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What is the probability distribution for the number of heads?

1 What is a Probability Distribution( 機率分配 )? Probability Distributions McGraw-Hill/Irwin Chapter 6 The McGraw-Hill Companies, Inc. 008 Experiment: Toss a coin three times. Observe the number of heads. The possible results are: zero heads, one head, two heads, and three heads. What is the probability distribution for the number of heads? Probability Distribution of Number of Heads Observed in 3 Tosses of a Coin Probability Distribution of Number of Heads Observed in 3 Tosses of a Coin The probability distribution for the number of heads is 1/8 a = 0 3/8 a = 1 PX ( = a) = 3/8 a = 1/8 a =

Random variable - a quantity resulting from an experiment that, by chance, can assume different values, associated with a probability value.

2 Characteristics of a Probability Distribution Random Variables ( 隨機變數 ) Experiment a coin is tossed 3 times. Outcomes eight outcomes Event one head occurs in the three tosses. Random variable - a quantity resulting from an experiment that, by chance, can assume different values, associated with a probability value. 5 6 Random Variables ( 隨機變數 ) Example: Random Variables 7 TTT TTH THT HTT THH HTH HHT HHH Random Variables W X Y Z W.P /8 3/8 3/8 1/8 8 Consider a random experiment of tossing three coins. X is a random variable, which assigns the number of heads to a number, associated with a probability value. 1/8 x = 0 3/8 x = 1 PX ( = x) = 3/8 x = 1/8 x = 3

3 Example: Random Variables Example : Random Variable 9 Consider the same random experiment of tossing three coins. Let Y be another random variable, which assigns the number of heads to a number, associated with a probability value. 1/8 y = 3 3/8 y = PY ( = y) = 3/8 y = 1 1/8 y = 0 10 Let X denote the number of spots of tossing two fair dice. The probability distribution for the number of spots as tossing two fair dice is 1/36 x = /36 x = 3 3/36 x = 4 4/36 x = 5 5/36 x = 6 P( X = x) = 6/36 x = 7 5/36 x = 8 4/36 x = 9 3/ 36 x = 10 / 36 x = 11 1/ 36 x = 1 Types of Random Variables Discrete Random Variables - Examples Discrete Random Variable ( 離散隨機變數 ) can assume only certain clearly separated values. It is usually the result of counting something. Continuous Random Variable ( 連續隨機變數 )can assume an infinite number of values within a given range. It is usually the result of some type of measurement The number of students in a class. The number of children in a family. The number of cars entering a carwash in a hour. Number of home mortgages approved by Coastal Federal Bank last week

4 Example: 離散隨機變數 Continuous Random Variables - Examples 已知某公司總機平均每 10 分鐘有 4 通電話, 則 5 分鐘內有 X 通電話的機率分配為 : x e PX ( = x) =, x= 0,1,,3,... x! The distance students travel to class. The time it takes an executive to drive to work. The length of an afternoon nap. The length of time of a particular phone call Example: 連續隨機變數期望值 (Expectation) 假設某股票的平均報酬率 5%, 報酬率標準差 30 % 假設今日股價 $10, 則一年後的股價 S 的機率分配為 : 假設 X是個隨機變數, 其機率分配函數為 f(x) 對任意給定的一個函數 g(x),g(x) 的期望值定義如下 : 1 lns ln PS ( = s) = e, 0< S< 0.3S π E[g(X)]= x x gxf ( ) ( x), X是離散隨機變數 gxf ( ) ( xdx ), X是連續隨機變數

5 Example: Random Variables Consider a random experiment of tossing three coins. X is a random variable, which assigns the number of heads to a number, associated with a probability value. 1/8 x = 0 3/8 1 3 x = PX ( = x) = E[X] = xpx ( = x) 3/8 x = x= 0 1/8 x = = = 期望值的性質假設 X 是一隨機變數, 其機率分配為 f(x),a, b 與 c 是常數, 則 : E[a] = a E[aX] = a E[X] E[aX + b] = ae[x]+b E[aX +bx+c]= ae[x ]+be[x]+c Features of a Discrete Distribution 離散機率分配 The main features of a discrete probability distribution are: The sum of the probabilities of the various outcomes is 1. The probability of a particular outcome is between 0 and 1. The outcomes are mutually exclusive. 一般來說, 離散機率分配可以寫成以下一般式 : PX ( = n) = p, j= 1,,3,... j 且滿足 j PX ( = n) = p = 1 j j= 1 j= 1 j

6 The Mean of a Probability Distribution MEAN The mean is a typical value used to represent the central location of a probability distribution. The mean of a probability distribution is also referred to as its expected value. Expected value of X is denoted by μ = E[X] The Variance, and Standard Deviation of a Probability Distribution Variance and Standard Deviation Measures the amount of spread in a distribution The variance is found by: 1. Subtract the mean from each value, and square this difference.. Multiply each squared difference by its probability. 3. Sum the resulting products. to arrive at. The standard deviation is found by taking the positive square root of the variance. σ = E[ (X-μ) ], the expected value of (X-μ). 1 Computation Form for Variance Mean, Variance, and Standard Deviation of a Probability Distribution - Example σ = E[( X μ) ] = E[ X μx + μ ] = EX [ ] μex [ ] + μ = EX [ ] μμ + μ = EX [ ] μ = EX ( EX) [ ] [ ] John Ragsdale sells new cars for Pelican Ford. John usually sells the largest number of cars on Saturday. He has developed the following probability distribution for the number of cars he expects to sell on a particular Saturday

7 Mean of a Probability Distribution - Example Variance and Standard Deviation of a Probability Distribution - Example 5 6 Bernoulli Distribution ( 百弩里分配 ) Binomial Probability Distribution ( 二項分配 ) 簡單隨機實驗只有兩種結果, 一種為成功, 機率為 π; 另一種失敗, 機率為 1- π 例如擲一公平銅板, 正面機率 0.5, 反面機率 0.5 生產線上抽取一產品樣本, 良品機率 π, 不良品機率 1- π 以隨機變數 X 表示簡單隨機實驗的結果 : X = 1 表示成功, 機率 π ; X = 0 表示失敗, 機率 1- π 隨機變數 X 的機率分配 : 丟擲一公平硬幣 10 次, 假設以 X 表示出現正面的次數, 則隨機變數 X 即是服從二項分配 (Binomial Distribution) X 的機率分配為 : 10 x 10 x PX ( = x) = C (0.5) (0.5), x= 0,1,,...,10 x 若丟擲 n 次, 則 X 的機率分配為 : ( ) n (0.5) x (0.5) n PX= x= C x, x= 0,1,,..., n x 7 f x x 1 x ( ) = π (1 π), X = 0,1, 0 π 1 8 丟擲一公平硬幣 n 次, 即是作 n 次 Bernoulli 實驗因此假設每次實驗結果為隨機變數 X i, 則 n X = X i i= 1 7

Binomial Probability Distribution ( 二項分配 ) Binomial Probability Formula Characteristics of a Binomial Probability Distribution The random variable is the result of

The outcomes are mutually exclusive. 9 30 Binomial Probability - Example Binomial Dist.

8 Binomial Probability Distribution ( 二項分配 ) Binomial Probability Formula Characteristics of a Binomial Probability Distribution The random variable is the result of counts. Each trial is independent and identical of any other trial. There are only two possible outcomes on a particular trial of an experiment. The outcomes are mutually exclusive Binomial Probability - Example Binomial Dist. Mean and Variance There are five flights daily from Pittsburgh via US Airways into the Bradford, Pennsylvania, Regional Airport. Suppose the probability that any flight arrives late is.0. What is the probability that none of the flights are late today?

Binomial Dist. Mean and Variance: Example Binomial Dist. Mean and Variance: Another Solution For the example regarding the number of late flights, recall that π =.

33 34 Binomial Distribution - Table Binomial Shapes for Varying π (n constant) Five percent of the worm gears produced by an automatic, highspeed Carter-Bell

9 Binomial Dist. Mean and Variance: Example Binomial Dist. Mean and Variance: Another Solution For the example regarding the number of late flights, recall that π =.0 and n = 5. What is the average number of late flights? What is the variance of the number of late flights? Binomial Distribution - Table Binomial Shapes for Varying π (n constant) Five percent of the worm gears produced by an automatic, highspeed Carter-Bell milling machine are defective. What is the probability that out of six gears selected at random none will be defective? Exactly one? Exactly two? Exactly three? Exactly four? Exactly five? Exactly six out of six?

Binomial Shapes for Varying n (π constant) Cumulative Distribution Function ( 累積分配函數 ) 假設 X 是一隨機變數, 其機率分配函數通常以函數形式表示, 以小寫的 f(x) 表示, f(x = x) = P(X = x) 以大寫的 F(x) 表示累積機率函數,F(x) 表示 X 發生小於 x 的機率, 定義如下 :

percent of front seat occupants used seat belts. A sample of 1 vehicles is selected. What is the probability the front seat occupants in at least 7 of the 1 vehicles are wearing seat belts?

10 Binomial Shapes for Varying n (π constant) Cumulative Distribution Function ( 累積分配函數 ) 假設 X 是一隨機變數, 其機率分配函數通常以函數形式表示, 以小寫的 f(x) 表示, f(x = x) = P(X = x) 以大寫的 F(x) 表示累積機率函數,F(x) 表示 X 發生小於 x 的機率, 定義如下 : F(x) = P(X x) Cumulative Binomial Probability Distributions Finite Population A study in June 003 by the Illinois Department of Transportation concluded that 76. percent of front seat occupants used seat belts. A sample of 1 vehicles is selected. What is the probability the front seat occupants in at least 7 of the 1 vehicles are wearing seat belts? A finite population is a population consisting of a fixed number of known individuals, objects, or measurements. Examples include: The number of students in this class. The number of cars in the parking lot. The number of homes built in Taipei

Characteristics of Hypergeometric Distribution ( 超幾何分配 ) Characteristics of Hypergeometric Distribution ( 超幾何分配 ) 41 假設 50 個燈泡只有 0 個品質合格試問任意抽取 0 個檢驗, 有 X 個合格的機率為何?

11 Characteristics of Hypergeometric Distribution ( 超幾何分配 ) Characteristics of Hypergeometric Distribution ( 超幾何分配 ) 41 假設 50 個燈泡只有 0 個品質合格試問任意抽取 0 個檢驗, 有 X 個合格的機率為何? 隨機變數 X 表示抽到的 0 個中, 品質合格的個數 X 的機率分配, 即是超幾合分配 4 There are only possible outcomes. The sample is selected from a finite population without replacement. The random variable is the number of successes in a fixed number of trials. The probability of a success is not the same on each trial. The trials are not independent. The size of the sample n is greater than 5% of the size of the population N (i.e. n/n.05). Hypergeometric Distribution Example 假設 50 個燈泡只有 0 個品質合格試問任意抽取 0 個檢驗, 有 X 個合格的機率為何? C C f( X) =, x= 0,1,,..., x 0 x 50 C0 抽到個合格的機率為 : C C C C f () = = C0 C

Hypergeometric Distribution - Example Hypergeometric Distribution - Example PlayTime Toys, Inc., employs 50 people in the Assembly Department. Forty of the employees belong to a union and ten do not.

12 Hypergeometric Distribution - Example Hypergeometric Distribution - Example PlayTime Toys, Inc., employs 50 people in the Assembly Department. Forty of the employees belong to a union and ten do not. Five employees are selected at random to form a committee to meet with management regarding shift starting times. What is the probability that four of the five selected for the committee belong to a union? Poisson Probability Distribution ( 卜瓦松 / 柏松機率分配 ) Poisson Probability Distribution ( 卜瓦松 / 柏松機率分配 ) 47 假設大台北地區, 下午 5 點 ~7 點, 平均發生 μ 件交通事故試問某日下午 5 點 ~ 7 點之間, 大台北地區發生 X 件車禍的機率為何? 一般以 X 服從 Poisson Probability Distribution 來描述這類事件的機率分配 x μ μ e f( x) =, x= 0,1,,..., x! 其中 μ 表示所探討時間單位內, 平均發生次數若改成下午 5 ~ 6 點間, 發生 Y 次事故的機率, 則 Y 的機率分配為 : y μ / ( μ /) e f( y) =, y = 0,1,,... y! 48 The Poisson probability distribution describes the number of times some event occurs during a specified interval. The interval may be time, distance, area, or volume. Assumptions of the Poisson Distribution (1) The probability is proportional to the length of the interval. () The intervals are independent. 1

Poisson Probability Distribution The Poisson distribution can be described mathematically using the formula: Poisson Probability Distribution - Example Assume baggage is rarely lost by Northwest

13 Poisson Probability Distribution The Poisson distribution can be described mathematically using the formula: Poisson Probability Distribution - Example Assume baggage is rarely lost by Northwest Airlines. Suppose a random sample of 1,000 flights shows a total of 300 bags were lost. Thus, the arithmetic mean number of lost bags per flight is 0.3 (300/1,000). If the number of lost bags per flight follows a Poisson distribution with u = 0.3, find the probability of not losing any bags for a flight Example Exercises 交通尖峰時間下午五點到七點, 全台北市平均事故 180 次假設發生機率服從 Poisson Dist. 請問, 在任一各三分鐘的區間內, 發生三次事故的機率為何? u = 180 次 /10 分 = 1.5 次 / 分 =4.5 次 /3 分 Poisson Distribution with u = 4.5 x -u ue 4.5e P(X=3)= = = x! 3! 1,5,9,11,13,15,19,5,7,31,33,41,43,45,47, 49,

14 End of Chapter

Stochastic Processes (XI) Hanjun Zhang School of Mathematics and Computational Science, Xiangtan University 508 YiFu Lou talk 06/

Stochastic Processes (XI) Hanjun Zhang School of Mathematics and Computational Science, Xiangtan University hjzhang001@gmail.com 508 YiFu Lou talk 06/04/2010 - Page 1 Outline 508 YiFu Lou talk 06/04/2010