{"id":1061,"date":"2023-06-22T01:45:29","date_gmt":"2023-06-22T01:45:29","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/normal-distribution-continued-dig-deeper\/"},"modified":"2024-02-13T01:35:48","modified_gmt":"2024-02-13T01:35:48","slug":"normal-distribution-continued-dig-deeper","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/normal-distribution-continued-dig-deeper\/","title":{"raw":"Normal Distribution (continued): Fresh Take","rendered":"Normal Distribution (continued): Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand the properties and characteristics of the standard normal distribution.<\/li>\r\n\t<li>Calculate probabilities and percentiles using normal distribution and [latex]z[\/latex]-scores with technology.<\/li>\r\n<\/ul>\r\n<\/section>\r\n<p>Recall that a normal distribution is a mathematical model with a smooth bell-shaped curve. Let's look at an example of how to calculate probabilities using normal distribution.<\/p>\r\n<p><iframe src=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\" width=\"100%\" height=\"850\"><\/iframe><\/p>\r\n<p><br \/>\r\n[<a href=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\" target=\"_blank\" rel=\"noopener\">Trouble viewing? Click to open in a new tab.<\/a>]<\/p>\r\n<section class=\"textbox example\">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 [latex]5.85[\/latex] cm and a standard deviation of [latex]0.24[\/latex] cm.<strong>(a)<\/strong> Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than [latex]6.0[\/latex] cm.[reveal-answer q=\"322532\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"322532\"]Go to the <strong>Find Probability<\/strong>\u00a0tab.<img class=\"aligncenter wp-image-5851 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003112\/NormalDist_Probability-1.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the probability that X &gt; 6, which is 26.6%.\" width=\"650\" height=\"350\" \/>[\/hidden-answer]<strong>(b) <\/strong>The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.[reveal-answer q=\"264658\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"264658\"] Go to the <strong>Find Percentile\/Quantile<\/strong>\u00a0tab.<img class=\"aligncenter wp-image-5852 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003457\/NormalDist_Percentile.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the middle 20% of the graph, which is P(5.789 &lt; X &lt; 5.911).\" width=\"650\" height=\"350\" \/>[\/hidden-answer]<strong>(c) <\/strong>Find the [latex]90[\/latex]<sup>th<\/sup> percentile for the diameters of mandarin oranges and interpret it in a complete sentence.[reveal-answer q=\"32411\"]Show answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"32411\"]Go to the <strong>Find Percentile\/Quantile<\/strong>\u00a0tab.<img class=\"aligncenter wp-image-5854 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003757\/NormalDist_Percentile-1.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the 90th percentile, which is 6.158.\" width=\"650\" height=\"350\" \/>\r\n<p>Interpretation: Ninety percent of the diameter of the mandarin oranges is at most [latex]6.158[\/latex] cm.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand the properties and characteristics of the standard normal distribution.<\/li>\n<li>Calculate probabilities and percentiles using normal distribution and [latex]z[\/latex]-scores with technology.<\/li>\n<\/ul>\n<\/section>\n<p>Recall that a normal distribution is a mathematical model with a smooth bell-shaped curve. Let&#8217;s look at an example of how to calculate probabilities using normal distribution.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\" width=\"100%\" height=\"850\"><\/iframe><\/p>\n<p>\n[<a href=\"https:\/\/lumen-learning.shinyapps.io\/normaldist\/\" target=\"_blank\" rel=\"noopener\">Trouble viewing? Click to open in a new tab.<\/a>]<\/p>\n<section class=\"textbox example\">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 [latex]5.85[\/latex] cm and a standard deviation of [latex]0.24[\/latex] cm.<strong>(a)<\/strong> Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than [latex]6.0[\/latex] cm.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q322532\">Show answer<\/button><\/p>\n<div id=\"q322532\" class=\"hidden-answer\" style=\"display: none\">Go to the <strong>Find Probability<\/strong>\u00a0tab.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5851 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003112\/NormalDist_Probability-1.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the probability that X &gt; 6, which is 26.6%.\" width=\"650\" height=\"350\" \/><\/div>\n<\/div>\n<p><strong>(b) <\/strong>The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q264658\">Show answer<\/button><\/p>\n<div id=\"q264658\" class=\"hidden-answer\" style=\"display: none\"> Go to the <strong>Find Percentile\/Quantile<\/strong>\u00a0tab.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5852 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003457\/NormalDist_Percentile.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the middle 20% of the graph, which is P(5.789 &lt; X &lt; 5.911).\" width=\"650\" height=\"350\" \/><\/div>\n<\/div>\n<p><strong>(c) <\/strong>Find the [latex]90[\/latex]<sup>th<\/sup> percentile for the diameters of mandarin oranges and interpret it in a complete sentence.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q32411\">Show answer<\/button><\/p>\n<div id=\"q32411\" class=\"hidden-answer\" style=\"display: none\">Go to the <strong>Find Percentile\/Quantile<\/strong>\u00a0tab.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-5854 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/10\/2022\/11\/04003757\/NormalDist_Percentile-1.png\" alt=\"Normal distribution with a mean of 5.85 and a standard deviation of 0.24. The graph marks the 90th percentile, which is 6.158.\" width=\"650\" height=\"350\" \/><\/p>\n<p>Interpretation: Ninety percent of the diameter of the mandarin oranges is at most [latex]6.158[\/latex] cm.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":8,"menu_order":25,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":3053,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1061"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/users\/8"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1061\/revisions"}],"predecessor-version":[{"id":5546,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1061\/revisions\/5546"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/3053"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1061\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1061"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1061"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1061"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1061"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}