/sub> = 25th percentile.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.25, 81, 15) = 70.9.\u00a0 Using <strong>TI 83\/84<\/strong> invNorm(0.25,81,15) = 70.9\r\n<em>Q<\/em><sub>3<\/sub> = 75th percentile.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.75, 81, 15) = 91.1.\u00a0 \u00a0Using <strong>TI 83\/84<\/strong> invNorm(0.75,81,15) = 91.1<\/li>\r\n \t<li>The middle 50% of the scores are between 70.9 and 91.1.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3><\/h3>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\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<ol>\r\n \t<li>Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.<\/li>\r\n \t<li>The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.<\/li>\r\n \t<li>Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"977845\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977845\"]\r\n<ol>\r\n \t<li>Using <strong>Excel<\/strong> 1 - NORM.DIST(6, 5.85, 0.24, 1) = 0.2660.\u00a0 Using <strong>TI 83\/84<\/strong> normalcdf(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.\" \/><\/li>\r\n \t<li>1 \u2013 0.20 = 0.80 The tails of the graph of the normal distribution each have an area of 0.40. Find <em>k1<\/em>, the 40th percentile, and <em>k2<\/em>, the 60th percentile (0.40 + 0.20 = 0.60). Using <strong>Excel<\/strong>\u00a0<em style=\"font-size: 0.9em;\">k1<\/em><span style=\"font-size: 0.9em;\"> = NORM.INV(0.40, 5.85, 0.24) = 5.79 cm <\/span><em style=\"font-size: 0.9em;\">k2<\/em><span style=\"font-size: 0.9em;\"> = NORM.INV(0.60, 5.85, 0.24) = 5.91 Using <strong>TI 83\/84\u00a0<\/strong><\/span><em>k1<\/em> = invNorm(0.40, 5.85, 0.24) = 5.79 cm <em>k2<\/em> = invNorm(0.60, 5.85, 0.24) = 5.91 cm<\/li>\r\n \t<li>Using <strong>Excel<\/strong> NORM.INV(0.90, 5.85, 0.24) = 6.16.\u00a0 Using <strong>TI 83\/84<\/strong> invNorm(0.90, 5.85, 0.24) = 6.16\u00a0 Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.<\/li>\r\n<\/ol>\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<\/h3>\r\nUsing the information from previous, answer the following:\r\n<ol>\r\n \t<li>The middle 40% of mandarin oranges from this farm are between ______ and ______.<\/li>\r\n \t<li>Find the 16th percentile and interpret it in a complete sentence.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"55342\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"55342\"]\r\n<ol>\r\n \t<li>The middle area = 0.40, so the two tails combined have an area of 0.60 (1 \u2013 0.40). The tails of the graph of the normal distribution each have an area of 0.30.\u00a0 Find <em>k1<\/em>, the 30th percentile and <em>k2<\/em>, the 70th percentile (0.40 + 0.30 = 0.70). Using <strong>Excel<\/strong> <em>k1<\/em> = NORM.INV(0.3, 5.85, 0.24) = 5.72 cm and<em> k2<\/em> = NORM.INV(0.7, 5.85, 0.24) = 5.98 cm Using <strong>TI 83\/84<\/strong>\u00a0<em>k1<\/em> = invNorm(0.30,5.85,0.24) = 5.72 cm\u00a0<em>k2<\/em> = invNorm(0.70,5.85,0.24) = 5.98 cm. The middle 40% of the mandarin oranges from this farm are between 5.72 and 5.98 cm.<\/li>\r\n \t<li>Using <strong>Excel<\/strong> NORM.INV(0.16, 5.85, 0.24) = 5.61.\u00a0 Using TI 83\/84 invNorm(0.16, 5.85, 0.24) = 5.61\u00a0 16% of the mandarin oranges have diameters of at most 5.61 cm.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>References<\/h2>\r\n\"Naegele's rule.\" Wikipedia. Available online at http:\/\/en.wikipedia.org\/wiki\/Naegele's_rule (accessed May 14, 2013).\r\n\r\n\"403: NUMMI.\" Chicago Public Media &amp; Ira Glass, 2013. Available online at http:\/\/www.thisamericanlife.org\/radio-archives\/episode\/403\/nummi (accessed May 14, 2013).\r\n\r\n\"Scratch-Off Lottery Ticket Playing Tips.\" WinAtTheLottery.com, 2013. Available online at http:\/\/www.winatthelottery.com\/public\/department40.cfm (accessed May 14, 2013).\r\n\r\n\"Smart Phone Users, By The Numbers.\" Visual.ly, 2013. Available online at http:\/\/visual.ly\/smart-phone-users-numbers (accessed May 14, 2013).\r\n\r\n\"Facebook Statistics.\" Statistics Brain. Available online at http:\/\/www.statisticbrain.com\/facebook-statistics\/(accessed May 14, 2013).\r\n<h2>Concept Review<\/h2>\r\nThe normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean\u00a0<em>\u00b5<\/em> and the standard deviation <em>\u03c3<\/em>. A special normal distribution, called the standard normal distribution is the distribution of <em>z<\/em>-scores. Its mean is zero, and its standard deviation is one.\r\n<h2>Formula Review<\/h2>\r\nNormal Distribution:\r\n<em>X<\/em> ~ <em>N<\/em>(<em>\u00b5<\/em>, <em>\u03c3<\/em>) where <em>\u00b5<\/em> is the mean and <em>\u03c3<\/em> is the standard deviation.\r\n\r\nStandard Normal Distribution:\r\n<em>Z<\/em> ~ <em>N<\/em>(0, 1).\r\n\r\nCalculator function for probability: normalcdf (lower\u00a0<em>x<\/em> value of the area, upper <em>x <\/em>value of the area, mean, standard deviation).\r\n\r\nExcel function for probability\/area to the left: NORM.DIST(<em>x<\/em> value, mean, standard deviation,1).\r\n\r\nCalculator function for the\u00a0<em>k<\/em>th percentile: <em>k<\/em> = invNorm (area to the left of <em>k<\/em>, mean, standard deviation)\r\n\r\nCalculator function for the\u00a0<em>k<\/em>th percentile: <em>k<\/em> = NORM.INV(area to the left of <em>k<\/em>, mean, standard deviation)","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul id=\"list4253\">\n<li>Recognize the normal probability distribution and apply it appropriately.<\/li>\n<li>Compare normal probabilities by converting to the standard normal distribution and directly using technology.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\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\" 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>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>.<\/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>) = <strong>Area to the right<\/strong> of the vertical line through <em>x<\/em>.<\/p>\n<p><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>\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 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=\"q337707\">Show Answer<\/span><\/p>\n<div id=\"q337707\" class=\"hidden-answer\" style=\"display: none\"> 1 \u2212 0.012 = 0.988 <\/div>\n<\/div>\n<\/div>\n<h2>Calculation of Probabilities and Finding Values from Probabilities<\/h2>\n<p>Probabilities are calculated using technology. We can also find values in the distribution (such as percentiles) from given probabilities using technology.<\/p>\n<p><strong>Regardless of the method used to find the probabilities or numbers, always draw a sketch of the normal distribution, shade the relevant area under the curve and label the known and unknown values.\u00a0<\/strong>You can use the fact that the mean is in the center of the distribution (50% of the area to its left and 50% of the area to its right) and the Empirical Rule percentages in the previous section to determine if your answers are reasonable.<\/p>\n<div class=\"textbox shaded\" style=\"text-align: left;\">\n<h3>Finding Probabilities FOR THE NORMAL DISTRIBUTION Using EXCEL<\/h3>\n<p>Access the norm.dist function (under\u00a0[latex]f_{x}[\/latex]). There are four inputs required [latex]x[\/latex], the mean, the standard deviation, and cumulative. <span style=\"color: #000000;\"><strong>For all work in this class, we are using this function to find P([latex]X < x[\/latex]), so we will only be using cumulative = 1 (or True).<\/strong>\u00a0<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2547 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/08203237\/norm.dist_-300x94.png\" alt=\"\" width=\"300\" height=\"94\" \/><\/p>\n<ul>\n<li><\/li>\n<li><\/li>\n<li><\/li>\n<li><\/li>\n<li><strong>Area\/Probability to the Left of a Number:<\/strong> To calculate <em>P<\/em>([latex]x <[\/latex] number),\u00a0 NORM.DIST(number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\n<li><strong>Area\/Probability Between Two Numbers:<\/strong> To calculate <em>P<\/em>([latex]number1 < x < number2[\/latex]), subtract the smaller area from the larger area. NORM.DIST(number2, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1) &#8211;\u00a0NORM.DIST(number1, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\n<li><strong>Area\/Probability to the Right of a Number:<\/strong> To calculate <em>P<\/em>([latex]x >[\/latex] number), subtract the area from one. 1 &#8211;\u00a0NORM.DIST(number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex], 1).<\/li>\n<\/ul>\n<p><em>If you convert your x values to z-scores, your distribution will be the Standard normal distribution. You can enter a mean of zero and a standard deviation of one into this NORM.DIST function or run the NORM.S.DIST function which only requires two inputs [the z-score and cumulative = 1 (or True)].<\/em><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Finding Probabilities FOR THE NORMAL DISTRIBUTION Using TI-83, 83+, 84, 84+<\/h3>\n<p>Go into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>. 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>. The syntax for the instructions are as follows:\u00a0 normalcdf(lower value, upper value, mean, standard deviation)<\/p>\n<p>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.\u00a0The number \u201310<sup>99<\/sup> is way out in the left tail of the normal curve.<\/p>\n<ul>\n<li><strong>Area\/Probability to the Left of a Number:<\/strong> To calculate <em>P<\/em>([latex]x <[\/latex] number),\u00a0 normcdf(-1E99, number, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex]).<\/li>\n<li><strong>Area\/Probability Between Two Numbers:<\/strong> To calculate <em>P<\/em>([latex]number1 < x< number2[\/latex]). Number2 is greater than number1\u00a0 normalcdf(number1,number2,[latex]\\displaystyle{\\mu},{\\sigma}[\/latex])<\/li>\n<li><strong>Area\/Probability to the Right of a Number:<\/strong> To calculate <em>P<\/em>([latex]x >[\/latex] number),\u00a0 normalcdf(number, 1E99, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex]).<\/li>\n<\/ul>\n<p><em>If the mean and standard deviation are left out of the function, the calculator will use the mean of zero and standard deviation of one for the Standard Normal Distribution.<\/em><\/p>\n<\/div>\n<div class=\"textbox shaded\" style=\"text-align: left;\">\n<h3>Finding VALUES FOR THE NORMAL DISTRIBUTION Using EXCEL<\/h3>\n<p>Access the norm.inv function (under\u00a0[latex]f_{x}[\/latex]). There are three inputs required (the area to the left, the mean and the standard deviation).<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2558 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5658\/2016\/04\/08210800\/norm.inv_-300x76.png\" alt=\"\" width=\"300\" height=\"76\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<ul>\n<li><strong>Number with a given probability\/area to its left:<\/strong> NORM.INV(probability\/area to the left, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex].<\/li>\n<\/ul>\n<p><em>&#8211; If the problem gives you an area to the right of your unknown number, convert it to the area to the left by subtracting it from one. Example: the x value that has area 0.015 its right also has 1 &#8211; 0.015 or 0.985 to its left<\/em><em>.<\/em><\/p>\n<p><em>&#8211; The NORM.S.INV function provides z-scores from the\u00a0<\/em><em>Standard normal distribution without having to enter the mean of zero and a standard deviation of one.<\/em><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Finding VALUES FOR THE NORMAL DISTRIBUTION Using TI-83, 83+, 84, 84+<code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><code style=\"line-height: 1.6em; text-align: justify;\" data-redactor-tag=\"code\"><\/code><\/code><\/h3>\n<p>Go into <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>. After pressing <code style=\"line-height: 1.6em; text-align: justify;\">2nd DISTR<\/code>, press<code style=\"line-height: 1.6em; text-align: justify;\">3:invNorm<\/code>. The syntax for the instructions are as follows: invNorm(probability\/area to the left, mean, standard deviation)<\/p>\n<ul>\n<li><strong style=\"font-size: 0.9em;\">Number with a given probability\/area to its left:<\/strong><span style=\"font-size: 0.9em;\">\u00a0invNorm(probability\/area to the left, [latex]\\displaystyle{\\mu},{\\sigma}[\/latex].<\/span><\/li>\n<\/ul>\n<p><em>&#8211; If the problem gives you an area to the right of your unknown number, convert it to the area to the left by subtracting it from one. Example: the x value that has area 0.015 its right also has 1 &#8211; 0.015 or 0.985 to its left<\/em><em>.<\/em><\/p>\n<p><em>&#8211; If the mean and standard deviation are left out of the function, the calculator will use the mean of zero and standard deviation of one for the Standard Normal Distribution.<\/em><\/p>\n<\/div>\n<h4>Note<\/h4>\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>The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.<\/p>\n<ol>\n<li>Find the probability that a randomly selected student scored more than 65 on the exam.<\/li>\n<li>Find the probability that a randomly selected student scored less than 85.<\/li>\n<li>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>).<\/li>\n<li>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>).<\/li>\n<\/ol>\n<p>Solution:\u00a0 Let\u00a0<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<ol>\n<li>Draw a graph. Then, find <em>P<\/em>(<em>x<\/em> &gt; 65).<br \/>\n<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.\" \/><\/li>\n<\/ol>\n<p>Using <strong>Excel <\/strong>1 &#8211; NORM.DIST(65, 63, 5, 1 or True)\u00a0The result is\u00a0[latex]P(x > 65)=0.3446[\/latex].<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(65,1E99,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">). The result is\u00a0[latex]P(x > 65)=0.3446[\/latex].\u00a0<\/span><\/p>\n<p><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">The probability that any student selected at random scores more than 65 is 0.3446.<\/span><\/p>\n<p>2. Draw a graph. Then find <em>P<\/em>(<em>x<\/em> &lt; 85), and shade the graph.<\/p>\n<p>Using <strong>Excel <\/strong>NORM.DIST(85, 63, 5, 1 or True)\u00a0The result is\u00a0[latex]P(x < 85)=1.0000[\/latex].\n\nUsing <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(-1E99,85,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">). The result is\u00a0[latex]P(x < 85)=1.0000[\/latex].\u00a0<\/span><\/p>\n<p>The probability that one student scores less than 85 is approximately one (or 100%).<\/p>\n<p>3. 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.<br \/>\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>.<br \/>\n<img decoding=\"async\" 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>Using <strong>Excel <\/strong>NORM.INV(0.90, 63, 5) The result is [latex]\\approx69.4[\/latex]<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 3:invNorm(0.90,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">)\u00a0The result is [latex]\\approx69.4[\/latex]<\/span><\/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.<\/p>\n<p>4. Find the 70th percentile. Draw a new graph and label it appropriately.<\/p>\n<p>Using <strong>Excel <\/strong>NORM.INV(0.70, 63, 5) The result is [latex]\\approx65.6[\/latex]<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 3:invNorm(0.70,63,5<\/code><span style=\"font-size: 1rem; text-align: initial;\">)\u00a0The result is [latex]\\approx65.6[\/latex]<\/span><\/p>\n<p><em>k<\/em> = 65.6 The 70th percentile is 65.6. This means that 70% of the test scores fall at or below 65.6 and 30% fall at or above 65.6.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/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=\"q315155\">Show Solution<\/span><\/p>\n<div id=\"q315155\" class=\"hidden-answer\" style=\"display: none\">\n<p>Draw a graph. Then find <em>P<\/em>(<em>x<\/em> &lt; 65), and shade the graph.<\/p>\n<p>Using <strong>Excel <\/strong>NORM.DIST(65, 68, 3, 1) = 0.1587<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong><code style=\"line-height: 1.6em; text-align: justify;\">2nd Distr 2:normalcdf(-1E99,65,68,3<\/code><span style=\"font-size: 1rem; text-align: initial;\">) = 0.1587\u00a0<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/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<ol>\n<li>Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.<\/li>\n<li>Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425730\">Show Solution<\/span><\/p>\n<div id=\"q425730\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Let <em>X<\/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. Find <em>P<\/em>(1.8 &lt; <em>x<\/em> &lt; 2.75).The probability for which you are looking is the area<br \/>\n<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\" class=\"alignleft\" 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.\" \/><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>Using <strong>Excel <\/strong>NORM.DIST(2.75, 2, 0.5, 1) &#8211; NORM.DIST(1.8, 2, 0.5, 1)= 0.5886<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong>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<p>2. 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>Using <strong>Excel <\/strong>NORM.INV(0.25,2,0.5) = 1.66<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong>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<\/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=\"q577099\">Show Answer<\/span><\/p>\n<div id=\"q577099\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using <strong>Excel <\/strong>NORM.DIST(70, 68, 3, 1) &#8211; NORM.DIST(66, 68, 3, 1)= 0.4950<\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong>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<\/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<ol>\n<li>Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.<\/li>\n<li>Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.<\/li>\n<li>Find the 80th percentile of this distribution, and interpret it in a complete sentence.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q588265\">Show Solution<\/span><\/p>\n<div id=\"q588265\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Using <strong>Excel <\/strong>NORM.DIST(64.7,36.9,13.9, 1) &#8211; NORM.DIST(23,36.9,13.9, 1)= 0.8186Using <strong>TI83\/84\u00a0<\/strong><span style=\"font-size: 0.9em;\">normalcdf(23,64.7,36.9,13.9) = 0.8186<\/span><\/li>\n<li>Using <strong>Excel <\/strong>NORM.DIST(50.8,36.9,13.9, 1) = 0.8413Using <strong>TI83\/84\u00a0<\/strong><span style=\"font-size: 1rem; text-align: initial;\">normalcdf(\u201310<\/span><sup style=\"text-align: initial;\">99<\/sup><span style=\"font-size: 1rem; text-align: initial;\">,50.8,36.9,13.9) = 0.8413<\/span><\/li>\n<li>Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Excel <\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">NORM.INV(0.8,36.9,13.9) = 48.6\u00a0 \u00a0<\/span>Using <strong>TI83\/84 <\/strong><span style=\"font-size: 1rem; text-align: initial;\">invNorm(0.80,36.9,13.9) = 48.6\u00a0<\/span>The 80th percentile is 48.6 years. 80% of the smartphone users in the age range 13 \u2013 55+ are 48.6 years old or less.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use the information in previous example to answer the following questions.<\/p>\n<ol>\n<li>Find the 30th percentile, and interpret it in a complete sentence.<\/li>\n<li>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.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q699940\">Show Solution<\/span><\/p>\n<div id=\"q699940\" 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>1.To find the 30th percentile, find\u00a0<em>k<\/em> such that <em>P<\/em>(<em>x<\/em> &lt; <em>k<\/em>) = 0.30. Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">Excel\u00a0<\/strong><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">NORM.INV(0.8,36.9,13.9) = 48.6\u00a0<\/span>Using <strong style=\"font-size: 1rem; orphans: 1; text-align: initial;\">TI83\/84\u00a0<\/strong>invNorm(0.30, 36.9, 13.9) = 29.6<\/p>\n<p>Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.<\/p>\n<p>2. 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>2.Using <strong>Excel <\/strong>NORM.DIST(27,36.9,13.9, 1) = 0.2382 <em>(ignore the slight discrepancy from the graph above)<\/em><\/p>\n<p>Using <strong>TI83\/84\u00a0<\/strong>normalcdf(\u201310<sup>99<\/sup>,27,36.9,13.9) = 0.2382<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/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<ol>\n<li>Calculate the interquartile range (<em>IQR<\/em>).<\/li>\n<li>Forty percent of the ages that range from 13 to 55+ are at least what age?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q845498\">Show Solution<\/span><\/p>\n<div id=\"q845498\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1\u00a0<\/sub>Calculate\u00a0<em>Q<\/em><sub>3<\/sub> = 75th percentile and <em>Q<\/em><sub>1<\/sub> = 25th percentile. Using <strong>Excel<\/strong> NORM.INV(0.75,36.9, 13.9) = 46.2754 and NORM.INV(0.25, 36.9,13.9) = 27.5246. Using the <strong>TI 83\/84<\/strong>\u00a0invNorm(0.75,36.9,13.9) = <em>Q<\/em><sub>3<\/sub> = 46.2754 invNorm(0.25,36.9,13.9) = <em>Q<\/em><sub>1<\/sub> = 27.5246<br \/>\n<em>IQR<\/em> = <em>Q<\/em><sub>3<\/sub> \u2013 <em>Q<\/em><sub>1<\/sub> = 18.7508 years<\/li>\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;) 0.40 = the area to the right. Area to the left = 1 \u2013 0.40 = 0.60. The area to the left of <em>k<\/em> = 0.60.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.60, 36.9, 13.9) using <strong>TI 83\/84<\/strong> invNorm(0.60,36.9,13.9) = 40.4215. <em>k<\/em> = 40.42. Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/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<ol>\n<li>Calculate the first- and third-quartile scores for this exam.<\/li>\n<li>The middle 50% of the exam scores are between what two values?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q889848\">Show Answer<\/span><\/p>\n<div id=\"q889848\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><em>Q<\/em><sub>1<\/sub> = 25th percentile.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.25, 81, 15) = 70.9.\u00a0 Using <strong>TI 83\/84<\/strong> invNorm(0.25,81,15) = 70.9<br \/>\n<em>Q<\/em><sub>3<\/sub> = 75th percentile.\u00a0 Using <strong>Excel<\/strong> NORM.INV(0.75, 81, 15) = 91.1.\u00a0 \u00a0Using <strong>TI 83\/84<\/strong> invNorm(0.75,81,15) = 91.1<\/li>\n<li>The middle 50% of the scores are between 70.9 and 91.1.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example<\/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<ol>\n<li>Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.<\/li>\n<li>The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.<\/li>\n<li>Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q977845\">Show Solution<\/span><\/p>\n<div id=\"q977845\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Using <strong>Excel<\/strong> 1 &#8211; NORM.DIST(6, 5.85, 0.24, 1) = 0.2660.\u00a0 Using <strong>TI 83\/84<\/strong> 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.\" \/><\/li>\n<li>1 \u2013 0.20 = 0.80 The tails of the graph of the normal distribution each have an area of 0.40. Find <em>k1<\/em>, the 40th percentile, and <em>k2<\/em>, the 60th percentile (0.40 + 0.20 = 0.60). Using <strong>Excel<\/strong>\u00a0<em style=\"font-size: 0.9em;\">k1<\/em><span style=\"font-size: 0.9em;\"> = NORM.INV(0.40, 5.85, 0.24) = 5.79 cm <\/span><em style=\"font-size: 0.9em;\">k2<\/em><span style=\"font-size: 0.9em;\"> = NORM.INV(0.60, 5.85, 0.24) = 5.91 Using <strong>TI 83\/84\u00a0<\/strong><\/span><em>k1<\/em> = invNorm(0.40, 5.85, 0.24) = 5.79 cm <em>k2<\/em> = invNorm(0.60, 5.85, 0.24) = 5.91 cm<\/li>\n<li>Using <strong>Excel<\/strong> NORM.INV(0.90, 5.85, 0.24) = 6.16.\u00a0 Using <strong>TI 83\/84<\/strong> invNorm(0.90, 5.85, 0.24) = 6.16\u00a0 Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Using the information from previous, answer the following:<\/p>\n<ol>\n<li>The middle 40% of mandarin oranges from this farm are between ______ and ______.<\/li>\n<li>Find the 16th percentile and interpret it in a complete sentence.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q55342\">Show Answer<\/span><\/p>\n<div id=\"q55342\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The middle area = 0.40, so the two tails combined have an area of 0.60 (1 \u2013 0.40). The tails of the graph of the normal distribution each have an area of 0.30.\u00a0 Find <em>k1<\/em>, the 30th percentile and <em>k2<\/em>, the 70th percentile (0.40 + 0.30 = 0.70). Using <strong>Excel<\/strong> <em>k1<\/em> = NORM.INV(0.3, 5.85, 0.24) = 5.72 cm and<em> k2<\/em> = NORM.INV(0.7, 5.85, 0.24) = 5.98 cm Using <strong>TI 83\/84<\/strong>\u00a0<em>k1<\/em> = invNorm(0.30,5.85,0.24) = 5.72 cm\u00a0<em>k2<\/em> = invNorm(0.70,5.85,0.24) = 5.98 cm. The middle 40% of the mandarin oranges from this farm are between 5.72 and 5.98 cm.<\/li>\n<li>Using <strong>Excel<\/strong> NORM.INV(0.16, 5.85, 0.24) = 5.61.\u00a0 Using TI 83\/84 invNorm(0.16, 5.85, 0.24) = 5.61\u00a0 16% of the mandarin oranges have diameters of at most 5.61 cm.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h2>References<\/h2>\n<p>&#8220;Naegele&#8217;s rule.&#8221; Wikipedia. Available online at http:\/\/en.wikipedia.org\/wiki\/Naegele&#8217;s_rule (accessed May 14, 2013).<\/p>\n<p>&#8220;403: NUMMI.&#8221; Chicago Public Media &amp; Ira Glass, 2013. Available online at http:\/\/www.thisamericanlife.org\/radio-archives\/episode\/403\/nummi (accessed May 14, 2013).<\/p>\n<p>&#8220;Scratch-Off Lottery Ticket Playing Tips.&#8221; WinAtTheLottery.com, 2013. Available online at http:\/\/www.winatthelottery.com\/public\/department40.cfm (accessed May 14, 2013).<\/p>\n<p>&#8220;Smart Phone Users, By The Numbers.&#8221; Visual.ly, 2013. Available online at http:\/\/visual.ly\/smart-phone-users-numbers (accessed May 14, 2013).<\/p>\n<p>&#8220;Facebook Statistics.&#8221; Statistics Brain. Available online at http:\/\/www.statisticbrain.com\/facebook-statistics\/(accessed May 14, 2013).<\/p>\n<h2>Concept Review<\/h2>\n<p>The normal distribution, which is continuous, is the most important of all the probability distributions. Its graph is bell-shaped. This bell-shaped curve is used in almost all disciplines. Since it is a continuous distribution, the total area under the curve is one. The parameters of the normal are the mean\u00a0<em>\u00b5<\/em> and the standard deviation <em>\u03c3<\/em>. A special normal distribution, called the standard normal distribution is the distribution of <em>z<\/em>-scores. Its mean is zero, and its standard deviation is one.<\/p>\n<h2>Formula Review<\/h2>\n<p>Normal Distribution:<br \/>\n<em>X<\/em> ~ <em>N<\/em>(<em>\u00b5<\/em>, <em>\u03c3<\/em>) where <em>\u00b5<\/em> is the mean and <em>\u03c3<\/em> is the standard deviation.<\/p>\n<p>Standard Normal Distribution:<br \/>\n<em>Z<\/em> ~ <em>N<\/em>(0, 1).<\/p>\n<p>Calculator function for probability: normalcdf (lower\u00a0<em>x<\/em> value of the area, upper <em>x <\/em>value of the area, mean, standard deviation).<\/p>\n<p>Excel function for probability\/area to the left: NORM.DIST(<em>x<\/em> value, mean, standard deviation,1).<\/p>\n<p>Calculator function for the\u00a0<em>k<\/em>th percentile: <em>k<\/em> = invNorm (area to the left of <em>k<\/em>, mean, standard deviation)<\/p>\n<p>Calculator function for the\u00a0<em>k<\/em>th percentile: <em>k<\/em> = NORM.INV(area to the left of <em>k<\/em>, mean, standard deviation)<\/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-239\">\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,Using the Normal Distribution. <strong>Authored by<\/strong>:  . <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:41\/Introductory_Statistics\">http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:41\/Introductory_Statistics<\/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>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":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"OpenStax, Statistics,Using the Normal Distribution\",\"author\":\" \",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.41:41\/Introductory_Statistics\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"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 License\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics \",\"author\":\"Barbara Illowski, Susan Dean\",\"organization\":\"Open Stax\",\"url\":\"http:\/\/cnx.org\/contents\/30189442-6998-4686-ac05-ed152b91b9de@17.44\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for 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