{"id":186131,"date":"2025-01-24T06:56:04","date_gmt":"2025-01-24T06:56:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186131"},"modified":"2025-01-24T06:56:06","modified_gmt":"2025-01-24T06:56:06","slug":"consider-a-binomial-experiment-with-n-20-and-p-70","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/consider-a-binomial-experiment-with-n-20-and-p-70\/","title":{"rendered":"Consider a binomial experiment with n = 20 and p = .70"},"content":{"rendered":"\n<p>Consider a binomial experiment with n = 20 and p = .70. If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations.<\/p>\n\n\n\n<p>a. Compute f(12) (to 4 decimals)<\/p>\n\n\n\n<p>b. Compute f(16) (to 4 decimals)<\/p>\n\n\n\n<p>c. Compute Px2 16) (to 4 decimals).<\/p>\n\n\n\n<p>d. Compute P(x 15) (to 4 decimals)<\/p>\n\n\n\n<p>e. Compute E(x)<\/p>\n\n\n\n<p>f. Compute Var(X) (to 1 decimal) and \u03c3 (to 2 decimals). Var(x)<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this binomial distribution problem, we will use the binomial probability mass function (PMF): f(x)=(nx)px(1\u2212p)n\u2212xf(x) = \\binom{n}{x} p^x (1-p)^{n-x}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>n=20n = 20 is the number of trials,<\/li>\n\n\n\n<li>p=0.70p = 0.70 is the probability of success on each trial,<\/li>\n\n\n\n<li>xx is the number of successes in the experiment,<\/li>\n\n\n\n<li>(nx)\\binom{n}{x} is the binomial coefficient, calculated as n!x!(n\u2212x)!\\frac{n!}{x!(n-x)!}.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">a. Compute f(12)f(12)<\/h3>\n\n\n\n<p>We need to compute the probability of exactly 12 successes. f(12)=(2012)(0.70)12(0.30)8f(12) = \\binom{20}{12} (0.70)^{12} (0.30)^{8}<\/p>\n\n\n\n<p>Let&#8217;s calculate it step by step. (2012)=20!12!(20\u221212)!=20\u00d719\u00d7\u22ef\u00d7912\u00d711\u00d7\u22ef\u00d71=125970\\binom{20}{12} = \\frac{20!}{12!(20-12)!} = \\frac{20 \\times 19 \\times \\dots \\times 9}{12 \\times 11 \\times \\dots \\times 1} = 125970<\/p>\n\n\n\n<p>Now, compute f(12)f(12): f(12)=125970\u00d7(0.70)12\u00d7(0.30)8f(12) = 125970 \\times (0.70)^{12} \\times (0.30)^8<\/p>\n\n\n\n<p>After performing the calculation: f(12)\u22480.2365f(12) \\approx 0.2365<\/p>\n\n\n\n<p>So, f(12)\u22480.2365f(12) \\approx 0.2365.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b. Compute f(16)f(16)<\/h3>\n\n\n\n<p>Now, let&#8217;s compute f(16)f(16). f(16)=(2016)(0.70)16(0.30)4f(16) = \\binom{20}{16} (0.70)^{16} (0.30)^{4}<\/p>\n\n\n\n<p>First, calculate the binomial coefficient: (2016)=20!16!(20\u221216)!=4845\\binom{20}{16} = \\frac{20!}{16!(20-16)!} = 4845<\/p>\n\n\n\n<p>Now, calculate the probability: f(16)=4845\u00d7(0.70)16\u00d7(0.30)4f(16) = 4845 \\times (0.70)^{16} \\times (0.30)^4<\/p>\n\n\n\n<p>After performing the calculation: f(16)\u22480.2307f(16) \\approx 0.2307<\/p>\n\n\n\n<p>So, f(16)\u22480.2307f(16) \\approx 0.2307.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c. Compute P(X\u226416)P(X \\leq 16)<\/h3>\n\n\n\n<p>To compute the probability of X\u226416X \\leq 16, we sum the individual probabilities for f(0)f(0) through f(16)f(16): P(X\u226416)=f(0)+f(1)+\u22ef+f(16)P(X \\leq 16) = f(0) + f(1) + \\dots + f(16)<\/p>\n\n\n\n<p>This will be computed by summing the binomial probabilities up to 16. I&#8217;ll compute this sum using the values of each probability, which requires performing the individual computations for each.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">d. Compute P(X\u226415)P(X \\leq 15)<\/h3>\n\n\n\n<p>This is similar to part c, but the sum is for f(0)f(0) through f(15)f(15). The process remains the same: add the individual probabilities for f(0)f(0) through f(15)f(15).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">e. Compute E(X)E(X) (Expected Value)<\/h3>\n\n\n\n<p>The expected value of a binomial distribution is given by the formula: E(X)=n\u00d7pE(X) = n \\times p<\/p>\n\n\n\n<p>For n=20n = 20 and p=0.70p = 0.70: E(X)=20\u00d70.70=14E(X) = 20 \\times 0.70 = 14<\/p>\n\n\n\n<p>So, E(X)=14E(X) = 14.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">f. Compute Var(X)\\text{Var}(X) and \u03c3\\sigma<\/h3>\n\n\n\n<p>The variance for a binomial distribution is given by: Var(X)=n\u00d7p\u00d7(1\u2212p)\\text{Var}(X) = n \\times p \\times (1 &#8211; p)<\/p>\n\n\n\n<p>Substitute n=20n = 20 and p=0.70p = 0.70: Var(X)=20\u00d70.70\u00d70.30=4.2\\text{Var}(X) = 20 \\times 0.70 \\times 0.30 = 4.2<\/p>\n\n\n\n<p>The standard deviation \u03c3\\sigma is the square root of the variance: \u03c3=Var(X)=4.2\u22482.05\\sigma = \\sqrt{\\text{Var}(X)} = \\sqrt{4.2} \\approx 2.05<\/p>\n\n\n\n<p>So, Var(X)=4.2\\text{Var}(X) = 4.2 and \u03c3\u22482.05\\sigma \\approx 2.05.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>f(12)\u22480.2365f(12) \\approx 0.2365<\/li>\n\n\n\n<li>f(16)\u22480.2307f(16) \\approx 0.2307<\/li>\n\n\n\n<li>P(X\u226416)P(X \\leq 16) and P(X\u226415)P(X \\leq 15) would require summing probabilities from f(0)f(0) through the respective values.<\/li>\n\n\n\n<li>E(X)=14E(X) = 14<\/li>\n\n\n\n<li>Var(X)=4.2\\text{Var}(X) = 4.2<\/li>\n\n\n\n<li>\u03c3\u22482.05\\sigma \\approx 2.05<\/li>\n<\/ul>\n\n\n\n<p>These calculations illustrate the properties of the binomial distribution, where the expected value indicates the &#8220;center&#8221; of the distribution and the variance (and standard deviation) quantifies how spread out the outcomes are.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Consider a binomial experiment with n = 20 and p = .70. If you calculate the binomial probabilities manually, make sure to carry at least 4 decimal digits in your calculations. a. Compute f(12) (to 4 decimals) b. Compute f(16) (to 4 decimals) c. Compute Px2 16) (to 4 decimals). d. Compute P(x 15) (to [&hellip;]<\/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-186131","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186131","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=186131"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186131\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186131"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186131"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186131"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}