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# Unit 4--probability and probability distribution (1).pptx

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# Unit 4--probability and probability distribution (1).pptx

STATS BY NAZIM

STATS BY NAZIM

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### Unit 4--probability and probability distribution (1).pptx

1. 1. Unit 4 : Probability concepts Probability distributions
2. 2. BASIC PROBABILITY & PROBABILITY DISTRIBUTIONS
3. 3. 3 Basic Concepts Random Experiment is a process leading to at least two possible outcomes with uncertainty as to which will occur. A coin is thrown A consumer is asked which of two products he or she prefers The daily change in an index of stock market prices is observed
4. 4. 4 Sample Spaces Collection of all possible outcomes e.g.: All six faces of a die: e.g.: All 52 cards a deck of bridge cards
5. 5. 5 Events and Sample Spaces An event is a set of basic outcomes from the sample space, and it is said to occur if the random experiment gives rise to one of its constituent basic outcomes. Simple Event Outcome With 1 Characteristic Joint Event 2 Events Occurring Simultaneously Compound Event One or Another Event Occurring
6. 6. 6 Simple Event A: Male B: Over age 20 C: Has 3 credit cards D: Red card from a deck of bridge cards E: Ace card from a deck of bridge cards
7. 7. 7 Joint Event D and E, (DE): Red, ace card from a bridge deck A and B, (AB): Male, over age 20 among a group of survey respondents
8. 8. 8 Intersection • Let A and B be two events in the sample space S. Their intersection, denoted AB, is the set of all basic outcomes in S that belong to both A and B. • Hence, the intersection AB occurs if and only if both A and B occur. • If the events A and B have no common basic outcomes, their intersection AB is said to be the empty set.
9. 9. 9 Union •Let A and B be two events in the sample space S. Their union, denoted AB, is the set of all basic outcomes in S that belong to at least one of these two events. •Hence, the union AB occurs if and only if either A or B or both occurs
10. 10. 10 Event Properties Mutually Exclusive Two outcomes that cannot occur at the same time E.g. flip a coin, resulting in head and tail Collectively Exhaustive One outcome in sample space must occur E.g. Male or Female
11. 11. Special Events Null Event Club & Diamond on 1 Card Draw Complement of Event For Event A, All Events Not In A: A' or A
12. 12. 12 What is Probability? 1. Numerical measure of likelihood that the event will occur Simple Event Joint Event Compound 2. Lies between 0 & 1 3. Sum of events is 1
13. 13. 13 Concept of Probability A Priori classical probability, the probability of success is based on prior knowledge of the process involved. i.e. the chance of picking a black card from a deck of bridge cards Empirical classical probability, the outcomes are based on observed data, not on prior knowledge of a process. i.e. the chance that individual selected at random from the Kalosha employee survey if satisfied with his or her job. (.89)
14. 14. 14 Concept of Probability Subjective probability, the chance of occurrence assigned to an event by a particular individual, based on his/her experience, personal opinion and analysis of a particular situation. i.e. The chance of a newly designed style of mobile phone will be successful in market.
15. 15. 15 (There are 2 ways to get one 6 and the other 4) e.g. P( ) = 2/36 Computing Probabilities • The probability of an event E: • Each of the outcomes in the sample space is equally likely to occur number of event outcomes ( ) total number of possible outcomes in the sample space P E X T  
16. 16. 16 Presenting Probability & Sample Space 1. Listing S = {Head, Tail} 2. Venn Diagram 3. Tree Diagram 4. Contingency Table
17. 17. 17 S Ā A Venn Diagram Example: ABC Employee Survey Event: A = Satisfied, Ā = Dissatisfied P(A) = 356/400 = .89, P(Ā) = 44/400 = .11
18. 18. 18 Kalosha Employee Tree Diagram Satisfied Not Satisfied Advanced Not Advanced Advanced Not Advanced P(A)=.89 P(Ā)=.11 .485 .405 .035 .075 Example: ABC Employee Survey Joint Probability
19. 19. 19 Event Event B1 B2 Total A1 P(A1  B1) P(A1  B2) P(A1) A2 P(A2  B1) P(A2  B2) P(A2) Total P(B1) P(B2) 1 Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability
20. 20. 20 Joint Probability Using Contingency Table ABC Employee Survey Satisfied Not Satisfied Total Advanced Not Advanced Total .485 .035 .405 .075 .52 .48 .89 .11 1.00 Joint Probability Simple Probability
21. 21. 21 Use of Venn Diagram Fig. 3.1: AB, Intersection of events A & B, mutually exclusive Fig. 3.2: AB, Union of events A & B Fig. 3.3: Ā, Complement of event A Fig. 3.4 and 3.5: The events AB and ĀB are mutually exclusive, and their union is B. (A  B)  (Ā  B) = B
22. 22. 22 Use of Venn Diagram Let E1, E2,…, Ek be K mutually exclusive and collective exhaustive events, and let A be some other event. Then the K events E1  A, E2  A, …, Ek  A are mutually exclusive, and their union is A. (E1  A)  (E2  A)  …  (Ek  A) = A
23. 23. 23 Compound Probability Addition Rule 1. Used to Get Compound Probabilities for Union of Events 2. P(A or B) = P(A  B) = P(A) + P(B)  P(A  B) 3. For Mutually Exclusive Events: P(A or B) = P(A  B) = P(A) + P(B) 4. Probability of Complement P(A) + P(Ā) = 1. So, P(Ā) = 1  P(A)
24. 24. 24 Addition Rule: Example A hamburger chain found that 75% of all customers use mustard, 80% use ketchup, 65% use both. What is the probability that a particular customer will use at least one of these? A = Customers use mustard B = Customers use ketchup AB = a particular customer will use at least one of these Given P(A) = .75, P(B) = .80, and P(AB) = .65, P(AB) = P(A) + P(B)  P(AB) = .75 + .80  .65= .90
25. 25. 25 Conditional Probability 1. Event Probability Given that Another Event Occurred 2. Revise Original Sample Space to Account for New Information Eliminates Certain Outcomes 3. P(A | B) = P(A and B) , P(B)>0 P(B)
26. 26. 26 Example Recall the previous hamburger chain example, what is the probability that a ketchup user uses mustard? P(A|B) = P(AB)/P(B) = .65/.80 = .8125 Please pay attention to the difference from the joint event in wording of the question.
27. 27. 27 S Black Ace Conditional Probability Black ‘Happens’: Eliminates All Other Outcomes and Thus Increase the Conditional Probability Event (Ace and Black) (S) Black Draw a card, what is the probability of black ace? What is the probability of black ace when black happens?
28. 28. 28 Color Type Red Black Total Ace 2 2 4 Non-Ace 24 24 48 Total 26 26 52 Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Black Ace Revised Sample Space P(Ace|Black) = P(Ace and Black) P(Black) = = 2/26 2/52 26/52
29. 29. 29 Statistical Independence 1. Event Occurrence Does Not Affect Probability of Another Event e.g. Toss 1 Coin Twice, Throw 3 Dice 2. Causality Not Implied 3. Tests For Independence P(A | B) = P(A), or P(B | A) = P(B), or P(A and B) = P(A)P(B)
30. 30. 30 Statistical Independence ABC Employee Survey Satisfied Not Satisfied Total Advanced Not Advanced Total .485 .035 .405 .075 .52 .48 .89 .11 1.00 P(A1) Note: (.52)(.89) = .4628  .485 P(B1) P(A1and B1)
31. 31. 31 Multiplication Rule 1. Used to Get Joint Probabilities for Intersection of Events (Joint Events) 2. P(A and B) = P(A  B) P(A  B) = P(A)P(B|A) = P(B)P(A|B) 3. For Independent Events: P(A and B) = P(AB) = P(A)P(B)
32. 32. 32 Practice of Randomized Response Question a) Is the second last digit of your office phone number odd? b) Recall your undergraduate course work, have you ever cheated in midterm or final exams? Respondents, please do as follows: 1. Flip a coin. 2. If the result is the national emblem answer question a); otherwise answer question b). Please circle “Yes” or “No” below as your answer. Yes No
33. 33. 33 Bayes’ Theorem 1. Permits Revising Old Probabilities Based on New Information 2. Application of Conditional Probability 3. Mutually Exclusive Events New Information Revised Probability Apply Bayes' Theorem Prior Probability
34. 34. 34 Bayes’s Theorem Example Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan? B1= repay, B2= default, A=college degree P(B1) = .5, P(A|B1) = .4, P(A|B2) = .1, P(B1|A) =? ) ( ) | ( ) ( ) | ( ) ( ) | ( ) | ( 2 2 1 1 1 1 1 B P B A P B P B A P B P B A P A B P   8 . 25 . 2 . ) 5 )(. 1 (. ) 5 )(. 4 (. ) 5 )(. 4 (.    
35. 35. 35 Event Prior Prob Cond. Prob Joint Prob Post. Prob Bi P(Bi) P(A|Bi) P(Bi  A) P(Bi |A) B1 .5 .4 .20 .20/.25 = .8 B2 .5 .1 .05 .05/.25 = .2 1.0 P(A) = 0.25 1.0 Bayes’ Theorem Example Table Solution Default Repay P(College) X =
36. 36. 36 Permutation and Combination Counting Rule 1 Example：In TV Series: ABC tossed 50 coins, the number of outcomes is 2·2 ·… ·2 ＝ 250. What is the probability of all coins with heads up? If any one of n different mutually exclusive and collectively exhaustive events can occur on each of r trials, the number of possible outcomes is equal to： n·n ·… ·n ＝ nr
37. 37. 37 Permutation and Combination Applicatin: If a license plate consists of 3 letters followed by 3 digits, the total number of outcomes would be? (most states in the US) Application: China License Plates How many licenses can be issued? Style 1992: one letter or digit plus 4 digits. Style 2002: 1) three letters + three digits 2) three digits + three letters 3) three digits + three digits
38. 38. 38 Permutation and Combination Counting Rule 2 Example: The number of ways that 5 books could be arranged on a shelf is: (5)(4)(3)(2)(1) = 120 The number of ways that all n objects can be arranged in order is： = n(n -1)(n -2)(2)(1) = n! Where n! is called factorial and 0! is defined as 1. n n P
39. 39. 39 Permutation and Combination Counting Rule 3: Permutation Example：What is the number of ways of arranging 3 books selected from 5 books in order? (5)(4)(3) = 60 The number of ways of arranging r objects selected from n objects in order is： )! ( ! r n n P n r  
40. 40. 40 Permutation and Combination Counting Rule 4: Combination Example：The number of combinations of 3 books selected from 5 books is (5)(4)(3)/[(3)(2)(1)] = 10 Note: 3! possible arrangements in order are irrelevant The number of ways that arranging r objects selected from n objects, irrespective of the order, is equal to )! ( ! ! r n r n r n Cn r           
41. 41. PROBABILITY DISTRIBUTIONS BINOMIAL POISSON 41
42. 42. Chap 5-42 Discrete Random Variables • Can only assume a countable number of values Examples: • Roll a die twice Let X be the number of times 4 comes up (then X could be 0, 1, or 2 times) • Toss a coin 5 times. Let X be the number of heads (then X = 0, 1, 2, 3, 4, or 5)
43. 43. Chap 5-43 • Variance of a discrete random variable • Standard Deviation of a discrete random variable where: E(X) = Expected value of the discrete random variable X Xi = the ith outcome of X P(Xi) = Probability of the ith occurrence of X Discrete Random Variable Summary Measures     N 1 i i 2 i 2 ) P(X E(X)] [X σ (continued)      N 1 i i 2 i 2 ) P(X E(X)] [X σ σ
44. 44. Chap 5-44 • Example: Toss 2 coins, X = # heads, compute standard deviation (recall E(X) = 1) Discrete Random Variable Summary Measures ) P(X E(X)] [X σ i 2 i    0.707 0.50 (0.25) 1) (2 (0.50) 1) (1 (0.25) 1) (0 σ 2 2 2         (continued) Possible number of heads = 0, 1, or 2
45. 45. Chap 5-45 The Covariance • The covariance measures the strength of the linear relationship between two variables • The covariance: ) Y X ( P )] Y ( E Y )][( X ( E X [ σ N 1 i i i i i XY      where: X = discrete variable X Xi = the ith outcome of X Y = discrete variable Y Yi = the ith outcome of Y P(XiYi) = probability of occurrence of the ith outcome of X and the ith outcome of Y
46. 46. Chap 5-46 The Sum of Two Random Variables • Expected Value of the sum of two random variables: • Variance of the sum of two random variables: • Standard deviation of the sum of two random variables: XY 2 Y 2 X 2 Y X σ 2 σ σ σ Y) Var(X       ) Y ( E ) X ( E Y) E(X    2 Y X Y X σ σ   
47. 47. Probability Distributions Continuous Probability Distributions Binomial Hypergeometric Poisson Probability Distributions Discrete Probability Distributions Normal Uniform Exponential
48. 48. Binomial Probability Distribution  A fixed number of observations, n  e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse  Two mutually exclusive and collectively exhaustive categories  e.g., head or tail in each toss of a coin; defective or not defective light bulb  Generally called “success” and “failure”  Probability of success is p, probability of failure is 1 – p  Constant probability for each observation  e.g., Probability of getting a tail is the same each time we toss the coin
49. 49. Binomial Probability Distribution (continued)  Observations are independent  The outcome of one observation does not affect the outcome of the other  Two sampling methods  Infinite population without replacement  Finite population with replacement
50. 50. Binomial Distribution Settings • A manufacturing plant labels items as either defective or acceptable • A firm bidding for contracts will either get a contract or not • A marketing research firm receives survey responses of “yes I will buy” or “no I will not” • New job applicants either accept the offer or reject it
51. 51. Rule of Combinations • The number of combinations of selecting X objects out of n objects is X)! (n X! n! Cx n   where: n! =(n)(n - 1)(n - 2) . . . (2)(1) X! = (X)(X - 1)(X - 2) . . . (2)(1) 0! = 1 (by definition)
52. 52. P(X) = probability of X successes in n trials, with probability of success p on each trial X = number of ‘successes’ in sample, (X = 0, 1, 2, ..., n) n = sample size (number of trials or observations) p = probability of “success” P(X) n X ! n X p (1-p) X n X ! ( )!    Example: Flip a coin four times, let x = # heads: n = 4 p = 0.5 1 - p = (1 - 0.5) = 0.5 X = 0, 1, 2, 3, 4 Binomial Distribution Formula
53. 53. Calculating a Binomial Probability What is the probability of one success in five observations if the probability of success is .1? X = 1, n = 5, and p = 0.1 0.32805 .9) (5)(0.1)(0 0.1) (1 (0.1) 1)! (5 1! 5! p) (1 p X)! (n X! n! 1) P(X 4 1 5 1 X n X           
54. 54. n = 5 p = 0.1 n = 5 p = 0.5 Mean 0 .2 .4 .6 0 1 2 3 4 5 X P(X) .2 .4 .6 0 1 2 3 4 5 X P(X) 0 Binomial Distribution • The shape of the binomial distribution depends on the values of p and n  Here, n = 5 and p = 0.1  Here, n = 5 and p = 0.5
55. 55. Binomial Distribution Characteristics • Mean  Variance and Standard Deviation np E(x) μ   p) - np(1 σ2  p) - np(1 σ  Where n = sample size p = probability of success (1 – p) = probability of failure
56. 56. Using Binomial Tables n = 10 x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50 0 1 2 3 4 5 6 7 8 9 10 … … … … … … … … … … … 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0055 0.0008 0.0001 0.0000 0.0000 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0000 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0368 0.0090 0.0014 0.0001 0.0000 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689 0.0212 0.0043 0.0005 0.0000 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115 0.0425 0.0106 0.0016 0.0001 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.1596 0.0746 0.0229 0.0042 0.0003 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0.2051 0.1172 0.0439 0.0098 0.0010 10 9 8 7 6 5 4 3 2 1 0 … p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x Examples: n = 10, p = 0.35, x = 3: P(x = 3|n =10, p = 0.35) = 0.2522 n = 10, p = 0.75, x = 2: P(x = 2|n =10, p = 0.75) = 0.0004
57. 57. The Poisson Distribution • Apply the Poisson Distribution when: • You wish to count the number of times an event occurs in a given area of opportunity • The probability that an event occurs in one area of opportunity is the same for all areas of opportunity • The number of events that occur in one area of opportunity is independent of the number of events that occur in the other areas of opportunity • The probability that two or more events occur in an area of opportunity approaches zero as the area of opportunity becomes smaller • The average number of events per unit is  (lambda)
58. 58. Poisson Distribution Formula where: X = number of events in an area of opportunity  = expected number of events e = base of the natural logarithm system (2.71828...) ! X e ) X ( P x    
59. 59. Poisson Distribution Characteristics • Mean  Variance and Standard Deviation λ μ  λ σ2  λ σ  where  = expected number of events
60. 60. Using Poisson Tables X  0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0 1 2 3 4 5 6 7 0.9048 0.0905 0.0045 0.0002 0.0000 0.0000 0.0000 0.0000 0.8187 0.1637 0.0164 0.0011 0.0001 0.0000 0.0000 0.0000 0.7408 0.2222 0.0333 0.0033 0.0003 0.0000 0.0000 0.0000 0.6703 0.2681 0.0536 0.0072 0.0007 0.0001 0.0000 0.0000 0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 0.0000 0.5488 0.3293 0.0988 0.0198 0.0030 0.0004 0.0000 0.0000 0.4966 0.3476 0.1217 0.0284 0.0050 0.0007 0.0001 0.0000 0.4493 0.3595 0.1438 0.0383 0.0077 0.0012 0.0002 0.0000 0.4066 0.3659 0.1647 0.0494 0.0111 0.0020 0.0003 0.0000 Example: Find P(X = 2) if  = 0.50 0.0758 2! (0.50) e X! e 2) P(X 2 0.50 X λ       λ
61. 61. Graph of Poisson Probabilities 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0 1 2 3 4 5 6 7 x P(x) X  = 0.50 0 1 2 3 4 5 6 7 0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 0.0000 P(X = 2) = 0.0758 Graphically:  = 0.50
62. 62. Poisson Distribution Shape • The shape of the Poisson Distribution depends on the parameter  : 0.00 0.05 0.10 0.15 0.20 0.25 1 2 3 4 5 6 7 8 9 10 11 12 x P(x) 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0 1 2 3 4 5 6 7 x P(x)  = 0.50  = 3.00
63. 63. 63 Continuous Probability Distributions •Continuous random variable • Values from an interval of numbers • Absence of gaps •Continuous probability distribution • Probability distribution of continuous random variable (probability of an interval) • The most important continuous probability distribution: The normal distribution
64. 64. 64 Normal Distribution Probability Density Function • ‘Bell-Shaped’ & Symmetrical • Mean & Median Are Equal • Random Variable Has Infinite Range Mean Median X f(X) Application: From discrete to continuous Working Years of BiMBA Classes
65. 65. 65 Normal Distribution Probability Density Function f(X) : probability density function of X  : mean  : standard deviation  = 3.14159; e = 2.71828 X = value of random variable (- < X < ) 2 2 1 2 1 ) (                              X e X f
66. 66. 66 Z=0 z=1 Z Standardize the Normal Distribution One table! Normal Distribution Standardized Normal Distribution X   Z = X –   f(X) f (Z)
67. 67. 67 Standardized Normal Distribution 6.2 5 0.12 10 X Z        Normal Distribution Standardized Normal Distribution Shaded Area Exaggerated 10   1 Z   5   6.2 X Z 0 Z   0.12 f(X)
68. 68. 68 Standardized Normal Distribution Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .5478 .02 0.1 .5478 Cumulative Standardized Normal Distribution Table (Portion) Probabilities Only One Table is Needed 0 1 Z Z     Z = 0.12 0
69. 69. 69 Normal Distribution Standardized Normal Distribution Shaded Area Exaggerated 10   1 Z   5   X Z 0 Z   2.9 5 8.1 5 .21 .31 10 10 X X Z Z                0.21  .31 8.1 2.9 P(2.9 < X < 8.1) = .2049
70. 70. 70 Z .00 .01 0.0 .5000 .5040 .5080 .5398 .5438 0.2 .5793 .5832 .5871 0.3 .6179 .6217 .6255 .6217 .02 0.1 .5478 Cumulative Standardized Normal Distribution Table (Portion) Shaded Area Exaggerated 0 1 Z Z     Z = 0.31 (continued) 0 P(2.9 < X < 8.1) = .2049
71. 71. 71 Z .00 .01 -03 .3821 .3783 .3745 .4207 .4168 -0.1.4602 .4562 .4522 0.0 .5000 .4960 .4920 .4168 .02 -02 .4129 Cumulative Standardized Normal Distribution Table (Portion) 0 1 Z Z     Z = -0.21 (continued) 0 .6217.4168 = .2049 P(2.9 < X < 8.1) = .2049
72. 72. 72 P(X > 8) = .3821 Normal Distribution Standardized Normal Distribution Shaded Area Exaggerated 10   1 Z   5   X Z 0 Z   8 5 .30 10 X Z        ? .30 8
73. 73. 73 Normal Distribution and  • Refer to Fig. 6.11, 6.12, and 6.13 (p. 204-205) 68.26% of normal values fall between 1 95.44% of normal values fall between 2 99.73% of normal values fall between 3  Check Standardized Normal Distribution Table and find the probability of normal values fall between 4, 5, and 6
74. 74. 74 Normal Distribution Thinking Challenge You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours &  = 200 hours. What’s the probability that a bulb will last? a. between 2000 & 2400 hours? b. less than 1470 hours
75. 75. 75 Z Z= 0 Z = 1 2.0 You work in Quality Control for GE. Light bulb life has a normal distribution with = 2000 hours &  = 200 hours. What’s the probability that a bulb will last? a. between 2000 & 2400 hours? b. less than 1470 hours Solution: P(2000  X  2400) Normal Distribution .4772 Standardized Normal Distribution Z X        2400 2000 200 2.0 X  = 2000  = 200 2400
76. 76. 76 .6217 Finding Z Values for Known Probabilities Z .00 0.2 0.0 .5000 .5040 .5080 0.1 .5398 .5438 .5478 0.2 .5793 .5832 .5871 .6179 .6255 .01 0.3 Cumulative Standardized Normal Distribution Table What is Z Given Probability = 0.6217 ? Shaded Area Exaggerated .6217 0 1 Z Z     .31 Z  0 Z
77. 77. 77 Uniform Distribution • The probability density function: Mean: Variance: 1 ( ) 0 for a x b b a f x othrewise            2 a b    2 2 ( ) 12 b a    2 1 1 2 1 2 ( ) , x x P x X x a x x b b a        
78. 78. 78 Exponential Distribution Density function: (X > 0) Mean and Standard Deviation: f(X) X  = 1.0  = 0.5   X f X e     1 1      
79. 79. 79 Exponential Distribution f (X) = eX P(a < X < b) = e -a – e -b a b X  0 f (X)
80. 80. 80 Exponential Distribution P( X < b ) =1 - e  : expected (average) number of “arrivals” X : time between successive “arrivals” a, b : value of the continuous random variable e = 2.71828 - b ( 0 < b <∞) The cumulative distribution function is : P(a < X < b) = e-a – e-b ( 0 < a < b) Or
81. 81. 81 Example Customers arrive at the check out line of a supermarket at the rate of 30 per hour. What is the probability that the arrival time between consecutive customers to be greater than 5 minutes?           .5 5 30/60 .5/minute, 5minutes > 1 1 1 .0821 b b P X b P X b e e               
82. 82. 82 Application If the number of times that a particular event occurs over an interval of time or space follows Poisson distribution. Then the time or space between successive occurrences of the event follows exponential distribution. The record shows the loss of a bike every 3 years. a) What is the probability that the time between successive losses of bikes is more than a year? b) What is the probability of less than one bike loss in a year?
83. 83. • Thank you