Answer
Standard deviation = 1.2083
Step-by-step explanation
[tex]\begin{gathered} \text{Mean = }\sum ^{}_{}xi\cdot\text{ p(xi)} \\ \text{Mean = }0\cdot\text{ 0.2 + 1 }\cdot\text{ }0.05\text{ + 2}\cdot\text{ }0.1\text{ + 3 }\cdot\text{ 0.65} \\ \text{Mean = 0 + 0.05 + 0.2 + 1.95} \\ \text{Mean = 2.2} \\ \text{Standard deviation = }\sqrt[]{\sum^{}_{}}(\text{ x - }\mu)^2\cdot\text{ p(xi)} \\ \text{let }\mu\text{ = 2.2} \\ \text{Standard deviation = }\sqrt[]{(0-2.2)^2\cdot0.2+(1-2.2)^2\cdot0.05+(2-2.2)^2\cdot0.1+(3-2.2)^2\cdot\text{ 0.65}} \\ \text{standard deviation = }\sqrt[]{0.968\text{ + 0.072 + 0.004 + 0.416}} \\ \text{Standard deviation = }\sqrt[]{1.46} \\ \text{Standard deviation = }1.2083 \\ \text{Hence, standard deviation is 1.2083} \end{gathered}[/tex]
