The first four raw moments of distribution are 2, 136, 320, and 40,000, The coefficient of skewness is:

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SSC CGL Tier-II ( JSO ) 2019 Official Paper ( Held On : 17 Nov 2020 )
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  1. \(\dfrac{40656}{(132)^2}\)
  2. \(\dfrac{(40656)^2}{(132)^3}\)
  3. \(\dfrac{-480}{(132)^2}\)
  4. \(\dfrac{(-480)^2}{(132)^3}\)

Answer (Detailed Solution Below)

Option 4 : \(\dfrac{(-480)^2}{(132)^3}\)
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Detailed Solution

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Given

μ'1, μ2, μ'3, μ'4 are raw moments

μ’1 = 2

μ’2 = 136

μ’3 = 320

μ’4 = 40,000

Calculation

We have to find the first our central moments

μ1 = μ’1 = 2

μ2 = μ’2 – (μ’1)2

⇒ 136 – (2)2

136 – 4

⇒ 132

μ3 = μ’3 – 3 × μ’2μ’1 + 2 × (μ’1)3

⇒ 320 – 3 × 136 × 2 + 2 × (2)3

⇒ 320 – 816 + 16

⇒ - 480

μ4 = μ’4 – 4 × μ’1 × μ’3 + 6 × μ’2 × (μ’1)2 – 3 × (μ’1)4

⇒ 40000 – 4 × 2 × 320 + 6 × 136 × 22 – 3 × 24

⇒ 40000 – 2560 + 3264 – 48

⇒ 40656

The coefficient of skewness  denoted by β1

⇒ β1 = μ2332

⇒ (-480)2/(132)3

∴ The coefficient of skewness is (-480)2/(132)3

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