{"id":265,"date":"2021-07-14T15:59:02","date_gmt":"2021-07-14T15:59:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-central-limit-theorem-for-sums\/"},"modified":"2023-12-05T09:22:49","modified_gmt":"2023-12-05T09:22:49","slug":"the-central-limit-theorem-for-sums","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/the-central-limit-theorem-for-sums\/","title":{"raw":"Distributions of Sums of Samples","rendered":"Distributions of Sums of Samples"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Calculate probabilities for the sampling distribution for a sum using technology<\/li>\r\n \t
  • Calculate percentiles for the sampling distribution for a sum using technology<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n
    \r\n\r\nSuppose\u00a0X<\/em> is a random variable with a distribution that may be known or unknown (it can be any distribution) and suppose:\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\nIf you draw random samples of size\u00a0n<\/em>, then as n<\/em> increases, the random variable [latex] \\sum X [\/latex]\u00a0<\/em>consisting of sums tends to be normally distributed<\/strong> and\r\n\r\n[latex] \\displaystyle {\\sum{X}{\\sim}{N}((n)(\\mu_X), (\\sqrt{n})(\\sigma_X))} [\/latex].\r\n\r\n<\/div>\r\nThe central limit theorem for sums<\/strong> says that if you keep drawing larger and larger samples and taking their sums, the sums form their own normal distribution (the sampling distribution), which approaches a normal distribution as the sample size increases. 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\r\nThe random variable \u03a3X<\/em> has the following z<\/em>-score associated with it:\r\n
        \r\n \t
      1. [latex] \\sum x [\/latex]\u00a0is one sum.<\/li>\r\n \t
      2. [latex]{z}=\\frac{{\\sum{x}-{({n})}{({\\mu}_{{X}})}}}{{{(\\sqrt{{n}})}{({\\mu}_{{X}})}}}[\/latex]\r\n
          \r\n \t
        1. [latex]{({n})}{({\\mu}_{{X}})}=\\text{the mean of }\\sum{X}[\/latex]<\/li>\r\n \t
        2. [latex]{\\left(\\sqrt{{n}}\\right)}{\\left({\\sigma}_{{X}}\\right)} =\\text{the standard deviation of }\\sum{X}[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n
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          USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header>To find probabilities for sums on the calculator, follow these steps:\r\n\r\n2nd\u00a0DISTR\r\n<\/code>2:normalcdf\r\n<\/code><\/code>normalcdf<\/code>(lower value of the area, upper value of the area, (n<\/em>)(mean), ([latex]\\displaystyle\\sqrt{{n}}[\/latex])(standard deviation))\r\n\r\nwhere:\r\n
            \r\n \t
          • mean<\/em> is the mean of the original distribution,<\/li>\r\n \t
          • standard deviation<\/em> is the standard deviation of the original distribution, and<\/li>\r\n \t
          • sample size<\/strong><\/em> = n<\/li>\r\n<\/ul>\r\n
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            Recall: Scientific Notation<\/h3>\r\nScientific notation is used by scientists, mathematicians, and engineers when they are working with very large or very small numbers. When a number is written in scientific notation, the exponent tells you if the term is a large or a small number. A positive exponent indicates a large number and a negative exponent indicates a small number that is between 0 and 1.\r\n\r\n \r\n
            \r\n

            Scientific Notation<\/h2>\r\nA positive number is written in scientific notation if it is written as [latex]a\\times10^{n}[\/latex]\u00a0where the coefficient a<\/i>\u00a0is [latex]1\\leq{a}<10[\/latex], and n <\/i>is an integer.\r\n\r\n<\/div>\r\nSome calculators and other devices use [latex]E[\/latex] instead of [latex]\\times 10[\/latex]. [latex]E[\/latex] represents an exponent of [latex]10[\/latex], then it is followed by a number which is the exponent used in scientific notation.\r\n\r\nFor example, [latex]1 \\times 10^5 = 1E5[\/latex].\r\n\r\n<\/div>\r\n
            \r\n

            Example 1<\/h3>\r\nAn 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\r\n1. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than 7,500.\r\n[reveal-answer q=\"609917\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"609917\"]\r\n\r\nLet 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>.\r\n\r\n[latex]\\displaystyle\\sum{X}[\/latex]\u00a0= the sum or total of 80 values. Since [latex]\\displaystyle{\\mu}_{x}=90,{\\sigma}_{x}=15[\/latex], and n<\/em> = 80, [latex]\\sum{X}[\/latex]~ N<\/em>((80)(90), ([latex]\\displaystyle\\sqrt{{80}}[\/latex])(15))\r\n
              \r\n \t
            • mean of the sums = (n<\/em>)(\u03bcX<\/sub><\/em>) = (80)(90) = 7,200<\/li>\r\n \t
            • standard deviation of the sums =[latex]\\displaystyle{(\\sqrt{{n}})}{({\\sigma}_{{X}})}={(\\sqrt{{80}})}{({15})}[\/latex]<\/li>\r\n \t
            • sum of 80 values = \u03a3x<\/em> = 7,500<\/li>\r\n<\/ul>\r\nFind P<\/em>(\u03a3x<\/em> > 7,500)\r\nP<\/em>(\u03a3x<\/em> > 7,500) = 0.0127\r\n\"This\r\n\r\n
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              USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header>
              \r\n
              <\/div>\r\n<\/section>normalcdf<\/code>(lower value, upper value, mean of sums, stdev<\/code> of sums)\r\n\r\nThe parameter list is abbreviated(lower, upper, (n<\/em>)(\u03bcX<\/sub><\/em>, [latex]\\displaystyle{(\\sqrt{{n}})}[\/latex](\u03c3X<\/sub><\/em>))\r\nnormalcdf<\/code> (7500, 1E99, (80)(90), [latex]\\displaystyle{(\\sqrt{{80}})}[\/latex](15)) = 0.0127\r\n\r\nReminder<\/strong>\r\n\r\n1E99 = 1099<\/sup>\r\n\r\nPress the EE<\/code> key for E.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Find the sum that is 1.5 standard deviations above the mean of the sums.\r\n[reveal-answer q=\"686260\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"686260\"]\r\n\r\nFind \u03a3x<\/em> where z<\/em> = 1.5.\r\n\r\n[latex]\\displaystyle{\\sum{x}}={(n)}{({\\mu}_{X})}+{(z)}{(\\sqrt{n})}{({\\sigma}_{X}})=(80)(90)+(1.5)(\\sqrt{80})(15)=7401.2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

              <\/h3>\r\n
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              Try it 1<\/h3>\r\nAn unknown distribution has a mean of 45 and a standard deviation of eight. A sample size of 50 is drawn randomly from the population. Find the probability that the sum of the 50 values is more than 2,400.\r\n[reveal-answer q=\"962102\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"962102\"]\r\n\r\n0.0040\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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              USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h3>\r\n<\/header>To find percentiles for sums on the calculator, follow these steps.\r\n\r\n2nd DIStR\r\n3:invNormk<\/em> = invNorm (area to the left of k<\/em>, (n<\/em>)(mean), (standard deviation))\r\n\r\nwhere:\r\n