{"id":264,"date":"2021-07-14T15:59:01","date_gmt":"2021-07-14T15:59:01","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-central-limit-theorem-for-sample-means-averages\/"},"modified":"2023-12-05T09:21:56","modified_gmt":"2023-12-05T09:21:56","slug":"the-central-limit-theorem-for-sample-means-averages","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-central-limit-theorem-for-sample-means-averages\/","title":{"raw":"Distributions of Sample Means","rendered":"Distributions of Sample Means"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<section>\r\n<ul id=\"list6234\">\r\n \t<li>Calculate the mean and standard deviation for the sampling distribution of a sample mean<\/li>\r\n \t<li>Calculate probabilities for the sampling distribution of a sample mean using technology<\/li>\r\n \t<li>Calculate percentiles for the sampling distribution of a sample mean using technology<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Subscript<\/h3>\r\nMathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable. In statistics we use subscripts to distinguish between random variables. For example, if [latex]X[\/latex] is a random variable, then [latex]\\mu _x[\/latex] is the population mean of the random variable [latex]X[\/latex] where [latex]\\overline{x}_x[\/latex] is the sample mean of the random variable [latex]X[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Central Limit Theorem<\/h3>\r\nSuppose\u00a0<em>X<\/em> is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:\r\n<ol>\r\n \t<li><em>\u03bc<\/em><em><sub data-redactor-tag=\"sub\">X<\/sub><\/em> = the mean of <em>X<\/em><\/li>\r\n \t<li><em>\u03c3<\/em><em><sub data-redactor-tag=\"sub\">X<\/sub><\/em> = the standard deviation of <em>X<\/em><\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Square Root Notation<\/h3>\r\n[latex]\\sqrt{m}[\/latex] is read as \"the square root of [latex]m[\/latex].\"\r\nIf [latex]m={n}^{2}[\/latex] then [latex]\\sqrt{m}=n[\/latex] for [latex]{n}\\ge 0[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221838\/CNX_BMath_Figure_05_07_003_img.png\" alt=\"A picture of an m inside a square root sign is shown. The sign is labeled as a radical sign and the m is labeled as the radicand.\" \/>\r\n\r\n<\/div>\r\nIf you draw random samples of size\u00a0<em>n<\/em>, then as <em>n<\/em> increases, the random variable [latex]\\displaystyle\\overline{{x}}[\/latex] which consists of sample means, tends to be\u00a0normally distributed and\u00a0[latex]\\displaystyle\\overline{{x}}[\/latex] ~\u00a0<em>N<\/em>\u00a0[latex]\\left(\\mu_x,\\frac{\\sigma X}{\\sqrt n}\\right)[\/latex].\r\n\r\n<\/div>\r\nThe\u00a0<strong><span id=\"term126\" data-type=\"term\">central limit theorem<\/span><\/strong>\u00a0for sample means says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate their means, those means tend to follow a normal distribution (the sampling distribution). As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of\u00a0<em data-effect=\"italics\">n<\/em>. The variable\u00a0<em data-effect=\"italics\">n<\/em>\u00a0is the number of values that are averaged together, not the number of times the experiment is done.\r\n\r\nTo put it more formally, if you draw random samples of size\u00a0<em>n<\/em>, the distribution of the random variable [latex]\\displaystyle\\overline{{x}}[\/latex], which consists of sample means, is called the <strong>sampling distribution of the mean<\/strong>. The sampling distribution of the mean approaches a normal distribution as <em>n<\/em>, the sample size, increases.\r\n\r\nThe random variable [latex]\\displaystyle\\overline{{x}}[\/latex] has a different z-score associated with it from that of the random variable <em>X<\/em>. The mean [latex]\\displaystyle\\overline{{x}}[\/latex] is the value of [latex]\\displaystyle\\overline{{x}}[\/latex] in one sample.\r\n<p style=\"text-align: center;\">[latex]z=\\displaystyle\\frac{{\\overline{x}-{\\mu}_{x}}}{{\\frac{{\\sigma{x}}}{{\\sqrt{n}}}}}[\/latex]<\/p>\r\n[latex]\\displaystyle{\\mu}_{x}[\/latex] is the average of both <em>X<\/em> and [latex]\\displaystyle\\overline{x}[\/latex]\r\n\r\n[latex]\\displaystyle{\\sigma}\\overline{x} = {{\\frac{{\\sigma{X}}}{{\\sqrt{n}}}}}[\/latex] = standard deviation of [latex]\\displaystyle\\overline{{x}}[\/latex] and is called the <strong>standard error of the mean<\/strong>.\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\nTo find probabilities for means on the calculator, follow these steps:\r\n2nd DISTR\r\n2:<code data-redactor-tag=\"code\">normalcdf<\/code>\r\n\r\n<code data-redactor-tag=\"code\">normalcdf<\/code> (Lower value of the area, upper value of the area, mean,[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex])\r\n\r\nwhere:\r\n<ul>\r\n \t<li><em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution,<\/li>\r\n \t<li><em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution, and<\/li>\r\n \t<li><em data-redactor-tag=\"em\">sample size<\/em> =\u00a0<em data-redactor-tag=\"em\">n<\/em><\/li>\r\n<\/ul>\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Using Square Roots<\/h3>\r\n[latex]n^2=m[\/latex] and [latex]\\sqrt{m}=n[\/latex] show the reverse relationship of numbers and their squares.\r\n\r\nIf [latex]n^2=m[\/latex], we say [latex]m[\/latex] is the square of [latex]n[\/latex]. If [latex]\\sqrt{m}=n[\/latex], then we can say [latex]n[\/latex] is the square root of [latex]m[\/latex].\r\n\r\n<strong>For example:\u00a0<\/strong>\r\n\r\nBecause [latex]1-^2=11[\/latex],\u00a0 we say [latex]100[\/latex] is the square of [latex]10[\/latex] and [latex]10[\/latex] is the square root of [latex]100[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 1<\/h3>\r\nAn unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size<em data-redactor-tag=\"em\">n<\/em> = 25 are drawn randomly from the population.\r\n\r\n1. Find the probability that the <strong data-redactor-tag=\"strong\">sample mean<\/strong> is between 85 and 92.\r\n\r\n[reveal-answer q=\"726810\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"726810\"]\r\n<p id=\"fs-idp14520352\" class=\" \">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= one value from the original unknown population. The probability question asks you to find a probability for the\u00a0<strong>sample mean<\/strong>.<\/p>\r\n<p id=\"element-126\" class=\" \">Let [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0= the mean of a sample of size 25. Since\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub><em data-effect=\"italics\">X<\/em><\/sub>\u00a0= 90,\u00a0<em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em>\u00a0= 15, and\u00a0<em data-effect=\"italics\">n<\/em>\u00a0= 25,<\/p>\r\n<p id=\"element-756\" class=\" \">[latex]\\displaystyle\\overline{{x}}[\/latex] ~\u00a0<em data-effect=\"italics\">N<\/em>\u00a0[latex]\\left ( 90, \\frac{15}{\\sqrt25} \\right )[\/latex].<\/p>\r\n<p id=\"element-181\" class=\" \">Find\u00a0<em data-effect=\"italics\">P<\/em>(85 &lt; [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&lt; 92). Draw a graph.<\/p>\r\n<p id=\"element-196\" class=\" \"><em data-effect=\"italics\">P<\/em>(85 &lt; [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&lt; 92) = 0.6997<\/p>\r\n<p id=\"element-594\" class=\" \">The probability that the sample mean is between 85 and 92 is 0.6997.<\/p>\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. Find the value that is two standard deviations above the expected value, 90, of the sample mean.\r\n[reveal-answer q=\"902931\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"902931\"]\r\n<p id=\"element-91\" class=\" \">To find the value that is two standard deviations above the expected value 90, use the formula:<\/p>\r\n<p id=\"eip-idm63061120\" class=\" \">value =\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>x<\/sub>\u00a0+ (#ofTSDEVs)[latex]\\left ( \\frac{\\sigma_x}{\\sqrt n} \\right )[\/latex]<\/p>\r\n<p id=\"eip-idp63452384\" class=\" \">value = 90 + 2[latex]\\left ( \\frac{15}{\\sqrt 25} \\right )[\/latex]\u00a0= 96<\/p>\r\n<p id=\"element-564\" class=\" \">The value that is two standard deviations above the expected value is 96.<\/p>\r\n<p id=\"fs-idp29133808\" class=\" \">The standard error of the mean is [latex]\\frac{\\sigma_x}{\\sqrt n}[\/latex]\u00a0=\u00a0[latex]\\frac{15}{\\sqrt 25}[\/latex]\u00a0= 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size\u00a0<em data-effect=\"italics\">n<\/em>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-08-at-8.51.05-AM.png\"><img class=\"aligncenter wp-image-516\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214600\/Screen-Shot-2015-06-08-at-8.51.05-AM.png\" alt=\"Graph of probability P(85&lt; x bar &lt; 92) where the region between 85 and 92 is shaded and there are tick marks at 85, 90 and 92.\" width=\"484\" height=\"236\" \/><\/a>\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<code data-redactor-tag=\"code\">normalcdf:\u00a0<\/code>(lower value, upper value, mean, standard error of the mean)\r\nThe parameter list is abbreviated (lower value, upper value, <em data-redactor-tag=\"em\">\u03bc<\/em>,[latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex]\r\n\r\n<code data-redactor-tag=\"code\">normalcdf:\u00a0<\/code>(85,92,90,\u00a0[latex]\\displaystyle\\frac{{15}}{{\\sqrt{25}}}[\/latex]\u00a0= 0.6997\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/embed\/FXZ2O1Lv-KE\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it 1<\/h3>\r\nAn unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size <em data-redactor-tag=\"em\">n<\/em> = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.\r\n\r\n[reveal-answer q=\"14694\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"14694\"]\r\n\r\nP(42 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 50) = 42, 50, 45,\u00a0[latex]\\displaystyle\\frac{{8}}{{\\sqrt{30}}}[\/latex] = 0.9797\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 2<\/h3>\r\nThe length of time, in hours, it takes an \"over 40\" group of people to play one soccer match is normally distributed with a <strong data-redactor-tag=\"strong\">mean of two hours<\/strong> and a <strong data-redactor-tag=\"strong\">standard deviation of 0.5 hours<\/strong>. A <strong data-redactor-tag=\"strong\">s<\/strong><strong data-redactor-tag=\"strong\">ample of size <em data-redactor-tag=\"em\">n<\/em> = 50<\/strong> is drawn randomly from the population. Find the probability that the <strong data-redactor-tag=\"strong\">sample mean<\/strong> is between 1.8 hours and 2.3 hours.\r\n\r\n[reveal-answer q=\"722778\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"722778\"]\r\n<p id=\"element-723\" class=\" \">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the time, in hours, it takes to play one soccer match.<\/p>\r\n<p id=\"element-302\" class=\" \">The probability question asks you to find a probability for the\u00a0<strong>sample mean time, in hours<\/strong>, it takes to play one soccer match.<\/p>\r\nLet [latex]\\displaystyle\\overline{x}[\/latex] =\u00a0the <strong data-redactor-tag=\"strong\">mean<\/strong> time, in hours, it takes to play one soccer match.\r\n\r\nIf [latex]{\\mu}_{x}[\/latex]= _________,\u00a0[latex]{\\sigma}_{x}[\/latex] = __________, and <em data-redactor-tag=\"em\">n<\/em> = ___________, then [latex]\\displaystyle\\overline{x}[\/latex] ~ <em data-redactor-tag=\"em\">N<\/em>(______, ______) by the <strong data-redactor-tag=\"strong\">central limit theorem for means<\/strong>.\r\n\r\n[latex]{\\mu}_{x}[\/latex] = 2\r\n[latex]{\\sigma}_{x}[\/latex]\u00a0 =\u00a0 0.5\r\nn\u00a0 =\u00a0 50 and\r\nX~N\u00a0 (2, [latex]\\frac{{0.5}}{{\\sqrt{50}}}[\/latex])\r\n\r\nFind\u00a0P(1.8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 2.3). Draw a graph.\r\n\r\nP(1.8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 2.3) = \u00a00.9977\r\n\r\n<code data-redactor-tag=\"code\">normalcdf<\/code>\u00a0:\u00a0(1.8,\u00a02.3,\u00a02,\u00a0[latex]\\displaystyle\\frac{{0.5}}{{\\sqrt{50}}}[\/latex]) = 0.9977\r\n\r\nThe probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.\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 2<\/h3>\r\nThe length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of <em data-redactor-tag=\"em\">n<\/em> = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.\r\n[reveal-answer q=\"469144\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"469144\"]\r\n\r\nP(2 &lt; [latex]\\overline{x}[\/latex] &lt; 3) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>( 2, 3, 2.5,\u00a0[latex]\\frac{0.25}{\\sqrt{60}}[\/latex] = 1\r\n\r\nTo find percentiles for means on the calculator, follow these steps.\r\n<ul>\r\n \t<li>2nd DIStR<\/li>\r\n \t<li>3:invNorm<\/li>\r\n \t<li><em data-redactor-tag=\"em\">k<\/em> = invNorm (area to the left of k, mean[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex]),\u00a0where: <em data-redactor-tag=\"em\">k<\/em> = the <em data-redactor-tag=\"em\">k<\/em>th percentile <em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution <em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution <em data-redactor-tag=\"em\">sample size<\/em> = <em data-redactor-tag=\"em\">n<\/em><\/li>\r\n<\/ul>\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\nTo find percentiles for means on the calculator, follow these steps.\r\n\r\n2nd DIStR\r\n3:invNorm\r\n\r\n<em data-redactor-tag=\"em\">k<\/em> = invNorm (area to the left of k, mean[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex])\r\n\r\nwhere:\r\n<ul>\r\n \t<li><em data-redactor-tag=\"em\">k<\/em> = the <em data-redactor-tag=\"em\">k<\/em>th percentile<\/li>\r\n \t<li><em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution<\/li>\r\n \t<li><em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution<\/li>\r\n \t<li><em data-redactor-tag=\"em\">sample size<\/em> = <em data-redactor-tag=\"em\">n<\/em><\/li>\r\n<\/ul>\r\n<div class=\"textbox exercises\">\r\n<h3>Example 3<\/h3>\r\nIn a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size <em data-redactor-tag=\"em\">n<\/em> = 100.\r\n\r\n1. What are the mean and standard deviation for the sample mean ages of tablet users?\r\n\r\n[reveal-answer q=\"245843\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"245843\"]\r\n\r\nSince the sample mean tends to target the population mean, we have <em data-redactor-tag=\"em\">\u03bc<sub data-redactor-tag=\"sub\">X<\/sub><\/em> = <em data-redactor-tag=\"em\">\u03bc<\/em> = 34. The sample standard deviation is given by [latex]\\displaystyle \\sigma_x = \\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{15}}{{\\sqrt{100}}}=\\frac{{15}}{{10}}={1.5}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. What does the distribution look like?\r\n\r\n[reveal-answer q=\"333899\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"333899\"]\r\n\r\nThe central limit theorem states that for large sample sizes (<em data-redactor-tag=\"em\">n<\/em>), the sampling distribution will be approximately normal.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).\r\n[reveal-answer q=\"923288\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"923288\"]\r\n\r\nThe probability that the sample mean age is more than 30 is given by <em data-redactor-tag=\"em\">P<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&gt; 30) = <code data-redactor-tag=\"code\">normalcdf<\/code>(30,E99,34,1.5) = 0.9962\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n4. Find the 95th percentile for the sample mean age (to one decimal place).\r\n[reveal-answer q=\"282327\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"282327\"]\r\n\r\nLet <em data-redactor-tag=\"em\">k<\/em> = the 95th percentile.\r\n\r\nk = invNorm(0.95, 34, [latex]\\displaystyle\\frac{{15}}{{\\sqrt{100}}}[\/latex]) = 36.5\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it 3<\/h3>\r\nIn an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article's data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29 to 35-year-olds, should you continue with your development strategy?\r\n\r\n[reveal-answer q=\"186202\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"186202\"]\r\n\r\nYou need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game.\u00a0Find\u00a0P(29 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 35) = <code data-redactor-tag=\"code\">normalcdf<\/code> = 0.0186\r\n\r\n(29, 35, 28,\u00a0[latex]\\displaystyle\\frac{{4.8}}{{\\sqrt{100}}}[\/latex] = 0.0186\r\n\r\nYou can conclude there is approximately a 2% chance that your game will be played by men whose mean age is between 29 and 35.\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\nThe mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.\r\n\r\n1. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user?\r\n[reveal-answer q=\"773220\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"773220\"]\r\n\r\n[latex]\\displaystyle{\\mu}_{\\overline{x}}={\\mu}=8.2 {\\sigma}_{\\overline{x}}=\\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{1}}{{\\sqrt{60}}} = 0.13[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. What is the standard error of the mean?\r\n[reveal-answer q=\"449047\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"449047\"]\r\n\r\nThis allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n3. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.\r\n\r\n[reveal-answer q=\"655350\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"655350\"]\r\n\r\nLet <em data-redactor-tag=\"em\">k<\/em> = the 90th percentile.\r\n\r\nk =\u00a0<code data-redactor-tag=\"code\">invNorm<\/code>(0.9, 8.2, \u00a0[latex]\\displaystyle\\frac{{1}}{{\\sqrt{60}}}[\/latex]) = 8.37.\u00a0This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.\r\n[reveal-answer q=\"610501\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"610501\"]\r\n\r\nP(8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 8.5) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>(8, 8.5, 8.2[latex]\\displaystyle\\frac{{1}}{{\\sqrt{60}}}[\/latex]) = 0.9293\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/embed\/J1twbrHel3o\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it 4<\/h3>\r\nCans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are <em data-redactor-tag=\"em\">n<\/em> = 34,[latex]\\displaystyle\\overline{x}[\/latex]\u00a0= 16.01 ounces. If the cans are filled so that <em data-redactor-tag=\"em\">\u03bc<\/em> = 16.00 ounces (as labeled) and <em data-redactor-tag=\"em\">\u03c3<\/em>= 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?\r\n[reveal-answer q=\"856728\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"856728\"]\r\n\r\nWe have P([latex]\\displaystyle\\overline{x}[\/latex] &gt; 16.01) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>(16.01, E99, 16,\u00a0[latex]\\displaystyle\\frac{{0.143}}{{\\sqrt{34}}}[\/latex]= 0.3417\r\n\r\nSince there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company's claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<section>\n<ul id=\"list6234\">\n<li>Calculate the mean and standard deviation for the sampling distribution of a sample mean<\/li>\n<li>Calculate probabilities for the sampling distribution of a sample mean using technology<\/li>\n<li>Calculate percentiles for the sampling distribution of a sample mean using technology<\/li>\n<\/ul>\n<\/section>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Recall: Subscript<\/h3>\n<p>Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable. In statistics we use subscripts to distinguish between random variables. For example, if [latex]X[\/latex] is a random variable, then [latex]\\mu _x[\/latex] is the population mean of the random variable [latex]X[\/latex] where [latex]\\overline{x}_x[\/latex] is the sample mean of the random variable [latex]X[\/latex].<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Central Limit Theorem<\/h3>\n<p>Suppose\u00a0<em>X<\/em> is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:<\/p>\n<ol>\n<li><em>\u03bc<\/em><em><sub data-redactor-tag=\"sub\">X<\/sub><\/em> = the mean of <em>X<\/em><\/li>\n<li><em>\u03c3<\/em><em><sub data-redactor-tag=\"sub\">X<\/sub><\/em> = the standard deviation of <em>X<\/em><\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Square Root Notation<\/h3>\n<p>[latex]\\sqrt{m}[\/latex] is read as &#8220;the square root of [latex]m[\/latex].&#8221;<br \/>\nIf [latex]m={n}^{2}[\/latex] then [latex]\\sqrt{m}=n[\/latex] for [latex]{n}\\ge 0[\/latex].<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24221838\/CNX_BMath_Figure_05_07_003_img.png\" alt=\"A picture of an m inside a square root sign is shown. The sign is labeled as a radical sign and the m is labeled as the radicand.\" \/><\/p>\n<\/div>\n<p>If you draw random samples of size\u00a0<em>n<\/em>, then as <em>n<\/em> increases, the random variable [latex]\\displaystyle\\overline{{x}}[\/latex] which consists of sample means, tends to be\u00a0normally distributed and\u00a0[latex]\\displaystyle\\overline{{x}}[\/latex] ~\u00a0<em>N<\/em>\u00a0[latex]\\left(\\mu_x,\\frac{\\sigma X}{\\sqrt n}\\right)[\/latex].<\/p>\n<\/div>\n<p>The\u00a0<strong><span id=\"term126\" data-type=\"term\">central limit theorem<\/span><\/strong>\u00a0for sample means says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate their means, those means tend to follow a normal distribution (the sampling distribution). As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has the same mean as the original distribution and a variance that equals the original variance divided by the sample size. Standard deviation is the square root of variance, so the standard deviation of the sampling distribution is the standard deviation of the original distribution divided by the square root of\u00a0<em data-effect=\"italics\">n<\/em>. The variable\u00a0<em data-effect=\"italics\">n<\/em>\u00a0is the number of values that are averaged together, not the number of times the experiment is done.<\/p>\n<p>To put it more formally, if you draw random samples of size\u00a0<em>n<\/em>, the distribution of the random variable [latex]\\displaystyle\\overline{{x}}[\/latex], which consists of sample means, is called the <strong>sampling distribution of the mean<\/strong>. The sampling distribution of the mean approaches a normal distribution as <em>n<\/em>, the sample size, increases.<\/p>\n<p>The random variable [latex]\\displaystyle\\overline{{x}}[\/latex] has a different z-score associated with it from that of the random variable <em>X<\/em>. The mean [latex]\\displaystyle\\overline{{x}}[\/latex] is the value of [latex]\\displaystyle\\overline{{x}}[\/latex] in one sample.<\/p>\n<p style=\"text-align: center;\">[latex]z=\\displaystyle\\frac{{\\overline{x}-{\\mu}_{x}}}{{\\frac{{\\sigma{x}}}{{\\sqrt{n}}}}}[\/latex]<\/p>\n<p>[latex]\\displaystyle{\\mu}_{x}[\/latex] is the average of both <em>X<\/em> and [latex]\\displaystyle\\overline{x}[\/latex]<\/p>\n<p>[latex]\\displaystyle{\\sigma}\\overline{x} = {{\\frac{{\\sigma{X}}}{{\\sqrt{n}}}}}[\/latex] = standard deviation of [latex]\\displaystyle\\overline{{x}}[\/latex] and is called the <strong>standard error of the mean<\/strong>.<\/p>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<p>To find probabilities for means on the calculator, follow these steps:<br \/>\n2nd DISTR<br \/>\n2:<code data-redactor-tag=\"code\">normalcdf<\/code><\/p>\n<p><code data-redactor-tag=\"code\">normalcdf<\/code> (Lower value of the area, upper value of the area, mean,[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex])<\/p>\n<p>where:<\/p>\n<ul>\n<li><em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution,<\/li>\n<li><em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution, and<\/li>\n<li><em data-redactor-tag=\"em\">sample size<\/em> =\u00a0<em data-redactor-tag=\"em\">n<\/em><\/li>\n<\/ul>\n<div class=\"textbox examples\">\n<h3>Recall: Using Square Roots<\/h3>\n<p>[latex]n^2=m[\/latex] and [latex]\\sqrt{m}=n[\/latex] show the reverse relationship of numbers and their squares.<\/p>\n<p>If [latex]n^2=m[\/latex], we say [latex]m[\/latex] is the square of [latex]n[\/latex]. If [latex]\\sqrt{m}=n[\/latex], then we can say [latex]n[\/latex] is the square root of [latex]m[\/latex].<\/p>\n<p><strong>For example:\u00a0<\/strong><\/p>\n<p>Because [latex]1-^2=11[\/latex],\u00a0 we say [latex]100[\/latex] is the square of [latex]10[\/latex] and [latex]10[\/latex] is the square root of [latex]100[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example 1<\/h3>\n<p>An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size<em data-redactor-tag=\"em\">n<\/em> = 25 are drawn randomly from the population.<\/p>\n<p>1. Find the probability that the <strong data-redactor-tag=\"strong\">sample mean<\/strong> is between 85 and 92.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q726810\">Show Answer<\/span><\/p>\n<div id=\"q726810\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-idp14520352\" class=\"\">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= one value from the original unknown population. The probability question asks you to find a probability for the\u00a0<strong>sample mean<\/strong>.<\/p>\n<p id=\"element-126\" class=\"\">Let [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0= the mean of a sample of size 25. Since\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub><em data-effect=\"italics\">X<\/em><\/sub>\u00a0= 90,\u00a0<em data-effect=\"italics\">\u03c3<sub>X<\/sub><\/em>\u00a0= 15, and\u00a0<em data-effect=\"italics\">n<\/em>\u00a0= 25,<\/p>\n<p id=\"element-756\" class=\"\">[latex]\\displaystyle\\overline{{x}}[\/latex] ~\u00a0<em data-effect=\"italics\">N<\/em>\u00a0[latex]\\left ( 90, \\frac{15}{\\sqrt25} \\right )[\/latex].<\/p>\n<p id=\"element-181\" class=\"\">Find\u00a0<em data-effect=\"italics\">P<\/em>(85 &lt; [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&lt; 92). Draw a graph.<\/p>\n<p id=\"element-196\" class=\"\"><em data-effect=\"italics\">P<\/em>(85 &lt; [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&lt; 92) = 0.6997<\/p>\n<p id=\"element-594\" class=\"\">The probability that the sample mean is between 85 and 92 is 0.6997.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. Find the value that is two standard deviations above the expected value, 90, of the sample mean.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q902931\">Show Answer<\/span><\/p>\n<div id=\"q902931\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"element-91\" class=\"\">To find the value that is two standard deviations above the expected value 90, use the formula:<\/p>\n<p id=\"eip-idm63061120\" class=\"\">value =\u00a0<em data-effect=\"italics\">\u03bc<\/em><sub>x<\/sub>\u00a0+ (#ofTSDEVs)[latex]\\left ( \\frac{\\sigma_x}{\\sqrt n} \\right )[\/latex]<\/p>\n<p id=\"eip-idp63452384\" class=\"\">value = 90 + 2[latex]\\left ( \\frac{15}{\\sqrt 25} \\right )[\/latex]\u00a0= 96<\/p>\n<p id=\"element-564\" class=\"\">The value that is two standard deviations above the expected value is 96.<\/p>\n<p id=\"fs-idp29133808\" class=\"\">The standard error of the mean is [latex]\\frac{\\sigma_x}{\\sqrt n}[\/latex]\u00a0=\u00a0[latex]\\frac{15}{\\sqrt 25}[\/latex]\u00a0= 3. Recall that the standard error of the mean is a description of how far (on average) that the sample mean will be from the population mean in repeated simple random samples of size\u00a0<em data-effect=\"italics\">n<\/em>.<\/p>\n<\/div>\n<\/div>\n<p><a href=\"https:\/\/courses.candelalearning.com\/introstats1xmaster\/wp-content\/uploads\/sites\/635\/2015\/06\/Screen-Shot-2015-06-08-at-8.51.05-AM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-516\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/132\/2016\/04\/21214600\/Screen-Shot-2015-06-08-at-8.51.05-AM.png\" alt=\"Graph of probability P(85&lt; x bar &lt; 92) where the region between 85 and 92 is shaded and there are tick marks at 85, 90 and 92.\" width=\"484\" height=\"236\" \/><\/a><\/p>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<p><code data-redactor-tag=\"code\">normalcdf:\u00a0<\/code>(lower value, upper value, mean, standard error of the mean)<br \/>\nThe parameter list is abbreviated (lower value, upper value, <em data-redactor-tag=\"em\">\u03bc<\/em>,[latex]\\displaystyle\\frac{{\\sigma}}{{\\sqrt{n}}}[\/latex]<\/p>\n<p><code data-redactor-tag=\"code\">normalcdf:\u00a0<\/code>(85,92,90,\u00a0[latex]\\displaystyle\\frac{{15}}{{\\sqrt{25}}}[\/latex]\u00a0= 0.6997<\/p>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Sampling distribution of the sample mean | Probability and Statistics | Khan Academy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/FXZ2O1Lv-KE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it 1<\/h3>\n<p>An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size <em data-redactor-tag=\"em\">n<\/em> = 30 are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14694\">Show Answer<\/span><\/p>\n<div id=\"q14694\" class=\"hidden-answer\" style=\"display: none\">\n<p>P(42 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 50) = 42, 50, 45,\u00a0[latex]\\displaystyle\\frac{{8}}{{\\sqrt{30}}}[\/latex] = 0.9797<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 2<\/h3>\n<p>The length of time, in hours, it takes an &#8220;over 40&#8221; group of people to play one soccer match is normally distributed with a <strong data-redactor-tag=\"strong\">mean of two hours<\/strong> and a <strong data-redactor-tag=\"strong\">standard deviation of 0.5 hours<\/strong>. A <strong data-redactor-tag=\"strong\">s<\/strong><strong data-redactor-tag=\"strong\">ample of size <em data-redactor-tag=\"em\">n<\/em> = 50<\/strong> is drawn randomly from the population. Find the probability that the <strong data-redactor-tag=\"strong\">sample mean<\/strong> is between 1.8 hours and 2.3 hours.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q722778\">Show Answer<\/span><\/p>\n<div id=\"q722778\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"element-723\" class=\"\">Let\u00a0<em data-effect=\"italics\">X<\/em>\u00a0= the time, in hours, it takes to play one soccer match.<\/p>\n<p id=\"element-302\" class=\"\">The probability question asks you to find a probability for the\u00a0<strong>sample mean time, in hours<\/strong>, it takes to play one soccer match.<\/p>\n<p>Let [latex]\\displaystyle\\overline{x}[\/latex] =\u00a0the <strong data-redactor-tag=\"strong\">mean<\/strong> time, in hours, it takes to play one soccer match.<\/p>\n<p>If [latex]{\\mu}_{x}[\/latex]= _________,\u00a0[latex]{\\sigma}_{x}[\/latex] = __________, and <em data-redactor-tag=\"em\">n<\/em> = ___________, then [latex]\\displaystyle\\overline{x}[\/latex] ~ <em data-redactor-tag=\"em\">N<\/em>(______, ______) by the <strong data-redactor-tag=\"strong\">central limit theorem for means<\/strong>.<\/p>\n<p>[latex]{\\mu}_{x}[\/latex] = 2<br \/>\n[latex]{\\sigma}_{x}[\/latex]\u00a0 =\u00a0 0.5<br \/>\nn\u00a0 =\u00a0 50 and<br \/>\nX~N\u00a0 (2, [latex]\\frac{{0.5}}{{\\sqrt{50}}}[\/latex])<\/p>\n<p>Find\u00a0P(1.8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 2.3). Draw a graph.<\/p>\n<p>P(1.8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 2.3) = \u00a00.9977<\/p>\n<p><code data-redactor-tag=\"code\">normalcdf<\/code>\u00a0:\u00a0(1.8,\u00a02.3,\u00a02,\u00a0[latex]\\displaystyle\\frac{{0.5}}{{\\sqrt{50}}}[\/latex]) = 0.9977<\/p>\n<p>The probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox key-takeaways\">\n<h3>Try it 2<\/h3>\n<p>The length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. A sample size of <em data-redactor-tag=\"em\">n<\/em> = 60 is drawn randomly from the population. Find the probability that the sample mean is between two hours and three hours.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q469144\">Show Answer<\/span><\/p>\n<div id=\"q469144\" class=\"hidden-answer\" style=\"display: none\">\n<p>P(2 &lt; [latex]\\overline{x}[\/latex] &lt; 3) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>( 2, 3, 2.5,\u00a0[latex]\\frac{0.25}{\\sqrt{60}}[\/latex] = 1<\/p>\n<p>To find percentiles for means on the calculator, follow these steps.<\/p>\n<ul>\n<li>2nd DIStR<\/li>\n<li>3:invNorm<\/li>\n<li><em data-redactor-tag=\"em\">k<\/em> = invNorm (area to the left of k, mean[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex]),\u00a0where: <em data-redactor-tag=\"em\">k<\/em> = the <em data-redactor-tag=\"em\">k<\/em>th percentile <em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution <em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution <em data-redactor-tag=\"em\">sample size<\/em> = <em data-redactor-tag=\"em\">n<\/em><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<\/div>\n<h3 class=\"os-title\" data-type=\"title\"><span class=\"os-title-label\">USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\n<p>To find percentiles for means on the calculator, follow these steps.<\/p>\n<p>2nd DIStR<br \/>\n3:invNorm<\/p>\n<p><em data-redactor-tag=\"em\">k<\/em> = invNorm (area to the left of k, mean[latex]\\displaystyle\\sqrt{\\frac{{\\text{standard deviation}}}{{\\text{sample size}}}}[\/latex])<\/p>\n<p>where:<\/p>\n<ul>\n<li><em data-redactor-tag=\"em\">k<\/em> = the <em data-redactor-tag=\"em\">k<\/em>th percentile<\/li>\n<li><em data-redactor-tag=\"em\">mean<\/em> is the mean of the original distribution<\/li>\n<li><em data-redactor-tag=\"em\">standard deviation<\/em> is the standard deviation of the original distribution<\/li>\n<li><em data-redactor-tag=\"em\">sample size<\/em> = <em data-redactor-tag=\"em\">n<\/em><\/li>\n<\/ul>\n<div class=\"textbox exercises\">\n<h3>Example 3<\/h3>\n<p>In a recent study reported Oct. 29, 2012 on the Flurry Blog, the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size <em data-redactor-tag=\"em\">n<\/em> = 100.<\/p>\n<p>1. What are the mean and standard deviation for the sample mean ages of tablet users?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245843\">Show Answer<\/span><\/p>\n<div id=\"q245843\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the sample mean tends to target the population mean, we have <em data-redactor-tag=\"em\">\u03bc<sub data-redactor-tag=\"sub\">X<\/sub><\/em> = <em data-redactor-tag=\"em\">\u03bc<\/em> = 34. The sample standard deviation is given by [latex]\\displaystyle \\sigma_x = \\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{15}}{{\\sqrt{100}}}=\\frac{{15}}{{10}}={1.5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. What does the distribution look like?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q333899\">Show Answer<\/span><\/p>\n<div id=\"q333899\" class=\"hidden-answer\" style=\"display: none\">\n<p>The central limit theorem states that for large sample sizes (<em data-redactor-tag=\"em\">n<\/em>), the sampling distribution will be approximately normal.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>3. Find the probability that the sample mean age is more than 30 years (the reported mean age of tablet users in this particular study).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923288\">Show Answer<\/span><\/p>\n<div id=\"q923288\" class=\"hidden-answer\" style=\"display: none\">\n<p>The probability that the sample mean age is more than 30 is given by <em data-redactor-tag=\"em\">P<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0&gt; 30) = <code data-redactor-tag=\"code\">normalcdf<\/code>(30,E99,34,1.5) = 0.9962<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>4. Find the 95th percentile for the sample mean age (to one decimal place).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q282327\">Show Answer<\/span><\/p>\n<div id=\"q282327\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em data-redactor-tag=\"em\">k<\/em> = the 95th percentile.<\/p>\n<p>k = invNorm(0.95, 34, [latex]\\displaystyle\\frac{{15}}{{\\sqrt{100}}}[\/latex]) = 36.5<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it 3<\/h3>\n<p>In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. You are researching a startup game targeted at the 35-year-old demographic. Your idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article&#8217;s data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. You take a sample of 100 randomly selected gamers. If your target market is 29 to 35-year-olds, should you continue with your development strategy?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q186202\">Show Answer<\/span><\/p>\n<div id=\"q186202\" class=\"hidden-answer\" style=\"display: none\">\n<p>You need to determine the probability for men whose mean age is between 29 and 35 years of age wanting to play a strategy game.\u00a0Find\u00a0P(29 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 35) = <code data-redactor-tag=\"code\">normalcdf<\/code> = 0.0186<\/p>\n<p>(29, 35, 28,\u00a0[latex]\\displaystyle\\frac{{4.8}}{{\\sqrt{100}}}[\/latex] = 0.0186<\/p>\n<p>You can conclude there is approximately a 2% chance that your game will be played by men whose mean age is between 29 and 35.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3><\/h3>\n<div class=\"textbox exercises\">\n<h3>Example 4<\/h3>\n<p>The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.<\/p>\n<p>1. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q773220\">Show Answer<\/span><\/p>\n<div id=\"q773220\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\displaystyle{\\mu}_{\\overline{x}}={\\mu}=8.2 {\\sigma}_{\\overline{x}}=\\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{1}}{{\\sqrt{60}}} = 0.13[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. What is the standard error of the mean?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q449047\">Show Answer<\/span><\/p>\n<div id=\"q449047\" class=\"hidden-answer\" style=\"display: none\">\n<p>This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>3. Find the 90th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q655350\">Show Answer<\/span><\/p>\n<div id=\"q655350\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let <em data-redactor-tag=\"em\">k<\/em> = the 90th percentile.<\/p>\n<p>k =\u00a0<code data-redactor-tag=\"code\">invNorm<\/code>(0.9, 8.2, \u00a0[latex]\\displaystyle\\frac{{1}}{{\\sqrt{60}}}[\/latex]) = 8.37.\u00a0This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q610501\">Show Answer<\/span><\/p>\n<div id=\"q610501\" class=\"hidden-answer\" style=\"display: none\">\n<p>P(8 &lt; [latex]\\displaystyle\\overline{x}[\/latex] &lt; 8.5) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>(8, 8.5, 8.2[latex]\\displaystyle\\frac{{1}}{{\\sqrt{60}}}[\/latex]) = 0.9293<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Standard error of the mean | Inferential statistics | Probability and Statistics | Khan Academy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/J1twbrHel3o?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try it 4<\/h3>\n<p>Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are <em data-redactor-tag=\"em\">n<\/em> = 34,[latex]\\displaystyle\\overline{x}[\/latex]\u00a0= 16.01 ounces. If the cans are filled so that <em data-redactor-tag=\"em\">\u03bc<\/em> = 16.00 ounces (as labeled) and <em data-redactor-tag=\"em\">\u03c3<\/em>= 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q856728\">Show Answer<\/span><\/p>\n<div id=\"q856728\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have P([latex]\\displaystyle\\overline{x}[\/latex] &gt; 16.01) =\u00a0<code data-redactor-tag=\"code\">normalcdf<\/code>(16.01, E99, 16,\u00a0[latex]\\displaystyle\\frac{{0.143}}{{\\sqrt{34}}}[\/latex]= 0.3417<\/p>\n<p>Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company&#8217;s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.<\/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-264\">\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>Introductory Statistics. <strong>Authored by<\/strong>: Barbara Illowsky, Susan Dean. <strong>Provided by<\/strong>: Open Stax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction\">https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/introductory-statistics\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/prealgebra\/pages\/1-introduction\">https:\/\/openstax.org\/books\/prealgebra\/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\/prealgebra\/pages\/1-introduction<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Sampling distribution of the sample mean | Probability and Statistics | Khan Academy. <strong>Authored by<\/strong>: Khan Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/embed\/FXZ2O1Lv-KE\">https:\/\/www.youtube.com\/embed\/FXZ2O1Lv-KE<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><li>Standard error of the mean | Inferential statistics | Probability and Statistics | Khan Academy. <strong>Authored by<\/strong>: Khan Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/embed\/J1twbrHel3o\">https:\/\/www.youtube.com\/embed\/J1twbrHel3o<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube LIcense<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169134,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Sampling distribution of the sample mean | Probability and Statistics | Khan Academy\",\"author\":\"Khan Academy\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/embed\/FXZ2O1Lv-KE\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"copyrighted_video\",\"description\":\"Standard error of the mean | Inferential statistics | Probability and Statistics | Khan Academy\",\"author\":\"Khan Academy\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/embed\/J1twbrHel3o\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube LIcense\"},{\"type\":\"cc\",\"description\":\"Introductory Statistics\",\"author\":\"Barbara Illowsky, Susan Dean\",\"organization\":\"Open 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