Applied Mathematics for Electrical Engineering (3130908)

Question-5c ( i )

BE | Semester-3 Winter-2019 | 26-11-2019

Q5) (c ( i ))

Attempt the following. In a distribution of two different groups the variances are 15 and 27. Whereas the third central moments are 32.4 and 67.56 respectively. Compare the skewness of two groups.

We know that,

Skewness =

β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}}

Here, for Group 1

μ_{2} = 15, μ_{3} = 32.4

Skewness =

β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = \frac{{(32.4)}^{2}}{{(15)}^{3}} = 0.31

Here, for Group 1

μ_{2} = 27, μ_{3} = 67.56

Skewness =

β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = \frac{{(67.56)}^{2}}{{(27)}^{3}} = 0.23

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