Question 7.9: The number K of parcels that the drivers of a parcel deliver......
The number K of parcels that the drivers of a parcel delivery service company can load in their trucks is a random variable with the
\,\displaystyle p_K\left (k\right )=\frac{40^k e^{-40}}{k !} \quad k=0,1,2, \ldotsThe weight W of a parcel in pounds is a continuous random variable with
\,\displaystyle f_W\left (w\right )= \begin{cases}\frac{1}{6} & 3 \leq w \leq 9 \\ 0 & \text { otherwise }\end{cases}Let X denote the weight of a randomly selected loaded truck.
a. What is the s-transform of the PDF of X?
b. What is the expected value of X?
c. What is the variance of X?
(a) Let the number of parcels in the truck be K=k, and let W_i denote the weight of parcel i, \,\displaystyle 1 \leq i \leq k. Then, since we have fixed K at k, we have that
\,\displaystyle \begin{aligned} & \left.X\right\vert _{K=k}=W_1+W_2+\cdots+W_k \\ & M_{X \mid K}\left (s \mid k\right )=\left[M_W\left (s\right )\right]^k \end{aligned}Thus, the s-transform of the PDF of \,\displaystyle X is
\,\displaystyle M_X\left (s\right )=\sum_{k=0}^{\infty} M_{X \mid K}\left (s \mid k\right ) p_K\left (k\right )=\sum_{k=0}^{\infty}\left[M_W\left (s\right )\right]^k p_K\left (k\right )=G_K\left (M_W\left (s\right )\right )where, G_K(z)=exp(-40{1-z}) and from equation (7.21), M_W(s) is given by:
\,\displaystyle M_X\left (s\right ) =\int_0^{\infty} e^{-s x} f_X\left (x\right ) d x=\frac{1}{b-a} \int_a^b e^{-s x} d x=\left[-\frac{e^{-s x}}{s\left (b-a\right )}\right]_a^b \\ =\frac{e^{-a s}-e^{-b s}}{s\left (b-a\right )}\qquad (7.21)
\,\displaystyle M_W\left (s\right )=\frac{e^{-a s}-e^{-b s}}{s\left (b-a\right )}=\frac{e^{-3 s}-e^{-9 s}}{6 s}(b) Since K is a Poisson random variable, its expected value and variance are given by
\,\displaystyle E[K]=\sigma_K^2=40Similarly, since W has a uniform distribution, its mean and variance are given by
\,\displaystyle \begin{gathered} E[W]=\frac{3+9}{2}=6 \\ \sigma_W^2=\frac{\left (9-3\right )^2}{12}=3 \end{gathered}Thus, \,\displaystyle E[X] =E[K]E[W]=240.
(c) The variance of X is given by
\,\displaystyle \sigma_X^2=E[K] \sigma_W^2+\{E[W]\}^2 \sigma_K^2=\left (40\right )\left (3\right )+\left (b^2\right )\left (40\right )=1560