This question was previously asked in

SSC CGL Tier-II ( JSO ) 2018 Official Paper ( Held On : 14 Sept 2019 )

- -0.20
- 0.20
- 1
- -1

Option 1 : -0.20

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298865

10 Questions
10 Marks
7 Mins

**Given**

Mean = x̅ = 25

Median = M_{d} = 26

Standard devation = σ = 5

**Formula**

Karl pearson coefficient of skewness = SK_{p} = (Mean – Mode)/Standard deviation

**Calculation**

SK_{P} = (25 – 26)/5

⇒ -1/5 = -0.2

**∴**** The karl pearson coefficient of skewness is -0.2**

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