1. Continuous Probability Distributions Chapter 8

2. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-2 8.1 Continuous Probability Distributions §Unlike a discrete random variable which we studied in Chapter 7, a continuous random variable （連續型隨機變數） is one that can assume an uncountable number of values in the interval (a,b). §We cannot list the possible values because there is an infinite number of them. §Because there is an infinite number of values, the probability that a continuous variable X will assume any particular value is 0. 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

3. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-3 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 8.1 Continuous Probability Distributions

4. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-4 Point Probabilities are Zero  Because there is an infinite number of values, the probability of each individual value is virtually 0.  Thus, we can determine the probability of a range of values only. l E.g. with a discrete random variable like tossing a die, it is meaningful to talk about P(X=5), say. l In a continuous setting (e.g. with time as a random variable), the probability the random variable of interest, say task length, takes exactly 5 minutes is infinitesimally small, hence P(X=5) = 0. § It is meaningful to talk about P(X ≤ 5).

5. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-5 §To calculate probabilities we define a probability density function f(x). §A function f(x) is called a probability density function （機 率密度函數） (over the range a ≤ x ≤ b if it meets the following requirements: 1) f(x) ≥ 0 for all x between a and b, and 2) The total area under the curve between a and b is 1.0 x1x1 x2x2 Area = 1 Probability Density Function 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. P(x 1 ≤X≤x 2 )

6. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-6 Uniform Distribution （均等分配） §Consider the uniform probability distribution (sometimes called the rectangular probability distribution （矩形分配） ). §It is described by the function: f(x) xba

7. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-7 §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. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-8 §The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2,000 gallons and a maximum of 5,000 gallons. Find the probability that sales are: l Between 2,500 and 3,000 gallons l More than 4,000 gallons l Exactly 2,500 gallons Example 8.1

9. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-9 §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 3,000 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 Example 8.1 There is about a 17% chance that between 2,500 and 3,000 gallons of gas will be sold on a given day.

10. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-10 §The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. §Find the probability that sales are more than 4,000 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 Example 8.1 There is a 33% chance the gas station will sell more than 4,000 gallons on any given day.

11. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-11 §The daily sale of gasoline is uniformly distributed between 2,000 and 5,000 gallons. §Find the probability that sales are 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 Example 8.1 The probability that the gas station will sell exactly 2,500 gallons is zero.

12. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-12 8.2 Normal Distribution （常態分配） §This is the most important continuous distribution. l Many distributions can be approximated by a normal distribution. l The normal distribution is the cornerstone distribution of statistical inference.

13. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-13  A random variable X with mean  and variance   is normally distributed if its probability density function is given by Normal Distribution Unlike the range of the uniform distribution (a ≤ x ≤ b), Normal distributions range from minus infinity to plus infinity.

14. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-14 The Normal Distribution § Important things to note: The normal distribution is fully defined by two parameters: its standard deviation and mean The normal distribution is bell shaped and symmetrical about the mean.

15. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-15 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

16. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-16 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)? Increasing the mean shifts the curve to the right. Increasing the standard deviation “ flattens ” the curve.

17. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-17 Standard Normal Distribution §A normal distribution whose mean is 0 and standard deviation is 1 is called the standard normal distribution （標準常態分配）. §As we shall see shortly, any normal distribution can be converted to a standard normal distribution with simple algebra. This makes calculations much easier. 0 1 1

18. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-18 §Two facts help calculate normal probabilities: l The normal distribution is symmetrical. l Any normal distribution can be transformed into a specific normal distribution called “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

19. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-19 §Solution l If X denotes the assembly time of a computer, we seek the probability P(45<X<60). l 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 This shifts the mean of X to zero… This changes the shape of the curve…

20. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-20 §Example - 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

21. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-21 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

22. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-22 §Example - 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

23. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-23 P(-.5<Z<0)+ P(0<Z<1) P(45<X<60) = P( < < ) 45X60  - 50  10 P(0<Z<1 §Example - continued = P(-.5<Z<1) = z=0 z 0 = 1 z 0 =-.5.3413 Finding Normal Probabilities

24. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-24 §The symmetry of the normal distribution makes it possible to calculate probabilities for negative values of Z using the table as follows: -z 0 +z 0 0 P(-z 0 <Z<0) = P(0<Z<z 0 ) Finding Normal Probabilities

25. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-25 §Example - continued Finding Normal Probabilities.3413.5-.5.1915

26. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-26 §Example - 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 “ Just over half the time, 53% or so, a computer will have an assembly time between 45 minutes and 1 hour ”

27. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-27 Using the Normal Table (Table 3)Table 3 §What is P(Z > 1.6) ? 0 1.6 P(0 < Z < 1.6) =.4452 P(Z > 1.6) =.5 – P(0 < Z < 1.6) =.5 –.4452 =.0548 z

28. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-28 Using the Normal Table (Table 3)Table 3 §What is P(Z < -2.23) ? 0 2.23 P(0 < Z < 2.23) P(Z 2.23) =.5 – P(0 < Z < 2.23) =.0129 z -2.23 P(Z > 2.23) P(Z < -2.23)

29. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-29 Using the Normal Table (Table 3)Table 3 §What is P(Z < 1.52) ? 0 1.52 P(Z < 0) =.5 P(Z < 1.52) =.5 + P(0 < Z < 1.52) =.5 +.4357 =.9357 z P(0 < Z < 1.52)

30. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-30 Using the Normal Table (Table 3)Table 3 §What is P(0.9 < Z < 1.9) ? 0 0.9 P(0 < Z < 0.9) P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9) =.4713 –.3159 =.1554 z 1.9 P(0.9 < Z < 1.9)

31. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-31 §Additional Example - 1 l Determine the following probability: P(Z>1.47) 1.47 0.5- P(0<Z<1.47) = P(Z>1.47) 0 P(Z>1.47) = 0.5 - 0.4292 = 0.0708 Finding Normal Probabilities

32. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-32 Additional Example - 2Additional Example - 2 –Determine the following probability: -2.251.85 P(-2.25<Z<1.85) P(-2.25<Z<0) = ? 0 2.25 P(0<Z<1.85) =.4678 P(0<Z<2.25) =.4878 P(-2.25<Z<1.85) = 0.4878 + 0.4678 = 0.9556 =.4878 Finding Normal Probabilities

33. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-33 Additional Example - 3Additional Example - 3 –Determine the following probability: P(.65<Z<1.36).65 1.36 0 P(0<Z<1.36) =.4131 P(0<Z<.65) =.2422 P(.65<Z<1.36) =.4131 -.2422 =.1709 Finding Normal Probabilities

34. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-34 10% 0% 2 0 -2 (i) P(X< 0 ) = P(Z< ) = P(Z< - 2) 0 - 10 5 =P(Z>2) = Z X §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 Example 8.2 Some advice: always draw a picture!

35. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-35 10% 0% (ii) P(X< 0 ) = P(Z< ) 0 - 10 10 = P(Z 1) = Z X §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(0.0228 ） Example 8.2 1 Increasing the standard deviation increases the probability of losing money, which is why the standard deviation (or the variance) is a measure of risk.

36. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-36 §Sometimes 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 ) = A Finding Values of Z zAzA A

37. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-37 §Example 8.3 & 8.4 l Determine z exceeded by 5% of the population l Determine z such that 5% of the population is below §Solution 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

38. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-38 Finding Values of Z §Other Z values are §Z.05 = 1.645 §Z.01 = 2.33

39. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-39 Using the values of Z §Because z.025 = 1.96 and - z.025 = -1.96, it follows that we can state P(-1.96 < Z < 1.96) =.95 §Similarly P(-1.645 < Z < 1.645) =.90

40. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-40 8.3 Exponential Distribution （指數分配） §The exponential distribution can be used to model l the length of time between telephone calls l the length of time between arrivals at a service station l 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.

41. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-41 §A random variable is exponentially distributed if its probability density function is given by Exponential Distribution §Note that x ≥ 0. Time (for example) is a non-negative quantity; the exponential distribution is often used for time related phenomena such as the length of time between phone calls or between parts arriving at an assembly station. Note also that the mean and standard deviation are equal to each other and to the inverse of the parameter of the distribution (lambda)

42. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-42 Exponential Distribution §The exponential distribution depends upon the value of.  Smaller values of “ flatten ” the curve. §E.g. exponential distributions for =.5, 1, 2

43. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-43 Exponential Distribution §If X is an exponential random variable, then we can calculate probabilities by: P(X > a) = e – a P(X < a) = 1 – e – a P(a 1 < X < a 2 ) = e –  a1) – e –  a2)

44. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-44 §Example 8.6 The lifetime of an alkaline battery is exponentially distributed with  =.05 per hour. l What is the mean and standard deviation of the battery’s lifetime? l Find the following probabilities: The battery will last between 10 and 15 hours. The battery will last for more than 20 hours. Exponential Distribution

45. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-45 §Solution The mean = standard deviation = 1/  1/.05 = 20 hours. l Let X denote the lifetime. P(X > 20) = e -.05(20) =.3679 Exponential Distribution “There is about a 13% chance a battery will only last 10 to 15 hours”

46. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-46 §Example 8.7 l The service rate at a supermarket checkout is 6 customers per hour. l If the service time is exponential, find the following probabilities: A service is completed in 5 minutes, A customer leaves the counter more than 10 minutes after arriving A service is completed between 5 and 8 minutes. Exponential Distribution

47. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-47 §Solution 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 l P(5 < X < 8) = e -.1(5) – e -.1(8) =.1572 Exponential Distribution

48. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-48 §Additional Example l Cars arrive randomly and independently to a tollbooth at an average of 360 cars per hour. l Use the exponential distribution to find the probability that the next car will not arrive within half a minute. §Solution Let X denote the time (in minutes) that elapses before the next car arrives. X is exponentially distributed with = 360/60 = 6 cars per minute. P(X>.5) = e -6(.5) =.0498. Exponential Distribution

49. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-49 l What is the probability that no car will arrive within the next half minute? l Solution If Y counts the number of cars that will arrive in the next half minute, then Y is a Poisson variable with  = (.5)(6) = 3 cars per half a minute. P(Y = 0) = e -3 (3 0 )/0! =.0498. Comment: If the first car will not arrive within the next half a minute then no car will arrives within the next half minute. Therefore, not surprisingly, the probability found here is the exact same probability found in the previous question. Exponential Distribution vs. Poisson Distribution

50. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-50 8.4 Other Continuous Distribution §Three other important continuous distributions which will be used extensively in later sections are introduced here: §Student t Distribution §Chi-Squared Distribution § F Distribution

51. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-51 Student t Distribution （ t 分配） §Here the letter t is used to represent the random variable, hence the name. The density function for the Student t distribution is as follows: § (nu) is called the degrees of freedom （自由度）   (Gamma function) is  (k)=(k-1)(k-2) … (2)(1)

52. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-52 Student t Distribution  Much like the standard normal distribution, the Student t distribution is “ mound ” shaped and symmetrical about its mean of zero: §The mean and variance of a Student t random variable are E(t) = 0 and

53. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-53 Student t Distribution §In much the same way that  and  define the normal distribution,, the degrees of freedom, defines the Student t Distribution: §As the number of degrees of freedom increases, the t distribution approaches the standard normal distribution.

54. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-54 Determining Student t Values §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. §We can do this using l t tables l Excel l Minitab

55. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-55 Determining Student t Values §The student t distribution is used extensively in statistical inference. Table 4 in Appendix B lists values of.Table 4 §That is, values of a Student t random variable with degrees of freedom such that:  The values for A are pre-determined “ critical ” values, typically in the 10%, 5%, 2.5%, 1% and 0.5% range.

56. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-56 tAtA t.100 t.05 t.025 t.01 t.005 A=.05 A -t A 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 ) = A Using the t Table tttt

57. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-57 Student t Probabilities and Values §Excel can calculate Student distribution probabilities and values.Excel

58. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-58 §The chi-squared density function is given by: §As before, the parameter is the number of degrees of freedom. Chi – Squared Distribution （卡方分配）

59. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-59 §The chi-squared distribution is not symmetrical §The square, i.e., forces non-negative values (e.g. finding is illogical). §Chi squared values can be found from the chi squared table, from Excel, or from Minitab. The Chi – Squared Distribution

60. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-60 §Table 5 in Appendix B makes it easy to look-up the right hand tail probability (A), for whichTable 5 The Chi – Squared Distribution

61. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-61 §For probabilities of this sort:  We use 1 – A, i.e. we are determining The Chi – Squared Distribution

62. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-62 =.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   

63. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-63 §To find the point in a chi-squared distribution with 8 degrees of freedom, such that the area to the right is.05, §Look up the intersection of the 8 d.f. row with the 0.05 column, yielding a value of 15.5073 For Example

64. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-64 §To find the point in a chi-squared distribution with 8 degrees of freedom, such that the area to the left is.05, §Look up the intersection of the 8 d.f. row with the 0.95 column, yielding a value of 2.73264 For Example

65. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-65 Chi-Squared Probabilities and Values §Excel can calculate chi-squared distribution probabilities and values.Excel

66. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-66 F Distribution （ F 分配） §The F density function is given by:  Two parameters define this distribution, and like we ’ ve already seen these are again degrees of freedom.  is the “ numerator ” degrees of freedom （分子的自由度）  is the “ denominator ” degrees of freedom （分母的自由度）

67. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-67 §The mean and variance of an F random variable are given by: §and §The F distribution is similar to the distribution in that its starts at zero (is non-negative) and is not symmetrical. F Distribution

68. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-68 Determining Values of F §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

69. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-69 Determining Values of F  For example, what is the value of F for 5% of the area under the right hand “ tail ” of the curve, with a numerator degree of freedom of 3 and a denominator degree of freedom of 7? §Solution: use the F look-up (Table 6)Table 6 Numerator Degrees of Freedom : COLUMN Denominator Degrees of Freedom : ROW F.05,3,7 There are different tables for different values of A. Make sure you start with the correct table!!

70. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-70 Determining Values of F §For areas under the curve on the left hand side of the curve, we can leverage the following relationship: Pay close attention to the order of the terms!

71. 2007 會計資訊系統計學 ( 一 ) 上課投影片. 8-71 F Probabilities and Values §Excel can calculate F distribution probabilities and values.Excel