Probability Distributions for Discrete RV

Transcription

1 Probability Distributions for Discrete RV

2 Probability Distributions for Discrete RV Definition The probability distribution or probability mass function (pmf) of a discrete rv is defined for every number x by p(x) = P(X = x) = P(all s S : X (s) = x). In words, for every possible value x of the random variable, the pmf specifies the probability of observing that value when the experiment is performed. (The conditions p(x) 0 and all possible x p(x) = 1 are required for any pmf.)

3 Probability Distributions for Discrete RV

4 Probability Distributions for Discrete RV Definition The cumulative distribution function (cdf) F (x) of a discrete rv X with pmf p(x) is defined for every number x by F (x) = P(X x) = y:y x p(y) For any number x, F(x) is the probability that the observed value of X will be at most x.

5 Probability Distributions for Discrete RV Definition The cumulative distribution function (cdf) F (x) of a discrete rv X with pmf p(x) is defined for every number x by F (x) = P(X x) = y:y x p(y) For any number x, F(x) is the probability that the observed value of X will be at most x. F (x) = P(X x) = P(X is less than or equal to x) p(x) = P(X = x) = P(X is exactly equal to x)

6 Probability Distributions for Discrete RV

7 Probability Distributions for Discrete RV pmf = cdf: F (x) = P(X x) = p(y) y:y x

8 Probability Distributions for Discrete RV pmf = cdf: F (x) = P(X x) = p(y) It is also possible cdf = pmf: y:y x

9 Probability Distributions for Discrete RV pmf = cdf: F (x) = P(X x) = It is also possible cdf = pmf: y:y x p(x) = F (x) F (x ) p(y) where x represents the largest possible X value that is strictly less than x.

10 Probability Distributions for Discrete RV

11 Probability Distributions for Discrete RV Proposition For any two numbers a and b with a b, P(a X b) = F (b) F (a ) where a represents the largest possible X value that is strictly less than a. In particular, if the only possible values are integers and if a and b are integers, then P(a X b) = P(X = a or a + 1 or... or b) = F (b) F (a 1) Taking a = b yields P(X = a) = F (a) F (a 1) in this case.

12 Expectations

13 Expectations Definition Let X be a discrete rv with set of possible values D and pmf p(x). The expected value or mean value of X, denoted by E(X ) or µ X, is E(X ) = µ X = x D x p(x)

14 Expectations Definition Let X be a discrete rv with set of possible values D and pmf p(x). The expected value or mean value of X, denoted by E(X ) or µ X, is E(X ) = µ X = x D x p(x) e.g (Problem 30) A group of individuals who have automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is y Then the expected value of p(y) moving violations for that group is µ Y = E(Y ) = = 0.60

15 Expectations

16 Expectations y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations.

17 Expectations y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = 0.60

18 Expectations y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = µ = = = 0.60

19 Expectations y p(y) Assume the total number of individuals in that group is 100, then there are 60 individuals without moving violation, 25 with 1 moving violation, 10 with 2 moving violations and 5 with 3 moving violations. The population mean is calculated as µ = = µ = = = 0.60 The population size is irrevelant if we know the pmf!

20 Expectations

21 Expectations Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1

22 Expectations Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1 Then the expected value for X is E(X ) = 0 p(0) + 1 p(1) = p

23 Expectations Examples: Let X be a Bernoulli rv with pmf 1 p x = 0 p(x) = p x = 1 0 x 0, or 1 Then the expected value for X is E(X ) = 0 p(0) + 1 p(1) = p We see that the expected value of a Bernoulli rv X is just the probability that X takes on the value 1.

24 Expectations

25 Expectations Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise

26 Expectations Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise The expected value for X is E(X ) = D x p(x) = xα(1 α) x 1 = α [ d dα (1 α)x ] x=1 x=1

27 Expectations Examples: Consider the cards drawing example again and assume we have infinitely many cards this time. Let X = the number of drawings until we get a. If the probability for getting a is α, then the pmf for X is { α(1 α) x 1 x = 1, 2, 3,... p(x) = 0 otherwise The expected value for X is E(X ) = D x p(x) = xα(1 α) x 1 = α [ d dα (1 α)x ] x=1 x=1 E(X ) = α{ d dα [ (1 α) x ]} = α{ d dα (1 α α )} = 1 α x=1

28 Expectations

29 Expectations Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 The expected value is NOT finite! x k x 2 = k 1 x =! x=1

30 Expectations Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 The expected value is NOT finite! Heavy Tail: x k x 2 = k 1 x =! x=1

31 Expectations Examples 3.20 Let X be the number of interviews a student has prior to getting a job. The pmf for X is { k x = 1, 2, 3,... p(x) = x 2 0 otherwise where k is chosen so that x=1 (k/x 2 ) = 1. (It can be showed that x=1 (1/x 2 ) <, which implies that such a k exists.) The expected value of X is µ = E(X ) = x=1 x k x 2 = k 1 x =! x=1 The expected value is NOT finite! Heavy Tail: distribution with a large amount of probability far from µ

32 Expectations

33 Expectations Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble?

34 Expectations Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x)

35 Expectations Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x) x

36 Expectations Example (Problem 38) Let X = the outcome when a fair die is rolled once. If before the 1 die is rolled you are offered either 3.5 dollars or 1 X dollars, would you accept the guaranteed amount or would you gamble? x p(x) x Then the expected dollars from gambling is E( 1 X ) = 6 x=1 1 x p( 1 x ) = = > 1 3.5

37 Expectations

38 Expectations Proposition If the rv X has a set of possible values D and pmf p(x), then the expected value of any function h(x ), denoted by E[h(X )] or µ hx, is computed by E[h(X )] = h(x) p(x) D

39 Expectations

40 Expectations Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece.

41 Expectations Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units,

42 Expectations Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900.

43 Expectations Example 3.23 A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900. The expected profit is E[h(X )] = h(0) p(0) + h(1) p(1) + h(2) p(2) + h(3) p(3) = ( 900)(0.1) + ( 100)(0.2) + (700)(0.3) + (1500)(0.4) = 700

44 Expectations

45 Expectations Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.)

46 Expectations Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.) e.g. for the previous example, E[h(X )] = E(800X 900) = 800 E(X ) 900 = 700

47 Expectations Proposition E(aX + b) = a E(X ) + b (Or, using alternative notation, µ ax +b = a µ X + b.) e.g. for the previous example, E[h(X )] = E(800X 900) = 800 E(X ) 900 = 700 Corollary 1. For any constant a, E(aX ) = a E(X ). 2. For any constant b, E(X + b) = E(X ) + b.

48 Expectations

49 Expectations Definition Let X have pmf p(x) and expected value µ. Then the variance of X, denoted by V (X ) or σ 2 X, or just σ2 X, is V (X ) = D (x µ) 2 p(x) = E[(X µ) 2 ] The stand deviation (SD) of X is σ X = σx 2

50 Expectations

51 Expectations Example: For the previous example, the pmf is given as x p(x)

52 Expectations Example: For the previous example, the pmf is given as x p(x) then the variance of X is V (X ) = σ 2 = 3 (x 2) 2 p(x) x=0 = (0 2) 2 (0.1) + (1 2) 2 (0.2) + (2 2) 2 (0.3) + (3 2) 2 (0.4) = 1

53 Expectations

54 Expectations Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n

55 Expectations Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n Proposition V (X ) = σ 2 = [ D x 2 p(x)] µ 2 = E(X 2 ) [E(X )] 2

56 Expectations Recall that for sample variance s 2, we have s 2 = S xx n 1 = x 2 i ( x i ) 2 n 1 n Proposition V (X ) = σ 2 = [ D x 2 p(x)] µ 2 = E(X 2 ) [E(X )] 2 e.g. for the previous example, the pmf is given as x p(x) Then V (X ) = E(X 2 ) [E(X )] 2 = (2) 2 = 1

57 Expectations

58 Expectations Proposition If h(x ) is a function of a rv X, then V [h(x )] = σ 2 h(x ) = D {h(x) E[h(X )]} 2 p(x) = E[h(X ) 2 ] {E[h(X )]} 2 If h(x ) is linear, i.e. h(x ) = ax + b for some nonrandom constant a and b, then V (ax + b) = σ 2 ax +b = a2 σ 2 X and σ ax +b = a σ X In particular, σ ax = a σ X, σ X +b = σ X

59 Expectations

60 Expectations Example 3.23 continued A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900.

61 Expectations Example 3.23 continued A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4. Let h(x ) denote the profit associated with selling X units, then h(x ) = revenue cost = 1000X + 200(3 X ) 1500 = 800X 900. The variance of h(x ) is V [h(x )] = V [800X 900] = V [X ] = 640, 000 And the SD is σ h(x ) = V [h(x )] = 800.

62 Binomial Distribution

63 Binomial Distribution 1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment;

64 Binomial Distribution 1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment; 2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which we denote by success (S) and failure (F );

65 Binomial Distribution 1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment; 2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which we denote by success (S) and failure (F ); 3. The trials are independent, so that the outcome on any particular trial dose not influence the outcome on any other trial;

66 Binomial Distribution 1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment; 2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which we denote by success (S) and failure (F ); 3. The trials are independent, so that the outcome on any particular trial dose not influence the outcome on any other trial; 4. The probability of success is constant from trial; we denote this probability by p.

67 Binomial Distribution 1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment; 2. Each trial can result in one of the same two possible outcomes (dichotomous trials), which we denote by success (S) and failure (F ); 3. The trials are independent, so that the outcome on any particular trial dose not influence the outcome on any other trial; 4. The probability of success is constant from trial; we denote this probability by p. Definition An experiment for which Conditions 1 4 are satisfied is called a binomial experiment.

68 Binomial Distribution

69 Binomial Distribution Examples: 1. If we toss a coin 10 times, then this is a binomial experiment with n = 10, S = Head, and F = Tail.

70 Binomial Distribution Examples: 1. If we toss a coin 10 times, then this is a binomial experiment with n = 10, S = Head, and F = Tail. 2. If we draw a card from a deck of well-shulffed cards with replacement, do this 5 times and record whether the outcome is or not, then this is also a binomial experiment. In this case, n = 5, S = and F = not.

71 Binomial Distribution Examples: 1. If we toss a coin 10 times, then this is a binomial experiment with n = 10, S = Head, and F = Tail. 2. If we draw a card from a deck of well-shulffed cards with replacement, do this 5 times and record whether the outcome is or not, then this is also a binomial experiment. In this case, n = 5, S = and F = not. 3. Again we draw a card from a deck of well-shulffed cards but without replacement, do this 5 times and record whether the outcome is or not. However this time it is NO LONGER a binomial experiment.

72 Binomial Distribution Examples: 1. If we toss a coin 10 times, then this is a binomial experiment with n = 10, S = Head, and F = Tail. 2. If we draw a card from a deck of well-shulffed cards with replacement, do this 5 times and record whether the outcome is or not, then this is also a binomial experiment. In this case, n = 5, S = and F = not. 3. Again we draw a card from a deck of well-shulffed cards but without replacement, do this 5 times and record whether the outcome is or not. However this time it is NO LONGER a binomial experiment. P( on second on first) = 12 = = P( on second) 51 We do not have independence here!

73 Binomial Distribution

74 Binomial Distribution Examples: 4. This time we draw a card from 100 decks of well-shulffed cards without replacement, do this 5 times and record whether the outcome is or not. Is it a binomial experiment?

75 Binomial Distribution Examples: 4. This time we draw a card from 100 decks of well-shulffed cards without replacement, do this 5 times and record whether the outcome is or not. Is it a binomial experiment? P( on second draw on first draw) = 1299 = P( on sixth draw on first five draw) = 1295 = P( on tenth draw not on first nine draw) = 1300 = Although we still do not have independence, the conditional probabilities differ so slightly that we can regard these trials as independent with P( ) = 0.25.

76 Binomial Distribution

77 Binomial Distribution Rule Consider sampling without replacement from a dichotomous population of size N. If the sample size (number of trials) n is at most 5% of the population size, the experiment can be analyzed as though it wre exactly a binomial experiment.

78 Binomial Distribution Rule Consider sampling without replacement from a dichotomous population of size N. If the sample size (number of trials) n is at most 5% of the population size, the experiment can be analyzed as though it wre exactly a binomial experiment. e.g. for the previous example, the population size is N = 5200 and the sample size is n = 5. We have n N 0.1%. So we can apply the above rule.

79 Binomial Distribution

80 Binomial Distribution Definition The binomial random variable X associated with a binomial experiment consisting of n trials is defined as X = the number of S s among the n trials

81 Binomial Distribution Definition The binomial random variable X associated with a binomial experiment consisting of n trials is defined as X = the number of S s among the n trials Possible values for X in an n-trial experiment are x = 0, 1, 2,..., n.

82 Binomial Distribution Definition The binomial random variable X associated with a binomial experiment consisting of n trials is defined as X = the number of S s among the n trials Possible values for X in an n-trial experiment are x = 0, 1, 2,..., n. Notation We use X Bin(n, p) to indicate that X is a binomial rv based on n trials with success probability p. We use b(x; n, p) to denote the pmf of X, and B(x; n, p) to denote the cdf of X, where B(x; n, p) = P(X x) = x b(x; n, p) y=0

83 Binomial Distribution

84 Binomial Distribution Example: Assume we toss a coin 3 times and the probability for getting a head for each toss is p. Let X be the binomial random variable associated with this experiment. We tabulate all the possible outcomes, corresponding X values and probabilities in the following table: Outcome X Probability Outcome X Probability HHH 3 p 3 TTT 0 (1 p) 3 HHT 2 p 2 (1 p) TTH 1 (1 p) 2 p HTH 2 p 2 (1 p) THT 1 (1 p) 2 p HTT 1 p (1 p) 2 THH 2 (1 p) p 2

85 Binomial Distribution Example: Assume we toss a coin 3 times and the probability for getting a head for each toss is p. Let X be the binomial random variable associated with this experiment. We tabulate all the possible outcomes, corresponding X values and probabilities in the following table: Outcome X Probability Outcome X Probability HHH 3 p 3 TTT 0 (1 p) 3 HHT 2 p 2 (1 p) TTH 1 (1 p) 2 p HTH 2 p 2 (1 p) THT 1 (1 p) 2 p HTT 1 p (1 p) 2 THH 2 (1 p) p 2 e.g. b(2; 3, p) = P(HHT ) + P(HTH) + P(THH) = 3p 2 (1 p).

86 Binomial Distribution

87 Binomial Distribution More generally, for the binomial pmf b(x; n, p), we have { } { } number of sequences of probability of any b(x; n, p) = length n consisting of x S s particular such sequence

88 Binomial Distribution More generally, for the binomial pmf b(x; n, p), we have { } { } number of sequences of probability of any b(x; n, p) = length n consisting of x S s particular such sequence { } ( ) number of sequences of n = and length n consisting of x S s x { } probability of any = p x (1 p) n x particular such sequence

89 Binomial Distribution More generally, for the binomial pmf b(x; n, p), we have { } { } number of sequences of probability of any b(x; n, p) = length n consisting of x S s particular such sequence Theorem { } ( ) number of sequences of n = and length n consisting of x S s x { } probability of any = p x (1 p) n x particular such sequence {( n ) b(x; n, p) = x p x (1 p) n x x = 0, 1, 2,..., n 0 otherwise

90 Binomial Distribution

91 Binomial Distribution Example: (Problem 55) Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 75% can be repaired, whereas the other 25% must be replaced with new units. if a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

92 Binomial Distribution Example: (Problem 55) Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 75% can be repaired, whereas the other 25% must be replaced with new units. if a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty? Let X = number of telephones which need replace. Then p = P(service and replace) = P(replace service) P(service) = =

93 Binomial Distribution Example: (Problem 55) Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 75% can be repaired, whereas the other 25% must be replaced with new units. if a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty? Let X = number of telephones which need replace. Then p = P(service and replace) = P(replace service) P(service) = = Now, P(X = 2) = b(2; 10, 0.05) = ( ) (1 0.05) 10 2 =

94 Binomial Distribution

95 Binomial Distribution Binomial Tables Table A.1 Cumulative Binomial Probabilities (Page 664) B(x; n, p) = x y=0 b(x; n, p)... b. n = 10 p

96 Binomial Distribution Binomial Tables Table A.1 Cumulative Binomial Probabilities (Page 664) B(x; n, p) = x y=0 b(x; n, p)... b. n = 10 p Then for b(2; 10, 0.05), we have b(2; 10, 0.05) = B(2; 10, 0.05) B(1; 10, 0.05) = =.074

97 Binomial Distribution

98 Binomial Distribution Mean and Variance Theorem If X Bin(n, p), then E(X ) = np, V (X ) = np(1 p) = npq, and σ X = npq (where q = 1 p).

99 Binomial Distribution Mean and Variance Theorem If X Bin(n, p), then E(X ) = np, V (X ) = np(1 p) = npq, and σ X = npq (where q = 1 p). The idea is that X = n i=1 Y + Y + + Y 1 2 n, where Y i s are independent Bernoulli random variable with probability p for one outcome, i.e. Y = { 1, with probabilityp 0, with probability1 p

100 Binomial Distribution Mean and Variance Theorem If X Bin(n, p), then E(X ) = np, V (X ) = np(1 p) = npq, and σ X = npq (where q = 1 p). The idea is that X = n i=1 Y + Y + + Y 1 2 n, where Y i s are independent Bernoulli random variable with probability p for one outcome, i.e. Y = { 1, with probabilityp 0, with probability1 p E(Y ) = p and V (Y ) = (1 p) 2 p + ( p) 2 (1 p) = p(1 p).

101 Binomial Distribution Mean and Variance Theorem If X Bin(n, p), then E(X ) = np, V (X ) = np(1 p) = npq, and σ X = npq (where q = 1 p). The idea is that X = n i=1 Y + Y + + Y 1 2 n, where Y i s are independent Bernoulli random variable with probability p for one outcome, i.e. Y = { 1, with probabilityp 0, with probability1 p E(Y ) = p and V (Y ) = (1 p) 2 p + ( p) 2 (1 p) = p(1 p). Therefore E(X ) = np and V (X ) = np(1 p) = npq.

102 Binomial Distribution

103 Binomial Distribution Example: (Problem 60) A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60% of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? What is the variance

104 Binomial Distribution Example: (Problem 60) A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60% of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? What is the variance Let X = the number of passenger cars and Y = revenue. Then Y = 1.00X (25 X ) = X.

105 Binomial Distribution Example: (Problem 60) A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60% of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? What is the variance Let X = the number of passenger cars and Y = revenue. Then Y = 1.00X (25 X ) = X. E(Y ) = E( X ) = E(X ) = (25 0.6) = 40

106 Binomial Distribution Example: (Problem 60) A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60% of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? What is the variance Let X = the number of passenger cars and Y = revenue. Then Y = 1.00X (25 X ) = X. E(Y ) = E( X ) = E(X ) = (25 0.6) = 40 V (Y ) = V ( X ) = ( 1.5) 2 V (X ) = 2.25 ( ) = 13.5