230. A discrete random variable $X$ is given by the distribution law:

$$
\begin{array}{lccc}
X & 1 & 2 & 4 \\
p & 0.1 & 0.3 & 0.6
\end{array}
$$

Find the central moments of the first, second, third, and fourth orders.
Solution. The first-order central moment is zero: $\boldsymbol{\mu}_{i}=\mathbf{0}$.

For the calculation of central moments, it is convenient to use formulas expressing central moments in terms of initial moments, so we will first find the initial moments:

$$
\begin{aligned}
& v_{1}=M(X)=1 \cdot 0.1+2 \cdot 0.3+4 \cdot 0.6=3.1 ; \\
& v_{2}=M\left(X^{2}\right)=1 \cdot 0.1+4 \cdot 0.3+16 \cdot 0.6=10.9 \\
& v_{3}=M\left(X^{3}\right)=1 \cdot 0.1+8 \cdot 0.3+64 \cdot 0.6=40.9 \\
& v_{4}=M\left(X^{4}\right)=1 \cdot 0.1+16 \cdot 0.3+256 \cdot 0.6=158.5 .
\end{aligned}
$$

Let's find the central moments:

$$
\begin{gathered}
\mu_{2}=v_{i}-v_{1}^{2}=10.9-3.1^{2}=1.29 ; \\
\mu_{3}=v_{3}-3 v_{1} v_{2}+2 v_{1}^{3}=40.9-3 \cdot 3.1 \cdot 10.9+2 \cdot 3.1^{3}=-0.888 ; \\
\mu_{4}=v_{4}-4 v_{3} v_{1}+6 v_{2} v_{1}^{2}-3 v_{1}^{4}= \\
=158.5-4 \cdot 40.9 \cdot 3.1+6 \cdot 10.9 \cdot 3.1^{2}-3 \cdot 3.1^{4}=2.7777
\end{gathered}
$$