UA MATH 115B - Sample Mean - D827969

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UA MATH 115B - Sample Mean

Course Math 115b- Business Mathematics II

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Ekstrom Math 115b Mathematics for Business Decisions, part IISample MeanMath 115bEkstrom Math 115b Sample MeanSamples give information about random variablesNote that and Also note that and Sample Mean xXE  2sXV   nXVxV    XExE Ekstrom Math 115b Sample MeanEx. If X is a finite R.V. that only assumes the values 0 and 1 with and , find and .Soln: Sample Mean 2.00 XP 8.01 XP XE XV     8.08.012.001100 XPXPXEEkstrom Math 115b Sample Mean Sample Mean Soln: The Excel file Sample Means.xls shows sets of 1000 samples taken, verifying that samples closely predict the behavior of the random variable              16.08.08.012.08.0011002222 XPXEXPXEXVEkstrom Math 115b Sample Mean Sample MeanThe shape of the approximate p.m.f. graphs change slightly for different sample sizesThe differences can be removed by plotting the standardizationThe graph of the standardized values appear to be approximately as followsEkstrom Math 115b Sample Mean Sample Mean: Finite R.V.Ekstrom Math 115b Sample Mean Sample Mean: Finite R.V.Ex. Let X be a finite R.V. that only assumes the values 0 and 1 with and . Consider the case when samples of size 4 are taken. Use BINOMDIST to find the values of the p.m.f. for B where B is the finite random variable that counts the number of 1’s. 2.00 XP 8.01 XPEkstrom Math 115b Sample Mean Sample MeanSoln: Note that the possible outcomes are:{(0000), (0001), (0010), (0011), (0100), (0101), (0110), (0111), (1000), (1001), (1010), (1011), (1100), (1101), (1110), (1111)}b0 0.00161 0.02562 0.15363 0.40964 0.4096 bfBEkstrom Math 115b Sample MeanEx. Find the variance and standard deviation of the finite random variable B from the previous example Sample Meanb0 0.0016 -3.2 10.24 0.0163841 0.0256 -2.2 4.84 0.1239042 0.1536 -1.2 1.44 0.2211843 0.4096 -0.2 0.04 0.0163844 0.4096 0.8 0.64 0.262144 bfB Bb 2Bb   bfbBB2Ekstrom Math 115b Sample MeanSo, the mean is 3.2, variance is 0.64, and standard deviation is 0.8 Sample Meanb0 0.0016 -3.2 10.24 0.0163841 0.0256 -2.2 4.84 0.1239042 0.1536 -1.2 1.44 0.2211843 0.4096 -0.2 0.04 0.0163844 0.4096 0.8 0.64 0.262144 bfB Bb 2Bb   bfbBB2Ekstrom Math 115b Sample MeanEx. Let S be the standardization of the finite random variable B from the previous example. Find the mean and standard deviation of S.Note that where , This means that the b-values must be adjusted Sample MeanBBBS2.3B8.0BEkstrom Math 115b Sample MeanEx. Let S be the standardization of the finite random variable B from the previous example. Find the mean and standard deviation of S. Sample Means-4 0.0016 -4 16 0.0256-2.75 0.0256 -2.75 7.5625 0.1936-1.5 0.1536 -1.5 2.25 0.3456-0.25 0.4096 -0.25 0.0625 0.02561 0.4096 1 1 0.4096 sfS Ss 2Ss   sfsSS2Ekstrom Math 115b Sample MeanSo, the mean is 0 and the standard deviation is 1.(This MUST be true for all standardized variables) Sample Means-4 0.0016 -4 16 0.0256-2.75 0.0256 -2.75 7.5625 0.1936-1.5 0.1536 -1.5 2.25 0.3456-0.25 0.4096 -0.25 0.0625 0.02561 0.4096 1 1 0.4096 sfS Ss 2Ss   sfsSS2Ekstrom Math 115b Sample MeanEx. Let X be a finite R.V. that only assumes the values 0 and 1 with and . Consider the case when samples of size 4 are taken. Use BINOMDIST to find the values of the p.m.f. for Sample Mean 2.00 XP 8.01 XPxEkstrom Math 115b Sample Mean Sample MeanSoln: We are finding the AVERAGE value in each sampleNote that the possible outcomes are:{(0000), (0001), (0010), (0011), (0100), (0101), (0110), (0111), (1000), (1001), (1010), (1011), (1100), (1101), (1110), (1111)}0.00 0.00160.25 0.02560.50 0.15360.75 0.40961.00 0.4096 xfXxEkstrom Math 115b Sample MeanEx. Standardize the values for from the previous example and find the mean and standard deviation of the standardized values. Sample MeanxxEkstrom Math 115b Sample MeanTo standardize the values, we must first find the mean and standard deviationThe mean is 0.8 and the standard deviation is 0.2 Sample Meanx0.00 0.0016 -0.80 0.64 0.0010240.25 0.0256 -0.55 0.3025 0.0077440.50 0.1536 -0.30 0.09 0.0138240.75 0.4096 -0.05 0.0025 0.0010241.00 0.4096 0.20 0.04 0.016384 xfX Xx 2Xx   xfxXX2Ekstrom Math 115b Sample MeanSo, the mean is 0 and the standard deviation is 1.(This MUST be true for all standardized variables) Sample Mean: Continuous R.V.s-4 0.0016 -4 16 0.0256-2.75 0.0256 -2.75 7.5625 0.1936-1.5 0.1536 -1.5 2.25 0.3456-0.25 0.4096 -0.25 0.0625 0.02561 0.4096 1 1 0.4096 sfS Ss 2Ss  

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UA MATH 115B - Sample Mean

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