Complete Question
The complete question is shown on the first uploaded image
Answer:
The standard deviation is
Explanation:
From the given data we can compute the expected mean for each random values as follows
![E(X) = \sum [ X * P(X = x )]\\\\ X \ \ \ \ \ \ X* P(X =x )\\ 0 \ \ \ \ \ \ \ \ \ \ 0* 0.4 = 0 \\ 2 \ \ \ \ \ \ \ \ \ \ 2 * 0.3 = 0.6 \\ 4 \ \ \ \ \ \ \ \ \ \ 4 * 0.2 = 0.8\\ 6 \ \ \ \ \ \ \ \ \ \ 6* 0.1 = 0.6](https://img.qammunity.org/2021/formulas/mathematics/college/owgy1x5kxmx8gscxndaqlzse8h6tv66hhk.png)
So


The
![E(X^2) = \sum [ X^2 * P(X = x )]\\\\ X \ \ \ \ \ \ \ \ \ \ X^2 * P(X=x ) \\ 0 \ \ \ \ \ \ \ \ \ \ 0^2 * 0.4 = 0 \\ 2 \ \ \ \ \ \ \ \ \ \ 2^2 * 0.3 = 12 \\ 4 \ \ \ \ \ \ \ \ \ \ 4^2 * 0.2 = 3.2 \\ 6 \ \ \ \ \ \ \ \ \ \ 6^2 * 0.1 = 3.6](https://img.qammunity.org/2021/formulas/mathematics/college/iiboinccx003cmptt7ukemcu8ipipu34qy.png)
So


Now the variance is mathematically evaluated as
![Var (X) = E(X^2 ) -[E(X]^2](https://img.qammunity.org/2021/formulas/mathematics/college/r49gei129nww8f05zy2wf1hl63maide56h.png)
Substituting value


The standard deviation is mathematically evaluated as


