{"id":63,"date":"2021-01-12T22:19:37","date_gmt":"2021-01-12T22:19:37","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/centrallimittheoremforsums\/"},"modified":"2024-03-31T22:10:01","modified_gmt":"2024-03-31T22:10:01","slug":"centrallimittheoremforsums","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/centrallimittheoremforsums\/","title":{"rendered":"Central Limit Theorem for Sums & Proportions"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

Central Limit Theorem for Sums<\/h2>\r\n\r\n
\r\n

Suppose X<\/em> is a random variable with a distribution that may be known or unknown<\/strong> (it can be any distribution) and suppose:<\/p>\r\n\r\n

    \r\n \t
  1. \u03bcX<\/sub><\/em> = the mean of \u03a7<\/em><\/li>\r\n \t
  2. \u03c3\u03a7<\/sub><\/em> = the standard deviation of X<\/em><\/li>\r\n<\/ol>\r\n

    If you draw random samples of size $n$, then as $n$ increases, the random variable \u03a3X<\/em> consisting of sums tends to be normally distributed<\/span> and $\\sum X \\sim N \\left( n\\cdot \\mu_x, \\sqrt{n}\\cdot \\sigma_x \\right)$<\/p>\r\nThe<\/strong> central limit theorem for sums<\/span> says that if you repeatedly draw samples of a given size (such as repeatedly rolling ten dice) and calculate the sum of each sample, these sums tend to follow a normal distribution. As sample sizes increase, the distribution of means more closely follows the normal distribution. The normal distribution has a mean equal to the original mean multiplied by the sample size and a standard deviation equal to the original standard deviation multiplied by the square root of the sample size.<\/strong>\r\n

    The random variable \u03a3X<\/em> has the following z<\/em>-score associated with it:<\/p>\r\n\r\n

      \r\n \t
    1. \u03a3x<\/em> is one sum.<\/li>\r\n \t
    2. $z = \\frac{\\sum x -n\\cdot \\mu_X}{\\sqrt{n}\\cdot \\sigma_x}$\r\n
        \r\n \t
      1. $n\\cdot \\mu_x = $ the mean of $\\sum X$<\/li>\r\n \t
      2. $\\sqrt{n}\\cdot \\sigma_X = $ the standard deviation of $\\sum X$<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
        \r\n

        Example <\/span>6.5<\/span><\/h3>\r\n<\/header>
        \r\n
        \r\n

        An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is drawn randomly from the population.\r\n<\/span><\/p>\r\n\r\n

        <\/header>
        \r\n
        \r\n
        \r\n
          \r\n \t
        1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.<\/li>\r\n \t
        2. Find the sum that is 1.5 standard deviations above the mean of the sums.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
          \r\n
          <\/div>\r\n
          \r\n

          Solution <\/span>6.5<\/span><\/h4>\r\n
          \r\n

          Let X<\/em> = one value from the original unknown population. The probability question asks you to find a probability for the sum (or total of) 80 values.<\/strong><\/p>\r\n

          \u03a3X<\/em> = the sum or total of 80 values. Since \u03bcX<\/sub><\/em> = 90, \u03c3X<\/sub><\/em> = 15, and n<\/em> = 80, $\\sum X \\sim N(80\\cdot 90, \\sqrt{80}\\cdot 15)$<\/span><\/span><\/p>\r\n\r\n