Again we have

$\begin{array}{ll}{\mu \prime }_{3}\hfill & =\sum _{x=0}^{n}{x}^{3}p\left(x\right)\hfill \\ =\sum _{x=0}^{n}\left[x\left(x-1\right)\left(x-2\right)+3x\left(x-1\right)+x\right]p\left(x\right)\hfill \\ \left(\mathrm{since}{x}^{3}=x\left(x-1\right)\left(x–2\right)+3x\left(x–1\right)+x\right)\hfill \\ =\sum _{x=0}^{n}\left[x\left(x-1\right)\left(x-2\right)+3\sum _{x=0}^{n}x\left(x-1\right)+p\left(x\right)+\sum _{x=0}^{n}xP\left(x\right)\hfill \\ =n\left(n–1\right)\left(n–2\right){p}^{3}{\left(q+p\right)}^{n–3}+3n\left(n–1\right){p}^{2}{\left(q+p\right)}^{n–2}+np{\left(q+p\right)}^{n–1}\hfill \\ =n\left(n–1\right)\left(n–2\right){p}^{3}+3n\left(n–1\right){p}^{2}+np\hfill \end{array}$

Since ${\mu }_{3}={\mu \prime }_{3}-3{\mu \prime }_{2}{\mu \prime }_{1}+2{\mu }_{1}^{\prime 3}$

We get

$\begin{array}{ll}{\mu }_{3}\hfill & =n\left(n–1\right)\left(n–2\right){p}^{3}+3n\left(n–1\right){p}^{2}+np–3\left({n}^{2}{p}^{2}–n{p}^{2}+np\right)\left(np\right)+2{n}^{3}{p}^{3}\hfill \\ =npq\left(q–p\right)\hfill \end{array}$

Similarly we get

${\mu \prime }_{4}=n\left(n-1\right)\left(n-2\right)$

Get Statistical Techniques for Transportation Engineering now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.