{"id":211335,"date":"2025-05-03T16:55:50","date_gmt":"2025-05-03T16:55:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=211335"},"modified":"2025-05-03T16:55:52","modified_gmt":"2025-05-03T16:55:52","slug":"at-a-particular-airport-81-of-the-flights-arrived-on-time-last-year-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/03\/at-a-particular-airport-81-of-the-flights-arrived-on-time-last-year-2\/","title":{"rendered":"At a particular airport, 81% of the flights arrived on time last year"},"content":{"rendered":"\n<p>At a particular airport, 81% of the flights arrived on time last year. If 15 flights are randomly selected, find the probability that 11 or fewer flights are on time.<\/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 solve this problem, we can model the number of flights arriving on time as a binomial random variable, because the outcome for each flight is either &#8220;on time&#8221; or &#8220;not on time,&#8221; and each flight is independent of the others.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Define the variables<\/h3>\n\n\n\n<p>We know that:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The probability of a flight arriving on time is $p = 0.81$.<\/li>\n\n\n\n<li>The probability of a flight <strong>not<\/strong> arriving on time is $q = 1 &#8211; p = 0.19$.<\/li>\n\n\n\n<li>The number of flights being considered is $n = 15$.<\/li>\n\n\n\n<li>We want to find the probability that 11 or fewer flights are on time, so we need to calculate $P(X \\leq 11)$, where $X$ is the number of flights that arrive on time.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Binomial probability formula<\/h3>\n\n\n\n<p>The probability mass function (PMF) for a binomial distribution is given by:<\/p>\n\n\n\n<p>$$<br>P(X = k) = \\binom{n}{k} p^k q^{n-k}<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\binom{n}{k}$ is the binomial coefficient, which calculates how many ways $k$ successes can occur in $n$ trials.<\/li>\n\n\n\n<li>$p^k$ is the probability of $k$ successes (flights on time).<\/li>\n\n\n\n<li>$q^{n-k}$ is the probability of $n-k$ failures (flights not on time).<\/li>\n<\/ul>\n\n\n\n<p>To calculate $P(X \\leq 11)$, we need to compute the sum of the individual probabilities from $X = 0$ to $X = 11$:<\/p>\n\n\n\n<p>$$<br>P(X \\leq 11) = \\sum_{k=0}^{11} P(X = k) = \\sum_{k=0}^{11} \\binom{15}{k} (0.81)^k (0.19)^{15-k}<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Using a normal approximation<\/h3>\n\n\n\n<p>Since the values of $n$ and $p$ are large enough, we can use the <strong>normal approximation<\/strong> to the binomial distribution for easier computation. The normal approximation to a binomial distribution $X$ is given by:<\/p>\n\n\n\n<p>$$<br>X \\sim N(\\mu, \\sigma^2)<br>$$<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$\\mu = np = 15 \\times 0.81 = 12.15$<\/li>\n\n\n\n<li>$\\sigma = \\sqrt{npq} = \\sqrt{15 \\times 0.81 \\times 0.19} \\approx 1.77$<\/li>\n<\/ul>\n\n\n\n<p>We can approximate $P(X \\leq 11)$ using the normal distribution with continuity correction. The continuity correction accounts for the fact that the binomial distribution is discrete, while the normal distribution is continuous. So, we will calculate:<\/p>\n\n\n\n<p>$$<br>P(X \\leq 11) \\approx P\\left( Z \\leq \\frac{11.5 &#8211; 12.15}{1.77} \\right)<br>$$<\/p>\n\n\n\n<p>Where $Z$ is the standard normal variable.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Z-score calculation<\/h3>\n\n\n\n<p>$$<br>Z = \\frac{11.5 &#8211; 12.15}{1.77} = \\frac{-0.65}{1.77} \\approx -0.37<br>$$<\/p>\n\n\n\n<p>Using the standard normal table or a calculator, we find the probability corresponding to a Z-score of $-0.37$ is approximately $0.355$.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>Thus, the probability that 11 or fewer flights are on time is approximately <strong>0.355<\/strong> or <strong>35.5%<\/strong>.<\/p>\n\n\n\n<p>This solution uses the normal approximation, which is valid for large $n$ and when $p$ is not too close to 0 or 1. The result shows that there is a 35.5% chance that 11 or fewer of the 15 randomly selected flights will be on time.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>At a particular airport, 81% of the flights arrived on time last year. If 15 flights are randomly selected, find the probability that 11 or fewer flights are on time. The correct answer and explanation is : To solve this problem, we can model the number of flights arriving on time as a binomial random [&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-211335","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211335","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=211335"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/211335\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=211335"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=211335"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=211335"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}