Statistical Techniques for Transportation Engineering

Again we have

$\begin{array}{ll}{\mu \prime }_3\hfill & =\sum _{x=0}^n{x}^{3}p(x)\hfill \\ =\sum _{x=0}^n[x(x-1)(x-2)+3x(x-1)+x]p(x)\hfill \\ (\mathrm{since}{x}^3=x(x-1)(x–2)+3x(x–1)+x)\hfill \\ =\sum _{x=0}^n[x(x-1)(x-2)+3\sum _{x=0}^{n}x(x-1)+p(x)+\sum _{x=0}^{n}xP(x)\hfill \\ =n(n–1)(n–2)p^3{(q+p)}^{n–3}+3n(n–1)p^2{(q+p)}^{n–2}+np{(q+p)}^{n–1}\hfill \\ =n(n–1)(n–2)p^3+3n(n–1)p^2+np\hfill \end{array}$

Since ${\mu }_3={\mu \prime }_3-3{\mu \prime }_2{\mu \prime }_1+2{\mu }_1^{\prime 3}$

We get

$\begin{array}{ll}{\mu }_3\hfill & =n(n–1)(n–2)p^3+3n(n–1)p^2+np–3(n^2{p}^2–n{p}^2+np)(np)+2n^3{p}^3\hfill \\ =npq(q–p)\hfill \end{array}$

Similarly we get

${\mu \prime }_4=n(n-1)(n-2)$