Answer to Question #201021 in Statistics and Probability for Rameen

Question #201021

1. Calculate the first four moments about the mean from the following frequency distribution. 

      Also calculate the Moment Ratios b1 and b2.                         

Groups

2------4

4-------6

6------8

8-----10

10----12

12----14

14----16

16----18

18----20


Frequency

18

24

47

80

102

66

40

21

15


                              


1
Expert's answer
2021-06-01T09:07:14-0400

Given data is:






Fist four moments about mean:

μ1=1nxifi=4445413=10.762\mu_1'=\dfrac{1}{n}\sum x_if_i=\dfrac{4445}{413}=10.762


μ2=1nxi2fi=48110413=116.489\mu_2'=\dfrac{1}{n}\sum x_i^2f_i=\dfrac{48110}{413}=116.489


μ3=1nxi3fi=699749413=1694.30\mu_3'=\dfrac{1}{n}\sum x_i^3f_i=\dfrac{699749}{413}=1694.30


μ4=1nxi4fi=9775376413=23669.19\mu_4'=\dfrac{1}{n}\sum x_i^4f_i=\dfrac{9775376}{413}=23669.19



Now,

μ1=μ1=10.762 μ2=μ2μ1=105.727 μ3=μ33μ2μ1+2μ13=1694.303760.96+2492.92=426.26\mu_1=\mu_1'=10.762\\\ \\\mu_2=\mu_2'-\mu_1'=105.727\\\ \\\mu_3=\mu_3'-3\mu_2'\mu_1'+2\mu_1'^3=1694.30-3760.96+2492.92=426.26


μ4=μ44μ3μ1+6μ2μ123μ14=23669.1972936.2264+80950.983739.38μ4=27944.563\mu_4=\mu_4'-4\mu_3'\mu_1'+6\mu_2'\mu_1'^2-3\mu_1'^4=23669.19-72936.2264+80950.98-3739.38\\\mu_4=27944.563


Now,


b1=μ32μ23=(426.26)2(105.727)3=181697.581181837.39=0.1537b_1=\dfrac{\mu_3^2}{\mu_2^3}=\dfrac{(426.26)^2}{(105.727)^3}=\dfrac{181697.58}{1181837.39}=0.1537



b2=μ4μ22=27944.56311178.19853=2.499b_2=\dfrac{\mu_4}{\mu_2^2}=\dfrac{27944.563}{11178.19853}=2.499


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