{"id":1068,"date":"2023-06-22T01:45:32","date_gmt":"2023-06-22T01:45:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/connection-between-binomial-and-normal-distributions-apply-it-2\/"},"modified":"2023-10-23T13:08:45","modified_gmt":"2023-10-23T13:08:45","slug":"connection-between-binomial-and-normal-distributions-apply-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/introstatstest\/chapter\/connection-between-binomial-and-normal-distributions-apply-it-2\/","title":{"raw":"Connection Between Binomial and Normal Distributions: Apply It 2","rendered":"Connection Between Binomial and Normal Distributions: Apply It 2"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Use normal probability distribution to calculate binomial probabilities&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Use normal probability distribution to calculate binomial probabilities<\/span><\/li>\r\n\t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Check the conditions for applying a normal distribution to approximate a binomial distribution&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Check the conditions for applying a normal distribution to approximate a binomial distribution<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h3>Continuity Correction<\/h3>\r\n<p>In addition to verifying that both the number of successes and the number of failures exceed [latex]10[\/latex], we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>continuity correction<\/h3>\r\n<p>In probability theory, a <strong>continuity correction<\/strong> is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex] X [\/latex], is represented in the normal distribution by the interval from [latex] X - 0.5 \\mbox{ to } X + 0.5[\/latex] or [latex]X \u00b10.5[\/latex].<\/p>\r\n<\/section>\r\n<p>Why is this necessary? Consider the following example.<\/p>\r\n<section class=\"textbox example\">The probability that George successfully makes exactly [latex]270[\/latex] free throws is [latex] P(X= 270) = 0.0766 [\/latex]. So, there is a [latex]7.66\\%[\/latex] chance that he will make exactly [latex]270[\/latex] free throws out of [latex]300[\/latex]. The [latex] P(X= 270.2) [\/latex] free throws is [latex]0[\/latex] because he cannot make [latex]0.2[\/latex] of a free throw. The number of free throws made is a discrete random variable.Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex] X [\/latex] is a continuous random variable. The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:\r\n\r\n<p style=\"text-align: center;\">[latex]-P(X &lt; a)[\/latex], [latex]P(X&gt;a)[\/latex] or [latex]P(a&lt; X&lt; b)[\/latex]<\/p>\r\n<p>For a continuous random variable, [latex] P(X = a) = 0 [\/latex] because there are an infinite number of possible numbers on any interval. Also, [latex] P( X \\leq a) = P(X&lt; a) [\/latex] for a continuous random variable.<\/p>\r\n<p>Thus, we will use [latex] P(269.5 &lt; X &lt; 270.5) [\/latex] to approximate [latex] P(X = 270)[\/latex].<\/p>\r\n<\/section>\r\n<p>Thus, provided that certain conditions ([latex]np \\ge 10[\/latex] and [latex]n(1-p) \\ge 10[\/latex]) are met, we can then use the continuity of correction to make the adjustment to the discrete random variable and use the normal distribution to approximate the binomial probabilities.<\/p>\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]1542[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Use normal probability distribution to calculate binomial probabilities&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Use normal probability distribution to calculate binomial probabilities<\/span><\/li>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Check the conditions for applying a normal distribution to approximate a binomial distribution&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Calibri&quot;}\">Check the conditions for applying a normal distribution to approximate a binomial distribution<\/span><\/li>\n<\/ul>\n<\/section>\n<h3>Continuity Correction<\/h3>\n<p>In addition to verifying that both the number of successes and the number of failures exceed [latex]10[\/latex], we need to adjust the discrete whole numbers used in a binomial distribution.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>continuity correction<\/h3>\n<p>In probability theory, a <strong>continuity correction<\/strong> is an adjustment that is made when a discrete distribution is approximated by a continuous distribution. We do this by adjusting the discrete whole numbers used in a binomial distribution so that any individual value, [latex]X[\/latex], is represented in the normal distribution by the interval from [latex]X - 0.5 \\mbox{ to } X + 0.5[\/latex] or [latex]X \u00b10.5[\/latex].<\/p>\n<\/section>\n<p>Why is this necessary? Consider the following example.<\/p>\n<section class=\"textbox example\">The probability that George successfully makes exactly [latex]270[\/latex] free throws is [latex]P(X= 270) = 0.0766[\/latex]. So, there is a [latex]7.66\\%[\/latex] chance that he will make exactly [latex]270[\/latex] free throws out of [latex]300[\/latex]. The [latex]P(X= 270.2)[\/latex] free throws is [latex]0[\/latex] because he cannot make [latex]0.2[\/latex] of a free throw. The number of free throws made is a discrete random variable.Now suppose we assume that the number of free throws made has a normal distribution. In other words, suppose that the random variable [latex]X[\/latex] is a continuous random variable. The probabilities for a continuous random variable are defined as the area under the curve, and we talk about the area under the curve over intervals using probability notation:<\/p>\n<p style=\"text-align: center;\">[latex]-P(X < a)[\/latex], [latex]P(X>a)[\/latex] or [latex]P(a< X< b)[\/latex]<\/p>\n<p>For a continuous random variable, [latex]P(X = a) = 0[\/latex] because there are an infinite number of possible numbers on any interval. Also, [latex]P( X \\leq a) = P(X< a)[\/latex] for a continuous random variable.<\/p>\n<p>Thus, we will use [latex]P(269.5 < X < 270.5)[\/latex] to approximate [latex]P(X = 270)[\/latex].<\/p>\n<\/section>\n<p>Thus, provided that certain conditions ([latex]np \\ge 10[\/latex] and [latex]n(1-p) \\ge 10[\/latex]) are met, we can then use the continuity of correction to make the adjustment to the discrete random variable and use the normal distribution to approximate the binomial probabilities.<\/p>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm1542\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=1542&theme=lumen&iframe_resize_id=ohm1542&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/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":2912,"module-header":"apply_it","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\/1068"}],"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":4,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1068\/revisions"}],"predecessor-version":[{"id":4092,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1068\/revisions\/4092"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/parts\/2912"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapters\/1068\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/media?parent=1068"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/pressbooks\/v2\/chapter-type?post=1068"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/contributor?post=1068"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/introstatstest\/wp-json\/wp\/v2\/license?post=1068"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}