Answer
(a) $\color{blue}{\ln 2 \approx 0.693}$
(b) $\color{blue}{\ln\left(\frac{5}{4}\right) \approx 0.223}$
(c) $\color{blue}{\ln\left(\frac{5}{4}\right) \approx 0.223}$
(d) $\color{blue}{f_Y(y) = \dfrac{1}{y},\ 1\lt y\lt e}$
Work Step by Step
(a)
$\begin{align*}
P(Y\lt 2) &= P(Y\le 2) - P(Y=2) & \\
&= P(Y\le 2) - 0 & [P(Y=2)=0\ \text{since}\ Y\ \text{is continuous}] \\
&= P(Y\le 2) & \\
&= F(2) & [\text{by the definition of a cdf}] \\
\color{blue}{P(Y\lt 2)} &\color{blue}{= \ln 2\approx 0.693} & [\text{since}\ 1\le 2 \le e]
\end{align*}$
(b)
$\begin{align*}
P(2\lt Y\le 2{\scriptsize \frac{1}{2}}) &= P(Y\le 2{\scriptsize \frac{1}{2}}) - P(Y\le 2) \\
&= F(2{\scriptsize \frac{1}{2}}) - F(2) & [\text{by the definition of a cdf}] \\
&= \ln (2{\scriptsize \frac{1}{2}}) - \ln 2 & [\text{since}\ 1\le 2,\ 2{\scriptsize \frac{1}{2}} \le e] \\
&= \ln \left(\frac{2{\scriptsize \frac{1}{2}}}{2}\right) \\
\color{blue}{P(2\lt Y\le 2{\scriptsize \frac{1}{2}})} &\color{blue}{= \ln \left({\scriptsize \frac{5}{4}}\right) \approx 0.223}
\end{align*}$
(c)
$\begin{align*}
P(2\le Y\le 2{\scriptsize \frac{1}{2}}) &= P(Y\le 2{\scriptsize \frac{1}{2}} - P(Y\lt 2) \\
&= P(Y\le 2{\scriptsize \frac{1}{2}} - (P(Y\lt 2) + 0) \\
&= P(Y\le 2{\scriptsize \frac{1}{2}} - (P(Y\lt 2) + P(Y=2))& [P(Y=2)=0\ \text{since}\ Y\ \text{is continuous}] \\
&= F(2{\scriptsize \frac{1}{2}}) - F(2) & [\text{by the definition of a cdf}] \\
&= \ln (2{\scriptsize \frac{1}{2}}) - \ln 2 & [\text{since}\ 1\le 2,\ 2{\scriptsize \frac{1}{2}} \le e] \\
&= \ln \left(\frac{2{\scriptsize \frac{1}{2}}}{2}\right) \\
\color{blue}{P(2\lt Y\le 2{\scriptsize \frac{1}{2}})} &\color{blue}{= \ln \left({\scriptsize \frac{5}{4}}\right) \approx 0.223}
\end{align*}$
(d)
$\begin{align*}
f_Y(y) &= \frac{dF_Y(y)}{dy} \\
&= \begin{cases} \dfrac{d}{dy}(0),& y\lt 1 \\ \dfrac{d}{dy}\ln y, & 1\lt y\lt e \\ \dfrac{d}{dy}(1), & y\gt e \end{cases} \\
&= \begin{cases} 0,& y\lt 1 \\ \dfrac{1}{y}, & 1\lt y\lt e \\ 0, & y\gt e \end{cases} \\
\color{blue}{f_Y(y)} &\color{blue}{= \frac{1}{y},\ 1\lt y\lt e}
\end{align*}$
