The first four moments of a distribution about the value 3 are -2,10,-25,50.
Calculate moment about mean, moment about origin, moment about the value 5

Solution:
The first four moments of a distribution about the value 3 are
`M_1=-2`

`M_2=10`

`M_3=-25`

`M_4=50`

Find Central moments using Moments about the value 3

First Central Moment
`m_1=0`

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Second Central Moment
`m_2=M_2-M_1^2`

`=10-(-2)^2`

`=10-4`

`=6`

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Third Central Moment
`m_3=M_3-3M_2M_1+2M_1^3`

`=(-25)-3*10*(-2)+2*(-2)^3`

`=(-25)+60-16`

`=19`

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Fourth Central Moment
`m_4=M_4-4M_3M_1+6M_2M_1^2-3M_1^4`

`=50-4*(-25)*(-2)+6*10*(-2)^2-3*(-2)^4`

`=50-200+240-48`

`=42`

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Now, find `bar x`

`M_1=bar x-A`

`-2=bar x-3`

`bar x=-2+3`

`bar x=1`

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Find Moment about origin using Central moments

First Moment about origin
`v_1=bar x`

`v_1=1`

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Second Moment about origin
`v_2=m_2+v_1^2`

`=6+1^2`

`=6+1`

`=7`

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Third Moment about origin
`v_3=m_3+3m_2v_1+v_1^3`

`=19+3*6*1+1^3`

`=19+18+1`

`=38`

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Fourth Moment about origin
`v_4=m_4+4m_3v_1+6m_2v_1^2+v_1^4`

`=42+4*19*1+6*6*1^2+1^4`

`=42+76+36+1`

`=155`

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Find Moment about the value 5 using Central moments

First Moment about the value 5
`M_1=bar x-A`

`=1-5`

`=-4`

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Second Moment about the value 5
`M_2=m_2+M_1^2`

`=6+(-4)^2`

`=6+16`

`=22`

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Third Moment about the value 5
`M_3=m_3+3m_2M_1+M_1^3`

`=19+3*6*(-4)+(-4)^3`

`=19-72-64`

`=-117`

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Fourth Moment about the value 5
`M_4=m_4+4m_3M_1+6m_2M_1^2+M_1^4`

`=42+4*19*(-4)+6*6*(-4)^2+(-4)^4`

`=42-304+576+256`

`=570`

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