The average retirement age of NFL players is 33 years with a standard deviation of 2 years.<\/p>\n\n\n\n
a) Find the probability that a randomly chosen NFL player retired at over 36 years old.<\/p>\n\n\n\n
b) What is the probability that a randomly chosen sample of 10NFL players has average retirement age between 34 and 35 years of age?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Given:<\/p>\n\n\n\n To find this probability, we will use the Z-score<\/strong> formula, which standardizes the given value: Substituting the values: Now, we look up the Z-score value of 1.5 in the standard normal distribution table, which gives the cumulative probability up to 36 years old: To find the probability that the player retired after<\/em> 36 years old, we subtract this value from 1: So, the probability that a randomly chosen NFL player retired over 36 years old is approximately 0.0668<\/strong> or 6.68%<\/strong>.<\/p>\n\n\n\n In this case, we are dealing with a sample of 10 NFL players, so we will use the sampling distribution<\/strong> of the sample mean. The mean of the sample mean remains the same as the population mean, but the standard deviation (standard error) of the sample mean is reduced by a factor of the square root of the sample size (n).<\/p>\n\n\n\n Given:<\/p>\n\n\n\n The standard error<\/strong> (SE) is: Next, we want to find the probability that the sample mean falls between 34 and 35 years. We will calculate the Z-scores for both 34 and 35 years, using the formula: For (X = 34): For (X = 35): Now, we find the cumulative probabilities for these Z-scores:<\/p>\n\n\n\n The probability that the sample mean is between 34 and 35 years old is the difference between these two probabilities: So, the probability that the average retirement age of 10 randomly chosen NFL players is between 34 and 35 years old is approximately 0.0563<\/strong> or 5.63%<\/strong>.<\/p>\n\n\n\n The average retirement age of NFL players is 33 years with a standard deviation of 2 years. a) Find the probability that a randomly chosen NFL player retired at over 36 years old. b) What is the probability that a randomly chosen sample of 10NFL players has average retirement age between 34 and 35 years […]<\/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-182015","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182015","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=182015"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182015\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182015"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182015"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182015"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part (a) Finding the probability that a randomly chosen NFL player retired over 36 years old<\/h3>\n\n\n\n
\n
[
Z = \\frac{X – \\mu}{\\sigma}
]
Where:<\/p>\n\n\n\n\n
[
Z = \\frac{36 – 33}{2} = \\frac{3}{2} = 1.5
]<\/p>\n\n\n\n
[
P(Z < 1.5) \\approx 0.9332
]<\/p>\n\n\n\n
[
P(X > 36) = 1 – P(Z < 1.5) = 1 – 0.9332 = 0.0668
]<\/p>\n\n\n\nPart (b) Finding the probability that the average retirement age of 10 randomly chosen NFL players is between 34 and 35 years old<\/h3>\n\n\n\n
\n
[
SE = \\frac{\\sigma}{\\sqrt{n}} = \\frac{2}{\\sqrt{10}} \\approx 0.6325
]<\/p>\n\n\n\n
[
Z = \\frac{X – \\mu}{SE}
]<\/p>\n\n\n\n
[
Z_{34} = \\frac{34 – 33}{0.6325} \\approx 1.58
]<\/p>\n\n\n\n
[
Z_{35} = \\frac{35 – 33}{0.6325} \\approx 3.16
]<\/p>\n\n\n\n\n
[
P(34 < \\bar{X} < 35) = P(Z < 3.16) – P(Z < 1.58) = 0.9992 – 0.9429 = 0.0563
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
\n