Let X be a r.v. has the probability function given by: Find the first four centeral moments

μ_1=E(X)=∑_x〖xP(x)〗=1.9
μ_2=E(X^2 )=∑_x〖x^2 P(x)〗=4
μ_3=E(X^3 )=∑_x〖x^3 P(x)〗=9.4
μ_4=E(X^4 )=∑_x〖x^4 P(x)〗=23.3
∴m_1=0
m_2=μ_2-μ_1^2=4-(1.9)^2=0.39
m_3=μ_3-3μ_1 μ_2+2μ_1^3=9.4-3(1.9)(4)+2(1.9)^3=0.318
m_4=μ_4-4μ_1 μ_3+6μ_1^2 μ_2-3μ_1^4
=23.3-4(1.9)(9.4)+6(1.9)^2 (4)-3(1.9)^4=-0.596