{"id":214234,"date":"2025-05-13T20:35:04","date_gmt":"2025-05-13T20:35:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214234"},"modified":"2025-05-13T20:35:06","modified_gmt":"2025-05-13T20:35:06","slug":"yard-work-lamar-and-hareesh-run-a-two-person-lawn-care-service","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/yard-work-lamar-and-hareesh-run-a-two-person-lawn-care-service\/","title":{"rendered":"Yard work Lamar and Hareesh run a two-person lawn care service"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Yard work Lamar and Hareesh run a two-person lawn care service. They have been caring for Mr. Johnson&#8217;s very large lawn for several years, and they have found that the time l, it takes L amar to mow the law a randomly selected day is approximately Normally distributed with a mean of 105 minutes and a standard deviation of 10 minutes. The time H it takes Hareesh to use the edger and string trimmer on a randomly elected day is approximately Normally distributed with a mean of 98 minutes and a standard deviation of 15 minutes. Assume that I, and H are independent random variables. F&#8217;ind the probability that Lamar and Hareesh finish their jobs within 5 minutes of each other on a randomly selected day<\/p>\n\n\n\n<ol start=\"67\" class=\"wp-block-list\"><\/ol>\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-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To find the probability that <strong>Lamar and Hareesh finish within 5 minutes of each other<\/strong>, we are essentially calculating the probability that the absolute difference between their completion times is less than or equal to 5 minutes.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>L\u223cN(105,102)L \\sim N(105, 10^2): Lamar&#8217;s mowing time is Normally distributed with mean 105 minutes and standard deviation 10 minutes.<\/li>\n\n\n\n<li>H\u223cN(98,152)H \\sim N(98, 15^2): Hareesh&#8217;s edging time is Normally distributed with mean 98 minutes and standard deviation 15 minutes.<\/li>\n\n\n\n<li>Let D=L\u2212HD = L &#8211; H: the difference in their times.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Distribution of D=L\u2212HD = L &#8211; H<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Since LL and HH are independent and Normally distributed, the difference DD is also Normally distributed. The mean and variance of DD are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03bcD=\u03bcL\u2212\u03bcH=105\u221298=7\\mu_D = \\mu_L &#8211; \\mu_H = 105 &#8211; 98 = 7<\/li>\n\n\n\n<li>\u03c3D2=\u03c3L2+\u03c3H2=102+152=100+225=325\\sigma_D^2 = \\sigma_L^2 + \\sigma_H^2 = 10^2 + 15^2 = 100 + 225 = 325<\/li>\n\n\n\n<li>\u03c3D=325\u224818.03\\sigma_D = \\sqrt{325} \\approx 18.03<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So, D\u223cN(7,18.032)D \\sim N(7, 18.03^2)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Probability that Lamar and Hareesh finish within 5 minutes<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">We want: P(\u2223L\u2212H\u2223\u22645)=P(\u22125\u2264D\u22645)P(|L &#8211; H| \\leq 5) = P(-5 \\leq D \\leq 5)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since D\u223cN(7,18.032)D \\sim N(7, 18.03^2), we standardize this: Z=D\u2212\u03bcD\u03c3D=D\u2212718.03Z = \\frac{D &#8211; \\mu_D}{\\sigma_D} = \\frac{D &#8211; 7}{18.03}<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Now convert the bounds:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Lower bound: \u22125\u2212718.03=\u22121218.03\u2248\u22120.665\\frac{-5 &#8211; 7}{18.03} = \\frac{-12}{18.03} \\approx -0.665<\/li>\n\n\n\n<li>Upper bound: 5\u2212718.03=\u2212218.03\u2248\u22120.111\\frac{5 &#8211; 7}{18.03} = \\frac{-2}{18.03} \\approx -0.111<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Now find the probability: P(\u22120.665\u2264Z\u2264\u22120.111)P(-0.665 \\leq Z \\leq -0.111)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Using standard normal tables or a calculator:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P(Z\u2264\u22120.111)\u22480.4557P(Z \\leq -0.111) \\approx 0.4557<\/li>\n\n\n\n<li>P(Z\u2264\u22120.665)\u22480.2524P(Z \\leq -0.665) \\approx 0.2524<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">P(\u22120.665\u2264Z\u2264\u22120.111)=0.4557\u22120.2524=0.2033P(-0.665 \\leq Z \\leq -0.111) = 0.4557 &#8211; 0.2524 = 0.2033<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">0.2033&nbsp;or&nbsp;20.33%\\boxed{0.2033} \\text{ or } \\boxed{20.33\\%}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (Approx. 300 words)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">In this problem, we are given two independent normally distributed variables representing the work times of Lamar and Hareesh. Lamar takes an average of 105 minutes to mow, with a standard deviation of 10 minutes. Hareesh takes an average of 98 minutes for edging, with a standard deviation of 15 minutes. Since both times are independent and Normally distributed, their difference (Lamar&#8217;s time minus Hareesh&#8217;s time) is also Normally distributed.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We define a new variable D=L\u2212HD = L &#8211; H, where LL and HH are the times taken by Lamar and Hareesh, respectively. The mean of this difference is 105\u221298=7105 &#8211; 98 = 7 minutes, and the variance is the sum of the individual variances because the variables are independent: 102+152=32510^2 + 15^2 = 325, giving a standard deviation of 325\u224818.03\\sqrt{325} \\approx 18.03.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">We are interested in the probability that the <strong>absolute difference<\/strong> between their times is <strong>less than or equal to 5 minutes<\/strong>, which is written mathematically as P(\u2223L\u2212H\u2223\u22645)P(|L &#8211; H| \\leq 5) or P(\u22125\u2264D\u22645)P(-5 \\leq D \\leq 5). We then convert these bounds to standard normal zz-scores and use the standard normal distribution to find the area between them, which gives us the desired probability.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The final answer is approximately 0.2033, meaning there is about a <strong>20.33% chance<\/strong> that Lamar and Hareesh finish their tasks within 5 minutes of each other on a randomly selected day.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Yard work Lamar and Hareesh run a two-person lawn care service. They have been caring for Mr. Johnson&#8217;s very large lawn for several years, and they have found that the time l, it takes L amar to mow the law a randomly selected day is approximately Normally distributed with a mean of 105 minutes and [&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-214234","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214234","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=214234"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214234\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214234"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214234"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214234"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}