Answer
$\color{blue}{\frac{\sqrt{179}}{12} \approx 1.115}$
Work Step by Step
From the given graph of the pdf, we have
$\begin{align*}
f_Y(y) = \begin{cases} 1-y, & 0 \le y \le 1 \\ \frac{1}{2}, & 2 \le y \le 3 \\ 0, & \text{otherwise.} \end{cases} \quad\quad\color{blue}{\text{(*** Eqn. 1 ***)}}
\end{align*}$
$\begin{align*}
E(Y) &= \int_{\mathbb{R}} y\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y)\ ] \\
&= \int_0^1 y\cdot (1-y)\ dy + \int_2^3 y\cdot \frac{1}{2}\ dy \qquad [\ \text{see Eqn. 1}\ ] \\
&= \int_0^1 (y-y^2)\ dy + \frac{1}{2}\int_2^3 y\ dy \\
&= \left( \frac{y^2}{2} - \frac{y^3}{3}\ \right\vert_0^1 + \frac{1}{2}\left( \frac{y^2}{2}\ \right\vert_2^3 \\
&= \left[\left( \frac{1^2}{2} - \frac{1^3}{3}\right) - \left( \frac{0^2}{2} - \frac{0^3}{3}\right)\right] + \frac{1}{2}\left( \frac{3^2}{2}- \frac{2^2}{2}\right) \\
&= \frac{1}{2} - \frac{1}{3} - 0 + \frac{1}{4}(3^2-2^2) \\
&= \frac{1}{6} + \frac{5}{4} \\
E(Y) &= \frac{17}{12} \\
\mu &= \frac{17}{12}
\end{align*}$
$\begin{align*}
E(Y^2) &= \int_{\mathbb{R}} y^2\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(Y^2)\ ] \\
&= \int_0^1 y^2\cdot (1-y)\ dy + \int_2^3 y^2\cdot \frac{1}{2}\ dy \qquad [\ \text{see Eqn. 1}\ ] \\
&= \int_0^1 (y^2-y^3)\ dy + \frac{1}{2}\int_2^3 y^2\ dy \\
&= \left( \frac{y^3}{3} - \frac{y^4}{4}\ \right\vert_0^1 + \frac{1}{2}\left( \frac{y^3}{3}\ \right\vert_2^3 \\ &= \left[\left( \frac{1^3}{3} - \frac{1^4}{4}\right) - \left(\frac{0^3}{3} - \frac{0^4}{4}\right)\right] + \frac{1}{2}\left( \frac{3^3}{3}- \frac{2^3}{3}\right) \\
&= \frac{1}{3} - \frac{1}{4} - 0 + \frac{1}{6}(27-8) \\
&= \frac{1}{12} + \frac{19}{6} \\
E(Y^2) &= \frac{13}{4}
\end{align*}$
By Theorem 3.6.1, since $\mu= \dfrac{17}{12}$ exists and $E(Y^2) = \dfrac{13}{4}$ is finite,
$\begin{align*}
\text{Var}(Y) &= E(Y^2) - \mu^2 \\
&= \frac{13}{4} - \left(\frac{17}{12}\right)^2 \\
&= \frac{468}{144} - \frac{289}{144} \\
\text{Var}(Y) &= \frac{179}{144} \\
\sqrt{\text{Var}(Y)} &= \frac{\sqrt{179}}{12} \\
\color{blue}{\sigma}\ &\color{blue}{= \frac{\sqrt{179}}{12} \;\approx\; 1.115} \\
\end{align*}$
