XP(X)40.460.1100.5\def\arraystretch{1.5} \begin{array}{c:c} X & P(X) \\ \hline 4 & 0.4 \\ \hdashline 6 & 0.1 \\ \hdashline 10 & 0.5 \end{array}X4610P(X)0.40.10.5
μ=∑i=13p(xi)xi=0.4×4+0.1×6+0.5×10=7.2\mu =\sum\limits ^3_{i=1}p(x_i)x_i=0.4\times 4+0.1\times 6+0.5\times 10=7.2μ=i=1∑3p(xi)xi=0.4×4+0.1×6+0.5×10=7.2
σ2=∑i=13p(xi)(xi−μ)2=0.4×(−3.2)2+0.1×(−1.2)2+0.5×2.82=8.16\sigma ^2= \sum\limits ^3_{i=1}p(x_i)(x_i-\mu)^2=0.4\times (-3.2)^2+0.1\times (-1.2)^2+0.5\times 2.8^2=8.16σ2=i=1∑3p(xi)(xi−μ)2=0.4×(−3.2)2+0.1×(−1.2)2+0.5×2.82=8.16
Answer: mean = 7.2, variance = 8.16.
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