Question

10. (a) For a distribution, the mean is 10, variance is 16, the skewness 4

sk is +1 and kurtosis 

b2

 is 4. Obtain the first four moments about the origin i.e. zero. Comment upon the 

nature of the distribution.

Answer 1

Mean $= m_1′ =10$ , variance =$m_2 =16$ , coefficient of skewness $= \gamma_1 = +1$ , coefficient of kurtosis is $\beta_2 = 4.$

First moment about the origin is

Mean $= m_1′ = 10$ .

Next, Variance $= 𝑚_2 = 16 =\dfrac{∑ x}{n}-(\dfrac{\sum x}{n})^2$

Second moment about the origin is $m_2=\dfrac{\sum x^2}{n}=m_2+\mu_1^2=Variance+(\dfrac{\sum x}{n})^2=16+10^2=116$

Coefficient of skewness is +1:

$\gamma1 = +1 =\dfrac{𝑚_3}{𝑚_2^{\frac{3}{2}}}\rightarrow 𝑚_3 = 𝑚_2^{\frac{3}{2}} = 16^{\frac{3}{2}} = 64$

    

Coefficient of kurtosis is 4:

$\beta_2 = 4\\ \dfrac{𝑚_4}{𝑚_2^2} = 4 \rightarrow 𝑚_4 = 4(16)^2 = 1024$

Third moment about the origin is

$𝑚_3′ = 𝑚_3 + 3𝑚_2𝑚_1′ + 𝑚_1′ ^3 = 64 + 3(16)(10) + 10^3 = 𝟏𝟓𝟒𝟒$

Fourth moment about the origin is

$𝑚_4′ = 𝑚_4 + 4𝑚_3𝑚_1′ + 6𝑚_2𝑚_2′ + 𝑚_1′ ^4 = 1024 + 4(64)(10) + 6(16)102 + 10^4 = 𝟐𝟑𝟏𝟖𝟒$

It is a positively skewed distribution.