Distance Breaking The Breaking Distance Of A Simple Random Sample Of Cars: 95% Confidence N = 32; ~ = 137ft And A Is Known To Be 7 Ft.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s first solve the problem correctly:<\/p>\n\n\n\n You are given:<\/p>\n\n\n\n Since the population standard deviation ( \\sigma ) is known<\/strong> and ( n > 30 ), we use the z-distribution<\/strong> to construct the confidence interval.<\/p>\n\n\n\n The formula for a confidence interval when ( \\sigma ) is known: Where:<\/p>\n\n\n\n At 95% confidence, ( z_{\\alpha\/2} = 1.96 ) (standard value).<\/p>\n\n\n\n Now, calculate the standard error (SE): Now compute the margin of error (ME): Finally, construct the confidence interval: Thus:<\/p>\n\n\n\n \u2705 The 95% confidence interval for the mean breaking distance is approximately (134.58 ft, 139.42 ft).<\/strong><\/p>\n\n\n\n In this problem, we are asked to calculate a 95% confidence interval for the mean breaking distance of cars based on a sample. The sample size provided is 32 cars, and the sample mean distance is 137 feet. Importantly, the standard deviation of the population is given as 7 feet.<\/p>\n\n\n\n Since the standard deviation of the population is known and the sample size is greater than 30, it is appropriate to use the z-distribution<\/strong> instead of the t-distribution<\/strong>. The z-distribution is ideal for large samples with a known standard deviation because it assumes that the sampling distribution of the sample mean is approximately normal.<\/p>\n\n\n\n The formula used to construct a confidence interval is: First, we calculate the standard error (SE)<\/strong>, which measures how much the sample mean is expected to vary from the true population mean. Using ( SE = \\sigma \/ \\sqrt{n} ), we find that SE is approximately 1.237. Then, we find the margin of error (ME)<\/strong> by multiplying the SE by the z-score corresponding to a 95% confidence level, which is 1.96.<\/p>\n\n\n\n The resulting margin of error is about 2.424. Adding and subtracting this margin of error from the sample mean gives the confidence interval. Thus, we are 95% confident that the true mean breaking distance for all cars falls between 134.58 feet and 139.42 feet.<\/p>\n","protected":false},"excerpt":{"rendered":" Distance Breaking The Breaking Distance Of A Simple Random Sample Of Cars: 95% Confidence N = 32; ~ = 137ft And A Is Known To Be 7 Ft. The correct answer and explanation is : Let’s first solve the problem correctly: You are given: Since the population standard deviation ( \\sigma ) is known and […]<\/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-209242","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209242","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=209242"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209242\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209242"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209242"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209242"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
[
\\bar{x} \\pm z_{\\alpha\/2} \\times \\frac{\\sigma}{\\sqrt{n}}
]<\/p>\n\n\n\n\n
[
SE = \\frac{\\sigma}{\\sqrt{n}} = \\frac{7}{\\sqrt{32}} \\approx \\frac{7}{5.6569} \\approx 1.237
]<\/p>\n\n\n\n
[
ME = z_{\\alpha\/2} \\times SE = 1.96 \\times 1.237 \\approx 2.424
]<\/p>\n\n\n\n
[
137 \\pm 2.424
]<\/p>\n\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
\n\n\n\n300-word Explanation:<\/h3>\n\n\n\n
[
\\bar{x} \\pm z_{\\alpha\/2} \\times \\frac{\\sigma}{\\sqrt{n}}
]
where ( \\bar{x} ) is the sample mean, ( \\sigma ) is the population standard deviation, and ( n ) is the sample size.<\/p>\n\n\n\n