{"id":217642,"date":"2025-05-22T09:09:41","date_gmt":"2025-05-22T09:09:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217642"},"modified":"2025-05-22T20:32:34","modified_gmt":"2025-05-22T20:32:34","slug":"what-is-the-probability-a-value-selected-at-random-from-this-distribution-is-greater-than-31","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/what-is-the-probability-a-value-selected-at-random-from-this-distribution-is-greater-than-31\/","title":{"rendered":". What is the probability a value selected at random from this distribution is greater than 31"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">A. Consider a normal distribution with mean 31 and standard deviation 6. What is the probability a value selected at random from this distribution is greater than 31? (Round your answer to two decimal places.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">B. Suppose 85% of the area under the standard normal curve lies to the right of z. Is z positive or negative?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">C. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u00b5 = 4; s = 2<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P(3 = x = 7) =<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">D. Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability. (Round your answer to four decimal places.)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u00b5 = 40; s = 14<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">P(50 = x = 70) =<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let&#8217;s go through each part of the question one by one, providing correct answers and detailed explanations.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>A. Probability that a value is greater than the mean in a normal distribution<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Given<\/strong>: Mean (\u03bc) = 31, Standard deviation (\u03c3) = 6<\/li>\n\n\n\n<li>We want to find: $$<br>P(X > 31)<br>$$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Since the distribution is <strong>normal<\/strong> and <strong>symmetric<\/strong>, the probability that a value is greater than the mean is always <strong>0.50<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\u2705 Answer: 0.50<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Explanation:<\/strong><br>In a normal distribution, the mean (\u03bc) divides the distribution into two equal halves. Because the distribution is symmetric about the mean, 50% of the data lies to the left of the mean and 50% lies to the right. Therefore, the probability that a randomly selected value from the distribution is <strong>greater than 31<\/strong> (i.e., greater than the mean) is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P(X &gt; \u03bc) = 0.50<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>B. If 85% of the area under the standard normal curve lies to the right of z, is z positive or negative?<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\u2705 Answer: Negative<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>Explanation:<\/strong><br>In the <strong>standard normal distribution (mean = 0)<\/strong>, the right-hand tail gets smaller as z increases. If <strong>85% of the area lies to the right<\/strong>, that means only 15% lies to the left. A high percentage of area to the right implies a <strong>z-value located to the left of the mean<\/strong>, because most of the distribution is <strong>after<\/strong> it. This can only occur when the z-score is <strong>negative<\/strong>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Mathematically, if $P(Z &gt; z) = 0.85$, then $z \\approx -1.036$. So z is negative.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>C. Find P(3 \u2264 x \u2264 7) for \u00b5 = 4, \u03c3 = 2<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We use the standardization formula:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>z = \\frac{x &#8211; \\mu}{\\sigma}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Convert x-values to z-scores:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For x = 3: $z = \\frac{3 &#8211; 4}{2} = -0.5$<\/li>\n\n\n\n<li>For x = 7: $z = \\frac{7 &#8211; 4}{2} = 1.5$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now, find area between z = -0.5 and z = 1.5 using a standard normal table or calculator:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$P(Z &lt; 1.5) = 0.9332$<\/li>\n\n\n\n<li>$P(Z &lt; -0.5) = 0.3085$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P(3 \\le x \\le 7) = 0.9332 &#8211; 0.3085 = 0.6247<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\u2705 Answer: 0.6247<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>D. Find P(50 \u2264 x \u2264 70) for \u00b5 = 40, \u03c3 = 14<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Convert to z-scores:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For x = 50: $z = \\frac{50 &#8211; 40}{14} = \\frac{10}{14} \\approx 0.7143$<\/li>\n\n\n\n<li>For x = 70: $z = \\frac{70 &#8211; 40}{14} = \\frac{30}{14} \\approx 2.1429$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Use a Z-table or calculator:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$P(Z &lt; 2.14) \\approx 0.9834$<\/li>\n\n\n\n<li>$P(Z &lt; 0.71) \\approx 0.7611$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>P(50 \\le x \\le 70) = 0.9834 &#8211; 0.7611 = 0.2223<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong>\u2705 Answer: 0.2223<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>A<\/strong>: 0.50<\/li>\n\n\n\n<li><strong>B<\/strong>: Negative<\/li>\n\n\n\n<li><strong>C<\/strong>: 0.6247<\/li>\n\n\n\n<li><strong>D<\/strong>: 0.2223<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">These calculations are based on the <strong>properties of the normal distribution<\/strong>, which is defined by its <strong>mean and standard deviation<\/strong>. Converting raw scores (x-values) into <strong>z-scores<\/strong> allows us to use standard tables or statistical software to compute exact probabilities<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-28.png\" alt=\"\" class=\"wp-image-217643\"\/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A. Consider a normal distribution with mean 31 and standard deviation 6. What is the probability a value selected at random from this distribution is greater than 31? (Round your answer to two decimal places.) B. Suppose 85% of the area under the standard normal curve lies to the right of z. Is z positive [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-217642","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217642","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=217642"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217642\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217642"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217642"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217642"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}