{"id":211623,"date":"2025-05-04T17:40:37","date_gmt":"2025-05-04T17:40:37","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=211623"},"modified":"2025-05-04T17:40:39","modified_gmt":"2025-05-04T17:40:39","slug":"locating-a-percentile-in-a-normal-distribution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/04\/locating-a-percentile-in-a-normal-distribution\/","title":{"rendered":"Locating a Percentile in a Normal Distribution"},"content":{"rendered":"\n<p>Locating a Percentile in a Normal Distribution (Level 1) Question 32 0\/0.333333333 points Identify the score corresponding to the indicated percentile. Cardiac pacemaker batteries will stop working if they are not properly calibrated and charged. Suppose that the length of time a battery operates follows a normal distribution with mean 25.0 days and standard deviation 2.0 days. At what point will 1% of the batteries fail?<\/p>\n\n\n\n<p>21.85 days<br>19 days<br>22.31 days<br>20.34 days<br>24.98 days<br>Locating a Percentile in a Normal Distribution (Level 2) 0.3340 333333333 15<\/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>To find the point at which <strong>1% of cardiac pacemaker batteries fail<\/strong>, we are essentially looking for the <strong>1st percentile<\/strong> of a <strong>normal distribution<\/strong> with a:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mean (\u03bc)<\/strong> = 25.0 days<\/li>\n\n\n\n<li><strong>Standard deviation (\u03c3)<\/strong> = 2.0 days<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step-by-Step Solution:<\/strong><\/h3>\n\n\n\n<p>To locate a percentile in a normal distribution, we use the <strong>z-score formula<\/strong>:<\/p>\n\n\n\n<p>$$<br>X = \\mu + z\\sigma<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$X$ is the value (battery life) at the desired percentile<\/li>\n\n\n\n<li>$\\mu$ is the mean<\/li>\n\n\n\n<li>$\\sigma$ is the standard deviation<\/li>\n\n\n\n<li>$z$ is the z-score corresponding to the desired percentile<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Find the z-score for the 1st percentile:<\/strong><\/h3>\n\n\n\n<p>The z-score corresponding to the 1st percentile (P = 0.01) is approximately:<\/p>\n\n\n\n<p>$$<br>z = -2.33<br>$$<\/p>\n\n\n\n<p>(This can be found using a <strong>z-table<\/strong> or calculator for the standard normal distribution.)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Plug into the formula:<\/strong><\/h3>\n\n\n\n<p>$$<br>X = 25.0 + (-2.33)(2.0) = 25.0 &#8211; 4.66 = 20.34<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Correct Answer: <em>20.34 days<\/em><\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (approx. 300 words):<\/strong><\/h3>\n\n\n\n<p>In problems involving normally distributed data, finding a percentile means determining the value below which a certain percentage of observations fall. In this case, we want to know the number of days at which only 1% of cardiac pacemaker batteries would have failed\u2014meaning they lasted that long or less.<\/p>\n\n\n\n<p>The data tells us that battery life is normally distributed with a mean of 25 days and a standard deviation of 2 days. These two parameters define the shape and spread of the normal curve.<\/p>\n\n\n\n<p>To solve this, we first find the z-score that corresponds to the 1st percentile in the standard normal distribution. Using statistical tables or a calculator, we find that the z-score for the 1st percentile is approximately -2.33. This negative value makes sense, as we&#8217;re looking far into the lower tail of the distribution.<\/p>\n\n\n\n<p>Next, we convert the z-score back into the actual time (X) using the formula:<\/p>\n\n\n\n<p>$$<br>X = \\mu + z\\sigma<br>$$<\/p>\n\n\n\n<p>Plugging in the numbers gives us:<\/p>\n\n\n\n<p>$$<br>X = 25.0 + (-2.33)(2.0) = 20.34 \\text{ days}<br>$$<\/p>\n\n\n\n<p>This tells us that 1% of the batteries will fail at or before 20.34 days. This is a crucial threshold for manufacturers and medical professionals to know, as they might want to replace or check devices before this point to ensure safety.<\/p>\n\n\n\n<p>Understanding percentiles in normal distributions helps in quality control, risk assessment, and designing safer products in real-world applications.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Locating a Percentile in a Normal Distribution (Level 1) Question 32 0\/0.333333333 points Identify the score corresponding to the indicated percentile. Cardiac pacemaker batteries will stop working if they are not properly calibrated and charged. Suppose that the length of time a battery operates follows a normal distribution with mean 25.0 days and standard deviation [&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-211623","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211623","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=211623"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211623\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211623"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211623"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211623"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}