{"id":266,"date":"2021-07-14T15:59:02","date_gmt":"2021-07-14T15:59:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/using-the-central-limit-theorem\/"},"modified":"2023-12-05T09:23:33","modified_gmt":"2023-12-05T09:23:33","slug":"using-the-central-limit-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/using-the-central-limit-theorem\/","title":{"raw":"Understanding the Central Limit Theorem","rendered":"Understanding the Central Limit Theorem"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Use the central limit theorem to calculate probabilities for normal and non-normal population distributions<\/li>\r\n \t
  • Use the central limit theorem to calculate percentiles for normal and non-normal population distributions<\/li>\r\n \t
  • Apply the continuity correction factor to find probabilities using technology<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n
    \r\n\r\nIt is important for you to understand when to use the\u00a0central limit theorem (CLT). If you are being asked to find the probability of the mean, use the CLT for the mean. If you are being asked to find the probability of a sum or total, use the\u00a0CLT for sums. This also applies to percentiles for means and sums.\r\n\r\nNote:<\/strong>\u00a0If you are being asked to find the probability of an\u00a0individual value, do not use the CLT. Use the distribution of its random variable.\r\n\r\n<\/div>\r\n
    \r\n

    Recall: Fraction Reasoning<\/h3>\r\nThe numerator is the top of a fraction while the denominator is the bottom of a fraction.\r\n\r\nAs the numerator gets larger, the fraction becomes larger.\r\n\r\nFor example, [latex]\\frac{3}{7}<\\frac{4}{7}<\\frac{5}{7}[\/latex],\u00a0this is because we are getting more parts of the same size as the denominator.\r\n\r\nAs the denominator gets larger, the fraction becomes smaller.\r\n\r\nFor example, [latex]\\frac{3}{4} > \\frac{3}{5} > \\frac{3}{6}[\/latex], this is because we are dividing by more parts as the denominator increases.\r\n\r\n<\/div>\r\n

    Examples of the Central Limit Theorem<\/h2>\r\n

    Law of Large Numbers<\/h3>\r\nThe\u00a0law of large numbers<\/span><\/strong>\u00a0says that if you take samples of larger and larger size from any population, then the mean [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0of the sample tends to get closer and closer to\u00a0\u03bc<\/em>. From the central limit theorem, we know that as\u00a0n<\/em>\u00a0gets larger and larger, the sample means follow a normal distribution. The larger\u00a0n<\/em>\u00a0gets, the smaller the standard deviation gets. (Remember that the standard deviation for [latex]\\displaystyle\\overline{{x}}[\/latex] is [latex]\\frac{\\sigma}{\\sqrt n}.) [\/latex]\u00a0This means that the sample mean [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0must be close to the population mean\u00a0\u03bc<\/em>. We can say that\u00a0\u03bc<\/em>\u00a0is the value that the sample means approach as\u00a0n<\/em>\u00a0gets larger. The central limit theorem illustrates the law of large numbers.\r\n

    Central Limit Theorem for the Mean and Sum Examples<\/h3>\r\n
    \r\n

    Example 1<\/h3>\r\nA study involving stress is conducted among the students on a college campus.\u00a0The stress scores follow a uniform distribution<\/strong> with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:\r\n
      \r\n \t
    1. The probability that the mean stress score<\/strong> for the 75 students is less than two.<\/li>\r\n \t
    2. The 90th percentile for the mean stress score<\/strong> for the 75 students.<\/li>\r\n \t
    3. The probability that the total of the 75 stress scores<\/strong> is less than 200.<\/li>\r\n \t
    4. The 90th percentile for the total stress score<\/strong> for the 75 students.<\/li>\r\n<\/ol>\r\nLet\u00a0X<\/em> = one stress score.\r\n\r\nProblems a. and b. ask you to find a probability or a percentile for a\u00a0mean<\/strong>. Problems c. and d. ask you to find a probability or a percentile for a total or sum<\/strong>. The sample size, n<\/em>, is equal to 75.\r\n\r\nSince the individual stress scores follow a uniform distribution,\u00a0X<\/em> ~ U<\/em>(1, 5) where a<\/em> = 1 and b<\/em> = 5\r\n\r\n[latex]\\displaystyle{\\mu}_{X}=\\frac{{a+b}}{{2}}=\\frac{{1+5}}{{2}}={3}[\/latex]\r\n\r\n[latex]\\displaystyle{\\sigma}_{X}=\\sqrt{\\frac{{{(b-a)}^{2}}}{{12}}}=\\sqrt{\\frac{{({5-1)}^{2}}}{{12}}}[\/latex]= 1.15\r\n\r\nFor problems a. and b., let\u00a0[latex]\\displaystyle\\overline{x}[\/latex] = the mean stress score for the 75 students. Then,\r\n\r\n[latex]\\displaystyle\\overline{x}\\sim{N}({3},\\frac{{1.15}}{{\\sqrt{75}}})\\text{ where } {n}={75}[\/latex].\r\n\r\na. Find P([latex]\\displaystyle\\overline{x }{<}{ 2}[\/latex]). Draw the graph.\r\n[reveal-answer q=\"497514\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"497514\"]\r\n\r\nP<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex] < 2) = 0 The probability that the mean stress score is less than two is about zero.\r\n\"This\r\n\r\n \r\n\r\n \r\n\r\nnormalcdf<\/code> [latex]\\displaystyle{({1},{2},{3},\\frac{{1.15}}{\\sqrt{{75}}})}={0}[\/latex]\r\n\r\nRemember<\/strong> that the smallest stress score is one.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nb. Find the 90th percentile for the mean of 75 stress scores. Draw a graph.\r\n[reveal-answer q=\"81823\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"81823\"]\r\n\r\nLet k<\/em> = the 90th percentile.\r\n\r\nFind\u00a0k<\/em>, where P<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex] < k<\/em>) = 0.90.\r\n\r\nk<\/em> = 3.2\r\n\r\n\"This\r\n\r\n \r\n\r\nThe 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.\r\n\r\ninvNorm<\/code>[latex]\\left (0.90,3,\\frac{1.15}{\\sqrt {75}} \\right ) = 3.2[\/latex]\r\n\r\nFor problems 3 and 4, \u03a3X = the sum of the 75 stress scores. Then, [latex]\\displaystyle\\sum{X}{\\sim}{N}{[{({75})}{({3})},{(\\sqrt{{75}})}{({1.15})}]}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nc. Find P<\/em>(\u03a3x<\/em> < 200). Draw the graph.\r\n[reveal-answer q=\"14477\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"14477\"]\r\n\r\nThe mean of the sum of 75 stress scores is (75)(3) = 225\r\n\r\nThe standard deviation of the sum of 75 stress scores is\r\n[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15) = 9.96\r\nP<\/em>(\u03a3x<\/em> < 200) = 0\r\n\"This\r\n\r\nThe probability that the total of 75 scores is less than 200 is about zero.<\/code>\r\n\r\nnormalcdf<\/code> (75,200,(75)(3),[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15)).\r\nRemember<\/strong>, the smallest total of 75 stress scores is 75, because the smallest single score is one.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nd. Find the 90th percentile for the total of 75 stress scores. Draw a graph.\r\n[reveal-answer q=\"102255\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"102255\"]\r\n\r\nLet k<\/em>= the 90th percentile.\r\n\r\nFind k<\/em> where P<\/em>(\u03a3x<\/em> < k<\/em>) = 0.90.\r\n\r\nk<\/em> = 237.8\r\n\"This\r\n\r\nThe 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.\r\n\r\ninvNorm<\/code>(0.90,(75)(3),[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15)) = 237.8\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h3>\r\n
      \r\n

      Try it 1<\/h3>\r\nUse the information in \"Central Limit Theorem for the Mean and Sum Examples,\"<\/span>\u00a0but use a sample size of 55 to answer the following questions.\r\n\r\n1. Find [latex]\\displaystyle{P}{(\\overline{{x }}{<}{ 7})}[\/latex].\r\n[reveal-answer q=\"477878\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"477878\"]\r\n\r\n0.0265\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Find [latex]\\displaystyle{P}{(\\sum{x}{>}{170})}[\/latex].\r\n[reveal-answer q=\"260959\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"260959\"]\r\n\r\n0.2789\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. Find the 80th percentile for the mean of 55 scores.\r\n[reveal-answer q=\"843205\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"843205\"]\r\n\r\n3.13\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n4. Find the 85th percentile for the sum of 55 scores.\r\n[reveal-answer q=\"585015\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"585015\"]\r\n\r\n173.84\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h3>\r\n
      \r\n

      Example 2<\/h3>\r\nSuppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used<\/strong> follows an exponential distribution<\/strong> with a mean of 22 minutes.\r\n\r\nConsider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.\r\n\r\nLet\u00a0X<\/em> = the excess time used by one individual<\/em> cell phone customer who exceeds his contracted time allowance.\r\n\r\n[latex]\\displaystyle{X}{\\sim}{E}{x}{p}{(\\frac{{1}}{{22}})}[\/latex]. From previous chapters, we know that \u03bc<\/em> = 22 and \u03c3<\/em> = 22.\r\n\r\nLet\u00a0[latex]\\displaystyle\\overline{{x}}[\/latex] = the mean excess time used by a sample of n<\/em> = 80 customers who exceed their contracted time allowance.\r\n\r\n[latex]\\displaystyle\\overline{{x}}{\\sim}{N}{({22},\\frac{{22}}{{\\sqrt{{80}}}})}[\/latex] by the central limit theorem for sample means\r\n\r\n \r\n\r\nUsing the CLT to find probability:<\/strong>\r\n\r\n1. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find [latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}[\/latex]. Draw the graph.\r\n[reveal-answer q=\"696931\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"696931\"]\r\n\r\nFind: [latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}[\/latex]\r\n\r\n[latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}={0.79199}[\/latex] using normalcdf<\/code> [latex]\\displaystyle{({20},{1}\\text{E99},{22},\\frac{{22}}{\\sqrt{{80}}})}[\/latex]\r\n\r\nThe probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.\r\n\r\n\"This\r\n\r\n \r\n\r\nRemember<\/strong>, 1E99 = 1099<\/sup> and \u20131E99 = \u20131099<\/sup>. Press the EE<\/code> key for E. Or just use 1099<\/sup> instead of 1E99.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P<\/em>(x<\/em> > 20).\r\n[reveal-answer q=\"193968\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"193968\"]\r\n\r\nFind P<\/em>(x > 20). Remember to use the exponential distribution for an individual: [latex]\\displaystyle{X}\\text{~}{E}{x}{p}{(\\frac{{1}}{{22}})}[\/latex].[latex]\\displaystyle{P}{({x}{>}{20})}={e}^{{{(-{({\\frac{1}{22}})}{({20})})}}}{\\quad\\text{or}\\quad}{e}^{{{(-{0.04545}{({20})})}}}={0.4029}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. Explain why the probabilities in parts 1 and 2 are different.\r\n[reveal-answer q=\"513351\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"513351\"]\r\n\r\n[latex]\\displaystyle{P}{({x}{>}{20})}={0.4029}\\text{ but } {P}{(\\overline{{x}}{>}{20})}={0.7919}[\/latex].\r\n\r\nThe probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.\r\n\r\nWhen asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the CLT. Use the CLT with the normal distribution when you are being asked to find the probability for a mean.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\nUsing the CLT to find percentiles:<\/strong>\r\n\r\n4. Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.\r\n[reveal-answer q=\"758671\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"758671\"]\r\n\r\nLet k<\/em> = the 95th percentile. Find k<\/em> where [latex]\\displaystyle{P}{(\\overline{{x}}{<}{k})}={0.95}[\/latex]\r\nk<\/em> = 26.0 using invNorm<\/code>[latex]\\left ( 0.95,22,\\frac{22}{\\sqrt {80}} \\right )[\/latex] = 26.0\r\n\"This\r\n\r\nThe 95th percentile for the sample mean excess time used<\/strong> is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.\r\n\r\nNinety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h3>\r\n
      \r\n

      Try it 2<\/h3>\r\nUse the information in Example 2, but change the sample size to 144.\r\n\r\n1. Find [latex]\\displaystyle{P}{({20}{<}\\overline{{x}}{<}{30})}[\/latex].\r\n[reveal-answer q=\"425056\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"425056\"]\r\n\r\n0.8623\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Find P<\/em>(\u03a3x<\/em> is at least 3,000).\r\n[reveal-answer q=\"201394\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"201394\"]\r\n\r\n0.7377\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. Find the 75th percentile for the sample mean excess time of [latex]144[\/latex] customers.\r\n[reveal-answer q=\"52807\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"52807\"]\r\n\r\n23.2\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n4. Find the 85th percentile for the sum of [latex]144[\/latex] excess times used by customers.\r\n[reveal-answer q=\"69005\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"69005\"]\r\n\r\n3,441.6\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h3>\r\n
      \r\n

      Example 3<\/h3>\r\nIn the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.\r\n\r\n1. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.\r\n[reveal-answer q=\"926094\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"926094\"]\r\n\r\nWe have, [latex]\\displaystyle{\\mu}_{x}={\\mu}=[\/latex] 2 and [latex]{\\sigma}_{x}=\\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{0.5}}{{10}}=0.05[\/latex]. Therefore:\r\n
        \r\n \t
      1. 50th percentile = \u03bcx<\/sub><\/em> = \u03bc<\/em> = 2<\/li>\r\n \t
      2. 25th percentile =\u00a0invNorm<\/code>(0.25,2,0.05) = 1.97<\/li>\r\n \t
      3. 75th percentile =\u00a0invNorm<\/code>(0.75,2,0.05) = 2.03<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.\r\n[reveal-answer q=\"64263\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"64263\"]\r\n\r\nWe have [latex]\\displaystyle{\\mu}_{\\sum{x}}=n({\\mu}_{x})={100}({2})= 200[\/latex] and [latex]{\\sigma}_{\\mu{x}}={\\sqrt n}({\\sigma}_{x})=10(0.5)=5[\/latex].\u00a0Therefore:\r\n
          \r\n \t
        1. 50th percentile =\u00a0[latex]\\displaystyle{\\mu}_{\\sum{x}}=n({\\mu}_{x})[\/latex] = 100(2) = 200<\/li>\r\n \t
        2. 25th percentile =\u00a0invNorm<\/code>(0.25,200,0.05) = 196.63<\/li>\r\n \t
        3. 75th percentile =\u00a0invNorm<\/code><\/i>(0.75,200,0.05) = 203.37<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.\r\n[reveal-answer q=\"651767\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"651767\"]\r\n\r\n[latex]{P}{(1.75 < {\\overline{x}} < {1.85})}[\/latex] =\u00a0normalcdf<\/code>(1.75,1.85,2,0.05) = 0.0013\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n4. Find the value that is two standard deviations above the sample mean.\r\n[reveal-answer q=\"663576\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"663576\"]\r\n\r\nUsing the z<\/em>-score equation, [latex]\\displaystyle{z}=\\frac{{\\overline{x}-{\\mu}_{\\overline{x}}}}{{{\\sigma}_{\\overline{x}}}}[\/latex].\r\n\r\nSolving for x,\u00a0we have x<\/em> = 2(0.05) + 2 = 2.1\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n5. Find the IQR<\/em> for the sum of the sample times.\r\n[reveal-answer q=\"242303\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"242303\"]\r\n\r\nThe IQR<\/em> is 75th percentile \u2013 25th percentile = 203.37 \u2013 196.63 = 6.74\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

          <\/h3>\r\n
          \r\n

          try it 3<\/h3>\r\nBased on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.\r\n\r\n1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.\r\n[reveal-answer q=\"785497\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"785497\"]\r\n\r\nP<\/em>(x > 120) =\u00a0normalcdf<\/code>(120, 1E99, 114.8, 13.1) [latex]\\approx [\/latex] 0.3457. There is about a 35% chance that the randomly selected woman will have a systolic blood pressure greater than 120.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.\r\n[reveal-answer q=\"503682\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"503682\"]\r\n\r\nP<\/em>( [latex]\\overline{x}[\/latex] > 120) = normalcdf<\/code>(120, 1E99, 114.8, [latex]\\dfrac{13.1}{\\sqrt{40}})\\approx[\/latex] 0.006.\r\nThere is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. If the sample were four women between the ages of [latex]18[\/latex] to [latex]24[\/latex] and we did not know the original distribution, could the central limit theorem be used?\r\n[reveal-answer q=\"963910\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"963910\"]\r\n\r\nThe central limit theorem could not be used if the sample size were four and we did not know the original distribution was normal. The sample size would be too small.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

          <\/h3>\r\n
          \r\n

          Example 4<\/h3>\r\nA study was done about violence against sex workers and the symptoms of the post-traumatic stress that they developed. The age range of the sex workers was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.\r\n\r\n1. In a sample of 25 sex workers, what is the probability that the mean age of the sex workers is less than 35?\r\n[reveal-answer q=\"687165\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"687165\"]\r\n\r\nP<\/em>([latex]\\overline{{x}}[\/latex] < 35) = normalcdf<\/code>(-E<\/em>99,35,30.9,1.8) = 0.9886\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.\r\n[reveal-answer q=\"685271\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"685271\"]\r\n\r\nP<\/em>([latex]\\overline{{x}}[\/latex] > 50) = normalcdf<\/code>(50, E<\/em>99,30.9,1.8) \u2248 0.\r\n\r\nFor this sample group, it is almost impossible for the group's average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. In a sample of 49 sex workers, what is the probability that the sum of the ages is no less than 1,600?\r\n[reveal-answer q=\"871464\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"871464\"]\r\n\r\nP<\/em>(\u03a3x<\/em> \u2265 1,600) = normalcdf<\/code>(1600,E99,1514.10,63) = 0.0864\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n4. Is it likely that the sum of the ages of the 49 sex workers is at most 1,595? Interpret the results.\r\n[reveal-answer q=\"897177\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"897177\"]\r\n\r\nP<\/em>(\u03a3x<\/em> \u2264 1,595) = normalcdf<\/code>(-E99,1595,1514.10,63) = 0.9005.\r\n\r\nThis means that there is a 90% chance that the sum of the ages for the sample group n<\/em> = 49 is at most 1,595.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n5. Find the 95th percentile for the sample mean age of 65 sex workers. Interpret the results.\r\n[reveal-answer q=\"630289\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"630289\"]\r\n\r\nThe 95th percentile = invNorm<\/code>(0.95,30.9,1.1) = 32.7.\r\n\r\nThis indicates that 95% of the sex workers in the sample of 65 are younger than 32.7 years, on average.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n6. Find the 90th percentile for the sum of the ages of 65 sex workers. Interpret the results.\r\n[reveal-answer q=\"808631\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"808631\"]\r\n\r\nThe 90th percentile = invNorm<\/code>(0.90,2008.5,72.56) = 2101.5.\r\n\r\nThis indicates that 90% of the sex workers in the sample of 65 have a sum of ages less than 2,101.5 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

          <\/h3>\r\n
          \r\n

          try it 4<\/h3>\r\nAccording to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.\r\n\r\n1. What mean doorway height would allow 95% of men to enter the aircraft without bending?\r\n[reveal-answer q=\"243773\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"243773\"]\r\n\r\nWe know that \u03bcx<\/sub><\/em> = \u03bc<\/em> = 69 and we have \u03c3x<\/sub><\/em> = 2.8. The height of the doorway is found to be\u00a0invNorm<\/code>(0.95,69,2.8) = 73.61.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?\r\n[reveal-answer q=\"460713\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"460713\"]\r\n\r\nWe know that \u03bcx<\/sub><\/em> = \u03bc<\/em> = 69 and we have \u03c3x<\/sub><\/em> = 0.28. So,\u00a0invNorm<\/code>(0.95,69,0.28) = 69.49.\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n3. For engineers designing the 757, which result is more relevant: the height from part 1 or part 2? Why?\r\n[reveal-answer q=\"981493\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"981493\"]\r\n\r\nWhen designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part 1.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

          Historical Note: Normal Approximation to the Binomial<\/h2>\r\nHistorically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n\u00a0<\/em>(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n<\/em>, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial<\/strong> distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution<\/strong>:\r\n
            \r\n \t
          • there are a certain number n<\/em> of independent trials<\/li>\r\n \t
          • the outcomes of any trial are success or failure<\/li>\r\n \t
          • each trial has the same probability of a success p<\/em><\/li>\r\n<\/ul>\r\nRecall that if X<\/em> is the binomial random variable, then X<\/em> ~ B<\/em>(n, p<\/em>). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np<\/em> and nq<\/em> must both be greater than five (np<\/em> > 5 and nq<\/em> > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean \u03bc<\/em> = np<\/em> and standard deviation [latex] \\sigma = {\\sqrt {npq}}[\/latex]. Remember that q <\/em>= 1 \u2013 p<\/em>. In order to get the best approximation, add 0.5 to x<\/em> or subtract 0.5 from x<\/em> (use x<\/em> + 0.5 or x<\/em> \u2013 0.5). The number 0.5 is called the continuity correction factor<\/strong> and is used in the following example.\r\n
            \r\n

            Example 5<\/h3>\r\nSuppose in a local kindergarten through 12th grade (K-12) school district, 53% of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.\r\n
              \r\n \t
            1. Find the probability that at least 150<\/strong> favor a charter school.<\/li>\r\n \t
            2. Find the probability that at most 160<\/strong> favor a charter school.<\/li>\r\n \t
            3. Find the probability that more than 155<\/strong> favor a charter school.<\/li>\r\n \t
            4. Find the probability that fewer than 147<\/strong> favor a charter school.<\/li>\r\n \t
            5. Find the probability that exactly 175<\/strong> favor a charter school.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"529\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"529\"]\r\n\r\nLet X<\/em> = the number that favor a charter school for grades K trough 5. X<\/em> ~B<\/em>(n, p<\/em>) where n<\/em> = 300 and p<\/em> = 0.53. Since np<\/em> > 5 and nq<\/em> > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are \u03bc<\/em> = np<\/em> and [latex]\\sigma = {\\sqrt {npq}}[\/latex]. The mean is 159 and the standard deviation is 8.6447. The random variable for the normal distribution is Y<\/em>. Y<\/em> ~ N<\/em>(159, 8.6447).\r\n\r\n \r\n\r\n1. You\u00a0include 150<\/strong> so P<\/em>(X<\/em> \u2265 150) has normal approximation P<\/em>(Y <\/em>\u2265 149.5) = 0.8641.\r\n\r\nnormalcdf<\/code>(149.5,10^99,159,8.6447) = 0.8641<\/span>\r\n\r\n \r\n\r\n2. You i<\/strong>nclude 160<\/strong> so <\/span>P<\/em>(<\/span>X<\/em> \u2264 160) has normal approximation <\/span>P<\/em>(<\/span>Y <\/em>\u2264 160.5) = 0.5689.\u00a0<\/span>\r\n\r\nnormalcdf<\/code>(0,160.5,159,8.6447) = 0.5689<\/span>\r\n\r\n \r\n\r\n3. You exclude 155 <\/strong>so <\/span>P<\/em>(<\/span>X<\/em> > 155) has normal approximation <\/span>P<\/em>(<\/span>y <\/em>> 155.5) = 0.6572.<\/span>\r\n\r\nnormalcdf<\/code>(155.5,10^99,159,8.6447) = 0.6572<\/span>\r\n\r\n \r\n\r\n4. You exclude 147<\/strong> so <\/span>P<\/em>(<\/span>X<\/em> < 147) has normal approximation\u00a0<\/span>P<\/em>(<\/span>Y<\/em> < 146.5) = 0.0741.<\/span>\r\n\r\nnormalcdf<\/code>(0,146.5,159,8.6447) = 0.0741<\/span>\r\n\r\n \r\n\r\n5. P<\/em>(<\/span>X<\/em> = 175) has normal approximation <\/span>P<\/em>(174.5 < <\/span>Y<\/em> < 175.5) = 0.0083.<\/span>\r\n\r\nnormalcdf<\/code>(174.5,175.5,159,8.6447) = 0.0083<\/span>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nBecause of calculators and computer software<\/strong> that let you calculate binomial probabilities for large values of <\/span>n<\/em> easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in \"binomial probability distribution calculation\" in an internet browser, you can find at least one online calculator for the binomial.<\/span><\/span>\r\n\r\nFor Example 3, the probabilities are calculated using the following binomial distribution: (n<\/em> = 300 and p<\/em> = 0.53). Compare the binomial and normal distribution answers.\r\n\r\nP<\/em>(X<\/em> \u2265 150) :1 - binomialcdf<\/code>(300,0.53,149) = 0.8641\r\n\r\nP<\/em>(X<\/em> \u2264 160) :binomialcdf<\/code>(300,0.53,160) = 0.5684\r\n\r\nP<\/em>(X<\/em> > 155) :1 - binomialcdf<\/code>(300,0.53,155) = 0.6576\r\n\r\nP<\/em>(X<\/em> < 147) :binomialcdf<\/code>(300,0.53,146) = 0.0742\r\n\r\nP<\/em>(X<\/em> = 175) :(You use the binomial pdf.)binomialpdf<\/code>(300,0.53,175) = 0.0083\r\n

              <\/h3>\r\n
              \r\n

              try it 5<\/h3>\r\nIn a city, 46% of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.\r\n[reveal-answer q=\"472946\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"472946\"]0.0401[\/hidden-answer]\r\n\r\n<\/div>","rendered":"
              \n

              Learning Outcomes<\/h3>\n
              \n
                \n
              • Use the central limit theorem to calculate probabilities for normal and non-normal population distributions<\/li>\n
              • Use the central limit theorem to calculate percentiles for normal and non-normal population distributions<\/li>\n
              • Apply the continuity correction factor to find probabilities using technology<\/li>\n<\/ul>\n<\/section>\n<\/div>\n
                \n

                It is important for you to understand when to use the\u00a0central limit theorem (CLT). If you are being asked to find the probability of the mean, use the CLT for the mean. If you are being asked to find the probability of a sum or total, use the\u00a0CLT for sums. This also applies to percentiles for means and sums.<\/p>\n

                Note:<\/strong>\u00a0If you are being asked to find the probability of an\u00a0individual value, do not use the CLT. Use the distribution of its random variable.<\/p>\n<\/div>\n

                \n

                Recall: Fraction Reasoning<\/h3>\n

                The numerator is the top of a fraction while the denominator is the bottom of a fraction.<\/p>\n

                As the numerator gets larger, the fraction becomes larger.<\/p>\n

                For example, [latex]\\frac{3}{7}<\\frac{4}{7}<\\frac{5}{7}[\/latex],\u00a0this is because we are getting more parts of the same size as the denominator.\n\nAs the denominator gets larger, the fraction becomes smaller.\n\nFor example, [latex]\\frac{3}{4} > \\frac{3}{5} > \\frac{3}{6}[\/latex], this is because we are dividing by more parts as the denominator increases.<\/p>\n<\/div>\n

                Examples of the Central Limit Theorem<\/h2>\n

                Law of Large Numbers<\/h3>\n

                The\u00a0law of large numbers<\/span><\/strong>\u00a0says that if you take samples of larger and larger size from any population, then the mean [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0of the sample tends to get closer and closer to\u00a0\u03bc<\/em>. From the central limit theorem, we know that as\u00a0n<\/em>\u00a0gets larger and larger, the sample means follow a normal distribution. The larger\u00a0n<\/em>\u00a0gets, the smaller the standard deviation gets. (Remember that the standard deviation for [latex]\\displaystyle\\overline{{x}}[\/latex] is [latex]\\frac{\\sigma}{\\sqrt n}.)[\/latex]\u00a0This means that the sample mean [latex]\\displaystyle\\overline{{x}}[\/latex]\u00a0must be close to the population mean\u00a0\u03bc<\/em>. We can say that\u00a0\u03bc<\/em>\u00a0is the value that the sample means approach as\u00a0n<\/em>\u00a0gets larger. The central limit theorem illustrates the law of large numbers.<\/p>\n

                Central Limit Theorem for the Mean and Sum Examples<\/h3>\n
                \n

                Example 1<\/h3>\n

                A study involving stress is conducted among the students on a college campus.\u00a0The stress scores follow a uniform distribution<\/strong> with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:<\/p>\n

                  \n
                1. The probability that the mean stress score<\/strong> for the 75 students is less than two.<\/li>\n
                2. The 90th percentile for the mean stress score<\/strong> for the 75 students.<\/li>\n
                3. The probability that the total of the 75 stress scores<\/strong> is less than 200.<\/li>\n
                4. The 90th percentile for the total stress score<\/strong> for the 75 students.<\/li>\n<\/ol>\n

                  Let\u00a0X<\/em> = one stress score.<\/p>\n

                  Problems a. and b. ask you to find a probability or a percentile for a\u00a0mean<\/strong>. Problems c. and d. ask you to find a probability or a percentile for a total or sum<\/strong>. The sample size, n<\/em>, is equal to 75.<\/p>\n

                  Since the individual stress scores follow a uniform distribution,\u00a0X<\/em> ~ U<\/em>(1, 5) where a<\/em> = 1 and b<\/em> = 5<\/p>\n

                  [latex]\\displaystyle{\\mu}_{X}=\\frac{{a+b}}{{2}}=\\frac{{1+5}}{{2}}={3}[\/latex]<\/p>\n

                  [latex]\\displaystyle{\\sigma}_{X}=\\sqrt{\\frac{{{(b-a)}^{2}}}{{12}}}=\\sqrt{\\frac{{({5-1)}^{2}}}{{12}}}[\/latex]= 1.15<\/p>\n

                  For problems a. and b., let\u00a0[latex]\\displaystyle\\overline{x}[\/latex] = the mean stress score for the 75 students. Then,<\/p>\n

                  [latex]\\displaystyle\\overline{x}\\sim{N}({3},\\frac{{1.15}}{{\\sqrt{75}}})\\text{ where } {n}={75}[\/latex].<\/p>\n

                  a. Find P([latex]\\displaystyle\\overline{x }{<}{ 2}[\/latex]). Draw the graph.\n\n\n

                  Show Answer<\/span><\/p>\n
                  \n

                  P<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex] < 2) = 0 The probability that the mean stress score is less than two is about zero.
                  \n\"This<\/p>\n

                   <\/p>\n

                   <\/p>\n

                  normalcdf<\/code> [latex]\\displaystyle{({1},{2},{3},\\frac{{1.15}}{\\sqrt{{75}}})}={0}[\/latex]<\/p>\n

                  Remember<\/strong> that the smallest stress score is one.<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  b. Find the 90th percentile for the mean of 75 stress scores. Draw a graph.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  Let k<\/em> = the 90th percentile.<\/p>\n

                  Find\u00a0k<\/em>, where P<\/em>([latex]\\displaystyle\\overline{{x}}[\/latex] < k<\/em>) = 0.90.<\/p>\n

                  k<\/em> = 3.2<\/p>\n

                  \"This<\/p>\n

                   <\/p>\n

                  The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of 75 stress scores are at most 3.2, and that 10% are at least 3.2.<\/p>\n

                  invNorm<\/code>[latex]\\left (0.90,3,\\frac{1.15}{\\sqrt {75}} \\right ) = 3.2[\/latex]<\/p>\n

                  For problems 3 and 4, \u03a3X = the sum of the 75 stress scores. Then, [latex]\\displaystyle\\sum{X}{\\sim}{N}{[{({75})}{({3})},{(\\sqrt{{75}})}{({1.15})}]}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  c. Find P<\/em>(\u03a3x<\/em> < 200). Draw the graph.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  The mean of the sum of 75 stress scores is (75)(3) = 225<\/p>\n

                  The standard deviation of the sum of 75 stress scores is
                  \n[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15) = 9.96
                  \nP<\/em>(\u03a3x<\/em> < 200) = 0
                  \n\"This<\/p>\n

                  The probability that the total of 75 scores is less than 200 is about zero.<\/code><\/p>\n

                  normalcdf<\/code> (75,200,(75)(3),[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15)).
                  \nRemember<\/strong>, the smallest total of 75 stress scores is 75, because the smallest single score is one.<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  d. Find the 90th percentile for the total of 75 stress scores. Draw a graph.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  Let k<\/em>= the 90th percentile.<\/p>\n

                  Find k<\/em> where P<\/em>(\u03a3x<\/em> < k<\/em>) = 0.90.<\/p>\n

                  k<\/em> = 237.8
                  \n\"This<\/p>\n

                  The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of 75 scores are no more than 237.8 and 10% are no less than 237.8.<\/p>\n

                  invNorm<\/code>(0.90,(75)(3),[latex]\\displaystyle{(\\sqrt{{75}})}[\/latex](1.15)) = 237.8<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                  <\/h3>\n
                  \n

                  Try it 1<\/h3>\n

                  Use the information in “Central Limit Theorem for the Mean and Sum Examples,”<\/span>\u00a0but use a sample size of 55 to answer the following questions.<\/p>\n

                  1. Find [latex]\\displaystyle{P}{(\\overline{{x }}{<}{ 7})}[\/latex].\n\n\n

                  Show Answer<\/span><\/p>\n
                  \n

                  0.0265<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  2. Find [latex]\\displaystyle{P}{(\\sum{x}{>}{170})}[\/latex].<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  0.2789<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  3. Find the 80th percentile for the mean of 55 scores.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  3.13<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  4. Find the 85th percentile for the sum of 55 scores.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  173.84<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                  <\/h3>\n
                  \n

                  Example 2<\/h3>\n

                  Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used<\/strong> follows an exponential distribution<\/strong> with a mean of 22 minutes.<\/p>\n

                  Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.<\/p>\n

                  Let\u00a0X<\/em> = the excess time used by one individual<\/em> cell phone customer who exceeds his contracted time allowance.<\/p>\n

                  [latex]\\displaystyle{X}{\\sim}{E}{x}{p}{(\\frac{{1}}{{22}})}[\/latex]. From previous chapters, we know that \u03bc<\/em> = 22 and \u03c3<\/em> = 22.<\/p>\n

                  Let\u00a0[latex]\\displaystyle\\overline{{x}}[\/latex] = the mean excess time used by a sample of n<\/em> = 80 customers who exceed their contracted time allowance.<\/p>\n

                  [latex]\\displaystyle\\overline{{x}}{\\sim}{N}{({22},\\frac{{22}}{{\\sqrt{{80}}}})}[\/latex] by the central limit theorem for sample means<\/p>\n

                   <\/p>\n

                  Using the CLT to find probability:<\/strong><\/p>\n

                  1. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find [latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}[\/latex]. Draw the graph.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  Find: [latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}[\/latex]<\/p>\n

                  [latex]\\displaystyle{P}{(\\overline{{x}}{>}{20})}={0.79199}[\/latex] using normalcdf<\/code> [latex]\\displaystyle{({20},{1}\\text{E99},{22},\\frac{{22}}{\\sqrt{{80}}})}[\/latex]<\/p>\n

                  The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.<\/p>\n

                  \"This<\/p>\n

                   <\/p>\n

                  Remember<\/strong>, 1E99 = 1099<\/sup> and \u20131E99 = \u20131099<\/sup>. Press the EE<\/code> key for E. Or just use 1099<\/sup> instead of 1E99.<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  2. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer’s excess time is longer than 20 minutes. This is asking us to find P<\/em>(x<\/em> > 20).<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  Find P<\/em>(x > 20). Remember to use the exponential distribution for an individual: [latex]\\displaystyle{X}\\text{~}{E}{x}{p}{(\\frac{{1}}{{22}})}[\/latex].[latex]\\displaystyle{P}{({x}{>}{20})}={e}^{{{(-{({\\frac{1}{22}})}{({20})})}}}{\\quad\\text{or}\\quad}{e}^{{{(-{0.04545}{({20})})}}}={0.4029}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  3. Explain why the probabilities in parts 1 and 2 are different.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  [latex]\\displaystyle{P}{({x}{>}{20})}={0.4029}\\text{ but } {P}{(\\overline{{x}}{>}{20})}={0.7919}[\/latex].<\/p>\n

                  The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.<\/p>\n

                  When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the CLT. Use the CLT with the normal distribution when you are being asked to find the probability for a mean.<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  Using the CLT to find percentiles:<\/strong><\/p>\n

                  4. Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances. Draw a graph.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  Let k<\/em> = the 95th percentile. Find k<\/em> where [latex]\\displaystyle{P}{(\\overline{{x}}{<}{k})}={0.95}[\/latex]\nk<\/em> = 26.0 using invNorm<\/code>[latex]\\left ( 0.95,22,\\frac{22}{\\sqrt {80}} \\right )[\/latex] = 26.0
                  \n\"This<\/p>\n

                  The 95th percentile for the sample mean excess time used<\/strong> is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.<\/p>\n

                  Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                  <\/h3>\n
                  \n

                  Try it 2<\/h3>\n

                  Use the information in Example 2, but change the sample size to 144.<\/p>\n

                  1. Find [latex]\\displaystyle{P}{({20}{<}\\overline{{x}}{<}{30})}[\/latex].\n\n\n

                  Show Answer<\/span><\/p>\n
                  \n

                  0.8623<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  2. Find P<\/em>(\u03a3x<\/em> is at least 3,000).<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  0.7377<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  3. Find the 75th percentile for the sample mean excess time of [latex]144[\/latex] customers.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  23.2<\/p>\n<\/div>\n<\/div>\n

                   <\/p>\n

                  4. Find the 85th percentile for the sum of [latex]144[\/latex] excess times used by customers.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  3,441.6<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                  <\/h3>\n
                  \n

                  Example 3<\/h3>\n

                  In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.<\/p>\n

                  1. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.<\/p>\n

                  Show Answer<\/span><\/p>\n
                  \n

                  We have, [latex]\\displaystyle{\\mu}_{x}={\\mu}=[\/latex] 2 and [latex]{\\sigma}_{x}=\\frac{{\\sigma}}{{\\sqrt{n}}}=\\frac{{0.5}}{{10}}=0.05[\/latex]. Therefore:<\/p>\n

                    \n
                  1. 50th percentile = \u03bcx<\/sub><\/em> = \u03bc<\/em> = 2<\/li>\n
                  2. 25th percentile =\u00a0invNorm<\/code>(0.25,2,0.05) = 1.97<\/li>\n
                  3. 75th percentile =\u00a0invNorm<\/code>(0.75,2,0.05) = 2.03<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                     <\/p>\n

                    2. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.<\/p>\n

                    Show Answer<\/span><\/p>\n
                    \n

                    We have [latex]\\displaystyle{\\mu}_{\\sum{x}}=n({\\mu}_{x})={100}({2})= 200[\/latex] and [latex]{\\sigma}_{\\mu{x}}={\\sqrt n}({\\sigma}_{x})=10(0.5)=5[\/latex].\u00a0Therefore:<\/p>\n

                      \n
                    1. 50th percentile =\u00a0[latex]\\displaystyle{\\mu}_{\\sum{x}}=n({\\mu}_{x})[\/latex] = 100(2) = 200<\/li>\n
                    2. 25th percentile =\u00a0invNorm<\/code>(0.25,200,0.05) = 196.63<\/li>\n
                    3. 75th percentile =\u00a0invNorm<\/code><\/i>(0.75,200,0.05) = 203.37<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      3. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      [latex]{P}{(1.75 < {\\overline{x}} < {1.85})}[\/latex] =\u00a0normalcdf<\/code>(1.75,1.85,2,0.05) = 0.0013<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      4. Find the value that is two standard deviations above the sample mean.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      Using the z<\/em>-score equation, [latex]\\displaystyle{z}=\\frac{{\\overline{x}-{\\mu}_{\\overline{x}}}}{{{\\sigma}_{\\overline{x}}}}[\/latex].<\/p>\n

                      Solving for x,\u00a0we have x<\/em> = 2(0.05) + 2 = 2.1<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      5. Find the IQR<\/em> for the sum of the sample times.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      The IQR<\/em> is 75th percentile \u2013 25th percentile = 203.37 \u2013 196.63 = 6.74<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                      <\/h3>\n
                      \n

                      try it 3<\/h3>\n

                      Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.<\/p>\n

                      1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>(x > 120) =\u00a0normalcdf<\/code>(120, 1E99, 114.8, 13.1) [latex]\\approx[\/latex] 0.3457. There is about a 35% chance that the randomly selected woman will have a systolic blood pressure greater than 120.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>( [latex]\\overline{x}[\/latex] > 120) = normalcdf<\/code>(120, 1E99, 114.8, [latex]\\dfrac{13.1}{\\sqrt{40}})\\approx[\/latex] 0.006.
                      \nThere is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      3. If the sample were four women between the ages of [latex]18[\/latex] to [latex]24[\/latex] and we did not know the original distribution, could the central limit theorem be used?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      The central limit theorem could not be used if the sample size were four and we did not know the original distribution was normal. The sample size would be too small.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                      <\/h3>\n
                      \n

                      Example 4<\/h3>\n

                      A study was done about violence against sex workers and the symptoms of the post-traumatic stress that they developed. The age range of the sex workers was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.<\/p>\n

                      1. In a sample of 25 sex workers, what is the probability that the mean age of the sex workers is less than 35?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>([latex]\\overline{{x}}[\/latex] < 35) = normalcdf<\/code>(-E<\/em>99,35,30.9,1.8) = 0.9886<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      2. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>([latex]\\overline{{x}}[\/latex] > 50) = normalcdf<\/code>(50, E<\/em>99,30.9,1.8) \u2248 0.<\/p>\n

                      For this sample group, it is almost impossible for the group’s average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      3. In a sample of 49 sex workers, what is the probability that the sum of the ages is no less than 1,600?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>(\u03a3x<\/em> \u2265 1,600) = normalcdf<\/code>(1600,E99,1514.10,63) = 0.0864<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      4. Is it likely that the sum of the ages of the 49 sex workers is at most 1,595? Interpret the results.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      P<\/em>(\u03a3x<\/em> \u2264 1,595) = normalcdf<\/code>(-E99,1595,1514.10,63) = 0.9005.<\/p>\n

                      This means that there is a 90% chance that the sum of the ages for the sample group n<\/em> = 49 is at most 1,595.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      5. Find the 95th percentile for the sample mean age of 65 sex workers. Interpret the results.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      The 95th percentile = invNorm<\/code>(0.95,30.9,1.1) = 32.7.<\/p>\n

                      This indicates that 95% of the sex workers in the sample of 65 are younger than 32.7 years, on average.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      6. Find the 90th percentile for the sum of the ages of 65 sex workers. Interpret the results.<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      The 90th percentile = invNorm<\/code>(0.90,2008.5,72.56) = 2101.5.<\/p>\n

                      This indicates that 90% of the sex workers in the sample of 65 have a sum of ages less than 2,101.5 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                      <\/h3>\n
                      \n

                      try it 4<\/h3>\n

                      According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.<\/p>\n

                      1. What mean doorway height would allow 95% of men to enter the aircraft without bending?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      We know that \u03bcx<\/sub><\/em> = \u03bc<\/em> = 69 and we have \u03c3x<\/sub><\/em> = 2.8. The height of the doorway is found to be\u00a0invNorm<\/code>(0.95,69,2.8) = 73.61.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      We know that \u03bcx<\/sub><\/em> = \u03bc<\/em> = 69 and we have \u03c3x<\/sub><\/em> = 0.28. So,\u00a0invNorm<\/code>(0.95,69,0.28) = 69.49.<\/p>\n<\/div>\n<\/div>\n

                       <\/p>\n

                      3. For engineers designing the 757, which result is more relevant: the height from part 1 or part 2? Why?<\/p>\n

                      Show Answer<\/span><\/p>\n
                      \n

                      When designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                      Historical Note: Normal Approximation to the Binomial<\/h2>\n

                      Historically, being able to compute binomial probabilities was one of the most important applications of the central limit theorem. Binomial probabilities with a small value for n\u00a0<\/em>(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n<\/em>, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial<\/strong> distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution<\/strong>:<\/p>\n