I have computed the statistics of both groups of students in a class, in which they are male and female students.How do you sort the data within each group to determine what proportion in each group is within one standard deviation of that group, and what is the sorted data look like base on the statistics? Are the proportions similar? a. For male studentsInterquartile rangeIQR = Qs – QIMean= 1.58 – 1.36 = 0.2220.41.457143N14GWA (x)dd?(Deviation from themean)-0.39714286MedianMed =1.42+1.451.4351.061.12360.1577221.331.7689-0.127142860.016165Minimum and maximum values1.341.361.7956-0.117142860.013722Min (x) = 1.06Маx (х) — 1.751.8496-0.097142860.0094371.381.9044-0.077142860.0059511.422.0164-0.037142860.001381.422.0164-0.037142860.001381.452.1025-0.007142860.0000511.522.31040.0628571430.0039511.562.43360.1028571430.010581.582.49640.1228571430.0150941.602.560.1428571430.0204081.632.65690.1728571430.02988First and third quartiles1.753.06250.2928571430.085765Q; = () • N• 14 = 3.5 or 4th quartileEx = 20.4, N= 14Ex = 30.0972Ed = -0.04Ed? = 0.371486Standard deviation and varianceQ=1.36Q = () .* N0.371486Population variance - g? = Ed²= 0.026514• 14 = 10.5 or 11th quartileQ = 1.58Population standard deviation - o = Vo? = V0.0265 = 0.1628• RangeRange = highest value – lowest value0.3716Sample variance - s² =N-10.028614-1= 1.75 – 1.06Sample standard deviation - s =V0.0286 = 0.1691= 0.69 GWA (x)xdd?b. For female students(Deviation from themean)• Mean1.48840.0830161.22-0.288125Ex1.241.5376-0.26812524.130.071891=1.508125161.271.6129-0.2381250.0567041.432.0449-0.0781250.006104• Median1.452.1025-0.0581250.0033791.492.2201-0.0181250.0003291.54+1.56Med1.551.492.2201-0.0181250.00032921.542.37160.0318750.0010161.562.43360.0518750.002691Minimum and maximum values1.562.43360.0518750.002691Min (x) = 1.22Маx (х) — 1.781.582.49640.0718750.0051661.592.52810.0818750.0067041.591.642.52810.0818750.0067042.68960.1318750.017391First and third quartiles1.702.890.1918750.036816Q1 = () • N1.783.16840.2718750.073916Ex = 24.13, N = 16Ex = 36.7659Ed = -0.007Ed = 0.374847| • 16 = 4th quartileQ1 = 1.43Q3 = )* N• 16 = 12th quartile= 1.59• Standard deviation and varianceRangeRange = highest value – lowest value= 1.78 – 1.22Population variance - g² = Ed*0.374847= 0.023416= 0.56Population standard deviation - o =0.0234 = 0.1530Σα"0.374847• Interquartile rangeSample variance - s² =0.0250N-115-1IQR = Qs – Q1= 1.59 – 1.44 = 0.16Sample standard deviation -s = Vs2 = V0.0249 = 0.1578%3D||

The proportions are not same for the sample that are within one standard deviation of the mean. For…

I have computed the statistics of both groups of students in a class, in which they are male and female students. How do you sort the data within each group to determine what proportion in each group is within one standard deviation of that group, and what is the sorted data look like base on the statistics? Are the proportions similar?

I have computed the statistics of both groups of students in a class, in which they are male and female students. How do you sort the data within each group to determine what proportion in each group is within one standard deviation of that group, and what is the sorted data look like base on the statistics? Are the proportions similar?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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