Continuous Probability Distributions Chapter 7 Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

Learning Objectives LO7-1 Describe the uniform probability distribution and use it to calculate probabilities. LO7-2 Describe the characteristics of a normal probability distribution. LO7-3 Describe the standard normal probability distribution and use it to calculate probabilities. LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate probabilities. LO7-5 Describe the exponential probability distribution and use it to calculate probabilities. 7-*

均等分配 The Uniform Distribution LO7-1 Describe the uniform probability distribution and use it to calculate probabilities. 均等分配 The Uniform Distribution 均等分配是最簡單的連續隨機變數分配。 The uniform probability distribution is perhaps the simplest distribution for a continuous random variable. 此分配為方形分配，由最大值與最小值定義其範圍，此方形的面積＝1（亦即：機率總和為1） This distribution is rectangular in shape and is defined by minimum (a) and maximum (b) values. 7-*

The Uniform Distribution – Mean and Standard Deviation LO7-1 The Uniform Distribution – Mean and Standard Deviation 在均等分配中，若知道最大值與最小值，我們就能定義均等分配的機率函數（可據此求機率），也能根據此函數求得平均數、變異數、標準差。 Knowing the minimum and maximum values of a uniform distribution, we can define the probability function, and calculate the mean, variance, and standard deviation of the distribution. 7-*

The Uniform Distribution – Example (p.208) LO7-1 The Uniform Distribution – Example (p.208) 西南亞利桑納州立大學提供通勤公車，週一至週五從早上6點到晚上11點，每半小時一班公車（由北主街到校園），學生等公車的時間從0分至30分呈均等分配。 Southwest Arizona State University provides bus service to students. On weekdays, a bus arrives at the North Main Street and College Drive stop every 30 minutes between 6 a.m. and 11 p.m. Students arrive at the bus stop at random times. The time that a student waits is uniformly distributed from 0 to 30 minutes. 1.請繪製此分配的圖形 2.請說明其總面積為1 3.學生等公車「一般」要等多久？換句話說，平均等候時間為？等候時間的標準差為？ 4.學生等公車超過25分鐘的機率？ 5.學生等公車介於10到20分鐘的機率？ Draw a graph of this distribution. Show that the area of this uniform distribution is 1.00. How long will a student “typically” have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times? What is the probability a student will wait more than 25 minutes? What is the probability a student will wait between 10 and 20 minutes? 7-*

The Uniform Distribution – Example (p.209) LO7-1 The Uniform Distribution – Example (p.209) 1. Graph of uniformly distributed waiting times between 0 and 30: P(X)= 1/(30-0) = 0.0333 7-*

The Uniform Distribution – Example (p.209) LO7-1 The Uniform Distribution – Example (p.209) 2. Show that the area of this distribution is 1.00. 7-*

The Uniform Distribution – Example (p.209) LO7-1 The Uniform Distribution – Example (p.209) 3. How long will a student “typically” have to wait for a bus? In other words what is the mean waiting time? What is the standard deviation of the waiting times? 7-*

The Uniform Distribution – Example (p.209-210) LO7-1 The Uniform Distribution – Example (p.209-210) 4. What is the probability a student will wait more than 25 minutes? 計算介於25到30間的面積＝等候時間大於25分鐘的機率 7-*

The Uniform Distribution – Example (p.210) LO7-1 The Uniform Distribution – Example (p.210) 5. What is the probability a student will wait between 10 and 20 minutes? 計算介於10到20分鐘之間的面積＝其機率 7-*

Self-review 7-1 (p.210) 澳洲的牧羊犬壽命相對較短，他們的壽命介於8到14歲之間，且呈現均等分配。問： (a). 請繪製此均等分配的圖形 (b). 說明此分配的面積為1 (c). 計算其平均值、標準差 (d). 某隻牧羊犬壽命介於10到14歲之機率？ (e). 牧羊犬壽命低於9歲的機率？ (f). 牧羊犬壽命正好等於9歲的機率？ What is the probability a dog will live exactly 9 years?

Self-review 7-1 (p.210) (a) a=8, b=14 縱軸截距 (機率): 1/(b-a)=1/(14-8)=0.167 (b) [1/(14-8)]*(14-8)=1 (c) mean=(a+b)/2=(14+8)/2=11 s.d. = √(14-8)2/12= 1.732 (d) P(10<X<14)= [1/(14-8)]*(14-10)=0.667 (e) P(X<9)= [1/(14-8)]*(9-a)=0.167*(9-8)=0.167 (f) P(X=9)=0.167*(9-9)=0

常態分配 (P.211) Normal Probability Distribution μ: 母體平均值 σ: 母體標準差 π是固定值＝3.1416 e 是固定值 =2.718 所以，常態分配基本上是由平均數、標準差來定義的 因此，常態分配可表示為：N(μ,σ2)

Characteristics of a Normal Probability Distribution LO7-2 Describe the characteristics of a normal probability distribution. Characteristics of a Normal Probability Distribution It is bell-shaped and has a single peak at the center of the distribution. It is symmetrical about the mean. It is asymptotic: The curve gets closer and closer to the X-axis but never actually touches it. To put it another way, the tails of the curve extend indefinitely in both directions. The location of a normal distribution is determined by the mean, . The dispersion or spread of the distribution is determined by the standard deviation, σ. The arithmetic mean, median, and mode are equal. As a probability distribution, the total area under the curve is defined to be 1.00. 因為對稱，平均值兩邊的機率值都是0.5 Because the distribution is symmetrical about the mean, half the area under the normal curve is to the right of the mean, and the other half to the left of it. 7-*

The Normal Distribution – Graphically LO7-2 The Normal Distribution – Graphically 7-*

The Family of Normal Distributions LO7-2 The Family of Normal Distributions Equal Means and Different Standard Deviations Different Means and Standard Deviations Different Means and Equal Standard Deviations 7-*

標準常態分配 The Standard Normal Probability Distribution LO7-3 Describe the standard normal probability distribution and use it to calculate probabilities. 標準常態分配 The Standard Normal Probability Distribution The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1. 亦即：N(1,0) It is also called the z distribution. A z-value is the signed distance between a selected value, designated x, and the population mean, , divided by the population standard deviation, σ. The formula is: 7-*

The Normal Distribution – Example (p.215) LO7-3 The Normal Distribution – Example (p.215)   玻璃產業的領班週薪呈現常態分配，平均值為＄1000，標準差為＄100。 The weekly incomes of shift foremen in the glass industry follow the normal probability distribution with a mean of $1,000 and a standard deviation of $100. 若某領班的週薪為＄1100，請問其薪資的z值為何？若另一領班的週薪為＄900，其z值為何？ What is the z value for the income, let’s call it x, of a foreman who earns $1,100 per week? For a foreman who earns $900 per week? 7-*

LO7-3 使用機率表：查第732頁的常態分配表 Areas Under the Normal Curve Using a Standard Normal Table 7-*

經驗法則（再測試）查第732頁的常態分配表 The Empirical Rule – Verification LO7-3 經驗法則（再測試）查第732頁的常態分配表 The Empirical Rule – Verification For z=1.00, the table’s value is 0.3413; times 2 is 0.6826. For z=2.00, the table’s value is 0.4772; times 2 is 0.9544. For z=3.00, the table’s value is 0.4987; times 2 is 0.9974. 7-*

課本範例：p.216 Autolite電池公司的品管部門測試電池壽命，1號(i.e. D-cell)電池平均壽命為19小時，電池壽命通常符合常態分配，標準差為1.2小時， 請問： 1.有68％的電池壽命介於哪兩個數值之間？ 2.有95％的電池壽命介於哪兩個數值之間？ 3.幾乎全部電池壽命介於哪兩個數值之間？

P.217 可用經驗法則來算： 1. 68％的電池壽命介於±1個標準差的範圍內：亦即 19 ±1*1.2  (17.8, 20.2) 小時 2. 95％的電池壽命介於±2個標準差的範圍內：亦即 19 ±2*1.2  (16.6, 21.4) 小時 3. 幾乎100%的電池壽命介於±3個標準差的範圍內：亦即 19 ±3*1.2  (15.4, 22.6) 小時

Normal Distribution – Finding Probabilities (Example 1) p.215, p.218 LO7-3 Normal Distribution – Finding Probabilities (Example 1) p.215, p.218 前例中領班週薪呈現平均值＄1,000，標準差＄100的常態分配。 In an earlier example we reported that the mean weekly income of a shift foreman in the glass industry is normally distributed with a mean of $1,000 and a standard deviation of $100. 請問隨便選個領班，他的週薪介於$1000到$1100之間的概率（機率）為何？ What is the likelihood of selecting a foreman whose weekly income is between $1,000 and $1,100? 7-*

Normal Distribution – Finding Probabilities (Example 1) LO7-3 Normal Distribution – Finding Probabilities (Example 1) 7-*

Finding Areas for z Using Excel p.219 LO7-3 Finding Areas for z Using Excel p.219 The Excel function: =NORM.DIST(x,Mean,Standard_dev,Cumu) =NORM.DIST(1100,1000,100,true) calculates the probability (area) for z=1. 7-*

Normal Distribution – Finding Probabilities (Example 2) p. 220 LO7-3 Normal Distribution – Finding Probabilities (Example 2) p. 220 再續前例 Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. 隨便選個領班，他的週薪介於$790到$1000之間的機率為何？ What is the probability of selecting a shift foreman in the glass industry whose income is between $790 and $1,000? 那麼剛剛選出的這個領班的週薪低於＄790的機率為何？ 7-*

Normal Distribution – Finding Probabilities (Example 3) p. 220 LO7-3 Normal Distribution – Finding Probabilities (Example 3) p. 220 續前例 Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. 剛剛抽出的領班，週薪低於＄790的機率為何？ What is the probability of selecting a shift foreman in the glass industry whose income is less than $790? The probability of selecting a shift foreman with income less than $790 is 0.5 - .4821 = .0179. 7-*

Normal Distribution – Finding Probabilities (Example 4) p.221 LO7-3 Normal Distribution – Finding Probabilities (Example 4) p.221 續前例 Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. 剛剛抽出的領班，週薪介於$840到＄1,200的機率為何？ What is the probability of selecting a shift foreman in the glass industry whose income is between $840 and $1,200? 7-*

Normal Distribution – Finding Probabilities (Example 5) p.222 LO7-3 Normal Distribution – Finding Probabilities (Example 5) p.222 續前例 Refer to the information regarding the weekly income of shift foremen in the glass industry. The distribution of weekly incomes follows the normal probability distribution with a mean of $1,000 and a standard deviation of $100. 剛剛抽出的領班，週薪介於$1,150到＄1,250的機率為何？ What is the probability of selecting a shift foreman in the glass industry whose income is between $1,150 and $1,250? 7-*

Using Z to Find X for a Given Probability – Example p.224 LO7-3 Using Z to Find X for a Given Probability – Example p.224 Layton 輪胎、橡皮公司想要對新胎MX100訂出最低保證的里程數，經測試後，平均里程數為67900英里，標準差為2050英里，且里程數服從常態分配。他們希望定下最低保證里程數後，換胎比例不超過4％ 請問他們該定最低保證多少英里？ Layton Tire and Rubber Company wishes to set a minimum mileage guarantee on its new MX100 tire. Tests reveal the mean mileage is 67,900 with a standard deviation of 2,050 miles and that the distribution of miles follows the normal probability distribution. Layton wants to set the minimum guaranteed mileage so that no more than 4 percent of the tires will have to be replaced. What minimum guaranteed mileage should Layton announce? 7-*

如何查表？如何找答案？P.225 因為凡輪胎壽命在最低里程數以下的，公司都保證免費換胎，所以必須查出低於某z值的機率必須≦4％

Using Z to Find X for a Given Probability – Example p.225 LO7-3 Using Z to Find X for a Given Probability – Example p.225 7-*

Using Z to Find X for a Given Probability – Example p.225 LO7-3 Using Z to Find X for a Given Probability – Example p.225 7-*

回顧第六章：Poisson vs Binomial Distribution Poisson分配乃是在連續區間內事件發生的次數，事件可在區間內任何一點發生，若將連續空間細分成無數個微小區間（亦即：n很大），每個微小區間內，事件發生就只有兩種可能：發生(x=1) 或不發生(x=0)，這樣，Poisson就成為n個白努利試驗(trials)，且成功的機率π=μ/n，或μ=nπ 所以，n很大（且π較小）的二項分配與普瓦松分配非常近似，故而，普瓦松分配在符合上述條件下，可作為二項分配的近似分配。 但若n很大，而π不夠小時，普瓦松分配與二項分配的機率不夠接近，反倒是二項分配比較接近常態分配，因此，可用常態分配來求二項分配的機率值（見下一章）。

回顧第六章：超幾何、二項與Poisson三種分配 之間的關係 當n/N≤0.05時可用二項分配取代超幾分配 當n>20且np≤7時，可用Poisson分配取代二項分配 超幾何分配 二項分配 Poisson分配 超幾何、二項與Poisson三種分配之間的關係

Normal Approximation to the Binomial (p.226) 當二項分配中的觀察點數目很大時，計算其隨機變量對應的機率，非常繁瑣！ e.g. 如果n =60, 求 P(x =33) = 60C33 (π)33 (1-π)27 但二項分配的觀察點數很大時，其分配趨近常態分配 觀察點數目要多大？ n > 5 and n(1-) >5 36

Normal Approximation to the Binomial LO7-4 Approximate the binomial probability distribution using the standard normal probability distribution to calculate probabilities. Normal Approximation to the Binomial The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n. The normal probability distribution is generally a good approximation to the binomial probability distribution when nπ and n(1-π ) are both greater than 5. 7-*

Normal Approximation to the Binomial LO7-4 Normal Approximation to the Binomial Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of n seems reasonable because, as n increases, a binomial distribution gets closer and closer to a normal distribution. 7-*

Continuity Correction Factor (p.228) 由於二項分配是間斷機率分配，常態分配是連續機率分配，若以常態分配來代替二項分配求其機率的話，必須做連續性調整。 也就是：將二項隨機變數之值加或減0.5 這0.5稱為連續性調整因子 (p.228) 二項機率: P(X=60) 常態分配: P (59.5<X<60.5) 39

Continuity Correction Factor LO7-4 Continuity Correction Factor The value .5 is subtracted or added, depending on the problem, to a selected value when a binomial probability distribution (a discrete probability distribution) is being approximated by a continuous probability distribution (the normal distribution). 7-*

How to Apply the Correction Factor LO7-4 How to Apply the Correction Factor Only one of four cases may arise: 1. For the probability at least X occurs, use the area above (X -0.5). 2. For the probability that more than X occurs, use the area above (X+0.5). 3. For the probability that X or fewer occurs, use the area below (X -0.5). 4. For the probability that fewer than X occurs, use the area below (X+0.5). 7-*

連續性修正因子 1/3 圖 連續性修正因子±1/2

連續性修正因子 2/3 設X～B(n,π)，當nπ≥5且n(1–π)≥5，則存在下列近似式 P(X≤d) P(X≥c) P(c≤X≤d)

連續性修正因子 3/3 設X～B(n,π)，當nπ≥5且n(1–π)≥5，則存在下列近似式 P(X<d) P(X>c) P(c<X<d) 式中μ=n．π ，σ=

Normal Approximation to the Binomial – Example p.227-229 LO7-4 Normal Approximation to the Binomial – Example p.227-229 若Santoni披薩餐廳發現70％的新顧客會再回來光顧，某週有80位新顧客光臨，其中60位或以上的顧客會再度光臨的機率為何？ Suppose the management of the Santoni Pizza Restaurant found that 70 percent of its new customers return for another meal. For a week in which 80 new (first-time) customers dined at Santoni’s, what is the probability that 60 or more will return for another meal? 7-*

Normal Approximation to the Binomial – Example （查二項分配表）p.228 LO7-4 Normal Approximation to the Binomial – Example （查二項分配表）p.228 If we use the binomial probabilities, we would compute: P(X ≥ 60) = 0.063+0.048+ … + 0.001) = 0.197 7-*

Normal Approximation to the Binomial – Example （用常態分配算機率）p.229 LO7-4 Normal Approximation to the Binomial – Example （用常態分配算機率）p.229 Step 1. Find the mean and the variance of a binomial distribution and find the z corresponding to an x of 59.5 (x - 0.5, the correction factor). Step 2: Determine the area from 59.5 and beyond. We now demonstrate here how the normal distribution is applied to the 7-*

例題 假設日本有12％家庭保地震險，某保險公司為推廣業務而作市調，隨機抽出900個家庭，令X為其中有保地震險的家庭數，問： a. X為何種分配？ b. 若用Poisson分配來計算（替代）此分配的機率合適嗎？Why? c. 求X的期望值與標準差 d. 以常態分配來求機率合適嗎？問至多126個家庭保地震險的機率為？ 解答： a. X為二項分配, n = 900, π= 0.12, E(X) = nπ= 108 b. 不合適，因為 nπ= 108 太大 c. E(X) = nπ= 108, σ= (900*0.12*0.88)0.5 = 9.75 d.合適，因為nπ= 108 > 5、且n(1-π)=792 > 5 P(x≦126) = P(z≦126+0.5-108/9.75) = P(z≦1.897)=0.9713

常態分配的加法定理 定理1：假設X~N(μ,σ2), 若W=a+bx, 則W呈何種機率分配？ W~N(a+bμ, b2σ2) 若X是常態分配，其線性函數W=a+bx 也是常態分配 例：ucc三合一咖啡每包重量X~N(12, 0.52), 亦知每包成本Y是每包重量的函數：Y=0.5+0.45X，請問：每包成本的平均數為何？變異數為何？ μy=a+bμx=0.5+0.45×12=5.9 σy=b2σx2=(0.45) 2 0.52=0.05

常態分配的加法定理 定理2：假設X~N(μx,σx2), Y~N(μy,σy2) 若W=aX+bY, 則W呈何種機率分配？ W~N(aμx+bμy, a2σx2+b2σy2)

常態分配的加法定理─範例 假設ucc咖啡每包成本與售價為一常態分配。每包成本平均為5.9元，變異數為0.05。每包售價平均10元，變異數為1。 批發給零售商的售價為75折。請問：ucc咖啡公司每包利潤的平均數與變異數為何？又每包利潤成何種分配？ 成本X~N (5.9, 0.05) 售價Y~N (10, 1) 利潤Z = 0.75Y-X= -X+0.75Y  Z ~N (aμx+bμy, a2σx2+b2σy2) Mean (z) = -1*5.9+0.75*10 =7.5-5.9=1.6 Variance (z)= (-1)2*(0.05)+(0.75)2*(1)=0.05+0.5625=0.6125 Standard deviation = 0.782 (元)

指數分配 The Family of Exponential Distributions p.231 LO7-5 Describe the characteristics and compute probabilities using the exponential distribution. 指數分配 The Family of Exponential Distributions p.231 Characteristics and uses: 正偏，類似普瓦松分配 Positively skewed, similar to the Poisson distribution (for discrete variables) Not symmetric like the uniform and normal distributions 僅用一個參數定義，λ：平均發生頻率（次數…） Described by only one parameter, which we identify as λ, often referred to as the “rate” of occurrence parameter 當λ遞減，分配則越來越不偏 As λ decreases, the shape of the distribution becomes “less skewed.” The exponential distribution usually describes inter-arrival situations such as: • The service times in a system • The time between “hits” on a web site • The lifetime of an electrical component • The time until the next phone call arrives in a customer service center 7-*

指數分配 vs 普瓦松分配 例如：餐廳每小時平均有6為顧客上門(λ=6)，普瓦松分配機率：求某特定時段，2位顧客上門的機率。E(X)＝V(X)=λ 若用指數分配，平均每小時6位顧客上門，代表平均10分鐘有1位顧客上門，或 μ=1/λ=1/6小時 P(x) = λ e-λx 、E(X) = μ=1/ λ、 V(X) =μ2= 1/ λ2 X:抵達時間 P(抵達時間< x) = 1 - e-λx 、 P(抵達時間> x) = e-λx 注意： P(抵達時間< x) = P(抵達時間≦x) 因為 P(抵達時間= x) = 0

Exponential Distribution – Example p.232-233 LO7-5 Exponential Distribution – Example p.232-233 某網路藥局的下單頻率服從指數分配，平均每20秒鐘接一個藥單。 Orders for prescriptions arrive at a pharmacy management website according to an exponential probability distribution at a mean of one every twenty seconds. 請問：下一個藥單 1) 不到5秒內可接到的機率是？ 2) 超過40秒才接到的機率是？ 3) 介於5到40秒接到的機率是？ Find the probability the next order arrives in: 1) less than 5 seconds, 2) more than 40 seconds, 3) or between 5 and 40 seconds. 7-*

P.232 例題如何解？ 先找λ= 1/20 = 0.05（因為平均每20秒接一個藥單,故μ=20, λ = 1/μ，每秒接0.05個藥單） 因P(下一藥單抵達時間< x) = 1 - e-λx 故 1) P(下一藥單抵達時間< 5) = 1 - e-(0.05)*5 =1-0.7788 = 0.2212 2) P(下一藥單抵達時間>40) = e-(0.05)*40 = 0.1353 3) P(5<下一藥單抵達時間< 40) = P(下一藥單抵達時間< 40) - P(下一藥單抵達時間<5) = (1-0.1353)-0.2212 = 0.6435

Exponential Distribution – Example p.232-233 LO7-5 Exponential Distribution – Example p.232-233 7-*

Exponential Distribution – Example p.234 LO7-5 Exponential Distribution – Example p.234 Compton電腦公司希望訂定該公司新電源設備的保證最低使用年限，品保測試發現設備使用壽命服從指數分配，平均可用4000小時，注意：4000小時為平均值，而非頻率，因此，λ＝1/4000＝0.00025(每小時的耗損率) Compton Computers wishes to set a minimum lifetime guarantee on its new power supply unit. Quality testing shows the time to failure follows an exponential distribution with a mean of 4000 hours. Note that 4000 hours is a mean and not a rate. Therefore, we must compute λ as 1/4000 or 0.00025 failures per hour. 該公司希望在保固期內僅5％電源設備耗損，他們該訂定多長的保固期？ Compton wants a warranty period such that only five percent of the power supply units fail during that period. What value should they set for the warranty period? 7-*

Exponential Distribution – Example p.235 因μ= 4000小時 （電源平均壽命）指數分配平均值 λ＝1/4000＝0.00025 (每小時電源壞掉的的頻率） 普瓦松分配平均值 且希望訂出x，使得電源壽命低於x的機率不大於0.05 0.05 = P(使用壽命、時間< x) = 1-e-λx ,  0.95 = e-λx = e-0.00025x ,  ln 0.95 = -0.00025x 上式可求出x（最低保用時數）

Exponential Distribution – Example p.235 LO7-5 Exponential Distribution – Example p.235 In this problem, we need to find the value of x or the average inter-arrival time so that the probability of a returned unit is 5%. 7-*

P.234 例題 若問：該公司生產的某個電源設備可以使用超過4500小時的機率為？ λ= 1/4000 = 0.00025 P(時間>4500) =1- P(時間<4500) = 1- (1-e-0.00025*4500) = e-0.00025*4500 = 0.3247

例題 某公司平均10天接3筆訂單，問：3天內接到下一筆訂單的機率為？至少需要5天才能接到下一筆訂單的機率為？ 解答： 此為指數分配 λ= 3/10 = 0.3（平均每天接0.3筆訂單）普瓦松分配的平均值 P(時間≦3) = 1-e-0.3*3 = 0.5934 P(時間≧5) = 1- P(時間≦5) = e-0.3*5 = 0.2231 注意：此指數分配的平均值為10/3= 3.33天，亦即：平均每3.33天接一筆訂單

例題 由A點開到B點的公車每隔15分鐘開一班車，某人沒看時刻表，就到站牌等車，請問他至少要等5分鐘的機率為？請用普瓦松分配與指數分配分別計算之。 解答： 普瓦松分配： （至少等5分鐘，所以5分鐘內來0班車） 而5分鐘區間平均來(5/15)班車 E(X) =μ= 1/3 普瓦松平均值(5分鐘區間) P(x=0) = (1/3)0 e-(1/3) /0! = = e-(1/3) = 0.7165 指數分配： E(X) =μ= 15 分鐘（平均15分鐘來一班車） λ= 1/15 = 0.067（平均每分鐘來0.067班車）普瓦松平均值(1分鐘區間) P(時間≧5) = = e-0.067*5 = e-(1/3) = 0.7165

表 Poisson隨機變數與指數隨機變數之比較 指數分配 1/3 表 Poisson隨機變數與指數隨機變數之比較 Poisson隨機變數 指數隨機變數 (a)20分鐘內，平均5部車子開進停車場(λ=5輛／20分鐘) (a)平均每隔4分鐘有一部車子開進停車場(μ=4分鐘／輛) (b)高速公路一每10公里平均有5個窪洞(λ=5個／10公里)。 (b)高速公路上，平均每2公里有1個洞(μ=2公里／個)。 (c)某一機器30分鐘內平故障3次(λ=3次／30分鐘)。 (c)某一機器平均每10分鐘故障1次(μ=10分鐘／次)

指數分配 2/3 設X為指數隨機變數，λ代表單位時間之平均次數，μ代表平均時間(每次) (6-21) (6-22)

指數分配 3/3 圖 指數分配機率之計算

各種分配的觀念與使用情形架構圖 各種分配的觀念與使用情形架構圖