Answer
$\color{blue}{e^{-3} \;\approx\; 0.05}$
Work Step by Step
Since $Y$ is an exponential random variable with $\lambda = 2$, then $f_Y(y) = \lambda e^{-\large \lambda y} = 2e^{-\large 2y},\ y \ge 0$ and
* $F_Y(y) = 1- e^{\large - \lambda y} = 1- e^{\large -2y}$ by Exercise 3.4.8 (p. 136),
* $\mu_Y = E(Y) = \dfrac{1}{\lambda} = 1/2$ by Exercise 3.5.11 (p. 146), and
* $\text{Var}(Y) = \dfrac{1}{\lambda^2} = 1/2^2$ by Exercise 3.6.11 (p. 158).
Thus,
$\begin{align*}
P\left(Y \gt E(Y) + 2\sqrt{\text{Var}(Y)} \right) &= P\left(Y \gt \frac{1}{2} + 2\sqrt{\left(\frac{1}{2}\right)^2} \right) \\
&= P\left(Y \gt \frac{1}{2} + 2\cdot \frac{1}{2}\right) \\
&= P\left(Y \gt \frac{3}{2}\right) \\
&= 1 - P\left(Y \le \frac{3}{2}\right) \\
&= 1 - F_Y(3/2) \\
&= 1 - \left(1 - e^{-\large 2(3/2)}\right) \\
&= e^{-\large 3} \\
\color{blue}{P\left(Y \gt E(Y) + 2\sqrt{\text{Var}(Y)} \right)}\ &\color{blue}{= e^{-3} \;\approx\; 0.0498 \;\approx\; 0.05}
\end{align*}$
