a. for value of k, it seems its a continuous PDF, so we know one thing about PDF as,
$\int_{0}^{infinite} PDF = 1$
but here PDF is divided in intervals,so
$\int_{0}^{6}kx2 + \int_{6}^{12} k(12-x)^{2} = 1$
solve it you will get k = $\frac{1}{144}$
b. probability of storm duration > 8hr
in this case our range would be 8 <= x <= 12
p = $\int_{8}^{12} \frac{1}{144} (12-x)^{2}$
solving this gives, p = 4/27....(i m not sure, please cross check)
c. in this case probability we have to find in interval 5 to 7 which is divided in two interval by 6
so our 2 intervals would be 5 to 6, and 6 to 7
p = $\int_{5}^{6} \frac{1}{144} x^{2} + \int_{6}^{7} \frac{1}{144} (12-x)^{2}$
solve it you will get, p = $\frac{91}{ 3 * 144}$ ..... please cross check
