ch13
CMOS模拟集成电路设计_ch13非线性与不匹配 23页PPT
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不匹配
• 2.3 失调消除技术
输出失调存储
需要专用的失调消除周期 S3,S4没有电荷注入失配 Av不能太大,否则AvVOS会使 放大器输出饱和, 通常Av<10
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不匹配
• 2.3 失调消除技术(续)
输入失调存储 需要专用的失调消除周期 较输出失调存储,可以用于较高电压增益的放大器中 S3,S4断开时,电荷注入失配可能会使放大器输出饱和
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小结
小结
• 1、非线性
– 非线性概念及度量
– 线性化技术
• 负反馈 • 工作在线性区的差动对 • 后校正
• 2、不匹配
– 大尺寸器件具有小的失配
– 失调消除技术
• 输入/输出失调存储 • 辅助放大级消除失调
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源级负反馈 单端电路,忽略体效应
Av GmRD1ggm mRs RD
线性度←→增益、噪声
差动对中的源级负反馈
a电路消耗ISSRS/2电压余度
b电路没有上述问题,但是产生 了较高噪声、失调
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非线性
• 1.5 线性化技术(续)
VDS恒定的工作在线性区输入管的差动对
ID 1 2C oW x L2 (V G S V T)H V D S V D 2 S
非线性
• 1.2 差动电路的非线性
差动电路的输入输出特性f(-x)=-f(x)
差动信号驱动的差动电路不会产生偶次谐波→↑线性度
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非线性
• 1.3 负反馈对非线性的影响
负反馈可以提高系统的线性度 考察一个“轻度非线性”系
Ch13 简单国民收入决定理论
第十三章 简单国民收入决定理论
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生命周期理论和永久收入理论的共同点
1、消费不只同现期收入相联系,而是以一生或永久 的收入作为消费决策的依据。
2、一次性暂时收入变化引起的消费支出变动甚小, 即其边际消费倾向很低,甚至近于零,但来自永久收 入变动的边际消费倾向很大,甚至近于1。 3、当政府想用税收政策影响消费时,如果减税或增 税只是临时性的,则消费并不会受到很大影响,只有 永久性税收变动,政策才会有明显效果。
响,这就是所谓“示范效应”。
第十三章 简单国民收入决定理论
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相对收入消费理论对短期消费函数的解释
CL=βy C CS2 Ct1 CS1 O Yt1 Y1 Y2 Y C2
C1
第十三章 简单国民收入决定理论
29
二、生命周期的消费理论
【美●弗朗科.莫迪利安】
弗朗科.莫迪利安的生命周期的消费理论强调人们
c dc c lim MPC y y 0 y dy
第十三章 简单国民收入决定理论
10
中国改革开放后居民消费倾向的变动
改革开放政策使居民的可支配收入迅速增加。随
着收入水平的快速提高,中国居民的消费倾向发生了 显著变化。在1988年中国居民平均消费倾向还在90%以 上,此后,平均消费倾向开始明显下降,1995年比 1988年平均消费倾向降低了16个百分点。与此同时,边 际消费倾向也明显下降,到1994年,边际消费倾向降 到64%。
平均储蓄倾向和边际储蓄倾向都是递增的, 且边际储蓄倾向大 简单国民收入决定理论
21
如果储蓄和收入之间存在线性关系, 因为 s=y-c , 且c=α+βy , 所以: s=y-(α+βy)= -α+(1–β)y
PMP知识点整理 CH13 项目相关方管理
13.1 如何来识别相关方?
识别相关方是定期识别项目相关方,分析和记录他们的利益、参与度、相互依赖性、影 响力和对项目成功的潜在影响的过程。
识别相关方的目的是什么?
识别相关方的目别相关方的输入有哪些?
输入:项目章程、商业文件、项目管理计划、项目文件、协议、事业环境因素、组织过程 资产。
相关方识别的工具是什么?
工具:专家判断、数据收集、数据分析、数据表现、会议。
13.3 相关方参与度评估矩阵是用来做什么的? 包含哪些要素?
相关方参与度评估矩阵用于将相关方当前参与水平与期望参与水平进行比较。对相 关方参与水平进行分类的方式之一,如图 13-6 所示。相关方参与水平可分为如 下:
①不了解型。 ②抵制型。 ③中立型。 ④支持型。 ⑤领导型。 领导型的相关方是领导的,支持型的相关方是支持的。
13.4 如何管理相关方参与?
管理相关方参与是与相关方进行沟通和协作以满足其需求与期望、处理问题,并促进相关方 合理参与的过程。在管理相关方参与过程中,需要开展多项活动,例如:
①在适当的项目阶段引导相关方参与,以便获取、确认或维持他们对项目成功的持续承诺; ②通过谈判和沟通管理相关方期望; ③处理与相关方管理有关的任何风险或潜在关注点,预测相关方可能在未来引发的问题; ④澄清和解决已识别的问题。 管理相关方参与有助于确保相关方明确了解项目目的、目标、收益和风险,以及他们的贡献
将如何促进项目成功。
13.5 如何控制相关方参与,有什么作用?
控制相关方参与--更改为监督相关方参与。监督相关方参与是监督项目相关方关系,并 通过修订参与策略和计划来引导相关方合理参与项目的过程。本过程的主要作用是, 随着项目进展和环境变化,维持或提升相关方参与活动的效率和效果。
ch13简单国民收入决定
同理,有Δy =Δc+ Δs;
两边同除Δy,则 Δy/Δy = Δc/Δy
+Δs/Δy
即: MPC + MPS = 1 推断:若MPC递减,那么MPS
必递增。
c,s Poor A
c=c(y)
Rich
a
450
s=s(y)
-a
y0
y
消费曲线与储蓄曲线
储蓄与消费的互补——函数曲线
c,s
y=c+s
2.生命周期假说 美国 莫迪利安尼(Modigliani)
假说: (1)人在一生中,收入有规律地波动。 (2)家庭会通过金融资产市场,以理性方式统筹安排一生
的资源,实现一生中最平稳的消费路径。
结论: 人们偏好平稳消费,工作年份储蓄,退休年份动用储蓄。 如果中年人比例增大,消费倾向会下降。 反之会增加。
消费中受收入水平影响
a
的部分。
45º
c= a + by y
3.消费倾向 propensity
消费倾向:消费与收入的比率。
平均消费倾向 APC average propensity to consume 平均每单位收入中消费所占比例。 计算:总消费在总收入中所占比例=消费/收入=c/y
APC如果<1,消费总量<收入总量(产生储蓄); APC如果=1,把全部收入都用于消费(储蓄为零); APC如果>1,消费总量大于收入总量(负债消费,即产生负储蓄)。
收入 消费 储蓄 MPS APS
1 9000 9110 -110
-0.01
2 10000 10000 0 0.11 0
3 11000 10850 150 0.15 0.01
CH13 Direct Foreign Investment(国际金融管理,英文版)
Motives for DFI
Cost-Related Motives
Use foreign technology.
React to exchange rate movements, such as when the foreign currency appears to be undervalued. DFI can also help reduce the MNC’s exposure to exchange rate fluctuations.
Asian crisis, an MNC that had diversified among the Asian countries might have fared better than if it had focused on one country. Even better would be diversification among the continents.
Diversify sales/production internationally.
C13 - 7
Motives for DFI
• The optimal method for a firm to penetrate
a foreign market is partially dependent on the characteristics of the market.
C13 - 9
Change in Distribution of DFI
By U.S. Firms Over Time
DFI Distribution in 1982
Asia & Pacific 15% Middle East 2% Africa 3%
13第三篇ch13信贷资产证券化
SPV自身风险的隔离
SPV的本质特性----破产隔离给予交易的安全 性以极大的保障
破产隔离,简言之,就是指实体一般不会遭受自 愿的或非自愿的破产
评级机构认为SPV是风险隔离的六点要求:
1.对SPV目标和能力的限制 2.对债务的限制 3.独立董事 4.不会合并或重组 5.独立契约 6.资产的担保权益
四.需要注意的问题
1.评级机构不对市场风险进行评判,因为市场 风险不是信用风险
2.评级机构也不会对投资者某项投资是否合适 发表任何见解
3.信用评级不表示某种证券的市场价格是否合 适
4.信用评级表明了对按照合同的期限偿还的 可能性的看法;信用评级既是非常精确的,也是 非常有限的
5.投资者的认可是评级机构成功的关键
信用评级概述
一.信用评级 信用评级是根据发行人.会计人员和其他
专业人士所提供的各种现在的信息和历史 纪录,依据对发行人的审查,以及在资产支撑 的证券中对发起人和基础应收款的服务商 的审查与对基础资产的分析的基础之上进 行的;当存在一种或多种形式的信用增级时, 评级也必须对信用增级的提供者进行分析
二.信用评级机构的作用
担保资产是一种权益资产,或者说是一种债权, 这种担保资产缺乏流动性,但是拥有稳定的, 可预期的未来现金流量
资产证券化具有信用提高功能,以达到证券市 场对这类证券的信用等级要求
资产证券化交易步骤
建立特设信托机构SPV 筛选可证券化的资产并组成资产池 原始权益人将资产“真实出售”给
3.资产证券化技术
证券化金融资产的特征
具有可预测的,稳定的未来现金流 在历史纪录中很少发生违约或损失事件 未来现金流比较均匀的分摊与资产的续存期内 所对应的债务人有广泛的地域分布和人口统计分布 原所有者持有该项资产已经有一段时间,在此期间
CH13固有免疫细胞免疫应答
第十三章 固有免疫细胞的免疫应答 Innate Immunity
目录
1 组织屏障及其作用 2 固有免疫细胞 3 固有免疫效应分子及其主要作用 4 固有免疫应答
固有免疫(innate immunity)
特点
遗传性(genetic;早,出生时已具备, 可稳定性遗传给后代)
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一、吞噬细胞(phagocyte)
吞噬细胞 phagocytes
单个核吞噬细胞(系统)
Mononuclear phagocyte system
中性粒细胞
neutrophils
单核细胞
巨噬细胞
monocyte
macrophage
小胶质细胞、库普弗细胞、破骨细胞
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(一)中性粒细胞----小吞噬细胞
➢树突状细胞:专职抗原提呈细胞。
➢NK细胞:ADCC作用(穿孔素、颗粒酶和FasL)。
➢γδT细胞:粘膜和皮肤组织,杀伤机制同CTL
➢NKT细胞:主要分布于骨髓、肝、胸腺,膜分子,杀伤机制同CTL。
➢B1细胞:发育较早,主要分布于胸腔、腹腔和肠壁的固有层,起抗早期感染和
维持自稳的作用。
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3. 微生物屏障 皮肤和粘膜表面的正常菌群。
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Secretions at epithelial surfaces
Site Source
Specific substances secreted
Eyes
Lacrimal glands (tears) Lysozyme, IgA and IgG
部位 皮肤 口腔 胃 肠道
Ch13英文课件 货币金融学 米什金 第七版
a financial institution has bought an asset, it is said to have taken a long position Short position(空头头寸) = if it has sold an asset that it has agreed to deliver to another party at a future date, it is said to have taken a short position
chapter 13
Financial Derivatives 衍生金融工具
Beginning in the 1970s and continuing into the 1980s and 1990s, interest rates and foreign exchange rates became more volatile, increasing the risk to financial institutions. To combat this, managers of financial institutions have demanded financial instruments to better manage risk.
远期(forwards)
Pros and Cons of forward contract
Pros : Flexible Cons: (1) Lack of liquidity: hard to find counterparty( 交易对手) (2) Subject to default risk: requires information to screen good from bad risk
多元微积分-ch13 更多积分
0
4
x
Draw a picture of the region.
7. Let f : D → R be a function defined on a nice subset D ⊂ R 2 . The average value A of f on D is defined to be A = 1 f ( x , y )dA . area of D ∫∫ D
2. Find the center of mass of the smaller of the two regions cut from the elliptical region x 2 + 4 y 2 = 12 by the parabola x = 4 y 2 if the density ρ ( x , y ) = 5x .
i =1
The three sums in the previous line are Riemann sums for two dimensional integrals! Thus as we take smaller and smaller rectangles, etc., we obtain for R, the location of the center of mass
Now choose a point ri = xi* i + y i* j in each rectangle. The mass of this rectangle will be approximately ρ ( x i* , yi* ) ∆Ai , where ∆Ai is the area of the rectangle. The equation for the center of mass of this system of rectangles is then
经济学原理ch13
E=Y
ED
I>S E>Y
ES
I<S E<Y
IS
Y
三、影响IS曲线移动的主要因素
r
Y=KA-Khr
纵轴截距A/h
横轴截距KA
斜率1/Kh
r0
0
Y0
Y1 KA
Y
影响IS曲线移动的主要因素(2)
KA0变动
乘数 K=1/(1-b),K大KA0大IS曲线右 移
厂商和消费者的信心: A=a+I0 信心大使自发性的消费支出增大 A 大 IS曲线右移
这种投资需求与 利率之间的反比 关系被定义为投 资的边际效率, 又称投资函数。
o
投资曲线
I
三、 投资曲线
I=I0-hr
I0 是自发性投资支出,是个常数。 h是利率对投资的影响程度大小,即
h ΔI Δr
r
I=I0-hr
o
I
投资曲线的斜率
在投资函数式I=I0-h r,h是一个重要参数,决定了 投资曲线的斜率。
三个模型的比较
类型
市场 状况
NI 决定模型
IS-LM模型
AD-AS模型
(简单的凯恩斯模型) (调整的凯恩斯模型) (完整的凯恩斯模型)
产品市场
产品市场和货币市场
产品市场、货币市场 和劳动力市场
假设
研究 内容
利率、价格不变, 不考虑货币因素
从总支出角度分析
价格不变, 考虑货币因素
利率、价格均变, 考虑 AS因素
如果用L2表示货币的投机需求,用r表示利率,
则这一货币需求量和利率的关系可表示为:
L2=L2(r) 或 L2=-hr 其中h是货币投机性需求的利率系数,负号表
Ch13 应用安全-网络与信息安全-安葳鹏-清华大学出版社
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解决电子邮件传播的攻击和易受攻击的电子邮件 系统的方法:
在电子邮件系统周围锁定电子邮件系统——电子邮件 系统周边控制开始于电子邮件网关的部署。 确保外部系统访问的安全性——电子邮件安全网关必 须负责处理来自所有外部系统的通信,并确保通过的 信息流量是合法的。通过确保外部访问的安全,可以 防止入侵者利用Web邮件等应用程序访问内部系统。 实时监视电子邮件流量——实时监视电子邮件流量对 于防止黑客利用电子邮件访问内部系统是至关重要的。
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13.2.2电子邮件安全保护技术和策略
漏洞 :
IMAP 和 POP 漏洞 拒绝服务(DoS)攻击 系统配置漏洞 利用软件问题 利用人为因素 特洛伊木马及自我传播
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典型的黑客攻击情况 :
步骤1:外部侦察 步骤 2:内部侦察 步骤 3:漏洞攻击 步骤 4:立足点 步骤 5:牟利
5
13.1.4几种典型应用服务安全的分析
⑴普通WWW浏览应用 ⑵DNS服务 ⑶BBS应用 ⑷E-Mail应用 ⑸MSN应用
都是基于B/S架构的网络应用服务
6
13.2 电子邮件安全
13.2.1 电子邮件安全概述
⑴ 电子邮件安全需求 基于 SMTP(Simple MailTransfer Protocol, 简单邮件 传输协议)的电子邮件系统被广泛应用,但邮件系统本身不 具备安全措施,邮件在收、发、存的过程中都是采用通用编 码方式,信息的发送和接收无鉴别和确认,信件内容容易被 篡改,不怀好意的人甚至可以冒名发信而被害者却丝毫不 知……显然传统的电子邮件不利于重要信息的传递。 资料显示,每年因毫不设防的电子邮件导致泄密、误解等造 成的经济损失至少在千亿美金以上。 更重要的是,电子邮件泄漏政治、军事秘密等恶性事件时常 发生,因此导致的损失更是难以估计。
Ch13交流-直流设备
第13章交流-直流设备(AC-DC Elements)单线图及地下电缆管道系统中的每一种设备类型均有其独立的编辑器。
除了设备的标识、母线联接以及状态之外,其它所有出现在各编辑器中的数据均视为工程电气属性数据。
单线图设备编辑器(One-line Diagram Element Editors)单线图工具条上的每一个设备均有其自己的定制编辑器。
本章讲述各个交流—直流设备的编辑器。
交流—直流设备(AC-DC Elements)后备电源变频器充电器逆变器13.1 后备电源(UPS(Uninterruptible Power Supply)可在该编辑器内输入与电力系统中的后备电源相关的各属性值。
后备电源由两个交流终端(输入与输出)和一个直流终端所组成。
直流终端位于其侧部,并且可被联接到一条直流母线(节点)。
后备电源的编辑器内包含有下列9个属性页:信息属性页额定值属性页负荷属性页短路阻抗属性页运行循环属性页谐波属性页可靠性属性页注释属性页评论属性页13.1.1 信息属性页(Info Page)在信息属性页中,指定后备电源的标识、所联接的母线的标识、投运/退出、设备馈线标签、名称和描述、数据类型、负荷优先级、配置状态、交流联接以及需求因子。
信息(Info)标识(ID)输入一个最多由25个包括文字和数字的字符组成的唯一的标识。
ETAP自动地为各后备电源分配一个唯一的标识。
这些分配的标识由默认标识加上一个整数组成,该整数从1开始随后备电源的数量增加而递增。
后备电源的默认标识可在菜单条中的默认菜单中更改,或者从工程视图中修改。
从…母线联接到…母线(In Bus and DC Bus)它们是后备电源的联接母线的标识。
如果其终端没有联接到任一条母线,那么该母线标识工程显示为空白。
为联接或重新联接一后备电流到一母线,请从该列表框中选择一母线。
在您点击OK 之后,单线图将被更新以显示新的联接。
注意:您可以将后备电源的终端和那些与该设备位于同一视图内的交流—直流母线,或者与位于其它视图内的母线进行联接(方法是联接复合网络的内部和外部引脚),您不能联接位于回收站内的母线。
ch13.简单国民收入决定理论
3.消费函数的线性表达式
C=α+βy α表示自发消费,不依存收 入变动而变动的消费,纵轴 截距;
c
C=α+βy
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α 0 y
MPC和APC的关系
消费函数为线性时, y C y APC MPC y y y 因此,APC MPC 随着收入的增加, 的值越来越小, y 所以,APC MPC
年轻时,c>y, s<0,负债;
青壮年期,c<y, s>0,s的用途有:还债,养老储蓄。 老年期,c>y, s<0,他有两件事要做:消费(用青年期的s), 讨债。
3.永久收入消费理论
提出者:米尔顿· 弗里德曼(美) 观点:暂时消费与暂时收入无关,而由永久收入决定。永久收入是指 消费者可以预计到的长期收入。 繁荣阶段,现实收入>永久收入,
第13章 简单国民收入决定理论
第一节 均衡产出
第二节 凯恩斯的消费理论 第三节 国民收入的基本模式 第四节 乘数理论
注 意
国民收入核算——解决GDP―是什么”、“是多少” 国民收入决定——解决GDP为什么会这样 “简单”——考察“两部门”、“产品市场”
第一节 均衡产出
一、简单国民收入决定理论的基本假设 两部门经济(居民户消费和储蓄,厂商生产和投资) 没有供给限制(存在资源闲置,物价不变,这是凯恩斯主 义的基本出发点) 折旧和公司未分配利润为零 投资为一常数:I=I0
400 300 200 100
200 200 200 200 200 200
4200 3900 3600 3300 3000 2700
CH13-工资成本核算及决策模型
目的和重要性
目的
通过建立科学的工资成本核算及决策模型,帮助企业合理规划和管理薪酬资源,提高员工的工作积极性和绩效, 增强企业的市场竞争力。
重要性
随着市场竞争的加剧和人力资源管理的日益重要,工资成本核算及决策模型对于企业的生存和发展具有重要意义。 一个合理的工资成本核算及决策模型可以帮助企业吸引和留住优秀人才,提高员工满意度和忠诚度,进而提升企 业的整体绩效和市场地位。
03 工资成本决策模型
决策模型一:成本加成模型
总结词
成本加成模型是一种基于企业生产成本和预期利润的定价策 略。
详细描述
成本加成模型首先确定产品的生产成本,然后根据预期利润 加成一定比例,计算出产品的售价。这种模型的优点在于简 单易行,能够快速确定价格,但缺点在于忽略了市场需求和 竞争状况,可能导致定价过高或过低。
决策模型二:目标成本模型
总结词
目标成本模型是一种基于市场竞争和客户需求的产品定价策略。
详细描述
目标成本模型首先分析市场需求和竞争状况,确定产品的目标售价,然后根据目标售价反向推算出产 品的目标成本,最后通过各种手段降低生产成本以达到目标成本。这种模型的优点在于能够更好地满 足市场需求和应对竞争,但需要企业具有较强的市场分析和成本控制能力。
从而降低工资成本。
推行弹性工作
推行弹性工作制度,如远程办公、弹 性工作时间等,提高员工的工作效率 和满意度,降低工资成本。
合理调整组织架构
通过合理调整组织架构,优化人力资 源配置,避免人力资源浪费,降低工 资成本。
05 结论
总结与回顾
工资成本核算及决策模型是企业管理中重要的环节,通过对工资成本的核 算和分析,可以更好地进行人力资源管理和决策2
Ch13矩形波导2
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P 0 max Emax 480 ab Er 1 2a
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温度℃ 饱和不 汽密度 克/米3
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10°
20°
30°
40°
50°
4.84
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Pmax Pmax 0 ( 1) 4
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根据电磁场理论
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C H A P T E R13Discrete ProbabilityDistributions andSimulationObjectivesTo demonstrate the basic ideas of discrete random variables.To introduce the concept of a probability distribution for a discrete random variable.To introduce and investigate applications of the binomial probability distribution.To show that simulation can be used to provide estimates of probability which areclose to exact solutions.To use simulation techniques to provide solutions to probability problems where anexact solution is too difficult to determine.To use coins and dice as simulation models.To introduce and use random number tables.13.1Discrete random variablesIn Chapter10the notion of the probability of an event occurring was explored,where an event was defined as any subset of a sample space.Sample spaces which were not sets of numberswere frequently encountered.For example,when a coin is tossed three times the samplespace is:ε={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}If it is only the number of heads that is of interest,however,a simpler sample space could beused whose outcomes are numbers.Let X represent the number of heads in the three tosses ofthe coin,then the possible values of X are0,1,2,and3.Since the actual value that X will takeis the outcome of a random experiment,X is called a random variable.Mathematically arandom variable is a function that assigns a number to each outcome in the sample spaceε.343344Essential Mathematical Methods1&2CASA random variable X is said to be discrete if it can assume only a countable number ofvalues.For example,suppose two balls are selected at random from a jar containing several white(W)and black balls(B).A random variable X is defined as the number of white balls obtained in the sample.Thus X is a discrete random variable which may take one of thevalues0,1,or2.The sample space of the experiment is:S={WW,WB,BW,BB}Each outcome in the sample space corresponds to a value of X,and vice-versa.Experimental outcome V alue of XWW X=2BW X=1WB X=1BB X=0Many events can be associated with a given experiment.Some examples are:Event Sample outcomesOne white ball:X=1{WB,BW}At least one white ball:X≥1{WW,WB,BW}No white balls:X=0{BB}A probability distribution can be thought of as the theoretical description of a randomexperiment.Consider tossing a die600times–we might obtain the following results:x123456Frequency981049397108100Experimental probability 986001046009360097600108600100600Theoretically,the frequencies will be equal,regardless of the number of trials(provided they are sufficiently large).This can most easily be expressed by giving probabilities–for instance,let X be‘the outcome of tossing a die’.x123456Theoretical frequency100100100100100100Pr(X=x)161616161616The table gives the probability distribution of X.Chapter13—Discrete Probability Distributions and Simulation345The probability distribution of X,p(x)=Pr(X=x)is a function that assigns probabilities to each value of X.It can be represented by a rule,a table or a graph,and must give a probability p(x)for every value x that X can take.For any discrete probability function the following must be true:1The minimum possible value of p(x)is zero,and the maximum possible value of p(x)is1. That is:0≤p(x)≤1for every value x that X can take.2All values of p(x)in every probability distribution must sum to exactly1.To determine the probability that X lies in an interval,we add together the probabilities that X takes all values included in that interval,as shown in the following example.Consider the function:x12345Pr(X=x)2c3c4c5c6ca For what value of c is this a probability distribution?b Find Pr(3≤X≤5).Solutiona To be a probability distribution we require2c+3c+4c+5c+6c=120c=1c=1 20b Pr(3≤X≤5)=Pr(X=3)+Pr(X=4)+Pr(X=5)=420+520+620=1520=34The table shows a probability distribution with random variable X.x123456Pr(X=x)0.20.20.070.170.130.23346Essential Mathematical Methods 1&2CASGive the following probabilities:a Pr(X >4)b Pr(2<X <5)c Pr(X ≥5|X ≥3)Solutiona Pr(X >4)=Pr(X =5)+Pr(X =6)=0.13+0.23=0.36b Pr(2<X <5)=Pr(X =3)+Pr(X =4)=0.07+0.17=0.24c Pr(X ≥5|X ≥3)=Pr(X ≥5)Pr(X ≥3)(as X ≥5and X ≥3implies X ≥5)=Pr(X =5)+Pr(X =6)Pr(X =3)+Pr(X =4)+Pr(X =5)+Pr(X =6)=0.13+0.230.07+0.17+0.13+0.23=0.360.6=35The following distribution table gives the probabilities for the number of people on a carnival ride at a paticular time of day.No.of people (t )012345Pr(T =t )0.050.20.30.20.10.15Find:a Pr(T >4)b Pr(1<T <5)c Pr(T <3|T <4)Solutiona Pr(T >4)=Pr(T =5)=0.15b Pr(1<T <5)=Pr(T =2)+Pr(T =3)+Pr(T =4)=0.6c Pr(T <3|T <4)=Pr (T <3)Pr (T <4)=0.550.75=1115Exercise 13A1A random variable X can take the values x =1,2,3,4.Indicate whether or not each of thefollowing is a probability function for such a variable,and if not,give reasons:a p (1)=0.05p (2)=0.35p (3)=0.55p (4)=0.15b p (1)=0.125p (2)=0.5p (3)=0.25p (4)=0.0625Chapter13—Discrete Probability Distributions and Simulation347c p(1)=13%p(2)=69%p(3)=1%p(4)=17%d p(1)=51p(2)=12p(3)=34p(4)=3e p(1)=0.66p(2)=0.32p(3)=−0.19p(4)=0.212For each of the following write a probability statement in terms of the discrete random variable X showing the probability that:a X is equal to2b X is greater than2c X is at least2d X is less than2e X is2or moref X is more than2g X is no more than2h X is greater than or equal to2i X is less than or equal to2j X is no less than2k X is greater than2and less than53A random variable X can take the values0,1,2,3,4,5.List the set of values that X can take for each of the following probability statements:a Pr(X=2)b Pr(X>2)c Pr(X≥2)d Pr(X<2)e Pr(X≤2)f Pr(2≤X≤5)g Pr(2<X≤5)h Pr(2≤X<5)i Pr(2<X<5)4Consider the following function:x12345Pr(X=x)k2k3k4k5ka For what value of k is this a probability distribution?b Find Pr(2≤X≤4).5The number of‘no-shows’on a scheduled airlineflight has the following probability distribution:r01234567p(r)0.090.220.260.210.130.060.020.01Find the probability that:a more than four people do not show up for theflightb at least two people do not show up for theflight.6Suppose Y is a random variable with the distribution given in the table.y0.20.30.40.50.60.70.80.9 Pr(Y=y)0.080.130.090.190.200.030.100.18 Find:a Pr(Y≤0.50)b Pr(Y>0.50)c Pr(0.30≤Y≤0.80)348Essential Mathematical Methods1&2CAS7The table shows a probability distribution with random variable X.x123456Pr(X=x)0.10.130.170.270.200.13Give the following probabilities:a Pr(X>3)b Pr(3<X<6)c Pr(X≥4|X≥2)8Suppose that a fair coin is tossed three times.a List the eight equally likely outcomes.b If X represents the number of heads shown,determine Pr(X=2).c Find the probability distribution of the random variable X.d Find Pr(X≤2).e Find Pr(X≤1|X≤2).9When a pair of dice is rolled,36equally likely outcomes are possible.Let Y denote the sum of the dice.a What are the possible values of the random variable Y?b Find Pr(Y=7).c Determine the probability distribution of the random variable Y.10When a pair of dice is rolled,36equally likely outcomes are possible.Let X denote the larger of the values showing on the dice.If both dice come up the same,then X denotesthe common value.a What are the possible values of the random variable X?b Find Pr(X=4).c Determine the probability distribution of the random variable X.11A dart is thrown at a circular board with a radius of10cm.The board has three rings:a bullseye of radius3cm,a second ring with an outer radius of7cm,and a third ring withouter radius10cm.Assume that the probability of the dart hitting a region R is given byPr(R)=area of R area of dartboarda Find the probability of scoring a bullseye.b Find the probability of hitting the middle ring.c Find the probability of hitting the outer ring.12Suppose that a fair coin is tossed three times.Y ou lose$3.00if three heads appear and $2.00if two heads appear.Y ou win$1.00if one head appears and$3.00if no heads appear;$Y is the amount you win or lose.a Find the probability distribution of the random variable Y.b Find Pr(Y≤1).Chapter 13—Discrete Probability Distributions and Simulation 34913.2Sampling without replacementConsider the sort of probability distribution that arises from the most common sampling situation.A jar contains three mints and four toffees,and Bob selects two (without looking).If the random variable of interest is the number of mints he selects,then this can take values of 0,1or 2.Suppose this was done experimentally many times (say 50).The following results may be obtained:Number of mints 012Number of times observed 18248Let X be the number of mints Bob selects.From the table,the probabilities can be estimated for each outcome as:Pr(X =0)≈1850=0.36Pr(X =1)≈2450=0.48Pr(X =2)≈850=0.16Obviously,it is not desirable that an experiment needs to be carried out every time a situation like this arises and it is not necessary.It is possible to work out the theoretical probability for each value of the random variable by using the knowledge of combinations from Chapter 12.Consider the situation in which the sample of two sweets contains no mints.Then it must contain two toffees.Thus Bob has selected no mints from the three available,and two toffees from the four available,which gives the number of favourable outcomes as:30 42Now the number of possible choices Bob has of choosing two sweets from seven is 72 andthus the probability of Bob selectingno mints is:Pr(X =0)= 30 4272=621=27350Essential Mathematical Methods 1&2CASSimilarly,we can determine:Pr(X =1)= 31 41 72 and Pr(X =2)= 32 4072=1221=321=47=17Thus the probability distribution for X is:x 012Pr(X =x )274717Note that the probabilities in the table add up to 1.If they did not add to 1it would be known that an error had been made.This problem can also easily becompleted with a tree diagram.MintTherefore Pr(X =0)=47×36=27Pr(X =1)=47×36+37×46=27+27=47and Pr(X =2)=37×26=17This can be considered as a sequence of two trials in which the second is dependent on the first.Instead of evaluating the probabilities for all the values of X and listing them in a table,the probability distribution could be given as a rule and,providing the values of X are specified for which the rule is appropriate,the same information as before is available.In this case the rule is:Pr(X =x )= 3x42−x72,x =0,1,2This example is an application of a distribution which is commonly called the hypergeometric distribution .Chapter 13—Discrete Probability Distributions and Simulation351Marine biologists are studying a group of dolphins which live in a small bay.They know there are 12dolphins in the group,four of which have been caught,tagged and released to mix back into the population.If the researchers return the following week and catch another group of three dolphins,what is the probability that two of these will already be tagged?SolutionLet X equal the number of tagged dolphins in the second sample.We wish to know the probability of selecting two of the four tagged dolphins,andone of the eight non tagged dolphins,when a sample of size 3is selected from apopulation of size 12.That is:Pr(X =2)= 42 81123=1255Exercise 13B1A company employs 30salespersons,12of whom are men and 18are women.Fivesalespersons are to be selected at random to attend an important conference.What is the probability of selecting two men and three women?2An electrical component is packaged in boxes of 20.A technician randomly selects threefrom each box for testing.If there are no faulty components,the whole box is passed.If there are any faulty components,the box is sent back for further inspection.If a box is known to contain four faulty components,what is the probability it will pass?3A pond contains seven gold and eight black fish.If three fish are caught at random in a net,find the probability that at least one of them is black.4A researcher has caught,tagged and released 10birds of a particular species into the forest.If there are known to be 25of this species of bird in the area,what is the probability that another sample of five birds will contain three tagged ones?5A tennis instructor has 10new and 10used tennis balls.If he selects six balls at random touse in a class,what is the probability that there will be at least two new balls?6A jury of six persons was selected from a group of 18potential jurors,of whom eight werefemale and 10male.The jury was supposedly selected at random,but it contained only one female.Do you have any reason to doubt the randomness of the selection?Explain your reasons.352Essential Mathematical Methods 1&2CAS 13.3Sampling with replacement:the binomial distributionSuppose a fair six-sided die is rolled four times and a random variable X is defined asthenumber of 3s observed.An approximate probability distribution for this random variable could be found by repeatedly rolling the die four times,and observing the outcomes.On one occasion the four rolls of the die was repeated 100times and the following resultsnoted:x 01234No.of times observed 5044510From this table the probabilities for each outcome could be estimated:Pr(X =0)≈50100=0.50Pr(X =1)≈40100=0.44Pr(X =2)≈5100=0.05Pr(X =3)≈1100=0.01Pr(X =4)≈0100=0The theoretical probability distribution can be determined in the following way.One possible outcome of the experiment is TTNN ,where T represents a 3and N represents not a 3.The probability of this particular outcome (that is,in this order)is16×16×56×56= 16 2562How many different arrangements of T,T,N ,and N are there?Listing we find that there are six:TTNN,TNTN,TNNT,NTTN,NTNT,NNTT .The number of arrangements could be found without listing them by recognising that this is equal to 42 ,the number of ways of placingthe two T s in the four available places.Thus,the probability of obtaining exactly two 3s when a fair die is tossed four times is:Pr(X =2)=42 16 2 562=25216Continuing in this way the entire probability distribution can be defined as given in the table.(Note that the probabilities shown do not add to exactly 1owing to rounding errors.)x 01234Pr(X =x )0.48220.38580.11570.01540.0008It would be convenient to be able to use a formula to summarise the probability distribution.In this case it is:Pr(X =x )= 4x 16x 56 n −xx =0,1,2,3,4This is an example of the binomial probability distribution ,which has arisen from abinomial experiment.A binomial experiment is one that possesses the following properties:The experiment consists of a number,n ,of identical trials.Each trial results in one of two outcomes,which are usually designated as either a success ,S ,or a failure ,F .The probability of success on a single trial,p say,is constant for all trials (and thus the probability of failure on a single trial is (1–p )).The trials are independent (so that the outcome on any trial is not affected by the outcome of any previous trial).The random variable of interest,X ,is the number of successes in n trials of a binomial experiment.Thus,X has a binomial distribution and the rule is:Pr(X=x )= n x (p )x(1−p )n −x x =0,1,...,nwhere n x =n !x !(n −x )!Rainfall records for the city of Melbourne indicate that,on average,the probability of rain falling on any one day in November is 0.4.Assuming that the occurrence of rain on any day is independent of whether or not rain falls on any other day,find the probability that rain will fall on any three days of a chosen week.SolutionSince there are only two possible outcomes on each day (rain or no rain),theprobability of rain on any day is constant (0.4)regardless of previous outcomes.The situation described is a binomial experiment.In this example occurrence of rain isconsidered as a success,and so define X as the number of days on which it rains in a given week.Thus X is a binomial random variable with p =0.4and n =7.Pr(X=x)=7x(0.4)x(0.6)7−x,x=0,1,...,7and Pr(X=3)=73(0.4)3(0.6)7−3=7!3!4!×0.064×0.13=0.290304(values held in calculator)the binomialcumulativegiven in Example6.Example6For the situation described calculator tofind the probabilitywhere n=7,p=0.4.shown.Once again it is found Pr(X=3)=0.2903.Here Pr(X ≤3)=0.7102.This time Pr(X ≥3)is required:Pr(X ≥3)=Pr(X =3)++Pr(X =5)+Pr(X =6)The CAS calculator does built-in function to calculate but it can be used to evaluate probability by recognising that:Pr(X ≥3)=1−(Pr(X ==1)+Pr(X =2))=1−Pr(X Here Pr(X ≤2)=and hence Pr(X ≥3)==0.5801CAS calculator can also display thebinomial probability distribution graphically.Thisshown in Example 7.Example 7Use the CAS calculator to plot the probability distribution functionPr(X =x )= n x p x (1x =0,1,...,nentry for Binomial Pdf without in List 1and then a scatterplotExample8The probability of winningExample9The probability of an archer from a shot is0.4.Find themaximum score:maximum score at least once.shots.b Pr(X=3|X>0)=Pr(X=3) Pr(X>0)=0.23041−Pr(X=0)=0.23041−0.65=0.23040.92224=0.2498(correct to4decimal places) Exercise13C1For the binomial distribution Pr(X=x)=6x(0.3)x(0.7)6−x,x=0,1,...,6,find:a Pr(X=3)b Pr(X=4)2For the binomial distribution Pr(X=x)=10x(0.1)x(0.9)10−x,x=0,1,...,10,find:a Pr(X=2)b Pr(X≤2)3A fair die is e your CAS calculator tofind the probability of observing:a exactly ten6sb fewer than ten6sc at least ten6s4Rainfall records for the city of Melbourne indicate that,on average,the probability of rain falling on any one day in November is0.35.Assuming that the occurrence of rain on any day is independent of whether or not rain falls on any other day,find the probability that:a rain will fall on thefirst three days of a given week,but not on the other fourb rain will fall on exactly three days of a given week,but not on the other fourc rain will fall on at least three days of a given week.5A die is rolled seven times and the number of2s that occur in the seven rolls is noted.Find the probability that:a thefirst roll is a2and the rest are notb exactly one of the seven rolls results in a2.6If the probability of a female child being born is0.5,use your CAS calculator tofind the probability that,if100babies are born on a certain day,more than60of them will be female.7A breakfast cereal manufacturer places a coupon in every tenth packet of cereal entitling the buyer to a free packet of cereal.Over a period of two months a family purchasesfive packets of cereal.a Find the probability distribution of the number of coupons in thefive packets.b What is the most probable number of coupons in thefive packets?8If the probability of a female child being born is0.48,find the probability that a family with exactly three children has at least one child of each sex.9An insurance company examines its records and notes that30%of accident claims are made by drivers aged under21.If there are100accident claims in the next12months,use your CAS calculator to determine the probability that40or more of them are made by drivers aged under21.10A restaurant is able to seat80customers inside,and many more at outside tables.Generally,80%of their customers prefer to sit inside.If100customers arrive one day,use your CAS calculator to determine the probability that the restaurant will seat inside all those who make this request.11A supermarket has four checkouts.A customer in a hurry decides to leave without making a purchase if all the checkouts are busy.At that time of day the probability of each checkout being free is0.25.Assuming that whether or not a checkout is busy isindependent of any other checkout,calculate the probability that the customer will make a purchase.12An aircraft has four engines.The probability that any one of them will fail on aflight is0.003.Assuming the four engines operate independently,find the probability that on aparticularflight:a no engine failure occursb not more than one engine failure occursc all four engines fail13A market researcher wishes to determine if the public has a preference for one of two brands of cheese,brand A or brand B.In order to do this,15people are asked to choose which cheese they prefer.If there is actually no difference in preference:a What is the probability that10or more people would state a preference for brand A?b What is the probability that10or more people would state a preference for brand A orbrand B?14It has been discovered that4%of the batteries produced at a certain factory are defective.A sample of10is drawn randomly from each hour’s production and the number ofdefective batteries is noted.In what percentage of these hourly samples would there be a least two defective batteries?Explain what doubts you might have if a particular sample contained six defective batteries.15An examination consists of10multiple-choice questions.Each question has four possible answers.At leastfive correct answers are required to pass the examination.a Suppose a student guesses the answer to each question.What is the probability thestudent will make:i at least three correct guesses?ii at least four correct guesses?iii at leastfive correct guesses?b How many correct answers do you think are necessary to decide that the student is notguessing each answer?Explain your reasons.16An examination consists of20multiple-choice questions.Each question has four possible answers.At least10correct answers are required to pass the examination.Suppose the student guesses the answer to each e your CAS calculator to determine theprobability that the student passes.17Plot the probability distribution function Pr(X=x)= nxp x(1−p)n−x,x=0,1,...,n,for n=10and p=0.318Plot the probability distribution function Pr(X=x)= nxp x(1−p)n−x,x=0,1,...,n,for n=15and p=0.6.19What is the least number of times a fair coin should be tossed in order to ensure that:a the probability of observing at least one head is more than0.95?b the probability of observing more than one head is more than0.95?20What is the least number of times a fair die should be rolled in order to ensure that:a the probability of observing at least one6is more than0.9?b the probability of observing more than one6is more than0.9?21Geoff has determined that his probability of hitting an ace when serving at tennis is0.1.What is the least number of balls he must serve to ensure that:a the probability of hitting at least one ace is more than0.8?b probability of hitting more than one ace is more than0.8?22The probability of winning in a game of chance is known to be0.05.What is the least number of times Phillip should play the game in order to ensure that:a the probability that he wins at least once is more than0.90?b the probability that he wins at least once is more than0.95?23The probability of a shooter obtaining a maximum score from a shot is0.7.Find the probability that out offive shots the shooter obtains the maximum score:a three timesb three times,given that it is known that he obtains the maximum score at least once.24Each week a securityfirm transports a large sum of money between two places.The day on which the journey is made is varied at random and,in any week,each of thefive daysfrom Monday to Friday is equally likely to be chosen.(In the following,give answerscorrect to4decimal places.)Calculate the probability that in a period of10weeks Friday will be chosen:a two timesb at least two timesc exactly three times,given it is chosen at least two times13.4Solving probability problems using simulationSimulation is a very powerful and widely used procedure which enables us tofind approximate answers to difficult probability questions.It is a technique which imitates the operation of thereal-world system being investigated.Some problems are not able to be solved directly andsimulation allows a solution to be obtained where otherwise none would be possible.In thissection some specific probability problems are looked at which may be solved by usingsimulation,a valuable and legitimate tool for the statistician.What is the probability that a family offive children will include at least four girls?SolutionThis problem could be simulated by tossing a coinfive times,once for each child,using a model based on the following assumptions:There is a probability of0.5of each child being female.The sex of each child is independent of the sex of the other children.That is,theprobability of a female child is always0.5.Since the probability of a female child is0.5,then tossing a fair coin is a suitablesimulation model.Let a head represent a female child and a tail a male child.A trialconsists of tossing the coinfive times to represent one complete family offivechildren and the result of the trial is the number of female children obtained in thetrial.To estimate the required probability several trials need to be conducted.Howmany trials are needed to estimate the probability?As we have already noted inSection8.2,the more repetitions of an experiment the better the estimate of theprobability.Initially about50trials could be considered.An example of the results that might be obtained from10trials is given in the tableon the next page:Trial number Simulation results Number of heads1THHTT 22HHHTH 43HHHTH 44HTTTH 25HTHHH 46HTTTH 27TTHHH 38HTHHT 39TTTHH 210HHTTT 2Continuing in this way,the following results were obtained for 50trials:Number of heads Number of times obtained011821731341051The results in the table can be used to estimate the required probability.Since atleast four heads were obtained in 11trials,estimate the probability of at least fourfemale children as 1150or 0.22.Of course,since this probability has been estimatedexperimentally,repeating the simulations would give a slightly different result,but we would expect to obtain approximately this value most of the time.Example 10can be recognised as a situation involving a binomial random variable,with n =5and p =0.5.Thus the exact answer to the question ‘What is the probability that a family of five children will include at least four girls?’is:Pr(X ≥4)= 54(0.5)4(0.5)1+ 55 (0.5)5(0.5)0=0.1875This is reasonably close to the answer obtained from the simulation.In Example 10simulation was used to provide an estimate of the value of a particularprobability.Simulation is also widely used to estimate the values of other quantities which are of interest in a probability problem.We may wish to know the average result,the largest result,the number of trials required to achieve a certain result,and so on.An example of this type of problem is given in Example 11.。