{"id":43200,"date":"2025-06-29T14:55:44","date_gmt":"2025-06-29T14:55:44","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43200"},"modified":"2025-06-29T14:55:45","modified_gmt":"2025-06-29T14:55:45","slug":"a-population-is-normally-distributed-with-a-mean-of-43-2-and-a-standard-deviation-of-5-7","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-population-is-normally-distributed-with-a-mean-of-43-2-and-a-standard-deviation-of-5-7\/","title":{"rendered":"A population is normally distributed with a mean of 43.2 and a standard deviation of 5.7."},"content":{"rendered":"\n<p>A population is normally distributed with a mean of 43.2 and a standard deviation of 5.7. Find the following probabilities. (Round your answers to four decimal places.) (a) P(43.2 &lt; x &lt; 48.9) (b) P(41.4 &lt; x &lt; 44.6) (c) P(x &lt; 50.0) (d) P(33.8 &lt; x &lt; 52.6) (e) P(x = 45.8) (f) P(x > 45.8)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve these problems, we need to use the Z-score formula and the standard normal distribution table. The Z-score formula is:Z=X\u2212\u03bc\u03c3Z = \\frac{X &#8211; \\mu}{\\sigma}Z=\u03c3X\u2212\u03bc\u200b<\/p>\n\n\n\n<p>where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>XXX is the value of interest,<\/li>\n\n\n\n<li>\u03bc\\mu\u03bc is the mean (43.2),<\/li>\n\n\n\n<li>\u03c3\\sigma\u03c3 is the standard deviation (5.7).<\/li>\n<\/ul>\n\n\n\n<p>We will then use the Z-scores to find the probabilities using the cumulative distribution function (CDF) of the standard normal distribution.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) P(43.2 &lt; x &lt; 48.9)<\/h3>\n\n\n\n<p>First, calculate the Z-scores for 43.2 and 48.9:<\/p>\n\n\n\n<p>For X=43.2X = 43.2X=43.2:Z1=43.2\u221243.25.7=0Z_1 = \\frac{43.2 &#8211; 43.2}{5.7} = 0Z1\u200b=5.743.2\u221243.2\u200b=0<\/p>\n\n\n\n<p>For X=48.9X = 48.9X=48.9:Z2=48.9\u221243.25.7=1Z_2 = \\frac{48.9 &#8211; 43.2}{5.7} = 1Z2\u200b=5.748.9\u221243.2\u200b=1<\/p>\n\n\n\n<p>Now, find the corresponding cumulative probabilities:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P(Z=0)=0.5000P(Z = 0) = 0.5000P(Z=0)=0.5000 (from standard normal table)<\/li>\n\n\n\n<li>P(Z=1)=0.8413P(Z = 1) = 0.8413P(Z=1)=0.8413<\/li>\n<\/ul>\n\n\n\n<p>Thus:P(43.2&lt;x&lt;48.9)=P(Z2)\u2212P(Z1)=0.8413\u22120.5000=0.3413P(43.2 &lt; x &lt; 48.9) = P(Z_2) &#8211; P(Z_1) = 0.8413 &#8211; 0.5000 = 0.3413P(43.2&lt;x&lt;48.9)=P(Z2\u200b)\u2212P(Z1\u200b)=0.8413\u22120.5000=0.3413<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(b) P(41.4 &lt; x &lt; 44.6)<\/h3>\n\n\n\n<p>For X=41.4X = 41.4X=41.4:Z1=41.4\u221243.25.7=\u22120.3158Z_1 = \\frac{41.4 &#8211; 43.2}{5.7} = -0.3158Z1\u200b=5.741.4\u221243.2\u200b=\u22120.3158<\/p>\n\n\n\n<p>For X=44.6X = 44.6X=44.6:Z2=44.6\u221243.25.7=0.2456Z_2 = \\frac{44.6 &#8211; 43.2}{5.7} = 0.2456Z2\u200b=5.744.6\u221243.2\u200b=0.2456<\/p>\n\n\n\n<p>From the standard normal table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P(Z=\u22120.3158)\u22480.3762P(Z = -0.3158) \\approx 0.3762P(Z=\u22120.3158)\u22480.3762<\/li>\n\n\n\n<li>P(Z=0.2456)\u22480.5968P(Z = 0.2456) \\approx 0.5968P(Z=0.2456)\u22480.5968<\/li>\n<\/ul>\n\n\n\n<p>Thus:P(41.4&lt;x&lt;44.6)=P(Z2)\u2212P(Z1)=0.5968\u22120.3762=0.2206P(41.4 &lt; x &lt; 44.6) = P(Z_2) &#8211; P(Z_1) = 0.5968 &#8211; 0.3762 = 0.2206P(41.4&lt;x&lt;44.6)=P(Z2\u200b)\u2212P(Z1\u200b)=0.5968\u22120.3762=0.2206<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(c) P(x &lt; 50.0)<\/h3>\n\n\n\n<p>For X=50.0X = 50.0X=50.0:Z=50.0\u221243.25.7=1.1263Z = \\frac{50.0 &#8211; 43.2}{5.7} = 1.1263Z=5.750.0\u221243.2\u200b=1.1263<\/p>\n\n\n\n<p>From the standard normal table:P(Z=1.1263)\u22480.8708P(Z = 1.1263) \\approx 0.8708P(Z=1.1263)\u22480.8708<\/p>\n\n\n\n<p>Thus:P(x&lt;50.0)=0.8708P(x &lt; 50.0) = 0.8708P(x&lt;50.0)=0.8708<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(d) P(33.8 &lt; x &lt; 52.6)<\/h3>\n\n\n\n<p>For X=33.8X = 33.8X=33.8:Z1=33.8\u221243.25.7=\u22121.6526Z_1 = \\frac{33.8 &#8211; 43.2}{5.7} = -1.6526Z1\u200b=5.733.8\u221243.2\u200b=\u22121.6526<\/p>\n\n\n\n<p>For X=52.6X = 52.6X=52.6:Z2=52.6\u221243.25.7=1.6579Z_2 = \\frac{52.6 &#8211; 43.2}{5.7} = 1.6579Z2\u200b=5.752.6\u221243.2\u200b=1.6579<\/p>\n\n\n\n<p>From the standard normal table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P(Z=\u22121.6526)\u22480.0495P(Z = -1.6526) \\approx 0.0495P(Z=\u22121.6526)\u22480.0495<\/li>\n\n\n\n<li>P(Z=1.6579)\u22480.9521P(Z = 1.6579) \\approx 0.9521P(Z=1.6579)\u22480.9521<\/li>\n<\/ul>\n\n\n\n<p>Thus:P(33.8&lt;x&lt;52.6)=P(Z2)\u2212P(Z1)=0.9521\u22120.0495=0.9026P(33.8 &lt; x &lt; 52.6) = P(Z_2) &#8211; P(Z_1) = 0.9521 &#8211; 0.0495 = 0.9026P(33.8&lt;x&lt;52.6)=P(Z2\u200b)\u2212P(Z1\u200b)=0.9521\u22120.0495=0.9026<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(e) P(x = 45.8)<\/h3>\n\n\n\n<p>Since the probability of a single point in a continuous distribution is always 0:P(x=45.8)=0P(x = 45.8) = 0P(x=45.8)=0<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(f) P(x &gt; 45.8)<\/h3>\n\n\n\n<p>For X=45.8X = 45.8X=45.8:Z=45.8\u221243.25.7=0.4561Z = \\frac{45.8 &#8211; 43.2}{5.7} = 0.4561Z=5.745.8\u221243.2\u200b=0.4561<\/p>\n\n\n\n<p>From the standard normal table:P(Z=0.4561)\u22480.6756P(Z = 0.4561) \\approx 0.6756P(Z=0.4561)\u22480.6756<\/p>\n\n\n\n<p>Thus:P(x&gt;45.8)=1\u2212P(Z=0.4561)=1\u22120.6756=0.3244P(x &gt; 45.8) = 1 &#8211; P(Z = 0.4561) = 1 &#8211; 0.6756 = 0.3244P(x&gt;45.8)=1\u2212P(Z=0.4561)=1\u22120.6756=0.3244<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Results:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) P(43.2 &lt; x &lt; 48.9) = 0.3413<\/li>\n\n\n\n<li>(b) P(41.4 &lt; x &lt; 44.6) = 0.2206<\/li>\n\n\n\n<li>(c) P(x &lt; 50.0) = 0.8708<\/li>\n\n\n\n<li>(d) P(33.8 &lt; x &lt; 52.6) = 0.9026<\/li>\n\n\n\n<li>(e) P(x = 45.8) = 0<\/li>\n\n\n\n<li>(f) P(x > 45.8) = 0.3244<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-438.jpeg\" alt=\"\" class=\"wp-image-43201\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-438.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-438-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-438-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-438-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A population is normally distributed with a mean of 43.2 and a standard deviation of 5.7. Find the following probabilities. (Round your answers to four decimal places.) (a) P(43.2 &lt; x &lt; 48.9) (b) P(41.4 &lt; x &lt; 44.6) (c) P(x &lt; 50.0) (d) P(33.8 &lt; x &lt; 52.6) (e) P(x = 45.8) (f) P(x [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-43200","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43200","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=43200"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43200\/revisions"}],"predecessor-version":[{"id":43202,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43200\/revisions\/43202"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43200"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43200"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}