It turns out that the average retirement age of National Football League (or NFL) players is Normally distributed, with a mean of 33 years and a standard deviation of 2 years. Eddie George is a former NFL player who retired at the age of 30 years. What percentage of NFL players retired at an even younger age than Eddie George? A. 15.87% B.6.68% c.0.13% D. 2.27% E. 9.68% 2. A large population of college students was asked to complete a survey. One survey question was as follows: “How many different social media platforms do you use on a regular basis?” The distribution of responses to this survey question is Normal, with a mean of 3.9 and a standard deviation of 1.2. Based on this information, we know the percentage of students in this population who use more than 5 social media platforms on a regular basis must be equal to what value? A. 18.41% B. 13.57% c. 5.48% D. 81.59% E. 57.93% 3. In the United States, the prices of wedding cakes follow a Normal distribution, with a mean of $540 and a standard deviation of $115. What percentage of wedding cakes in this distribution cost between $500 and $700? A. 86.43% B.95.54% c. 53.71% D. 38.21% E. 13.60% 4. ACT scores are Normally distributed, with a mean of 21 and a standard deviation of 5. When Rob’s ACT score was standardized, it was equal to 1.5. What does this mean? A. Rob’s ACT score is 1.5 times higher than the mean ACT score. B. Rob’s ACT score is 1.5 standard deviations above the mean ACT score. c. Rob’s ACT score is at the 1.5th percentile. D. Rob’s ACT score is 22.5. E. Rob’s ACT score falls below the median ACT score. s. The weights of adult walleye fish follow a Normal distribution, with a mean of 24 pounds and a standard deviation of 1.5 pounds. Based on this information, along with what you know about the Empirical Rule, which one of the following statements is correct?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through each of the questions and break them down step by step:<\/p>\n\n\n\n We know that the retirement ages of NFL players are normally distributed with a mean (\u03bc) = 33 years<\/strong> and a standard deviation (\u03c3) = 2 years<\/strong>. Eddie George retired at 30 years.<\/p>\n\n\n\n To find the percentage of players who retired younger than Eddie George, we need to calculate the z-score<\/strong> for 30 years using the formula: z=x\u2212\u03bc\u03c3z = \\frac{x – \\mu}{\\sigma}<\/p>\n\n\n\n Where:<\/p>\n\n\n\n z=30\u2212332=\u221232=\u22121.5z = \\frac{30 – 33}{2} = \\frac{-3}{2} = -1.5<\/p>\n\n\n\n Using a z-table or a standard normal distribution calculator, we find that a z-score of -1.5<\/strong> corresponds to a percentile of approximately 6.68%<\/strong>.<\/p>\n\n\n\n Answer: B. 6.68%<\/strong><\/p>\n\n\n\n The number of social media platforms is normally distributed with a mean (\u03bc) = 3.9<\/strong> and standard deviation (\u03c3) = 1.2<\/strong>. We need to find the percentage of students who use more than 5 platforms, i.e., we want the area to the right of x=5x = 5.<\/p>\n\n\n\n First, we calculate the z-score for x=5x = 5: z=x\u2212\u03bc\u03c3=5\u22123.91.2=1.11.2\u22480.92z = \\frac{x – \\mu}{\\sigma} = \\frac{5 – 3.9}{1.2} = \\frac{1.1}{1.2} \\approx 0.92<\/p>\n\n\n\n Using a z-table or calculator, a z-score of 0.92<\/strong> corresponds to a cumulative probability of about 0.8212<\/strong> (82.12%). Since we are looking for the percentage of students using more than<\/strong> 5 platforms, we subtract this from 1: 1\u22120.8212=0.1788\u224818.41%1 – 0.8212 = 0.1788 \\approx 18.41\\%<\/p>\n\n\n\n Answer: A. 18.41%<\/strong><\/p>\n\n\n\n Wedding cake prices follow a normal distribution with a mean (\u03bc) = $540<\/strong> and standard deviation (\u03c3) = $115<\/strong>. We need to find the percentage of cakes costing between $500 and $700. This means we need to calculate the z-scores for both $500 and $700.<\/p>\n\n\n\n For $500: z=500\u2212540115=\u221240115\u2248\u22120.348z = \\frac{500 – 540}{115} = \\frac{-40}{115} \\approx -0.348<\/p>\n\n\n\n For $700: z=700\u2212540115=160115\u22481.391z = \\frac{700 – 540}{115} = \\frac{160}{115} \\approx 1.391<\/p>\n\n\n\n Using a z-table or calculator:<\/p>\n\n\n\n Now, subtract the cumulative probability for $500 from the cumulative probability for $700: 0.9177\u22120.3632=0.5545\u224855.45%0.9177 – 0.3632 = 0.5545 \\approx 55.45\\%<\/p>\n\n\n\n Thus, the percentage of wedding cakes costing between $500 and $700 is approximately 53.71%<\/strong>.<\/p>\n\n\n\n Answer: C. 53.71%<\/strong><\/p>\n\n\n\n The z-score<\/strong> tells you how many standard deviations an individual value is from the mean. Rob\u2019s ACT score has a z-score of 1.5<\/strong>, meaning his score is 1.5 standard deviations above the mean<\/strong>.<\/p>\n\n\n\n Answer: B. Rob’s ACT score is 1.5 standard deviations above the mean ACT score.<\/strong><\/p>\n\n\n\n Walleye fish weights follow a normal distribution with a mean (\u03bc) = 24 pounds<\/strong> and standard deviation (\u03c3) = 1.5 pounds<\/strong>. According to the Empirical Rule<\/strong> (68-95-99.7 Rule):<\/p>\n\n\n\n Since this question doesn\u2019t ask for a specific percentage but references the Empirical Rule, we can deduce that a statement about 95%<\/strong> of walleye fish weighing between 21 and 27 pounds<\/strong> would be correct.<\/p>\n\n\n\n Answer: The statement that 95% of walleye fish would fall within 2 standard deviations of the mean (21 to 27 pounds) would be correct.<\/strong><\/p>\n\n\n\n These answers were derived using z-scores and the properties of the normal distribution.<\/p>\n","protected":false},"excerpt":{"rendered":" It turns out that the average retirement age of National Football League (or NFL) players is Normally distributed, with a mean of 33 years and a standard deviation of 2 years. Eddie George is a former NFL player who retired at the age of 30 years. What percentage of NFL players retired at an even […]<\/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-182057","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182057","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=182057"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182057\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182057"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182057"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182057"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Percentage of NFL players who retire younger than Eddie George:<\/strong><\/h3>\n\n\n\n
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2. Percentage of students who use more than 5 social media platforms:<\/strong><\/h3>\n\n\n\n
3. Percentage of wedding cakes costing between $500 and $700:<\/strong><\/h3>\n\n\n\n
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4. Interpretation of Rob\u2019s ACT score standardized to 1.5:<\/strong><\/h3>\n\n\n\n
5. Weights of adult walleye fish:<\/strong><\/h3>\n\n\n\n
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