31+k+61+41=1⇒k=41μ=3α+41−43μ=3α−21σ=(α231+41+941)−(3α−21)2σ=92α2+3α+49σ=μ+2σ2=(μ+2)2⇒92α2+3α+49=9α2+49+α9α2−32α=0α=0,( reject ) or α=6(∵x=0 is already given) ⇒σ+μ=2μ+2=5
Let the mean and the standard deviation of the probability distribution $\begin{array}{|c|c|c|c|c|}
\hline \mathrm{X} & \mathrm{\alpha} & \mathrm{1} & 0 & -3 \
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{3} & \mathrm{~K} & \frac{1}{6} & \frac{1}{4} \
\hline
\end{array}be\muand\sigma,respectively.If\sigma-\mu=2,then\sigma+\mu$ is equal to________
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