Question

The first four moments of a frequency distribution about the valuo sare 1, 22, !17 and 560 Determine the corresponding movements ☺ about the mean and cli) about zero. -0.55, 4.46,-0.43 and 68.58 fing B, and ß2.​

Answer 1

Step-by-step explanation:

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Given moments about an arbitary origin 5.

μ1' = -4 μ 2' = 22 μ 3' = - 117 μ4'=560

Moments about mean:

μ2 = μ2' - (μ1')^2

μ1'μ2' + 2(μ1')^3

μ3=μ3' - 32

μ4 = μ4' - 4μ1'μ3' + 6μ2'(μ1)^2 - 3(μ1')^4

substituting the values,

μ2 = 22 - (-4)^2 = 22-16=6

μ3 = -117-3(-4)(22)+2(-4)^3=-117+264-128=19

μ4 = 560-4(-4)(117)+6(22)(-4)^2-3(-4)^4

=560-1872+2112-768=32

Moments about mean are μ1=0,μ2 = 6,μ3=19,μ4=32

Moments about zero::

Let the moments about zero be denoted by v1,v2,v3,v4

First moment about zero = v1 or mean

Second moment about zero= v3=μ1 + (v1)^2

Third moment about zero = v2=μ2 + 3v1v2 - 2v1^2

fourth moment about zero = v4 = μ4 + 4v1v2 - 6(v1)^2v2 + 3v1^4

Hope it helps!!!!!