{"id":197941,"date":"2025-03-07T19:48:55","date_gmt":"2025-03-07T19:48:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197941"},"modified":"2025-03-07T19:48:58","modified_gmt":"2025-03-07T19:48:58","slug":"a-population-of-values-has-a-normal-distribution-with-%c2%b5-15-7-and-s-1-5-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/a-population-of-values-has-a-normal-distribution-with-%c2%b5-15-7-and-s-1-5-3\/","title":{"rendered":"A population of values has a normal distribution with \u00b5 = 15.7 and s = 1.5"},"content":{"rendered":"\n<p>A population of values has a normal distribution with \u00b5 = 15.7 and s = 1.5. You intend to draw a random sample of size n = 18. First calculate z, round it to two (2) decimal places, then use the rounded z-score to determine the required probability accurate to four (4) decimal places. Find the probability that a single randomly selected value is less than 16.7. P(x &lt; 16.7) = ?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>We are asked to find the probability that a single randomly selected value from a population that follows a normal distribution with a mean ((\\mu)) of 15.7 and a standard deviation ((\\sigma)) of 1.5 is less than 16.7. This is essentially asking us to find the probability (P(x &lt; 16.7)), where (x = 16.7).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the Z-score<\/h3>\n\n\n\n<p>To standardize the value of 16.7 in terms of the Z-score, we can use the formula:<\/p>\n\n\n\n<p>[<br>z = \\frac{x &#8211; \\mu}{\\sigma}<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(x = 16.7) is the value we are interested in,<\/li>\n\n\n\n<li>(\\mu = 15.7) is the population mean, and<\/li>\n\n\n\n<li>(\\sigma = 1.5) is the population standard deviation.<\/li>\n<\/ul>\n\n\n\n<p>Plugging in the values:<\/p>\n\n\n\n<p>[<br>z = \\frac{16.7 &#8211; 15.7}{1.5} = \\frac{1.0}{1.5} = 0.6667<br>]<\/p>\n\n\n\n<p>Thus, the Z-score is approximately <strong>0.67<\/strong> when rounded to two decimal places.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find the Probability<\/h3>\n\n\n\n<p>Once we have the Z-score, we can find the probability that a randomly selected value from this normal distribution is less than 16.7 by using the standard normal distribution table (Z-table) or a calculator.<\/p>\n\n\n\n<p>The Z-table provides the cumulative probability for a given Z-score. For (z = 0.67), the cumulative probability (i.e., the area under the standard normal curve to the left of (z = 0.67)) is approximately <strong>0.7486<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>Thus, the probability that a randomly selected value is less than 16.7 is:<\/p>\n\n\n\n<p>[<br>P(x &lt; 16.7) \\approx 0.7486<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>In a normal distribution, the Z-score represents how many standard deviations a value is away from the mean. By transforming the original value into a Z-score, we can use the standard normal distribution, which has a mean of 0 and a standard deviation of 1, to calculate the probability. The Z-score of 0.67 tells us that 16.7 is 0.67 standard deviations above the mean of 15.7. The Z-table then provides the cumulative probability to the left of this Z-score, which corresponds to the probability of selecting a value less than 16.7 from the population.<\/p>\n\n\n\n<p>In summary, the probability is approximately <strong>0.7486<\/strong> or <strong>74.86%<\/strong>. This means there is a 74.86% chance that a randomly selected value from this population will be less than 16.7.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-585.png\" alt=\"\" class=\"wp-image-197942\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A population of values has a normal distribution with \u00b5 = 15.7 and s = 1.5. You intend to draw a random sample of size n = 18. First calculate z, round it to two (2) decimal places, then use the rounded z-score to determine the required probability accurate to four (4) decimal places. Find [&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-197941","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197941","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=197941"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197941\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197941"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197941"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197941"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}