[tex]f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}[/tex]
a. 9:00 AM is the 60 minute mark:
[tex]f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248[/tex]
b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is
[tex]\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173[/tex]
c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is
[tex]\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982[/tex]
The probability of doing so for at least 2 of 5 days is
[tex]\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1[/tex]
i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.
d. Integrate the PDF to obtain the CDF:
[tex]F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<0\\1-0.1e^{-0.1x}&\text{for }x\ge0\end{cases}[/tex]
Then the desired probability is
[tex]F_X(30)-F_X(15)\approx0.950-0.777=0.173[/tex]
