The joint probability distribution of the number $X$ of cars and the number $Y$ of buses per signal cycle at a proposed left turn lane is displayed in the accompanying joint probability table.

%%Table in LaTeX

\begin{table}[htbp]
\centering
\begin{tabular}{cc|p{0.5in}p{0.5in}p{0.5in}}
 & & & $y$ & \\
$p(x,y)$ &  & 0 & 1 & 2 \\\hline
\multirow{3}{*}{$x$}& 0 & .025 & .015 & .010\\
~& 1 & .050 & .030 & .020\\
~& 2 & .125 & .075 & .050\\
~& 3 & .150 & .090 & .060\\
~& 4 & .100 & .060 & .040\\
~& 5 & .050 & .030 & .020\\
\hline
\end{tabular}
\end{table}

• (a) What is the probability that there is exactly one car and exactly one bus during a cycle?
• (b) What is the probability that there is at most one car and at most one bus during a cycle?
• (c) What is the probability that there is exactly one car during a cycle? Exactly one bus?
• (d) Suppose the left turn lane is to have a capacity of five cars and one bus is equivalent to three cars. What is the probability of an overflow during a cycle?
• (e) Are $X$ and $Y$ independent rv's? Explain.

The number of customers waiting for gift-wrap service at a department store is an rv $X$ with possible values $0,1,2,3,4$ and corresponding probabilities $.1,.2,.3,.25,.15$. A randomly selected customer will have $1,2$, or $3$ packages for wrapping with probabilities $.6,.3$, and $.1$, respectively. Let $Y$= the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer).

• (a) Determine $P(X=3,\ Y=3)$, i.e., $p(3,3)$.
• (b) Determine $p(4,11)$.
   

A certain market has both an express checkout line and a superexpress checkout line. Let $X_1$ denote the number of customers in line at the express checkout at a particular time of day and let $X_2$ denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of $X_1$ and $X_2$ is as given in the accompanying table.

%%Table in LaTeX

\begin{table}[htbp]
\centering
\begin{tabular}{cc|p{0.5in}p{0.5in}p{0.5in}p{0.5in}}
 & & & $x_2$ & & \\
 & & 0 & 1 & 2 & 3\\\hline
\multirow{5}{*}{$x_1$}& 0 & .08 & .07 & .04 & .00\\
~& 1 & .06 & .15 & .05 & .04\\
~& 2 & .05 & .04 & .10 & .06\\
~& 3 & .00 & .03 & .04 & .07\\
~& 4 & .00 & .01 & .05 & .06\\
\hline
\end{tabular}
\end{table}

• (a) What is $P(X_1=1,\ X_2=1)$, that is, the probability that there is exactly one customer in each line?
• (b) What is $P(X_1=X_2)$, that is, the probability that the numbers of customers in the two lines are identical?
• (c) Let $A$ denote the event that there are at least two more customers in one line than in the other line. Express $A$ in terms of $X_1$ and $X_2$, and calculate the probability of this event.
• (d) What is the probability that the total number of customers in the two lines is exactly four? At least four?
   

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let $X$ denote the number of hoses being used on the self-service island at a particular time, and let $Y$ denote the number of hoses on the full-service island in use at that time. The joint pmf of $X$ and $Y$ appears in the accompanying tabulation.

%%Table in LaTeX

\begin{table}[htbp]
\centering
\begin{tabular}{cc|p{0.5in}p{0.5in}p{0.5in}}
 & & & $y$ & \\
$p(x,y)$ &  & 0 & 1 & 2 \\\hline
\multirow{3}{*}{$x$}& 0 & .10 & .04 & .02\\
~& 1 & .08 & .20 & .06\\
~& 2 & .06 & .14 & .30\\
\hline
\end{tabular}
\end{table}

• (a) What is $P(X=1\ \text{and}\ Y=1)$?
• (b) Compute $P(X\leqslant 1\ \text{and}\ Y\leqslant 1)$.
• (c) Give a word description of the event $\{X\neq 0\ \text{and}\ Y\neq 0\}$ and compute the probability of this event.
• (d) Compute the marginal pmf of $X$ and of $Y$. Using $p_{X}(x)$, what is $P(X\leqslant 1)$?
• (e) Are $X$ and $Y$ independent rv's? Explain.
   

随机变量 $X$ 被称为服从带参数 $\alpha,\beta$(都是正的), 以及参数 $A$ 和 $B$ 的 $\beta$-分布(beta distribution), 如果其 pdf (概率密度函数)定义为

\[
f(x;\alpha,\beta,A,B)=\begin{cases}
\frac{1}{B-A}\cdot\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\cdot\Gamma(\beta)}\biggl(\frac{x-A}{B-A}\biggr)^{\alpha-1}\biggl(\frac{B-x}{B-A}\biggr)^{\beta-1}, & x\in[A,B]\\
0, & \text{其他}
\end{cases}
\]

特别的, 当 $A=0$, $B=1$ 时, 该分布称为标准 $\beta$-分布 (standard beta distribution).

首先, 可以证明, 这样定义的 pdf 是合理的, 即

\[
\int_{A}^{B}f(x;\alpha,\beta,A,B)dx=1.
\]

其次, 可以计算服从 $\beta$-分布的随机变量 $X$, 其均值和方差为:

\[
\mu=A+(B-A)\cdot\frac{\alpha}{\alpha+\beta},\qquad\sigma^2=\frac{(B-A)^2\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}
\]

A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength(延展强度) $X$ of a material has a lognormal distribution.

[基於某種材料失效機制的理論論證是以假定材料的延展強度 $X$ 具有對數正態分佈為基礎。]

Suppose the parameters are $\mu=5$ and $\sigma=.1$.

• (a) Compute $E(X)$ and $V(X)$.
• (b) Compute $P(X > 120)$.
• (c) Compute $P(110\leqslant X\leqslant 130)$.
• (d) What is the value of median ductile strength?
• (e) If ten different samples of an alloy steel(合金鋼) of this type were subjected to a strength test, how many would you expect to have strength at least $120$?
• (f) If the smallest $5\%$ of strength values were unacceptable, what would the minimum acceptable strength be?
   

Let $X$= the hourly median power (in decibels (分贝/décibels/Dezibel)) of received radio signals transmitted between two cities. The authors of the article "Families of Distributions for Hourly Median Power and Instantaneous Power of Received Radio Signals" (J. Research National Bureau of Standards, vol. 67D, 1963: 753--762) argue that the lognormal distribution provides a reasonable probability model for $X$. If the parameter values are $\mu=3.5$ and $\sigma=1.2$, calculate the following:

• (a) The mean value and standard deviation of received power
• (b) The probability that received power is between $50$ and $250$ dB.
• (c) The probability that $X$ is less than its mean value. Why is this probability not $.5$?
   

Let $X$ have a Weibull distribution with the pdf from Expression (see alse Question2480)
\[
f(x;\alpha,\beta)=\begin{cases}
\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

Verify that

\[\mu=\beta\Gamma(1+\frac{1}{\alpha}).\] 

(Hint: In the integral for $E(X)$, make the change of variable $y=(\frac{x}{\beta})^{\alpha}$, so that $x=\beta y^{\frac{1}{\alpha}}$.)

(The proof can be found in the solution of Question2480.)
 

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985:519--522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials(固体绝缘材料) subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens(样本) subjected to AC voltage(交流电压). The values of the parameters depend on the voltage and temperature; suppose $\alpha=2.5$ and $\beta=200$ (values suggested by data in the article).

• (a) What is the probability that a specimen's lifetime is at most $200$? Less than $200$? More than $300$?
• (b) What is the probability that a specimen's lifetime is between $100$ and $200$?
• (c) What value is such that exactly $50\%$ of all specimens have lifetimes exceeding the value?
   

Weibull 分布是由瑞典物理学家 Waloddi Weibull 于1939年引进的.

他的1951年的一篇文章 "A Statistical Distribution Function of Wide Applicability"(J. Applied Mechanics, vol. 18: 293--297) 中讨论了一些应用.

定义: 

A random variable $X$ is said to have a Weilbull distribution with parameters $\alpha$ and $\beta$ ($\alpha > 0$, $\beta > 0$) if the pdf of $X$ is

\[
f(x;\alpha,\beta)=\begin{cases}
\frac{\alpha}{\beta^{\alpha}}x^{\alpha-1}e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

与之十分相近的是 $\Gamma$-分布(Gamma distribution). 回顾其 pdf 定义为

\[
f_2(x;\alpha,\beta)=\begin{cases}
\frac{1}{\beta^{\alpha}\Gamma{\alpha}}x^{\alpha-1}e^{-\frac{x}{\beta}}, & x\geqslant 0,\\
0, & x < 0
\end{cases}
\]

令 Weibull 分布和 $\Gamma$-分布中的参数 $\alpha=1$, $\lambda=\frac{1}{\beta}$, 我们都可以得到指数分布(exponential distribution)

\[
f(x;\lambda)=\begin{cases}
\lambda e^{-\lambda x}, & x\geqslant 0,\\
0, & x < 0.
\end{cases}
\]

也就是说, 指数分布同是 $\Gamma$-分布和 Weibull 分布的特殊情形. 但是, 存在 $\Gamma$-分布, 其不属于 Weibull 分布; 也存在 Weibull 分布, 不属于 $\Gamma$-分布.

对于所定义的 Weibull 分布, 证明其 cdf 为

\[
F(x;\alpha,\beta)=\begin{cases}
1-e^{-(\frac{x}{\beta})^{\alpha}}, & x\geqslant 0,\\
0, & x < 0.\\
\end{cases}
\]

Prop. 若 $X$ 是服从 Weibull 分布的随机变量, 则其均值和方差为

\[
\begin{eqnarray}
\mu=E(X)&=&\beta\cdot\Gamma(1+\frac{1}{\alpha})\\
\sigma^2=V(X)&=&\beta^2\cdot\biggl[\Gamma(1+\frac{2}{\alpha})-\Bigl(\Gamma(1+\frac{1}{\alpha})\Bigr)^2\biggr]
\end{eqnarray}
\]