Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more?<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n To find the probability that a randomly selected stock will close up \\$0.75 or more<\/strong>, we need to calculate the probability that a value from a normal distribution<\/strong> with:<\/p>\n\n\n\n is greater than or equal to \\$0.75<\/strong>.<\/p>\n\n\n\n The Z-score formula is:<\/p>\n\n\n\n $$ Where:<\/p>\n\n\n\n $$ Now, we use the standard normal distribution table<\/strong> (Z-table) or a calculator to find the probability that Z is greater than 1.21<\/strong>.<\/p>\n\n\n\n From the Z-table:<\/p>\n\n\n\n $$ So,<\/p>\n\n\n\n $$ $$ This problem is a classic example of using the normal distribution<\/strong> to find probabilities related to real-world data. Here, we are told that the change in daily closing stock prices is normally distributed \u2014 this implies that most changes will cluster around the mean (\\$0.35), with fewer observations occurring as we move further away from this average.<\/p>\n\n\n\n In statistics, when data follow a normal distribution, we can standardize any value using the Z-score<\/strong>, which tells us how many standard deviations a particular value is from the mean. In this case, we are asked to find the probability that a stock will close up by \\$0.75 or more<\/strong>. First, we calculate how far \\$0.75 is from the mean (\\$0.35), measured in standard deviations. The result, a Z-score of approximately 1.21<\/strong>, means that \\$0.75 is 1.21 standard deviations above the mean<\/strong>.<\/p>\n\n\n\n To find the probability of this occurring, we look up the Z-score in the standard normal distribution table, which provides the probability that a value is less than<\/strong> a given Z-score. Because we want the probability of being greater than \\$0.75<\/strong>, we subtract that value from 1.<\/p>\n\n\n\n The final result, 0.1131 or 11.31%<\/strong>, tells us that there’s about an 11.31% chance<\/strong> that a randomly selected stock will have a price increase of \\$0.75 or more on a given day. This relatively low probability reflects the fact that \\$0.75 is significantly above the average change of \\$0.35, making it a less common outcome.<\/p>\n","protected":false},"excerpt":{"rendered":" Assuming that the change in daily closing prices for stocks on the New York Stock Exchange is a random variable that is normally distributed with a mean of $.35 and a standard deviation of $.33. Based on this information, what is the probability that a randomly selected stock will close up $.75 or more? The […]<\/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-216722","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216722","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=216722"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216722\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n\n\n\nStep 1: Use the Z-score formula<\/strong><\/h3>\n\n\n\n
Z = \\frac{X – \\mu}{\\sigma}
$$<\/p>\n\n\n\n\n
Z = \\frac{0.75 – 0.35}{0.33} = \\frac{0.40}{0.33} \\approx 1.21
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Use the Z-score to find the probability<\/strong><\/h3>\n\n\n\n
P(Z < 1.21) \\approx 0.8869
$$<\/p>\n\n\n\n
P(Z > 1.21) = 1 – 0.8869 = 0.1131
$$<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/strong><\/h3>\n\n\n\n
\\boxed{0.1131} \\text{ or } \\boxed{11.31\\%}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h3>\n\n\n\n