{"id":253839,"date":"2025-07-12T07:38:16","date_gmt":"2025-07-12T07:38:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253839"},"modified":"2025-07-12T07:38:18","modified_gmt":"2025-07-12T07:38:18","slug":"the-life-expectancy-of-a-particular-brand-of-tire-is-normally-distributed-with-a-mean-of-50000-and-a-standard-deviation-of-4000-miles","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/12\/the-life-expectancy-of-a-particular-brand-of-tire-is-normally-distributed-with-a-mean-of-50000-and-a-standard-deviation-of-4000-miles\/","title":{"rendered":"The life expectancy of a particular brand of tire is normally distributed with a mean of 50,000 and a standard deviation of 4,000 miles."},"content":{"rendered":"\n<p>The life expectancy of a particular brand of tire is normally distributed with a mean of 50,000 and a standard deviation of 4,000 miles. What percentage of tires will have a life of 42,600 to 57,400 miles? a. 46.78% b. 93.57% c. 50% d. 23.39%<\/p>\n\n\n\n<p><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>To solve this problem, we need to use the concept of the <strong>normal distribution<\/strong> and <strong>z-scores<\/strong> to find the percentage of tires that will have a life expectancy between 42,600 and 57,400 miles. The mean life expectancy of the tires is 50,000 miles, and the standard deviation is 4,000 miles.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Standardize the Values<\/h3>\n\n\n\n<p>We first convert the given values (42,600 and 57,400 miles) into <strong>z-scores<\/strong>, which measure how many standard deviations each value is away from the mean. The formula for a z-score 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 we&#8217;re interested in,<\/li>\n\n\n\n<li>\u03bc\\mu\u03bc is the mean,<\/li>\n\n\n\n<li>\u03c3\\sigma\u03c3 is the standard deviation.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">For 42,600 miles:<\/h4>\n\n\n\n<p>z1=42,600\u221250,0004,000=\u22127,4004,000=\u22121.85z_1 = \\frac{42,600 &#8211; 50,000}{4,000} = \\frac{-7,400}{4,000} = -1.85z1\u200b=4,00042,600\u221250,000\u200b=4,000\u22127,400\u200b=\u22121.85<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">For 57,400 miles:<\/h4>\n\n\n\n<p>z2=57,400\u221250,0004,000=7,4004,000=1.85z_2 = \\frac{57,400 &#8211; 50,000}{4,000} = \\frac{7,400}{4,000} = 1.85z2\u200b=4,00057,400\u221250,000\u200b=4,0007,400\u200b=1.85<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Look Up the Z-Scores in the Standard Normal Distribution Table<\/h3>\n\n\n\n<p>Using the z-table, we find the probabilities corresponding to the z-scores of -1.85 and 1.85.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For z=\u22121.85z = -1.85z=\u22121.85, the cumulative probability is approximately 0.0322.<\/li>\n\n\n\n<li>For z=1.85z = 1.85z=1.85, the cumulative probability is approximately 0.9678.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the Percentage<\/h3>\n\n\n\n<p>To find the percentage of tires between 42,600 and 57,400 miles, we subtract the cumulative probability for z=\u22121.85z = -1.85z=\u22121.85 from the cumulative probability for z=1.85z = 1.85z=1.85: P(42,600\u2264X\u226457,400)=P(z2)\u2212P(z1)=0.9678\u22120.0322=0.9356P(42,600 \\leq X \\leq 57,400) = P(z_2) &#8211; P(z_1) = 0.9678 &#8211; 0.0322 = 0.9356P(42,600\u2264X\u226457,400)=P(z2\u200b)\u2212P(z1\u200b)=0.9678\u22120.0322=0.9356<\/p>\n\n\n\n<p>Thus, approximately <strong>93.57%<\/strong> of the tires will have a life expectancy between 42,600 and 57,400 miles.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Answer: <strong>b. 93.57%<\/strong><\/h3>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>By using z-scores and the cumulative probability from the normal distribution table, we can find that about 93.57% of the tires will last between 42,600 and 57,400 miles. This is a direct application of probability theory involving the normal distribution.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-166.jpeg\" alt=\"\" class=\"wp-image-253849\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The life expectancy of a particular brand of tire is normally distributed with a mean of 50,000 and a standard deviation of 4,000 miles. What percentage of tires will have a life of 42,600 to 57,400 miles? a. 46.78% b. 93.57% c. 50% d. 23.39% The Correct Answer and Explanation is: To solve this problem, [&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-253839","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253839","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=253839"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253839\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}