| X | $\alpha$ | 1 | 0 | - 3 |
| $\mathrm { P } ( \mathrm { X } )$ | $\frac { 1 } { 3 }$ | K | $\frac { 1 } { 6 }$ | $\frac { 1 } { 4 }$ |
Let the mean and the standard deviation of the probability distribution
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\begin{tabular}{ | c | c | c | c | c | }
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X & $\alpha$ & 1 & 0 & - 3 \\
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$\mathrm { P } ( \mathrm { X } )$ & $\frac { 1 } { 3 }$ & K & $\frac { 1 } { 6 }$ & $\frac { 1 } { 4 }$ \\
\hline
\end{tabular}
\end{center}
be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to $\_\_\_\_$