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# Probability Arunesh Chand Mankotia 2005

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### Probability Arunesh Chand Mankotia 2005

2. 2. Sample SpaceThe possible outcomes of a random experimentare called the basic outcomes, and the set of allbasic outcomes is called the sample space. The symbol S will be used to denote the sample space.
3. 3. Sample Space - An Example -What is the sample space for a roll of a single six-sided die? S = [1, 2, 3, 4, 5, 6]
4. 4. Mutually Exclusive If the events A and B have no common basic outcomes,they are mutually exclusive and their intersection A ∩ B is said to be the empty set indicating that A ∩ B cannot occur. More generally, the K events E1, E2, . . . , EK are said to be mutually exclusive if every pair of them is a pair of mutually exclusive events.
5. 5. Venn DiagramsVenn Diagrams are drawings, usually using geometric shapes, used to depict basicconcepts in set theory and the outcomes of random experiments.
6. 6. Intersection of Events A and BS S A A∩B B A B(a) A∩B is the striped area (b) A and B are Mutually Exclusive
7. 7. Collectively Exhaustive Given the K events E1, E2, . . ., EK in thesample space S. If E1 ∪ E2 ∪ . . . ∪EK = S, these events are said to be collectively exhaustive. exhaustive
8. 8. ComplementLet A be an event in the sample space S. Theset of basic outcomes of a random experiment belonging to S but not to A is called the complement of A and is denoted by A.
9. 9. Venn Diagram for theComplement of Event AS A A
10. 10. Unions, Intersections, and ComplementsA die is rolled. Let A be the event “Number rolled is even”and B be the event “Number rolled is at least 4.” Then A = [2, 4, 6] and B = [4, 5, 6] A = [1, 3, 5] and B = [1, 2, 3] A ∩ B = [4, 6] A ∪ B = [2, 4, 5, 6] A ∪ A = [1, 2, 3, 4, 5, 6] = S
11. 11. Classical Probability The classical definition of probability is the proportion of times that an event will occur,assuming that all outcomes in a sample space are equally likely to occur. The probability of an event is determined by counting the number of outcomes in the sample space that satisfy theevent and dividing by the number of outcomes in the sample space.
12. 12. Classical Probability The probability of an event A is NA P(A) = N where NA is the number of outcomes that satisfy thecondition of event A and N is the total number of outcomes in the sample space. The important idea here is that one can develop a probability from fundamental reasoning about the process.
13. 13. CombinationsThe counting process can be generalized by using the following equation to compare the number of combinations of n things taken k at a time. n! C = n k 0!= 1 k!(n − k )!
14. 14. Relative FrequencyThe relative frequency definition of probability isthe limit of the proportion of times that anevent A occurs in a large number of trials, n, nA P(A) = nwhere nA is the number of A outcomes and n isthe total number of trials or outcomes in thepopulation. The probability is the limit as nbecomes large.
15. 15. Subjective Probability The subjective definition of probabilityexpresses an individual’s degree of belief about the chance that an event will occur. These subjective probabilities are used in certain management decision procedures.
16. 16. Probability PostulatesLet S denote the sample space of a random experiment, Oi, the basic outcomes, and A, an event. For each event A of the sample space S, we assume that a number P(A) is defined and we have the postulatesq If A is any event in the sample space S 0 ≤ P ( A) ≤ 1q Let A be an event in S, and let Oi denote the basic outcomes. Then P ( A) = ∑ P (Oi ) A where the notation implies that the summation extends over all the basic outcomes in A. 3. P(S) = 1
17. 17. Probability RulesLet A be an event and A its complement. The the complement rule is: is P ( A ) = 1 − P ( A)
18. 18. Probability Rules The Addition Rule of Probabilities: ProbabilitiesLet A and B be two events. The probability of their union isP ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B )
19. 19. Probability RulesVenn Diagram for Addition Rule P ( A ∪ B ) = P ( A) + P ( B ) − P ( A ∩ B ) P(A∪B) A B =P(A) P(B) P(A∩B)A B + A B - A B
20. 20. Probability Rules Conditional Probability: Probability Let A and B be two events. The conditional probability ofevent A, given that event B has occurred, is denoted by the symbol P(A|B) and is found to be: P( A ∩ B) P( A | B) = P( B) provided that P(B > 0).
21. 21. Probability Rules Conditional Probability: Probability Let A and B be two events. The conditional probability ofevent B, given that event A has occurred, is denoted by the symbol P(B|A) and is found to be: P( A ∩ B) P ( B | A) = P ( A) provided that P(A > 0).
22. 22. Probability Rules The Multiplication Rule of Probabilities: ProbabilitiesLet A and B be two events. The probability of their intersection can be derived from the conditional probability as P( A ∩ B) = P( A | B) P( B) Also, P ( A ∩ B ) = P ( B | A) P ( A)
23. 23. Statistical IndependenceLet A and B be two events. These events are said to be statistically independent if and only if P ( A ∩ B) = P( A) P ( B) From the multiplication rule it also follows that P(A | B) = P(A) (if P(B) > 0) P(B | A) = P(B) (if P(A) > 0)More generally, the events E1, E2, . . ., Ek are mutually statistically independent if and only if P(E1 ∩ E 2 ∩  ∩ E K ) = P(E1 ) P(E 2 )  P(E K )
24. 24. Bivariate Probabilities B1 B2 ... BkA1 P(A1∩B1) P(A1∩B2) ... P(A1∩Bk)A2 P(A2∩B1) P(A2∩B2) ... P(A2∩Bk) . . . . . . . . . . . . . . .Ah P(Ah∩B1) P(Ah∩B2) ... P(Ah∩Bk) Outcomes for Bivariate Events
25. 25. Joint and Marginal Probabilities In the context of bivariate probabilities, the intersection probabilities P(Ai ∩ Bj) are called jointprobabilities. The probabilities for individual events P(Ai) and P(Bj) are called marginal probabilities. probabilities Marginal probabilities are at the margin of abivariate table and can be computed by summing the corresponding row or column.
26. 26. Probabilities for the Television Viewing and Income Example Viewing High Middle Low TotalsFrequency Income Income Income Regular 0.04 0.13 0.04 0.21 Occasional 0.10 0.11 0.06 0.27 Never 0.13 0.17 0.22 0.52 Totals 0.27 0.41 0.32 1.00
27. 27. Tree Diagrams P(A1 ∩ B1) = .04 P(A1 ∩ B2) = .13 P(A1 ∩ B3) = .04 1 .2 )= P(A2 ∩ B1) = .10 1 A P( P(A2) = .27P(S) = 1 P(A2 ∩ B2) = .11 P( A P(A2 ∩ B3) = .06 3 )= .5 P(A3∩ B1) = .13 2 P(A3 ∩ B2) = .17 P(A3 ∩ B3) = .22
28. 28. Probability Rules Rule for Determining the Independence of AttributesLet A and B be a pair of attributes, each broken intomutually exclusive and collectively exhaustive event categories denoted by labels A1, A2, . . ., Ah andB1, B2, . . ., Bk. If every Ai is statistically independent of every event Bj, then the attributes A and B are independent.
29. 29. Bayes’ TheoremLet A and B be two events. Then Bayes’ Theorem states that: P(A | B)P(B) P( A | B) = P(A) and P(B | A)P(A) P( A | B) = P(B)
30. 30. Bayes’ Theorem (Alternative Statement)Let E1, E2, . . . , Ek be mutually exclusive and collectively exhaustive events and let A be some other event. Theconditional probability of Ei given A can be expressed as Bayes’ Theorem: Theorem P(A | E i )P(E i )P(E i | A) = P(A | E1 )P(E1 ) + P(A | E 2 )P(E 2 ) +  + P(A | E K )P(E K )
31. 31. Bayes’ Theorem - Solution Steps -1. Define the subset events from the problem.2. Define the probabilities for the events defined in step 1.3. Compute the complements of the probabilities.4. Apply Bayes’ theorem to compute the probability for the problem solution.
32. 32. Discrete Random Variables and Probability Distributions©
33. 33. Random VariablesA random variable is a variable that takes onnumerical values determined by the outcome of a random experiment.
34. 34. Discrete Random Variables A random variable is discrete if it can take on no more than a countable number of values.
35. 35. Discrete Random Variables (Examples)1. The number of defective items in a sample of twenty items taken from a large shipment.2. The number of customers arriving at a check-out counter in an hour.3. The number of errors detected in a corporation’s accounts.4. The number of claims on a medical insurance policy in a particular year.
36. 36. Continuous Random Variables A random variable is continuous if it can take any value in an interval.
37. 37. Continuous Random Variables (Examples)1. The income in a year for a family.2. The amount of oil imported into the U.S. in a particular month.3. The change in the price of a share of IBM common stock in a month.4. The time that elapses between the installation of a new computer and its failure.5. The percentage of impurity in a batch of chemicals.
38. 38. Discrete Probability DistributionsThe probability distribution function (DPF), P(x), of a discrete random variable expresses the probability that X takes the value x, as a function of x. That is P ( x) = P ( X = x), for all values of x.
39. 39. Discrete Probability Distributions Graph the probability distribution function for the roll of a single six-sided die. P(x) 1/6 1 2 3 4 5 6 x
40. 40. Required Properties of Probability Distribution Functions of Discrete Random Variables Let X be a discrete random variable with probability distribution function, P(x). Thenq P(x) ≥ 0 for any value of xq The individual probabilities sum to 1; that is ∑ P( x) = 1 x Where the notation indicates summation over all possible values x.
41. 41. Cumulative Probability FunctionThe cumulative probability function, F(x0), of arandom variable X expresses the probability that X does not exceed the value x0, as a function of x0. That is F ( x0 ) = P ( X ≤ x0 ) Where the function is evaluated at all values x0
42. 42. Derived Relationship Between Probability Function and Cumulative Probability Function Let X be a random variable with probability functionP(x) and cumulative probability function F(x0). Then it can be shown that F ( x0 ) = ∑ P( X ) x ≤ x0 Where the notation implies that summation is over all possible values x that are less than or equal to x0.
43. 43. Derived Properties of CumulativeProbability Functions for Discrete Random Variables Let X be a discrete random variable with a cumulative probability function, F(x0). Then we can show thatq 0 ≥ F(x0) ≥ 1 for every number x0q If x0 and x1 are two numbers with x0 < x1, then F(x0) ≤ F(x1)
44. 44. Expected Value The expected value, E(X), of a discrete random variable X is defined E ( X ) = ∑ xP( x) xWhere the notation indicates that summation extends over all possible values x.The expected value of a random variable is called its mean and is denoted µx.
45. 45. Expected Value: Functions of Random Variables Let X be a discrete random variable with probability function P(x) and let g(X) be some function of X. Then the expected value, E[g(X)], of that function is defined as E[ g ( X )] = ∑ g ( x) P ( x) x
46. 46. Variance and Standard Deviation Let X be a discrete random variable. The expectation of the squared discrepancies about the mean, (X - µ)2, is called the variance, denoted σ2x and is given by variance σ x = E ( X − µ x ) 2 = ∑ ( x − µ x ) 2 P( x) 2 x The standard deviation, σx , is the positive square root deviation of the variance.
47. 47. Variance (Alternative Formula)The variance of a discrete random variable X can beExpressed as σ = E( X ) − µx 2 2 2 x = ∑ x P( x) − µ x 2 2 x
48. 48. Expected Value and Variance forDiscrete Random Variable Using Microsoft Excel Sales P(x) Mean Variance 0 0.15 0 0.570375 1 0.3 0.3 0.27075 2 0.2 0.4 0.0005 3 0.2 0.6 0.2205 4 0.1 0.4 0.42025 5 0.05 0.25 0.465125 1.95 1.9475 Expected Value = 1.95 Variance = 1.9475
49. 49. Summary of Properties for Linear Function of a Random VariableLet X be a random variable with mean µx , and variance σ2x; and let a and b be any constant fixed numbers. Define therandom variable Y = a + bX. Then, the mean and variance of Y are µY = E (a + bX ) = a + bµ X and σ 2 Y = Var (a + bX ) = b σ 2 2 X so that the standard deviation of Y is σY = bσ X
50. 50. Summary Results for the Mean and Variance of Special Linear Functionsq Let b = 0 in the linear function, W = a + bX. Then W = a (for any constant a). E (a) = a and Var (a ) = 0If a random variable always takes the value a, it will have a mean a and a variance 0.q Let a = 0 in the linear function, W = a + bX. Then W = bX. E (bX ) = bµ X and Var (a ) = b 2σ X 2
51. 51. Mean and Variance of ZLet a = -µX/σX and b = 1/ σX in the linear function Z = a + bX. Then, X − µX Z = a + bX = σX so that  X − µX  µX 1 E  σ =−  + µX = 0  X  σX σX and  X − µX  1 2 Var   σ  = 2 σ X =1  σ  X  X
52. 52. Bernoulli DistributionA Bernoulli distribution arises from a random experimentwhich can give rise to just two possible outcomes. These outcomes are usually labeled as either “success” or“failure.” If π denotes the probability of a success and the probability of a failure is (1 - π ), the the Bernoulli probability function is P (0) = (1 − π ) and P (1) = π
53. 53. Mean and Variance of a Bernoulli Random Variable The mean is: µ X = E ( X ) = ∑ xP( x) = (0)(1 − π ) + (1)π = π X And the variance is: σ = E[( X − µ X ) ] = ∑ ( x − µ X ) P( x) 2 X 2 2 X = (0 − π ) 2 (1 − π ) + (1 − π ) 2 π = π (1 − π )
54. 54. Sequences of x Successes in n TrialsThe number of sequences with x successes in n independent trials is: n! C = n x x!(n − x)! Where n! = n x (x – 1) x (n – 2) x . . . x 1 and 0! = 1. n These C x sequences are mutually exclusive, since no two of them can occur at the same time.
55. 55. Binomial DistributionSuppose that a random experiment can result in two possible mutuallyexclusive and collectively exhaustive outcomes, “success” and “failure,”and that π is the probability of a success resulting in a single trial. If n independent trials are carried out, the distribution of the resulting number of successes “x” is called the binomial distribution. Its distributionprobability distribution function for the binomial random variable X = x is: P(x successes in n independent trials)= n! ( n− x ) P( x) = π (1 − π ) x x!(n − x)! for x = 0, 1, 2 . . . , n
56. 56. Mean and Variance of a Binomial Probability DistributionLet X be the number of successes in n independent trials, each with probability of success π. The x follows a binomial distribution with mean, mean µ X = E ( X ) = nπ and variance, variance σ = E[( X − µ ) ] = nπ (1 − π ) 2 X 2
57. 57. Binomial Probabilities - An Example –An insurance broker, has five contracts, and he believesthat for each contract, the probability of making a sale is0.40. What is the probability that he makes at most one sale?P(at most one sale) = P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.078 + 0.259 = 0.337 5! P(no sales) = P(0) = (0.4) 0 (0.6) 5 = 0.078 0!5! 5! P(1 sale) = P(1) = (0.4)1 (0.6) 4 = 0.259 1!4!
58. 58. Binomial Probabilities, n = 100, π =0.40 Sample size 100 Probability of success 0.4 Mean 40 Variance 24 Standard deviation 4.898979 Binomial Probabilities Table X P(X) P(<=X) P(<X) P(>X) P(>=X) 36 0.059141 0.238611 0.179469 0.761389 0.820531 37 0.068199 0.30681 0.238611 0.69319 0.761389 38 0.075378 0.382188 0.30681 0.617812 0.69319 39 0.079888 0.462075 0.382188 0.537925 0.617812 40 0.081219 0.543294 0.462075 0.456706 0.537925 41 0.079238 0.622533 0.543294 0.377467 0.456706 42 0.074207 0.69674 0.622533 0.30326 0.377467 43 0.066729 0.763469 0.69674 0.236531 0.30326
59. 59. Poisson Probability DistributionAssume that an interval is divided into a very large number of subintervals so that the probability of the occurrence of an event in any subinterval is very small. The assumptions of a Poisson probability distribution are:2) The probability of an occurrence of an event is constant for all subintervals.3) There can be no more than one occurrence in each subinterval.4) Occurrences are independent; that is, the number of occurrences in any non-overlapping intervals in independent of one another.
60. 60. Poisson Probability Distribution The random variable X is said to follow the Poissonprobability distribution if it has the probability function: e − λ λx P( x) = , for x = 0, 1,2,... x! where P(x) = the probability of x successes over a given period of time or space, given λ λ = the expected number of successes per time or space unit; λ > 0 e = 2.71828 (the base for natural logarithms) The mean and variance of the Poisson probability distribution are: are µ x = E ( X ) = λ and σ x2 = E[( X − µ ) 2 ] = λ
61. 61. Partial Poisson Probabilities for λ = 0.03 Obtained Using Microsoft ExcelPoisson Probabilities Table X P(X) P(<=X) P(<X) P(>X) P(>=X) 0 0.970446 0.970446 0.000000 0.029554 1.000000 1 0.029113 0.999559 0.970446 0.000441 0.029554 2 0.000437 0.999996 0.999559 0.000004 0.000441 3 0.000004 1.000000 0.999996 0.000000 0.000004 4 0.000000 1.000000 1.000000 0.000000 0.000000
62. 62. Poisson Approximation to the Binomial Distribution Let X be the number of successes resulting from n independenttrials, each with a probability of success, π. The distribution of thenumber of successes X is binomial, with mean nπ. If the number oftrials n is large and nπ is of only moderate size (preferably nπ ≤ 7), this distribution can be approximated by the Poisson distribution with λ = nπ. The probability function of the approximating distribution is then: e − nπ (nπ ) x P( x) = , for x = 0, 1,2,... x!
63. 63. Covariance Let X be a random variable with mean µ X , and let Y be arandom variable with mean, µ Y . The expected value of (X - µ X )(Y - µ Y ) is called the covariance between X and Y, denoted Cov(X, Y). For discrete random variablesCov ( X , Y ) = E[( X − µ X )(Y − µY )] = ∑∑ ( x − µ x )( y − µ y ) P ( x, y ) x y An equivalent expression is Cov ( X , Y ) = E ( XY ) − µ x µ y = ∑∑ xyP( x, y ) − µ x µ y x y
64. 64. CorrelationLet X and Y be jointly distributed random variables. The correlation between X and Y is: Cov ( X , Y ) ρ = Corr ( X , Y ) = σ XσY
65. 65. Covariance and Statistical Independence If two random variables are statisticallyindependent, the covariance between them is 0.independent However, the converse is not necessarily true.
66. 66. Portfolio Analysis The random variable X is the price for stock A and the random variable Y is the price for stock B. The marketvalue, W, for the portfolio is given by the linear function, W = aX + bYWhere, a, is the number of shares of stock A and, b, is the number of shares of stock B.
67. 67. Portfolio Analysis The mean value for W is, µW = E[W ] = E[aX + bY ] = aµ X + bµY The variance for W is, σ = a σ + b σ + 2abCov ( X , Y ) 2 W 2 2 X 2 2 Y or using the correlation,σ = a σ + b σ + 2abCorr ( X , Y )σ X σ Y 2 W 2 2 X 2 2 Y
68. 68. Continuous Random Variables and Probability Distributions©
69. 69. Continuous Random Variables A random variable is continuous if it can take any value in an interval.
70. 70. Cumulative Distribution Function The cumulative distribution function, F(x), for a function continuous random variable X expresses the probability that X does not exceed the value of x, as a function of x F ( x) = P( X ≤ x)
71. 71. Cumulative Distribution Function F(x) 1 0 1 Cumulative Distribution Function for a Random variable Over 0 to 1
72. 72. Cumulative Distribution Function Let X be a continuous random variable with a cumulative distribution function F(x), and let a and b be two possible values of X, with a < b. The probability that X lies between a and b is P(a < X < b) = F (b) − F (a )
73. 73. Probability Density Function Let X be a continuous random variable, and let x be any number lying in the range of values this random variable can take. The probability density function, f(x), of the random variable is a function with the following function properties:q f(x) > 0 for all values of xq The area under the probability density function f(x) over all values of the random variable X is equal to 1.0q Suppose this density function is graphed. Let a and b be two possible values of the random variable X, with a<b. Then the probability that X lies between a and b is the area under the density function between these points.q The cumulative density function F(x0) is the area under the probability density function f(x) up to x0 x0 f ( x0 ) = ∫ f ( x)dx xm
74. 74. Shaded Area is the Probability That X is Between a and b0 a b x
75. 75. Probability Density Function for aUniform 0 to 1 Random Variable f(x) 1 0 1 x
76. 76. Areas Under Continuous Probability Density Functions  Let X be a continuous random variable with the probability density function f(x) and cumulative distribution F(x). Then the following properties hold: q The total area under the curve f(x) = 1. q The area under the curve f(x) to the left of x0 is F(x0), where x0 is any value that the random variable can take.
77. 77. Properties of the Probability Density Functionf(x) Comments 1 Total area under the uniform probability density function is 1. 0 0 x0 1 x
78. 78. Properties of the Probability Density Function Commentsf(x) Area under the uniform probability density function to the left of 1 x0 is F(x0), which is equal to x0 for this uniform distribution because f(x)=1. 0 0 x0 1 x
79. 79. Rationale for Expectations ofContinuous Random Variables Suppose that a random experiment leads to an outcome that can be represented by a continuous random variable. If N independent replications of this experiment are carried out, then the expected value of the random variable is the average of thevalues taken, as the number of replications becomes infinitely large. The expected value of a random variable is denoted by E(X).
80. 80. Rationale for Expectations ofContinuous Random Variables (continued) Similarly, if g(x) is any function of the randomvariable, X, then the expected value of this function isthe average value taken by the function over repeated independent trials, as the number of trials becomes infinitely large. This expectation is denoted E[g(X)]. By using calculus we can define expected values forcontinuous random variables similarly to that used for discrete random variables. E[ g ( x)] = ∫ g ( x) f ( x)dx x
81. 81. Mean, Variance, and Standard DeviationLet X be a continuous random variable. There are two important expected values that are used routinely to define continuous probability distributions.q The mean of X, denoted by µX, is defined as the expected value of X. X µ X = E( X )q The variance of X, denoted by σX2, is defined as the expectation of the X squared deviation, (X - µX)2, of a random variable from its mean σ = E[( X − µ X ) ] 2 X 2Or an alternative expression can be derived σ X = E( X 2 ) − µ X 2 2q The standard deviation of X, σX, is the square root of the variance. X
82. 82. Linear Functions of VariablesLet X be a continuous random variable with mean µ X andvariance σ X2, and let a and b any constant fixed numbers. Define the random variable W as W = a + bX Then the mean and variance of W are µW = E (a + bX ) = a + bµ X and σ = Var (a + bX ) = b σ 2 W 2 2 X and the standard deviation of W is σW = bσ X
83. 83. Linear Functions of Variable (continued)An important special case of the previous results is the standardized random variable X − µX Z= σX which has a mean 0 and variance 1.
84. 84. Reasons for Using the Normal Distribution1. The normal distribution closely approximates the probability distributions of a wide range of random variables.2. Distributions of sample means approach a normal distribution given a “large” sample size.3. Computations of probabilities are direct and elegant.4. The normal probability distribution has led to good business decisions for a number of applications.
85. 85. Probability Density Function for a Normal Distribution (Figure 6.8) 0.4 0.3 0.2 0.1 0.0 µ x
86. 86. Probability Density Function of the Normal Distribution The probability density function for a normally distributed random variable X is 1 − ( x − µ ) 2 / 2σ 2 f ( x) = e for - ∞ < x < ∞ 2πσ 2Where µ and σ 2 are any number such that -∞ < µ < ∞ and -∞ < σ 2 < ∞ and where e and π are physical constants, e = 2.71828. . . and π = 3.14159. . .
87. 87. Properties of the Normal DistributionSuppose that the random variable X follows a normal distribution with parameters µ and σ2. Then the following properties hold:q The mean of the random variable is µ, E( X ) = µq The variance of the random variable is σ2, E[( X − µ X ) 2 ] = σ 2q The shape of the probability density function is a symmetric bell- shaped curve centered on the mean µ as shown in Figure 6.8.vii. By knowing the mean and variance we can define the normal distribution by using the notation X ~ N (µ ,σ ) 2
88. 88. Effects of µ on the Probability Density Function of a Normal Random Variable0.40.3 Mean = 5 Mean = 60.20.10.0 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 x
89. 89. Effects of σ2 on the Probability DensityFunction of a Normal Random Variable 0.4 Variance = 0.0625 0.3 0.2 Variance = 1 0.1 0.0 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 x
90. 90. Cumulative Distribution Function of the Normal DistributionSuppose that X is a normal random variable with mean µ and variance σ 2 ; that is X~N(µ, σ 2). Then the cumulative distribution function is F ( x0 ) = P ( X ≤ x0 ) This is the area under the normal probability densityfunction to the left of x0, as illustrated in Figure 6.10. Asfor any proper density function, the total area under the curve is 1; that is F(∞) = 1.
91. 91. Shaded Area is the Probability that X does not Exceed x0 for a Normal Random Variablef(x) x0 x
92. 92. Range Probabilities for Normal Random VariablesLet X be a normal random variable with cumulative distribution function F(x), and let a and b be two possible values of X, with a < b. Then P (a < X < b) = F (b) − F (a )The probability is the area under the corresponding probability density function between a and b.
93. 93. Range Probabilities for Normal Random Variablesf(x) a µ b x
94. 94. The Standard Normal Distribution Let Z be a normal random variable with mean 0 and variance 1; that is Z ~ N (0,1)We say that Z follows the standard normal distribution.Denote the cumulative distribution function as F(z), and a and b as two numbers with a < b, then P (a < Z < b) = F (b) − F (a)
95. 95. Standard Normal Distribution with Probability for z = 1.25 0.8944 z 1.25
96. 96. Finding Range Probabilities for Normally Distributed Random Variables Let X be a normally distributed random variable with mean µ and variance σ 2. Then the random variable Z = (X - µ)/σ has a standard normal distribution: Z ~ N(0, 1) It follows that if a and b are any numbers with a < b, then a−µ b−µ  P ( a < X < b) = P <Z<   σ σ  b−µ  a−µ  = F  − F   σ   σ where Z is the standard normal random variable and F(z) denotes its cumulative distribution function.
97. 97. Computing Normal Probabilities A very large group of students obtains test scores that are normally distributed with mean 60 and standard deviation 15. What proportion of the students obtained scores between 85 and 95?  85 − 60 95 − 60  P (85 < X < 95) = P <Z<   15 15  = P (1.67 < Z < 2.33) = F (2.33) − F (1.67) = 0.9901 − 0.9525 = 0.0376That is, 3.76% of the students obtained scores in the range 85 to 95.
98. 98. Approximating Binomial Probabilities Using the Normal Distribution Let X be the number of successes from n independent Bernoullitrials, each with probability of success π. The number of successes, X, is a Binomial random variable and if nπ(1 - π) > 9 a good approximation is  a − nπ b − nπ  P ( a < X < b) = P ≤Z≤   nπ (1 − π ) nπ (1 − π )   Or if 5 < nπ(1 - π) < 9 we can use the continuity correction factor to obtain  a − 0.5 − nπ b + 0.5 − nπ  P ( a ≤ X ≤ b) = P  ≤Z≤   nπ (1 − π ) nπ (1 − π )    where Z is a standard normal variable.
99. 99. Covariance Let X and Y be a pair of continuous random variables,with respective means µ x and µ y. The expected value of (x - µ x)(Y - µ y) is called the covariance between X and Y. That is Cov( X , Y ) = E[( X − µ x )(Y − µ y )]An alternative but equivalent expression can be derived as Cov( X , Y ) = E ( XY ) − µ x µ yIf the random variables X and Y are independent, then the covariance between them is 0. However, the converse is not true.
100. 100. CorrelationLet X and Y be jointly distributed random variables. The correlation between X and Y is Cov( X , Y ) ρ = Corr ( X , Y ) = σ XσY
101. 101. Sums of Random VariablesLet X1, X2, . . .Xk be k random variables with means µ1, µ2,. . . µk and variances σ12, σ22,. . ., σk2. The following properties hold:ii. The mean of their sum is the sum of their means; that is E ( X 1 + X 2 +  + X k ) = µ1 + µ 2 +  + µ kiv. If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances; that is Var ( X 1 + X 2 +  + X k ) = σ 12 + σ 2 +  + σ k2 2 However, if the covariances between pairs of random variables are not 0, the variance of their sum is K −1 K Var ( X 1 + X 2 +  + X k ) = σ 12 + σ 2 +  + σ k2 + 2∑ 2 ∑ Cov( X , X i j ) i =1 j =i +1
102. 102. Differences Between a Pair of Random VariablesLet X and Y be a pair of random variables with means µX and µY and variances σX2 and σY2. The following properties hold:ii. The mean of their difference is the difference of their means; that is E ( X − Y ) = µ X − µYiv. If the covariance between X and Y is 0, then the variance of their difference is Var ( X − Y ) = σ X + σ Y 2 2vi. If the covariance between X and Y is not 0, then the variance of their difference is Var ( X − Y ) = σ X + σ Y − 2Cov ( X , Y ) 2 2
103. 103. Linear Combinations of Random Variables The linear combination of two random variables, X and Y, is W = aX + bY Where a and b are constant numbers. The mean for W is, µW = E[W ] = E[aX + bY ] = aµ X + bµY The variance for W is, σ W = a 2σ X + b 2σ Y + 2abCov ( X , Y ) 2 2 2 Or using the correlation, σ W = a 2σ X + b 2σ Y + 2abCorr ( X , Y )σ X σ Y 2 2 2 If both X and Y are joint normally distributed random variablesthen the resulting random variable, W, is also normally distributed with mean and variance derived above.