1. 1 Continuous Probability Distributions Chapter 8

2. 2 在隨機實驗中，所觀察的隨機變數，如：人的 身高、體重、家庭所得、地區的氣溫、股票的 價格、工人的工資、醫生看病的時間，都是連 續的隨機變數。這些連續隨機變數具有共同的 特性，其機率分配的形態頗為類似，比如人的 身高，大部份的人都很接近，而且集中在平均 身高左右，只有少部份的人較高或較矮。同樣 的，氣溫、體重也類似此一形態。又如醫生看 病的時間，大部份的人看病時間不長，只有少 數病人需要花較長的時間去診斷，而形成右偏 的分配。 在隨機實驗中，所觀察的隨機變數，如：人的 身高、體重、家庭所得、地區的氣溫、股票的 價格、工人的工資、醫生看病的時間，都是連 續的隨機變數。這些連續隨機變數具有共同的 特性，其機率分配的形態頗為類似，比如人的 身高，大部份的人都很接近，而且集中在平均 身高左右，只有少部份的人較高或較矮。同樣 的，氣溫、體重也類似此一形態。又如醫生看 病的時間，大部份的人看病時間不長，只有少 數病人需要花較長的時間去診斷，而形成右偏 的分配。

3. 3 均等分佈 (Uniform Distribution) 均等分佈 (Uniform Distribution) 常態分佈 (Normal Distribution) 常態分佈 (Normal Distribution) 指數分佈 (Exponential Distribution) 指數分佈 (Exponential Distribution)

4. 4 A continuous random variable has an uncountably infinite number of values in the interval (a,b).A continuous random variable has an uncountably infinite number of values in the interval (a,b). 8.2 Continuous Probability Distributions The probability that a continuous variable X will assume any particular value is zero. Why?The probability that a continuous variable X will assume any particular value is zero. Why? 01 1/2 1/3 2/3 1/2 + 1/2 = 1 1/3 + 1/3 + 1/3 = 1 1/4 + 1/4 + 1/4 + 1/4 = 1 The probability of each value

5. 5 01 1/2 1/3 2/3 1/2 + 1/2 = 1 1/3 + 1/3 + 1/3 = 1 1/4 + 1/4 + 1/4 + 1/4 = 1 As the number of values increases the probability of each value decreases. This is so because the sum of all the probabilities remains 1. When the number of values approaches infinity (because X is continuous) the probability of each value approaches 0. The probability of each value 8.2 Continuous Probability Distributions

6. 6 To calculate probabilities we define a probability density function f(x).To calculate probabilities we define a probability density function f(x). The density function satisfies the following conditionsThe density function satisfies the following conditions –f(x) is non-negative, –The total area under the curve representing f(x) equals 1. x1x1 x2x2 Area = 1 P(x 1 <=X<=x 2 ) Probability Density Function (pdf) The probability that X falls between x 1 and x 2 is found by calculating the area under the graph of f(x) between x 1 and x 2.The probability that X falls between x 1 and x 2 is found by calculating the area under the graph of f(x) between x 1 and x 2.

7. 7 –A random variable X is said to be uniformly distributed if its density function is –The expected value and the variance are Uniform Distribution

8. 8 均等分佈 均等分佈是指隨機變數在某一連續 的區間，有同等的機率密度，如 ：等車的時間，顧客排隊買票時 間。 均等分佈是指隨機變數在某一連續 的區間，有同等的機率密度，如 ：等車的時間，顧客排隊買票時 間。

9. 9 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 25003000 P(2500  X  3000) = (3000-2500)(1/3000) =.1667 Uniform Distribution

10. 10 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 4000 P(X  4000) = (5000-4000)(1/3000) =.333 Uniform Distribution

11. 11 Example 8.1 Example 8.1 –The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. Find the probability that sales are: –Between 2,500 and 3,500 gallons –More than 4,000 gallons –Exactly 2,500 gallons 20005000 1/3000 f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000] x 2500 P(X=2500) = (2500-2500)(1/3000) = 0 Uniform Distribution

12. 12 常態分佈 許多連續的隨機變數，如：某一年齡層的身高、體重， 人的智商，某公司的產品與標準規格的差異，測量的 誤差等。大都會集中於平均數的附近，其中特別大的 數值或特別小的數值不多，而且對稱的分散於平均數 的兩邊，亦即其次數分配係一個鐘形 (bell-shaped) 的分 配，而且數值大部份集中於離平均數三個標準差之內 ，此種隨機變數，我們稱為常態隨機變數，其分佈稱 為常態分佈，該分配是由德國人 Karl F. Gauss 所提出的 ，故又稱為高斯分佈 (Gaussian Distribution) 。常態分佈 是一個最重要、最常用的分配，因為社會及自然界的 現象以常態分佈的情形最為普遍。 許多連續的隨機變數，如：某一年齡層的身高、體重， 人的智商，某公司的產品與標準規格的差異，測量的 誤差等。大都會集中於平均數的附近，其中特別大的 數值或特別小的數值不多，而且對稱的分散於平均數 的兩邊，亦即其次數分配係一個鐘形 (bell-shaped) 的分 配，而且數值大部份集中於離平均數三個標準差之內 ，此種隨機變數，我們稱為常態隨機變數，其分佈稱 為常態分佈，該分配是由德國人 Karl F. Gauss 所提出的 ，故又稱為高斯分佈 (Gaussian Distribution) 。常態分佈 是一個最重要、最常用的分配，因為社會及自然界的 現象以常態分佈的情形最為普遍。

13. 13 常態分佈的重要性 因為社會與自然界中，常態分佈是最為常見的 現象，因此可以做為我們統計推論程序中的基 本模式。 因為社會與自然界中，常態分佈是最為常見的 現象，因此可以做為我們統計推論程序中的基 本模式。 間斷機率分佈在某些條件下可利用常態分佈， 求其近似值。 間斷機率分佈在某些條件下可利用常態分佈， 求其近似值。 常態分佈在大樣本時，是推論統計的基礎。 常態分佈在大樣本時，是推論統計的基礎。 在統計推論上，許多樣本統計量的抽樣分佈， 如： t 分佈，卡方分佈， F 分佈必須假設母體為 常態分佈方可獲得。 在統計推論上，許多樣本統計量的抽樣分佈， 如： t 分佈，卡方分佈， F 分佈必須假設母體為 常態分佈方可獲得。

14. 14 8.3 Normal Distribution This is the most important continuous distribution.This is the most important continuous distribution. –Many distributions can be approximated by a normal distribution. –The normal distribution is the cornerstone distribution of statistical inference.

15. 15 A random variable X with mean  and variance   is normally distributed if its probability density function is given byA random variable X with mean  and variance   is normally distributed if its probability density function is given by Normal Distribution

16. 16 The Shape of the Normal Distribution The normal distribution is bell shaped, and symmetrical around   Why symmetrical? Let  = 100. Suppose x = 110. Now suppose x = 90 11090

17. 17 The effects of  and  How does the standard deviation affect the shape of f(x)?  = 2  =3  =4  = 10  = 11  = 12 How does the expected value affect the location of f(x)?

18. 18 標準常態分佈 (Standard Normal Distribution) 不同的常態分佈因其平均數與標準差互不相同，因此 有不同的常態曲線，而要計算某一常態分佈，其在某 一區間的機率，必須利用積分的方法。 不同的常態分佈因其平均數與標準差互不相同，因此 有不同的常態曲線，而要計算某一常態分佈，其在某 一區間的機率，必須利用積分的方法。 求出該區間常態曲線以下的面積。這種計算方法相當麻 煩且費時，有沒有比較容易省事的方法呢？ 求出該區間常態曲線以下的面積。這種計算方法相當麻 煩且費時，有沒有比較容易省事的方法呢？

19. 19 設 X~ ，令 Z ＝ ，則 Z 為一標準 常態變數，一般以 Z~ N(0,1) 來表示。標 準常態分佈的任何值域之機率可查標準 常態機率值表而獲得，如：求 Z 落在 (a,b) 間的機率 可查表而獲得。 設 X~ ，令 Z ＝ ，則 Z 為一標準 常態變數，一般以 Z~ N(0,1) 來表示。標 準常態分佈的任何值域之機率可查標準 常態機率值表而獲得，如：求 Z 落在 (a,b) 間的機率 可查表而獲得。

20. 20 Two facts help calculate normal probabilities:Two facts help calculate normal probabilities: –The normal distribution is symmetrical. –Any normal distribution can be transformed into a specific normal distribution called… “STANDARD NORMAL DISTRIBUTION” “STANDARD NORMAL DISTRIBUTION” Example The amount of time it takes to assemble a computer is normally distributed, with a mean of 50 minutes and a standard deviation of 10 minutes. What is the probability that a computer is assembled in a time between 45 and 60 minutes? Finding Normal Probabilities

21. 21 SolutionSolution –If X denotes the assembly time of a computer, we seek the probability P(45<X<60). –This probability can be calculated by creating a new normal variable the standard normal variable. E(Z) =  = 0 V(Z) =  2 = 1 Every normal variable with some  and , can be transformed into this Z. Therefore, once probabilities for Z are calculated, probabilities of any normal variable can be found. Finding Normal Probabilities

22. 22 Example - continuedExample - continued P(45<X<60) = P( < < ) 45X60  - 50  10 = P(-0.5 < Z < 1) To complete the calculation we need to compute the probability under the standard normal distribution Finding Normal Probabilities

23. 23 Standard normal probabilities have been calculated and are provided in a table. The tabulated probabilities correspond to the area between Z=0 and some Z = z 0 >0 Z = 0 Z = z 0 P(0<Z<z 0 ) Using the Standard Normal Table

24. 24 Example - continuedExample - continued P(45<X<60) = P( < < ) 45X60  - 50  10 = P(-.5 < Z < 1) z 0 = 1 z 0 = -.5 We need to find the shaded area Finding Normal Probabilities

25. 25 P(-.5<Z<0)+ P(0<Z<1) P(45<X<60) = P( < < ) 45X60  - 50  10 P(0<Z<1 Example - continuedExample - continued = P(-.5<Z<1) = z=0 z 0 = 1 z 0 =-.5.3413 Finding Normal Probabilities

26. 26 The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows:The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows: -z0-z0 +z0+z0 0 P(-z 0 <Z<0) = P(0<Z<z 0 ) Finding Normal Probabilities

27. 27 Example - continuedExample - continued Finding Normal Probabilities.3413.5-.5.1915

28. 28 Example - continuedExample - continued Finding Normal Probabilities.1915.3413.5-.5 P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) =.1915 +.3413 =.5328 1.0

29. 29 10% 0% 2 0 -2 (i) P(X< 0 ) = P(Z< ) = P(Z< - 2) 0 - 10 5 =P(Z>2) = Z X Example 8.2Example 8.2 –The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. –What is the probability of losing money?.4772 0.5 - P(0<Z<2) = 0.5 -.4772 =.0228 Finding Normal Probabilities

30. 30 10% 0% (ii) P(X< 0 ) = P(Z< ) 0 - 10 10 = P(Z 1) = Z X Example 8.2Example 8.2 –The rate of return (X) on an investment is normally distributed with mean of 10% and standard deviation of (i) 5%, (ii) 10%. –What is the probability of losing money?.3413 0.5 - P(0<Z<1) = 0.5 -.3413 =.1587 Finding Normal Probabilities Find Normal Probabilities 1

31. 31 Sometimes we need to find the value of Z for a given probabilitySometimes we need to find the value of Z for a given probability We use the notation z A to express a Z value for which P(Z > z A ) = AWe use the notation z A to express a Z value for which P(Z > z A ) = A Finding Values of Z zAzA A

32. 32 Example 8.3 & 8.4Example 8.3 & 8.4 –Determine z exceeded by 5% of the population –Determine z such that 5% of the population is below SolutionSolution z.05 is defined as the z value for which the area on its right under the standard normal curve is.05. 0.05 Z 0.05 0 0.45 1.645 Finding Values of Z 0.05 -Z 0.05

33. 33 二項分佈，波氏分佈與常態分佈 n 很大， p 很小時，二項分佈  波氏分佈。n 很大， p 很小時，二項分佈  波氏分佈。 n 很大， p 不是很小時，二項分佈  常態分佈。n 很大， p 不是很小時，二項分佈  常態分佈。 np  5 and nq  5 時，則可以常態分佈代替 二項分佈，求其機率值，np  5 and nq  5 時，則可以常態分佈代替 二項分佈，求其機率值，  ＝ np ，  ＝ np ，

34. 34 指數分佈 (Exponential Distribution) 指數分佈是另一個常用的連續機率分佈，通常 應用於間隔或等待的時間，如：燈泡被使用至 壞掉的時間，顧客排隊等待服務的時間，機器 發生故障的間隔時間等。 指數分佈是另一個常用的連續機率分佈，通常 應用於間隔或等待的時間，如：燈泡被使用至 壞掉的時間，顧客排隊等待服務的時間，機器 發生故障的間隔時間等。 指數分佈的隨機變數為非負之值，且通常該隨 機變數超過其平均數的機率較小，如服務時間 、發生事件間隔時間等，通常具有這樣的性質 。 指數分佈的隨機變數為非負之值，且通常該隨 機變數超過其平均數的機率較小，如服務時間 、發生事件間隔時間等，通常具有這樣的性質 。

35. 35 Exponential Distribution The exponential distribution can be used to modelThe exponential distribution can be used to model –the length of time between telephone calls –the length of time between arrivals at a service station –the life-time of electronic components. When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.When the number of occurrences of an event follows the Poisson distribution, the time between occurrences follows the exponential distribution.

36. 36 A random variable is exponentially distributed if its probability density function is given by f(x) = e - x, x>=0.  is the distribution parameter ( >0). is the distribution parameter ( >0). E(X) = 1/ V(X) = (1/   Exponential Distribution

37. 37 f(x) = 2e -2x f(x) = 1e -1x f(x) =.5e -.5x 0 1 2 3 4 5 Exponential distribution for =.5, 1, 2 a b P(a<x<b) = e - a - e - b

38. 38 Finding exponential probabilities is relatively easy:Finding exponential probabilities is relatively easy: –P(X > a) = e – a. –P(X < a) = 1 – e – a –P(a 1 < X < a 2 ) = e –  a1) – e –  a2) Exponential Distribution

39. 39 Example 8.5Example 8.5 –The lifetime of an alkaline battery is exponentially distributed with  =.05 per hour. –What is the mean and standard deviation of the battery’s lifetime? –Find the following probabilities: The battery will last between 10 and 15 hours.The battery will last between 10 and 15 hours. The battery will last for more than 20 hours?The battery will last for more than 20 hours? Exponential Distribution

40. 40 SolutionSolution –The mean = standard deviation = 1/  1/.05 = 20 hours. –Let X denote the lifetime. P(10<X<15) = e -.05(10) – e -.05(15) =.1341P(10<X<15) = e -.05(10) – e -.05(15) =.1341 P(X > 20) = e -.05(20) =.3679P(X > 20) = e -.05(20) =.3679 Exponential Distribution

41. 41 Example 8.6Example 8.6 –The service rate at a supermarket checkout is 6 customers per hour. –If the service time is exponential, find the following probabilities: A service is completed in 5 minutes, A service is completed in 5 minutes, A customer leaves the counter more than 10 minutes after arrivingA customer leaves the counter more than 10 minutes after arriving A service is completed between 5 and 8 minutes.A service is completed between 5 and 8 minutes. Exponential Distribution

42. 42 SolutionSolution –A service rate of 6 per hour = A service rate of.1 per minute ( =.1/minute). –P(X < 5) = 1-e - x = 1 – e -.1(5) =.3935 –P(X >10) = e - x = e -.1(10) =.3679 –P(5 < X < 8) = e -.1(5) – e -.1(8) =.1572 Exponential Distribution Compute Exponential probabilities

43. 43 8.5 Other Continuous Distribution Three new continuous distributions:Three new continuous distributions: –Student t distribution –Chi-squared distribution –F distribution

44. 44 The Student t Distribution The Student t density functionThe Student t density function  is the parameter of the student t distribution E(t) = 0 V(t) =  (  – 2) (for n > 2)

45. 45 The Student t Distribution = 3 = 10

46. 46 Determining Student t Values The student t distribution is used extensively in statistical inference.The student t distribution is used extensively in statistical inference. Thus, it is important to determine values of t A associated with a given number of degrees of freedom.Thus, it is important to determine values of t A associated with a given number of degrees of freedom. We can do this usingWe can do this using –t tables –Excel –Minitab

47. 47 tAtA t.100 t.05 t.025 t.01 t.005 A=.05 A -tA-tA The t distribution is symmetrical around 0 =1.812 =-1.812 The table provides the t values (t A ) for which P(t > t A ) = AThe table provides the t values (t A ) for which P(t > t A ) = A Using the t Table tttt

48. 48 The Chi – Squared Distribution The Chi – Squared density function:The Chi – Squared density function: The parameter is the number of degrees of freedom.The parameter is the number of degrees of freedom.

49. 49 The Chi – Squared Distribution

50. 50 Chi squared values can be found from the chi squared table, from Excel, or from Minitab.Chi squared values can be found from the chi squared table, from Excel, or from Minitab. The  2 -table entries are the   values of the right hand tail probability (A), for which P(       = A.The  2 -table entries are the   values of the right hand tail probability (A), for which P(       = A. Determining Chi-Squared Values A 2A2A

51. 51 =.05 A =.99 Using the Chi-Squared Table                A  To find  2 for which P(  2 <  2 )=.01, lookup the column labeled  2 1-.01 or  2.99   

52. 52 The F Distribution The density function of the F distribution: 1 and 2 are the numerator and denominator degrees of freedom.The density function of the F distribution: 1 and 2 are the numerator and denominator degrees of freedom. ! ! !

53. 53 This density function generates a rich family of distributions, depending on the values of 1 and 2This density function generates a rich family of distributions, depending on the values of 1 and 2 The F Distribution 1 = 5, 2 = 10 1 = 50, 2 = 10 1 = 5, 2 = 10 1 = 5, 2 = 1

54. 54 Determining Values of F The values of the F variable can be found in the F table, Excel, or from Minitab.The values of the F variable can be found in the F table, Excel, or from Minitab. The entries in the table are the values of the F variable of the right hand tail probability (A), for which P(F 1, 2 >F A ) = A.The entries in the table are the values of the F variable of the right hand tail probability (A), for which P(F 1, 2 >F A ) = A.

55. 55 Homework for Chapter 8 Chapter 85, 11, 13, 21, 37, 53, 83, 85, 93, 99, 105. 93, 99, 105.