Gamma disttribution

Exercise12

1. Suppose that the amount of gasoline $X$ kiloliter sold per week in a gas station follows a distribution with the following density function.

$$f(x) = \left\{\begin{array}{lc} ce^{-1}\{-(x-2)^2 + 2\}, & 1 < x < 3\\ ce^{-(x-2)}, & x \geq 3\\ 0, & \mbox{その他} \end{array}\right.$$

(a)

Find the value of the constant $c$.

(b)

Find the probability that $2.5 \leq X \leq 3.5$．

2. The blood types of Japanese are said to be A type $35\%$, B type $25\%$, AB type $10\%$, and O type $30\%$. Now, when there are four people, find the probability that all four people have different blood types.

3.. It is assumed that the telephone talk time follows an exponential distribution of 2 minutes on average. Find the probability that you will have to wait at least 10 minutes when there are 3 people in the phone box, including those who are already busy.

Answer

I. To make $f(x)$ a density function,

$$\int_{-\infty}^{\infty}f(x) dx = 1$$

(a)

$$\int_{\infty}^{\infty}f(x)dx = \int_{-\infty}^{1}f(x)dx + \int_{1}^{3}f(x)dx + \int_{3}^{\infty}f(x)dx$$

$$= 0 + \int_{1}^{3}ce^{-1}\{-(x-2)^2 + 2\} dx + \int_{3}^{\infty}ce^{-(x-2)}dx$$

$$= ce^{-1}\left[-\frac{1}{3}(x-2)^{3} + 2x \right]_{1}^{3} - \left[ce^{-(x-2)}\right]_{3}^{\infty}$$

$$= ce^{-1}(-\frac{1}{3} + 6 - (\frac{1}{3} + 2)) - ce^{-1}(0 - (-1))$$

$$=$$

2. Let $X_{1}$ be the number of people with blood type A，$X_{2}$ be the numbeBr of people with blood type ，$X_{3}$ be the number of people with blood type AB，$X_{4}$ be the number of people with blood type O. Then ${\vec X} = (X_{1},X_{2}.X_{3}.X_{4}) \sim M(4, 0.35, 0.25, 0.1, 0.3)$．Thus,

$$P_{r}(X_{1} = 1, X_{2} = 1, X_{3} = 1, X_{4} = 1) = \frac{4!}{1! 1! 1! 1!}(0.35)(0.25)(0.1)(0.3) = 0.063 .$$

3. Let $X_{1}, X_{2}, X_{3}$ be the talk time of 3 people. Then $X_{1} \sim E_{x}(\lambda) = \Gamma(1, 1/\lambda), X_{2} \sim E_{x}(\lambda) = \Gamma(1, 1/\lambda), X_{3} \sim E_{x}(\lambda) = \Gamma(1, 1/\lambda)$．Now let $X = X_{1} + X_{2} + X_{3}$. Then $X$ represents the total talk time of 3 people．Thus, $X \sim \Gamma(3, 1/\lambda)$．Also, $E(X_{1}) = 1/\lambda = 2$ implies $X \sim \Gamma(3, 2)$．Thus, the probability of having to wait more than 10 minutes is

$$P_{r}(X \geq 10) = 1 - P_{r}(0 \leq X \leq 10) = \int_{10}^{\infty} \frac{1}{ \Gamma(3) 2^{3}}x^2 e^{-x/2} dx$$

$$= 1 - \frac{1}{16}\int_{0}^{10}x^2 e^{-x/2}dx = 1 - [-2x^2 e^{-x/2}\mid_{0}^{10} + 4\int_{0}^{10}xe^{-x/2}dx ]$$

$$= 1 - \frac{1}{16}[-200e^{-5} + 4(-2xe^{-x/2}\mid_{0}^{10} + 2\int_{0}^{10}e^{-x/2}dx)]$$

$$= 1 - \frac{1}{16}[-200e^{-5} -80e^{-5} - 16e^{-x/2}\mid_{0}^{10}] = 1 - \frac{1}{16}[-296e^{-5} + 16] = \frac{296 e^{-5}}{16} = 0.1246$$