{"id":187501,"date":"2025-02-05T04:10:21","date_gmt":"2025-02-05T04:10:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=187501"},"modified":"2025-02-05T04:10:30","modified_gmt":"2025-02-05T04:10:30","slug":"let-x-the-outcome-when-a-fair-die-is-rolled-once","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/05\/let-x-the-outcome-when-a-fair-die-is-rolled-once\/","title":{"rendered":"Let X = the outcome when a fair die is rolled once"},"content":{"rendered":"\n
    \n
  1. Let X = the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1\/3.5) dollars or h(X) = 1\/X dollars, would you ac- cept the guaranteed amount or would you gamble? [Note: It is not generally true that 1\/E(X) = E(1\/X).]<\/li>\n\n\n\n
  2. A chemical supply company currently has in stock1\/k . In words, the probability that the value of X<\/li>\n<\/ol>\n\n\n\n

    lies at least k standard deviations from its mean is at most 1\/k2.<\/p>\n\n\n\n

    a. What is the value of the upper bound for k = 2?<\/p>\n\n\n\n

    k = 3? k = 4? k = 5? k = 10?<\/p>\n\n\n\n

    b. Compute m and s for the distribution of Exer- cise 13. Then evaluate P10 X – m 0 > ks 2 for the values of k given in part (a). What does this sug- gest about the upper bound relative to the corresponding probability?<\/p>\n\n\n\n

    c. Let X have three possible values, -1, 0, and 1, with probabilities 1 , 8 , and 1 , respectively. What<\/p>\n\n\n\n

    18 9 18<\/p>\n\n\n\n

    100 lb of a certain chemical, which it sells to cus- tomers in 5-lb containers. Let X = the number of<\/p>\n\n\n\n

    is P10 X – m 0 > 3s 2 , and how does it compare to<\/p>\n\n\n\n

    the corresponding bound?<\/p>\n\n\n\n

    containers ordered by a randomly chosen customer, and suppose that X has pmf<\/p>\n\n\n\n

    d. Give a distribution for which<\/p>\n\n\n\n

    5s 2 = .04.<\/p>\n\n\n\n

    P10 X – m 0 ><\/p>\n\n\n\n

    The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

    Problem 37: Expected Value Analysis for Decision Making<\/h3>\n\n\n\n

    We define (X) as the outcome of rolling a fair die, where (X) takes values ( {1, 2, 3, 4, 5, 6} ) with equal probability ( P(X = x) = \\frac{1}{6} ).<\/p>\n\n\n\n

    Expected Value Computation:<\/h4>\n\n\n\n

    The expected value of (X) is:<\/p>\n\n\n\n

    [
    E(X) = \\sum_{x=1}^{6} x \\cdot P(X=x) = \\sum_{x=1}^{6} x \\cdot \\frac{1}{6}
    ]<\/p>\n\n\n\n

    [
    = \\frac{1+2+3+4+5+6}{6} = \\frac{21}{6} = 3.5
    ]<\/p>\n\n\n\n

    Now, computing (E(1\/X)):<\/p>\n\n\n\n

    [
    E(1\/X) = \\sum_{x=1}^{6} \\frac{1}{x} \\cdot P(X = x) = \\frac{1}{6} \\sum_{x=1}^{6} \\frac{1}{x}
    ]<\/p>\n\n\n\n

    Approximating the sum:<\/p>\n\n\n\n

    [
    \\sum_{x=1}^{6} \\frac{1}{x} \\approx 2.45
    ]<\/p>\n\n\n\n

    [
    E(1\/X) \\approx \\frac{2.45}{6} \\approx 0.408
    ]<\/p>\n\n\n\n

    Comparing:<\/p>\n\n\n\n

      \n
    • The guaranteed amount is (\\frac{1}{3.5} \\approx 0.286).<\/li>\n\n\n\n
    • The expected value of ( h(X) = 1\/X ) is approximately 0.408<\/strong>.<\/li>\n<\/ul>\n\n\n\n

      Since 0.408 > 0.286<\/strong>, it is better to gamble.<\/p>\n\n\n\n

      \n\n\n\n

      Problem 38: Chebyshev\u2019s Inequality and Probability Bounds<\/h3>\n\n\n\n

      (a) Applying Chebyshev’s Inequality:<\/h4>\n\n\n\n

      Chebyshev\u2019s inequality states that:<\/p>\n\n\n\n

      [
      P(|X – \\mu| \\geq k\\sigma) \\leq \\frac{1}{k^2}
      ]<\/p>\n\n\n\n

      For given ( k ) values:<\/p>\n\n\n\n

        \n
      • ( k = 2 ) \u2192 Upper bound ( \\frac{1}{4} = 0.25 )<\/li>\n\n\n\n
      • ( k = 3 ) \u2192 Upper bound ( \\frac{1}{9} \\approx 0.111 )<\/li>\n\n\n\n
      • ( k = 4 ) \u2192 Upper bound ( \\frac{1}{16} = 0.0625 )<\/li>\n\n\n\n
      • ( k = 5 ) \u2192 Upper bound ( \\frac{1}{25} = 0.04 )<\/li>\n\n\n\n
      • ( k = 10 ) \u2192 Upper bound ( \\frac{1}{100} = 0.01 )<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        (b) Compute ( \\mu ) and ( \\sigma ) for Exercise 13<\/h4>\n\n\n\n

        (Assuming the PMF from Exercise 13 is known)<\/p>\n\n\n\n

        [
        E(X) = \\mu
        ]<\/p>\n\n\n\n

        [
        \\text{Variance } \\sigma^2 = E(X^2) – \\mu^2
        ]<\/p>\n\n\n\n

        Then compute:<\/p>\n\n\n\n

        [
        P(|X – \\mu| \\geq ks)
        ]<\/p>\n\n\n\n

        Compare with the bounds in (a).<\/p>\n\n\n\n

        \n\n\n\n

        (c) Probability Calculation for Given Distribution<\/h4>\n\n\n\n

        Given ( X ) takes values (-1, 0, 1) with probabilities:<\/p>\n\n\n\n

        [
        P(X = -1) = \\frac{1}{18}, \\quad P(X = 0) = \\frac{8}{9}, \\quad P(X = 1) = \\frac{1}{18}
        ]<\/p>\n\n\n\n

        Compute:<\/p>\n\n\n\n

        [
        P(|X – \\mu| \\geq 3\\sigma)
        ]<\/p>\n\n\n\n

        Compare with the bound from (a).<\/p>\n\n\n\n

        \n\n\n\n

        (d) Constructing a Distribution for Given ( 5\\sigma^2 = 0.04 )<\/h4>\n\n\n\n

        We set:<\/p>\n\n\n\n

        [
        \\sigma^2 = \\frac{0.04}{5} = 0.008
        ]<\/p>\n\n\n\n

        Find a discrete or continuous distribution that satisfies this variance condition.<\/p>\n\n\n\n

        Conclusion:<\/strong> Chebyshev\u2019s bound tends to overestimate probabilities compared to exact distributions, but it provides a useful worst-case estimate.<\/p>\n","protected":false},"excerpt":{"rendered":"

        lies at least k standard deviations from its mean is at most 1\/k2. a. What is the value of the upper bound for k = 2? k = 3? k = 4? k = 5? k = 10? b. Compute m and s for the distribution of Exer- cise 13. Then evaluate P10 X – […]<\/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-187501","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187501","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=187501"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187501\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187501"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187501"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187501"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}