Cereal A company\u2019s cereal boxes advertise 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean \u00b5 = 9.70 ounces and standard deviation s = 0.03 ounces. <\/p>\n\n\n\n
(a) What is the probability that a randomly selected box of the cereal contains less than 9.65 ounces of cereal? Show your work. <\/p>\n\n\n\n
(b) Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal x \u2013 in these boxes is 9.65 ounces or less? Show your work.
<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n We are given that the weight of cereal in a randomly selected box follows a Normal distribution<\/strong> with:<\/p>\n\n\n\n We will calculate the probabilities for each part using the Standard Normal Distribution (Z-score formula)<\/strong>:<\/p>\n\n\n\n [ We need to find:<\/p>\n\n\n\n [ First, compute the Z-score:<\/p>\n\n\n\n [ Using a Z-table, the probability corresponding to (Z = -1.67) is 0.0475<\/strong>.<\/p>\n\n\n\n Thus, the probability that a randomly selected box contains less than 9.65 ounces is 0.0475 (or 4.75%)<\/strong>.<\/p>\n\n\n\n For a sample of size n = 5<\/strong>, the sampling distribution of the sample mean follows:<\/p>\n\n\n\n [ Now, compute the Z-score for (X\u0304 = 9.65):<\/p>\n\n\n\n [ Using a Z-table, the probability corresponding to (Z = -3.73) is 0.0001<\/strong>.<\/p>\n\n\n\n Thus, the probability that the mean weight of 5 boxes<\/strong> is 9.65 ounces or less is 0.0001 (or 0.01%)<\/strong>.<\/p>\n\n\n\n This problem is a classic application of the Normal distribution<\/strong> and sampling distribution of the mean<\/strong>. In part (a), we calculate the probability of a single<\/strong> box having less than 9.65 ounces by standardizing the value using the Z-score formula. The result of 0.0475<\/strong> means that only 4.75%<\/strong> of individual boxes contain less than 9.65 ounces, indicating that most boxes meet or exceed the advertised weight.<\/p>\n\n\n\n In part (b), the situation changes because we are dealing with an SRS (Simple Random Sample) of 5 boxes<\/strong>. When working with sample means, the variability decreases because the standard deviation is adjusted by the square root of the sample size ((\\sqrt{n})). This results in a smaller standard error (SE = 0.0134<\/strong>), making extreme deviations from the mean (such as 9.65 ounces) much less likely.<\/p>\n\n\n\n The probability of the sample mean being less than 9.65 ounces is extremely small (0.0001 or 0.01%<\/strong>), meaning it is almost impossible to randomly pick 5 boxes and get an average weight of 9.65 ounces or less. This illustrates the law of large numbers<\/strong>, where larger sample sizes tend to produce means closer to the population mean.<\/p>\n\n\n\n In summary, while some individual boxes<\/strong> may weigh less than 9.65 ounces, the chance of a group of 5 boxes<\/strong> averaging that weight is negligible.<\/p>\n","protected":false},"excerpt":{"rendered":" Cereal A company\u2019s cereal boxes advertise 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean \u00b5 = 9.70 ounces and standard deviation s = 0.03 ounces. (a) What is the probability that a randomly selected box of the cereal contains less than 9.65 […]<\/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-204470","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204470","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=204470"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/204470\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=204470"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=204470"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=204470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Z = \\frac{X – \\mu}{\\sigma}
]<\/p>\n\n\n\n(a) Probability of a single box containing less than 9.65 ounces<\/strong><\/h3>\n\n\n\n
P(X < 9.65)
]<\/p>\n\n\n\n
Z = \\frac{9.65 – 9.70}{0.03} = \\frac{-0.05}{0.03} = -1.67
]<\/p>\n\n\n\n
\n\n\n\n(b) Probability that the mean weight of an SRS of 5 boxes is 9.65 ounces or less<\/strong><\/h3>\n\n\n\n
\n
SE = \\frac{\\sigma}{\\sqrt{n}} = \\frac{0.03}{\\sqrt{5}} = \\frac{0.03}{2.236} \\approx 0.0134
]<\/p>\n\n\n\n
Z = \\frac{9.65 – 9.70}{0.0134} = \\frac{-0.05}{0.0134} \\approx -3.73
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n