{"id":763,"date":"2020-08-04T20:21:21","date_gmt":"2020-08-05T00:21:21","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&#038;p=763"},"modified":"2025-05-07T17:55:52","modified_gmt":"2025-05-07T21:55:52","slug":"6-3-central-limit-theorem","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/6-3-central-limit-theorem\/","title":{"raw":"6.3 Central Limit Theorem (CLT)","rendered":"6.3 Central Limit Theorem (CLT)"},"content":{"raw":"The central limit theorem is one of the most important theorems in statistics.\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Fact: The Central Limit Theorem<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nWhen a random sample of size <em>n<\/em> is drawn from any population with mean [latex]\\mu[\/latex] and standard deviation [latex]\\sigma[\/latex] , the distribution of the sample mean [latex]\\bar{X}[\/latex] will be (approximately) normally distributed if the sample size <em>n<\/em> is large enough. In general, [latex]n \\geq 30[\/latex]\u00a0is large enough if the population distribution is not too extremely skewed.\r\n\r\n<\/div>\r\n<\/div>\r\nNote that:\r\n<ul>\r\n \t<li>The central limit theorem is about the <strong>shape of the distribution of the sample mean <\/strong>[latex]\\bar{X}[\/latex]. It is the distribution of the random variable [latex]\\bar{X}[\/latex]\u00a0that will be normally distributed if the sample size <em>n<\/em> is large enough.<\/li>\r\n \t<li>The required sample size <em>n<\/em> depends on how skewed the population distribution is. If the population distribution, the distribution of [latex]X[\/latex], is symmetric, [latex]n \\geq 5[\/latex]\u00a0might be large enough to claim that the sample mean [latex]\\bar{X}[\/latex]\u00a0is approximately normally distributed; if the distribution of [latex]X[\/latex] is not too extremely skewed, [latex]n \\geq 30[\/latex]\u00a0should be enough; if the population is very skewed, we might need \u00a0[latex]n \\geq 100[\/latex]\u00a0(see the central limit theorem for proportion in Chapter 10).<\/li>\r\n<\/ul>\r\nIn addition to the results on the mean and standard deviation of [latex]\\bar{X}[\/latex], we can claim that:\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Fact: The Distribution of the Sample Mean [latex]\\color{white} \\bar{X}[\/latex]<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nFor a normal population or large sample, the sample mean [latex]\\bar{X}[\/latex] follows a normal distribution with mean [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu[\/latex] and standard deviation [latex]\\sigma_{\\scriptsize \\bar{X}} = \u00a0\\frac{\\sigma}{\\sqrt{n}}[\/latex]. That is [latex]\\bar{X} \\sim N(\\mu, \\frac{\\sigma}{\\sqrt{n}})[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Distribution of the sample mean [latex]\\color{white} \\bar{X}[\/latex]<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<ol start=\"1\" type=\"1\">\r\n \t<li>Let [latex]X[\/latex] denote student grades in a particular class, and suppose [latex]X[\/latex] is normally distributed with a mean of 70 and a standard deviation of 10, i.e., [latex]X \\sim N(\\mu=70, \\sigma = 10)[\/latex].\r\n<ol type=\"a\">\r\n \t<li>If I randomly pick four students, determine the distribution of their average grade. Indicate the mean, standard deviation, and shape.\r\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 70[\/latex].\r\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{10}{\\sqrt{4}} = 5[\/latex].\r\nShape: normal since the population is normal. Recall that when the population distribution is normal, the distribution of the sample mean [latex]\\bar{X}[\/latex] is also normal regardless of the sample size.\r\nTherefore, the average grade of four randomly selected students [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=70, \\sigma_{\\scriptsize \\bar{X}} = 5)[\/latex].<\/li>\r\n \t<li>If I randomly pick 100 students, determine the distribution of their average grade. Indicate the mean, standard deviation, and shape.\r\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 70[\/latex].\r\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{10}{\\sqrt{100}} = 1[\/latex].\r\nShape: normal since the population is normal. Recall that when the population distribution is normal, the distribution of the sample mean [latex]\\bar{X}[\/latex] is also normal regardless of the sample size.\r\nTherefore, the average grade of 100 randomly selected students [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=70, \\sigma_{\\scriptsize \\bar{X}} = 1)[\/latex].<\/li>\r\n \t<li>If I randomly pick four students, find the probability that their average is between 60 and 90.\r\nBy part (a), for [latex]n = 4[\/latex], average grade [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}} = 70, \\sigma_{\\scriptsize \\bar{X}} = 5)[\/latex] . Therefore,\r\n[latex]\\begin{align*}P(60 \\leq \\bar{X} \\leq 90)&amp;= P \\left( \\frac{60 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\leq \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\leq \\frac{90 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right)\\\\ &amp;= P \\left( \\frac{60 - 70}{5} \\leq Z \\leq \\frac{90 - 70}{5} \\right)\\\\\r\n&amp;=P(-2 \\leq Z \\leq 4)\\\\\r\n&amp;= P( Z \\leq 4) - P(Z \\leq -2) \\\\\r\n&amp;= 1 - 0.0228 = 0.9772.\r\n\\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Suppose the lifetime of a brand of laptops follows an extremely right-skewed distribution with a mean [latex]\\mu = 5[\/latex] years and a standard deviation [latex]\\sigma = 5[\/latex].\r\n<ol type=\"a\">\r\n \t<li>If I randomly pick four laptops, determine the distribution of their average lifetime. Indicate the mean, standard deviation, and shape.\r\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 5[\/latex] years.\r\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{5}{\\sqrt{4}} = 2.5[\/latex] years.\r\nShape: Not normal, still right-skewed. The population is extremely right-skewed, and the sample size [latex]n=4[\/latex] is too small to apply the central limit theorem.<\/li>\r\n \t<li>If I randomly pick 100 laptops, determine the distribution of their average lifetime. Indicate the mean, standard deviation, and shape.\r\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 5[\/latex] years.\r\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{5}{\\sqrt{100}} = 0.5[\/latex] years.\r\nShape: approximately normal. The population is extremely right-skewed, but the sample size [latex]n = 100 \\: &gt; \\: 30[\/latex] is large enough to apply the central limit theorem. Therefore, [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=5, \\sigma_{\\scriptsize \\bar{X}} = 0.5)[\/latex].<\/li>\r\n \t<li>If I randomly pick 100 laptops, find the probability that their average lifetime is at least four years.\r\nBy part (b), for [latex]n=100[\/latex], the average lifetime [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=5, \\sigma_{\\scriptsize \\bar{X}} = 0.5)[\/latex]. Hence,\r\n[latex]\\begin{align*}P(\\bar{X} \\geq 4)&amp;= P \\left( \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\geq \\frac{4 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right)\\\\\r\n&amp;= P \\left( Z \\geq \\frac{4 - 5}{0.5} \\right)= P(Z \\geq -2) \\\\\r\n&amp;=P(Z \\leq 2) = 0.9722. \\end{align*}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Distribution of the Sample Mean<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nLet [latex]X=[\/latex] the rent of a one-bedroom apartment in Edmonton, and suppose that [latex]X[\/latex] follows a distribution with a mean of $700 and a standard deviation of $400. The distribution of [latex]X[\/latex] (the population distribution) is given by the density curve below.<a id=\"retex6.1\"><\/a>\r\n\r\n[caption id=\"attachment_718\" align=\"aligncenter\" width=\"458\"]<img class=\"wp-image-718 size-full\" src=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve.png\" alt=\"A density curve of rent. The curve is right skewed. Image description available.\" width=\"458\" height=\"465\" \/> <strong>Exercise 6.1<\/strong> [<a href=\"https:\/\/openbooks.macewan.ca\/introstats\/back-matter\/image-description\/#ex6.1\">Image Description (See Appendix D Exercise 6.1)<\/a>][\/caption]\r\n<ol type=\"a\">\r\n \t<li>Describe the population distribution of the rent of a one-bedroom apartment in Edmonton, i.e., the distribution of [latex]X[\/latex]. Comment on modality, center, spread, and shape.<\/li>\r\n \t<li>If you randomly pick four one-bedroom apartments, describe the sampling distribution of their average rent. Indicate the mean, standard deviation, and shape.<\/li>\r\n \t<li>If you randomly pick 100 one-bedroom apartments, describe the sampling distribution of their average rent. Indicate the mean, standard deviation, and shape.<\/li>\r\n \t<li>If you randomly pick 100 one-bedroom apartments, find the probability that their average rent is above $800.<\/li>\r\n<\/ol>\r\n<details><summary>Show\/Hide Answer<\/summary>\r\n<ol type=\"a\">\r\n \t<li>Unimodal, right skewed, centered at the mean 700 with a spread of 400 as the standard deviation.<\/li>\r\n \t<li>Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = $700 [\/latex].\r\nStandard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{400}{\\sqrt{4}} = $200 [\/latex].\r\nShape: Not normal, still right-skewed. The population is right-skewed, and the sample size [latex]n=4[\/latex] is too small to apply the central limit theorem.<\/li>\r\n<\/ol>\r\n<ol start=\"3\" type=\"a\">\r\n \t<li>Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = $700 [\/latex].\r\nStandard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{400}{\\sqrt{100}} = $40 [\/latex].\r\nShape: approximately normal. The population is right-skewed, but the sample size [latex]n = 100 \\: &gt; \\: 30[\/latex], so it is large enough to apply the central limit theorem.<\/li>\r\n<\/ol>\r\n<ol start=\"4\" type=\"a\">\r\n \t<li>By part (c), [latex]n = 100[\/latex] for, the average rent [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=700, \\sigma_{\\scriptsize \\bar{X}}=40)[\/latex]. Hence,<\/li>\r\n<\/ol>\r\n<p style=\"padding-left: 40px;\">[latex]\\begin{align*} P(\\bar{X} \\geq 800) &amp;= P \\left( \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\geq \\frac{800 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right) \\\\ &amp;= P \\left( Z \\geq \\frac{800-700}{40} \\right) \\\\ &amp;= P(Z \\geq 2.5) \\\\ &amp;= P(Z \\leq -2.5) \\\\ &amp;=0.0062. \\end{align*}[\/latex]<\/p>\r\n\r\n<\/details><\/div>\r\n<\/div>","rendered":"<p>The central limit theorem is one of the most important theorems in statistics.<\/p>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Fact: The Central Limit Theorem<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>When a random sample of size <em>n<\/em> is drawn from any population with mean [latex]\\mu[\/latex] and standard deviation [latex]\\sigma[\/latex] , the distribution of the sample mean [latex]\\bar{X}[\/latex] will be (approximately) normally distributed if the sample size <em>n<\/em> is large enough. In general, [latex]n \\geq 30[\/latex]\u00a0is large enough if the population distribution is not too extremely skewed.<\/p>\n<\/div>\n<\/div>\n<p>Note that:<\/p>\n<ul>\n<li>The central limit theorem is about the <strong>shape of the distribution of the sample mean <\/strong>[latex]\\bar{X}[\/latex]. It is the distribution of the random variable [latex]\\bar{X}[\/latex]\u00a0that will be normally distributed if the sample size <em>n<\/em> is large enough.<\/li>\n<li>The required sample size <em>n<\/em> depends on how skewed the population distribution is. If the population distribution, the distribution of [latex]X[\/latex], is symmetric, [latex]n \\geq 5[\/latex]\u00a0might be large enough to claim that the sample mean [latex]\\bar{X}[\/latex]\u00a0is approximately normally distributed; if the distribution of [latex]X[\/latex] is not too extremely skewed, [latex]n \\geq 30[\/latex]\u00a0should be enough; if the population is very skewed, we might need \u00a0[latex]n \\geq 100[\/latex]\u00a0(see the central limit theorem for proportion in Chapter 10).<\/li>\n<\/ul>\n<p>In addition to the results on the mean and standard deviation of [latex]\\bar{X}[\/latex], we can claim that:<\/p>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Fact: The Distribution of the Sample Mean [latex]\\color{white} \\bar{X}[\/latex]<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>For a normal population or large sample, the sample mean [latex]\\bar{X}[\/latex] follows a normal distribution with mean [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu[\/latex] and standard deviation [latex]\\sigma_{\\scriptsize \\bar{X}} = \u00a0\\frac{\\sigma}{\\sqrt{n}}[\/latex]. That is [latex]\\bar{X} \\sim N(\\mu, \\frac{\\sigma}{\\sqrt{n}})[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Distribution of the sample mean [latex]\\color{white} \\bar{X}[\/latex]<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<ol start=\"1\" type=\"1\">\n<li>Let [latex]X[\/latex] denote student grades in a particular class, and suppose [latex]X[\/latex] is normally distributed with a mean of 70 and a standard deviation of 10, i.e., [latex]X \\sim N(\\mu=70, \\sigma = 10)[\/latex].\n<ol type=\"a\">\n<li>If I randomly pick four students, determine the distribution of their average grade. Indicate the mean, standard deviation, and shape.<br \/>\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 70[\/latex].<br \/>\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{10}{\\sqrt{4}} = 5[\/latex].<br \/>\nShape: normal since the population is normal. Recall that when the population distribution is normal, the distribution of the sample mean [latex]\\bar{X}[\/latex] is also normal regardless of the sample size.<br \/>\nTherefore, the average grade of four randomly selected students [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=70, \\sigma_{\\scriptsize \\bar{X}} = 5)[\/latex].<\/li>\n<li>If I randomly pick 100 students, determine the distribution of their average grade. Indicate the mean, standard deviation, and shape.<br \/>\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 70[\/latex].<br \/>\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{10}{\\sqrt{100}} = 1[\/latex].<br \/>\nShape: normal since the population is normal. Recall that when the population distribution is normal, the distribution of the sample mean [latex]\\bar{X}[\/latex] is also normal regardless of the sample size.<br \/>\nTherefore, the average grade of 100 randomly selected students [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=70, \\sigma_{\\scriptsize \\bar{X}} = 1)[\/latex].<\/li>\n<li>If I randomly pick four students, find the probability that their average is between 60 and 90.<br \/>\nBy part (a), for [latex]n = 4[\/latex], average grade [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}} = 70, \\sigma_{\\scriptsize \\bar{X}} = 5)[\/latex] . Therefore,<br \/>\n[latex]\\begin{align*}P(60 \\leq \\bar{X} \\leq 90)&= P \\left( \\frac{60 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\leq \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\leq \\frac{90 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right)\\\\ &= P \\left( \\frac{60 - 70}{5} \\leq Z \\leq \\frac{90 - 70}{5} \\right)\\\\  &=P(-2 \\leq Z \\leq 4)\\\\  &= P( Z \\leq 4) - P(Z \\leq -2) \\\\  &= 1 - 0.0228 = 0.9772.  \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<li>Suppose the lifetime of a brand of laptops follows an extremely right-skewed distribution with a mean [latex]\\mu = 5[\/latex] years and a standard deviation [latex]\\sigma = 5[\/latex].\n<ol type=\"a\">\n<li>If I randomly pick four laptops, determine the distribution of their average lifetime. Indicate the mean, standard deviation, and shape.<br \/>\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 5[\/latex] years.<br \/>\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{5}{\\sqrt{4}} = 2.5[\/latex] years.<br \/>\nShape: Not normal, still right-skewed. The population is extremely right-skewed, and the sample size [latex]n=4[\/latex] is too small to apply the central limit theorem.<\/li>\n<li>If I randomly pick 100 laptops, determine the distribution of their average lifetime. Indicate the mean, standard deviation, and shape.<br \/>\n<span style=\"text-align: initial; font-size: 1em;\">Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = 5[\/latex] years.<br \/>\n<\/span>Standard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{5}{\\sqrt{100}} = 0.5[\/latex] years.<br \/>\nShape: approximately normal. The population is extremely right-skewed, but the sample size [latex]n = 100 \\: > \\: 30[\/latex] is large enough to apply the central limit theorem. Therefore, [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=5, \\sigma_{\\scriptsize \\bar{X}} = 0.5)[\/latex].<\/li>\n<li>If I randomly pick 100 laptops, find the probability that their average lifetime is at least four years.<br \/>\nBy part (b), for [latex]n=100[\/latex], the average lifetime [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=5, \\sigma_{\\scriptsize \\bar{X}} = 0.5)[\/latex]. Hence,<br \/>\n[latex]\\begin{align*}P(\\bar{X} \\geq 4)&= P \\left( \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\geq \\frac{4 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right)\\\\  &= P \\left( Z \\geq \\frac{4 - 5}{0.5} \\right)= P(Z \\geq -2) \\\\  &=P(Z \\leq 2) = 0.9722. \\end{align*}[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Distribution of the Sample Mean<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Let [latex]X=[\/latex] the rent of a one-bedroom apartment in Edmonton, and suppose that [latex]X[\/latex] follows a distribution with a mean of $700 and a standard deviation of $400. The distribution of [latex]X[\/latex] (the population distribution) is given by the density curve below.<a id=\"retex6.1\"><\/a><\/p>\n<figure id=\"attachment_718\" aria-describedby=\"caption-attachment-718\" style=\"width: 458px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-718 size-full\" src=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve.png\" alt=\"A density curve of rent. The curve is right skewed. Image description available.\" width=\"458\" height=\"465\" srcset=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve.png 458w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve-295x300.png 295w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve-65x66.png 65w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve-225x228.png 225w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/07\/m06_Rent_DensityCurve-350x355.png 350w\" sizes=\"auto, (max-width: 458px) 100vw, 458px\" \/><figcaption id=\"caption-attachment-718\" class=\"wp-caption-text\"><strong>Exercise 6.1<\/strong> [<a href=\"https:\/\/openbooks.macewan.ca\/introstats\/back-matter\/image-description\/#ex6.1\">Image Description (See Appendix D Exercise 6.1)<\/a>]<\/figcaption><\/figure>\n<ol type=\"a\">\n<li>Describe the population distribution of the rent of a one-bedroom apartment in Edmonton, i.e., the distribution of [latex]X[\/latex]. Comment on modality, center, spread, and shape.<\/li>\n<li>If you randomly pick four one-bedroom apartments, describe the sampling distribution of their average rent. Indicate the mean, standard deviation, and shape.<\/li>\n<li>If you randomly pick 100 one-bedroom apartments, describe the sampling distribution of their average rent. Indicate the mean, standard deviation, and shape.<\/li>\n<li>If you randomly pick 100 one-bedroom apartments, find the probability that their average rent is above $800.<\/li>\n<\/ol>\n<details>\n<summary>Show\/Hide Answer<\/summary>\n<ol type=\"a\">\n<li>Unimodal, right skewed, centered at the mean 700 with a spread of 400 as the standard deviation.<\/li>\n<li>Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = $700[\/latex].<br \/>\nStandard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{400}{\\sqrt{4}} = $200[\/latex].<br \/>\nShape: Not normal, still right-skewed. The population is right-skewed, and the sample size [latex]n=4[\/latex] is too small to apply the central limit theorem.<\/li>\n<\/ol>\n<ol start=\"3\" type=\"a\">\n<li>Mean: [latex]\\mu_{\\scriptsize \\bar{X}} = \\mu = $700[\/latex].<br \/>\nStandard deviation: [latex]\\sigma_{\\scriptsize \\bar{X}} = \\frac{\\sigma}{\\sqrt{n}} = \\frac{400}{\\sqrt{100}} = $40[\/latex].<br \/>\nShape: approximately normal. The population is right-skewed, but the sample size [latex]n = 100 \\: > \\: 30[\/latex], so it is large enough to apply the central limit theorem.<\/li>\n<\/ol>\n<ol start=\"4\" type=\"a\">\n<li>By part (c), [latex]n = 100[\/latex] for, the average rent [latex]\\bar{X} \\sim N(\\mu_{\\scriptsize \\bar{X}}=700, \\sigma_{\\scriptsize \\bar{X}}=40)[\/latex]. Hence,<\/li>\n<\/ol>\n<p style=\"padding-left: 40px;\">[latex]\\begin{align*} P(\\bar{X} \\geq 800) &= P \\left( \\frac{\\bar{X} - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\geq \\frac{800 - \\mu_{\\scriptsize \\bar{X}}}{\\sigma_{\\scriptsize \\bar{X}}} \\right) \\\\ &= P \\left( Z \\geq \\frac{800-700}{40} \\right) \\\\ &= P(Z \\geq 2.5) \\\\ &= P(Z \\leq -2.5) \\\\ &=0.0062. \\end{align*}[\/latex]<\/p>\n<\/details>\n<\/div>\n<\/div>\n","protected":false},"author":19,"menu_order":3,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-763","chapter","type-chapter","status-publish","hentry"],"part":671,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/763","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":35,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/763\/revisions"}],"predecessor-version":[{"id":5293,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/763\/revisions\/5293"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/763\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=763"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=763"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=763"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=763"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}