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### chap3-4

Course: STAT 3717, Fall 2008
School: Youngstown
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(Chapter Probability 3) Lecture Notes 3-1 Examples: - flip a coin - toss a die Whats the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing). So toss a coin twice. Do it! Did you get one head &amp; one tail? Whats it all mean? STAT 3717: Probability - 2 STAT 3717: Probability - 1 Random Experiment: an experiment whose outcomes depend on...

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(Chapter Probability 3) Lecture Notes 3-1 Examples: - flip a coin - toss a die Whats the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing). So toss a coin twice. Do it! Did you get one head & one tail? Whats it all mean? STAT 3717: Probability - 2 STAT 3717: Probability - 1 Random Experiment: an experiment whose outcomes depend on chance. Sample Point or basic Possible Outcome of an experiment. Sample Space (S): collection (or set) of all possible outcomes in random experiment. Event: a collection of outcomes. Possible Outcomes (Sample point): 1 2 3 4 5 6 Sample Space, S : { 1, 2, 3, 4, 5, 6 } STAT 3717: Probability - 4 STAT 3717: Probability - 3 Possible Outcomes: -HH -HT -TH -TT Sample Space, S : { HH, HT, TH, TT } Sample Space: S = {Head, Tail} S = {Life span of a human} = {x | x0 ,xR} Event: E = {Head} E = {Life span of a human is less than 3 years} STAT 3717: Probability - 5 STAT 3717: Probability - 6 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-2 Probability is always a value between 0 and 1. Total probability of all possible outcomes equals 1. A rough definition:(frequentist definition) Probability of event A, denoted P(A), is the proportion of times that the event A would occur in a very long series of repetitions of a random experiment. STAT 3717: Probability - 7 STAT 3717: Probability - 8 ! Total Heads / Number of Tosses 1.00 0.75 0.50 0.25 0.00 0 25 50 75 " # $100 125 STAT 3717: Probability - 10 Number of Tosses STAT 3717: Probability - 9$ If outcomes are equally likely to occur, the distribution is P(1) = 1/6, P(3) = 1/6, P(5) = 1/6, P(2) = 1/6, P(4) = 1/6, P(6) = 1/6, % $Probability Density (Mass) Distribution 0.2 0.15 0.1 0.05 0 1 2 3 4 5 6 and total probability is 1. STAT 3717: Probability - 11 STAT 3717: Probability - 12 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-3 Possible Outcomes: -HH -HT -TH -TT Probability (observed): P(HH) = 4/9 P(HT) = 2/9 P(TH) = 2/9 P(TT) = 1/9 !$ Event A : { observe at least one head } Event B : { observe exactly one head } P(A) = P(HH) + P(HT) + P(TH) = 4/9 + 2/9 + 2/9 = 8/9 P(B) = 2/9 + 2/9 = 4/9 STAT 3717: Probability - 13 STAT 3717: Probability - 14 & Intersection of events: A B <=> A and B A B A B Union of events: A B <=> A or B A Sample Space B AB STAT 3717: Probability - 15 STAT 3717: Probability - 16 Relative Frequency and Probability Number of children per household from a sample of 300 households Class 0 1 2 3 4 5 Total STAT 3717: Probability - 17 ' If a household is randomly selected from these 300 households, consider events: A = a household has less than 3 children. B = a household has between 2 to 4 children P(A) = .81, P(B) = .42 P(A and B) = P(2 children) = .24 STAT 3717: Probability - 18 Frequency 54 117 72 42 12 3 300 Relative Frequency .18 .39 .24 .14 .04 .01 1.00 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-4 & ( ) !$A B P(A or B) = P(A) + P(B) - P(A and B) = .81 + .42 - .24 = .99 Sample Space P(A B) = P(A) + P(B) - P(A B) STAT 3717: Probability - 19 STAT 3717: Probability - 20$$*$ Relative Frequency and Probability Number of children per household from a sample of 300 households Class 0 1 2 3 4 5 Total STAT 3717: Probability - 22 A and B are mutually exclusive (disjointed) events Frequency 54 117 72 42 12 3 300 A B P(A B) = P(A) + P(B) STAT 3717: Probability - 21 Relative Frequency .18 .39 .24 .14 .04 .01 1.00 ) !$Ac is the complement of event A Ac A In the 300 household example, if event A is selecting a household with 0 children, and event B is selecting a household with 5 children, then, A and B are mutually exclusive, and P(A) = .18, P(B) = 0.01 P(A B) = P(A) + P(B) = .19 STAT 3717: Probability - 23 Sample Space P(A) + P(Ac) = 1 STAT 3717: Probability - 24 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-5 In the 300 household example, if event A is selecting a household that has 0 children and event Ac is selecting a household that has at least one child, then A and Ac are also mutually exclusive and complementary. P(A) = .18, P(Ac) = 1 P(A) = 1 .18 = .82 STAT 3717: Probability - 25 How to determine probability? Empirical Probability Theoretical Probability (Subjective approach) STAT 3717: Probability - 26 ) Empirical study: Outcome Head Tail Total Frequency 512 488 1000 ) Empirical probability assignment for event E: P(E) = Number of times event E occurs Number of times experiment is repeated m = n Probability of Head: P(Head) = 512/1000 = .512 = 51.2% STAT 3717: Probability - 27 STAT 3717: Probability - 28 Relative Frequency and Probability Distribution Number of children per household from a sample of 300 households Class 0 1 2 3 4 5 Total STAT 3717: Probability - 29 Discrete Distribution Relative Frequency Distribution 0.5 0.4 0.3 0.2 0.1 0 0 1 2 3 4 5 Frequency 54 117 72 42 12 3 300 Relative Frequency .18 .39 .24 .14 .04 .01 1.00 STAT 3717: Probability - 30 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-6 ) If a household is randomly selected from the 300 households, what is the probability that it has 3 children? P(3 children) = (42)/300 = .14 or 14% Make a reasonable assumption: We have a balanced coin! Hence, all possible outcomes are equally likely. STAT 3717: Probability - 31 STAT 3717: Probability - 32 ) Theoretical probability assignment for event E: P(E) = Number of equally likely outcomes in event E Size of the sample space n( E ) = n( S ) )$ In a randomly observed sample of 100 individuals, 25 have cancer (event A), and 50 are smokers (event B). Also, 20 are smokers and have cancer. STAT 3717: Probability - 34 Probability of Head: P(Head) = 1/2 = .50 = 50% STAT 3717: Probability - 33 & 25 & 50 A A B B Sample Space n(S) = 100 STAT 3717: Probability - 35 n(A) = 25 Sample Space n(S) = 100 STAT 3717: Probability - 36 n(A) = 25 n(B) = 50 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-7 +$25 ? & 50 20 ? ? +$ 25 5 & 50 20 30 45 A B A B Sample Space Sample Space n(A B) = 20 STAT 3717: Probability - 37 STAT 3717: Probability - 38 AB +$& 45 5 A 20 30 B .05 A & .45 .30 .20 B Sample Space n(S) = 100 STAT 3717: Probability - 39 Sample Space AB STAT 3717: Probability - 40 P(A B) = .20 , Smoke B Cancer A No cancer c A 50 100 20 Not Smoke c B 25 Cancer A No cancer c A Smoke B 20 (.2) 30 50 Not Smoke c B 5 45 50 25 75 100 n(A B) = 20 (Smoke and have cancer) n(A) = 25, n(B) = 50, n(S) = 100 STAT 3717: Probability - 41 n(A B) = 20 (Smoke and have cancer) P(A B) = n(A B) / n(S) = 20/100 = .2 STAT 3717: Probability - 42 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-8 , Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 P(A ) = .75 c & A B = 1 or 100% 100 n(A B) = 20 (Smoke and have cancer) P(A B) = n(A B) / n(S) = 20/100 = .2 STAT 3717: Probability - 43 Sample Space STAT 3717: Probability - 44 % % Smoke B Cancer A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 c P(A ) = .75 = 1 or 100% A B No cancer c A 100 Sample Space P(A B) = P(A) + P(B) - P(A B) STAT 3717: Probability - 45 n(A B) = 55 (Smoke or have cancer) P(A B) = n(A B) / n(S) = 55/100 = .55 STAT 3717: Probability - 46 ( ) !$ The conditional probability of A to occur given B has occurred is denoted as P(A|B) and is, P(A | B) = P(A) = .25 P(A B) = P(A) + P(B) - P(A B) = .25 + .50 - .20 = .55 Smoke B Cancer A No cancer c A P(B) =.5 STAT 3717: Probability - 47 Not Smoke c B (.2) 20 25 if P(B) is not zero, 50 = 1 or 100% n(A B) n(B) P(A B) = P(B) 100 n(E) = number of equally likely outcomes in event E. STAT 3717: Probability - 48 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-9 Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 45 (.45) 50 c P(B ) = .5 25 P(A) = .25 75 c P(A ) = .75 = 1 or 100% Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 25 P(A) = .25 100 75 45 c (.45) P(A ) = .75 50 100 c P(B ) = .5 = 1 or 100% P(A|B) = ? P(A|Bc) = ? STAT 3717: Probability - 49 P(A|B) = P(A B)/P(B) = .2/.5 = .4 P(A|Bc) = ? STAT 3717: Probability - 50 , Smoke B Cancer A No cancer c A 20 (.2) 30 (.3) 50 P(B) =.5 Not Smoke c B 5 (.05) 25 P(A) = .25 75 45 c (.45) P(A ) = .75 50 100 c P(B ) = .5 = 1 or 100% 4 times If two events A and B are independent then P(A|B) = P(A) P(B|A) = P(B) and P(A B) = P(A) P(B) . Reason: Since P(A|B) = P(A B) / P(B), P(A B) = P(A|B) P(B) = P(A) P(B) STAT 3717: Probability - 52 P(A|B) = P(A B)/P(B) = .2/.5 = .4 P(A|Bc) = P(A B)/P(Bc) = .05/.5 = .1 STAT 3717: Probability - 51 ) Ten percent of the people in a large population have disease H. If one person is randomly selected what is the probability that this person has disease H? P(H) = 10% ?. ., Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = ? . H1 denotes the event that the first people chosen has disease H. H2 denotes the event that the 2nd people chosen has disease H. STAT 3717: Probability - 53 STAT 3717: Probability - 54 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-10 , Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = P(H1| H2) P(H2) = P(H1) P(H2) (H1 and H2 are almost independent.) STAT 3717: Probability - 55 , Ten percent of the people in a large population have disease H. If two people are randomly selected what is the probability that both of them have disease H? P(H1 H2) = P(H1) P(H2) = .10 x .10 = .01 = 1% . STAT 3717: Probability - 56 , Ten percent of the people in a large population have disease H. If three people are randomly selected what is the probability that both of them have disease H? P(H1 H2 H3) = P(H1) P(H2) P(H3) = .10 x .10 x .10 = .001 = 0.1% . STAT 3717: Probability - 57 $Tree diagram Multiplication principle Permutation rule Combination rule +$ STAT 3717: Probability - 58 Multiplication Principle Coin Spinner 1 4 1 2 3 4 1 2 3 4 2 3 Counting Rule: Multiplication Principle: In a sequence of k events in which the first one has n1 possibilities and the second event has n2 and the third has n3, and so forth, the total number of possible outcomes of the sequence of events will be n1 n2 n3 nk Example: H T STAT 3717: Probability - 59 2x4=8 possible outcomes STAT 3717: Probability - 60 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-11 $How many ways can a three-letter code be formed by using A, B, and C without repeating use of the same digit? n1 = 3, n2 = 2, n3 = 1 number of ways = 3 x 2 x 1 = 3! = 6 n! = 123... n Example: 3! = 123 = 6 STAT 3717: Probability - 61 STAT 3717: Probability - 62$ How many ways can a four-digit code be formed by selecting 4 distinct digits from the nine digits, 1 through 9, without repeating the use of any digit? 9876 = 9! / 5! = 9! / (9-4)! (formula) Permutation Rule Permutation Rule: The number of possible permutations of selecting r objects from a collection of n distinct objects is Prn = n! (n r )! Note: 1. No repetitions. 2. Order of selection is important. STAT 3717: Probability - 63 STAT 3717: Probability - 64 Combination Rule How many ways can a combination of 3 distinct letters be selected from the three letters A, B, C and D? Permutations = 4! / (4 3)! = 24 Order is not important, the possible permutations of 3 letters = 3! = 6 Combination Rule: The number of possible combinations of r objects from a collection of n distinct objects is Crn = n Prn n! = = r r! r!(n r )! Combinations = 24 / 6 = 4. STAT 3717: Probability - 65 Note: 1. No repetition. 2. Order is not important. STAT 3717: Probability - 66 STAT 3717 Yates / Chang Probability 3) Lecture (Chapter Notes 3-12 How many ways can a combination of 4 distinct digits be selected from nine digits, 1 through 9? 9876 / 4! = 9! / (4! 5!) = 9! / [4! (9-4)!] = 126 STAT 3717: Probability - 67 How many ways can 6 distinct numbers be selected from a set of 47 distinct numbers? 47! / (6! 41!) = 10,737,573 (formula) STAT 3717: Probability - 68 ! If n elements are selected from a population so that every set of n elements selected from the population has a equal probability of being selected, the n elements are called a (simple) random sample. A random sample is likely to be representative of the population. STAT 3717: Probability - 69 ! Selecting a random sample: - picking numbers from a hat - using a table of random numbers (random number generator) STAT 3717: Probability - 70 ! - & / . ! & What is Random Variable? A random variable assumes a numerical value associated with the outcome of a random experiment (by chance). Usually is denoted by a capital letter. X, Y, Z, ... Very useful for mathematically modeling the distribution of variables. STAT 3717: Probability - 71 STAT 3717: Probability - 72 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-13  ! & ! & Continuous Random Variable: A random variable that assumes continuous values in one or more intervals. Discrete Random Variable: A random variable that assumes discrete values by chance. STAT 3717: Probability - 73 STAT 3717: Probability - 74 ! X = 1, if Head occurs, and X = 0, if Tail occurs. & $If a balanced coin is tossed, Head and Tail are equally likely to occur, P(X=1) = .5 = 1/2 and P(X=0) = .5 = 1/2 1/2 0 1 Example: (Toss a balanced coin) P(Head) = P(X=1) = P(1) = .5 P(Tail) = P(X=0) = P(0) = .5 Probability distribution function: p(X) = .5 , if X = 0, 1, and p(X) = 0 , otherwise. STAT 3717: Probability - 75 P(all possible outcomes) = P(X=1 or 0) = P(X=1) + P(X=0) = 1/2 + 1/2 = 1.0 Total probability is 1. STAT 3717: Probability - 76 &$ $* Table 0$ Count p(X) Experiment: Toss 2 Coins. Count # Tails. Probability Distribution Values, X Probabilities, P(X) 0 1 2 1984-1994 T/Maker Co. # Tails # times 0 1 2 1 2 1 .25 .50 .25 1/4 = .25 2/4 = .50 1/4 = .25 Graph p(X) .50 .25 .00 STAT 3717: Probability - 78 0 1 2 X STAT 3717: Probability - 77 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-14 $1. List of All possible [x, p(x)] pairs x = Value of Random Variable (Outcome) p(x) = Probability Associated with Value ! Example: Toss of a single die. & 1/6 1 2 3 4 5 6 Probability distribution function: p(X) = 1/6, if X = 1, 2, 3, 4, 5, or 6, p(X) = 0 otherwise. 2. 0 p(x) 1, for all possible values of x. 3. p(x) = 1, sum over all possible values of x. Example: What is probability of getting a number less than 3 when roll a balanced die? p( X < 3 ) = p( X 2) = ? Answer: 2/6 = 1/3 STAT 3717: Probability - 79 STAT 3717: Probability - 80 ! &$ The probability distribution of a discrete random variable X gives the probability associated with each possible X value. Each probability is the limiting relative frequency of occurrence of the corresponding X value when the experiment is repeatedly performed. The probability distribution may be given by means of a table, formula, or graph. STAT 3717: Probability - 82 A simple mathematical notation to describe an event. e.g.: X < 3, X = 0, ... Mathematical function can be used to model the distribution through the use of random variable. e.g.: Binomial, Poisson, Normal, STAT 3717: Probability - 81 ! Histogram for categorical variable x. (n = 1000) 500 250 0 1 2 x & p(x) frequency x 0 1 2 p(x) .25 .50 .25 .50 .25 0 1 2 x Mean = xi / n = (250 x 0 + 500 x 1 + 250 x 2) / 1000 = 1 STAT 3717: Probability - 83 (250x0 + 500x1 + 250x2)/1000 = 1 Mean of the (.25x0 + .5x1 + .25x2) = 1 distribution STAT 3717: Probability - 84 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-15  The mean value (expected value) of a discrete random variable (distribution) X, denoted by X or just , or E[X], is defined as = x p(x) Example: Toss a balanced coin and interested in number of heads that turn up. {x = 1 implies Head and x = 0 implies not Head, and p(1) = .5, p(0) = .5.} * $.50 .25 0 1 2 So, = 1 p(1) + 0 p(0) = 1 .5 + 0 .5 = .5 STAT 3717: Probability - 85 If the distribution is p(0) = .25, p(1) = .5, p(2) = .25, then the mean of the distribution is = (.25x0 + .5x1 + .25x2) = 1 STAT 3717: Probability - 86 * x -4 -1 11 p(x) .25 .50 .25 p(x)$ Probability distribution (or line chart) x -100 -1 11 -4 -1 11 x * p(x) .10 .25 .65 p(x) .75 & $.50 .25 .50 .25 -100 -1 11 x The mean of the distribution is = (.25x(-4) + .5x(-1) + .25x11) = 1.25 STAT 3717: Probability - 87 The mean of the distribution is = (-100) x.1+ (-1)x.25+ 11x.65 = -3.1 STAT 3717: Probability - 88$ $The variance of a discrete random variable (distribution) X, denoted by X 2 or just 2 (or V[X]) is defined as$ $Example: Toss a balanced coin and interested in number of heads that turn up. { x = 1 implies head and x = 0 implies not head, and p(1) = .5, p(0) = .5.} The mean value, = 1p(1) + 0 p(0) = .5 2 = (x - )2 p(x) The standard deviation of a discrete random variable (distribution) X, denoted by X or just is defined as 2 2 = (1-.5)2 p(1) + (0-.5)2 p(0) = .125 + .125 = .25 STAT 3717: Probability - 90 The variance, = STAT 3717: Probability - 89 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-16 x (x-)2 0 1 1 0 2 1 p(x) .25 .50 .25 p(x) .50 .25 0 1 2 x$ ! & . Mean = = 0x0.25 + 1x.5 + 2x.25 = 1 2 = (0-1)2x.25+ (1-1)2x.5+ (2-1)2x.25 = .5 STAT 3717: Probability - 91 STAT 3717: Probability - 92 Bernoulli Trial Definition: Bernoulli trial is a random experiment whose outcomes are classified as one of the two categories. (S , F) or (Success, Failure) or (1, 0) P(S) = P(X=1) = p, P(F) = P(X=0) = 1 p. Example: (Head, Tail), (Died, Survived) 1 $Example: In a random experiment of tossing an unbalanced coin, the probability of Head is 0.3, what is the probability distribution? P(Head) P(Tail) = P(X=1) = 0.3, = P(X=0) = 1 0.3 = 0.7. (A discrete distribution.) STAT 3717: Probability - 94 STAT 3717: Probability - 93 1 * 1 How to model the number successes, x, in a sequence of n independent and identical Bernoulli trials? A random experiment involving a sequence of independent and identical Bernoulli trials. Example: Toss a coin ten times and observing Head or Tail turns up. Roll a die 3 times and observing a 6 or not 6 turns up. A random sample of 5 people and observing 2 of them have disease H. (Approximately binomial.) STAT 3717: Probability - 96 STAT 3717: Probability - 95 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-17 , If two events A and B are independent then P(A|B) = P(A) P(B|A) = P(B) and P(A B) = P(A) P(B) . Reason: Since P(A|B) = P(A B) / P(B), P(A B) = P(A|B) P(B) = P(A) P(B) STAT 3717: Probability - 97 , The probability of getting all 6s in roll a balanced die 2 times experiment. P(S1 S2) = P(S1) P(S2 | S1) = P(S1) P(S2) = (1/6)2 STAT 3717: Probability - 98 , If events A1, A2, , Ak are independent, then P(A1 and A2 and and Ak) = P(A1) P(A2) P(Ak) , The probability of getting all 6s in rolling a balanced die 5 times experiment. P(S1 S2 S3 S4 S5) = P(S1) P(S2) P(S3) P(S4) P(S5) = (1/6)5 STAT 3717: Probability - 99 STAT 3717: Probability - 100 , The probability of getting 6s in the first two trials and non-6s from the last three trials from rolling a balanced die 5 times experiment. P(S1 S2 F3 F4 F5) = P(S1) P(S2) P(F3) P(F4) P(F5) = (1/6)2 (5/6)3 STAT 3717: Probability - 101 1 The probability of getting two 6s in rolling a balanced die 5 times experiment. P(SSFFF) = (1/6)2 x (5/6)3 P(SFSFF) = (1/6)2 x (5/6)3 P(SFFSF) = (1/6)2 x (5/6)3 5 5! How many of them? = = 10 2 2! 3! STAT 3717: Probability - 102 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-18 1 The probability of getting two 6s in rolling a balanced die 5 times experiment is 1 In a binomial experiment involving n independent and identical Bernoulli trials each with probability of success p, the probability of having x successes can be calculated with the binomial probability distribution function, and it is, for x = 0, 1, , n, P(2 Ss) = ( 2 ) x (1/6) x (5/6) 2 5 3 P (X = x ) = = STAT 3717: Probability - 104 n x p x (1 p) n x = 5!/(2!3!) x (1/6)2 x (1 1/6)3 STAT 3717: Probability - 103 n! p x (1 p ) n x x!(n x )! 1$ Parameters of the distribution: Mean of the distribution, = np Variance of the distribution, 2 = np(1 p) Standard deviation, , is the square root of variance. 1 Example: A balanced die is rolled three times (or three balanced dice are rolled), what is the probability to see two 6s? [ = 3(1/6) = 1/2, 2 = 3(1/6)(5/6) = 5/12 ] Identify n = 3, p = 1/6, x = 2 P(X=2) = { 3! / [2!1!] } (1/6)2(5/6)3-2 = 3(1/6)2(5/6)1 = .069 P(X = x) = n! p x (1 p) n x x!(n x )! = np(1 p) STAT 3717: Probability - 105 STAT 3717: Probability - 106 Random Sample A (Simple) Random Sample of size n consists of n individuals from the population chosen in such a way that every set of n individuals has an equal chance to be the sample actually selected. 1 Example: If there are10% of the population in a community have a certain disease, what is the probability that 4 people in a random sample of 5 people from this community has the disease? (Approximately Binomial) Identify n = 5, x = 4, p = .10 P(X=4) = { 5! / [4!(5-4)!] }(.10)4(1.10)5-4 = 5(.10) 4(.90)1 = .00045 P(X = x) = STAT 3717: Probability - 108 STAT 3717: Probability - 107 n! p x (1 p) n x x!(n x )! STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-19 1 Example: In the previous problem, what is the probability that 4 or more people have the disease? Identify n = 5, x = 4, p = .10 P(X4) = P(X=4) + P(X=5) = .00045 + { 5! / [5!(5-5)!] }(.10)5(1 .10)5-5 = .00045 + .00001 = .00046 $The Poisson distribution is used to model discrete events that occur infrequently in time or space. Model the number of successes in a given time period or in a given unit space. STAT 3717: Probability - 110 (What is this number telling us?) STAT 3717: Probability - 109$ Let X represents the number of occurrences of some event of interest over a given interval from a Poisson process, and the is the mean of the distribution, the probability of X assumes the value x is, for x = 0, 1, 2, , P( X = x) = e x! STAT 3717: Probability - 111 x The probability that a single event occurs within an interval is proportional to the length of the interval. Within a single interval, an infinite number of occurrences is possible. The events occurs independently both within the same interval and between consecutive non-overlapping intervals. It can be used to approximate Binomial prob, with large n. STAT 3717: Probability - 112 If on average there are 4 people catch flu in a given week in a community during a certain season, what is the probability of observing 2 people catch flu in this community in a given week period during the season? (Assume the number of people catching flu in a given period of time follow a Poisson Process.) =4 x=2 P(X=2) =(e-442)/2! = .1465 STAT 3717: Probability - 113 If on average there are 4 people catch flu in a given week in a community during a certain season, what is the probability of observing less than 2 people catch flu in this community in a given week period during the season? (Assume the number of people catching flu in a given period of time follow a Poisson Process.) =4 x = 0, 1 P(X<2) = P(X=0) + P(X=1) = (e-440)/0! + (e-441)/1! = .0916 STAT 3717: Probability - 114 STAT 3717 Yates / Chang Probability (Chapter 3) Lecture Notes 3-20 P ( X = x) = n! p x (1 p) n x x!(n x)! x 2 $e-5 =? 4! = ?$ "" " $! STAT 3717: Probability - 115 STAT 3717: Probability - 116 P ( X = x) = e x x! x " " " " STAT 3717: Probability - 117 STAT 3717 Yates / Chang Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education. Below is a small sample set of documents: Youngstown - STAT - 3717 Normal DistributionContinuous Random Variable Continuous Random VariablesChapter 51. Random VariableA Numerical Outcome of an Experiment Weight of a Student (e.g., 115, 156.8, etc.)2. Continuous Random VariableWhole or Fractional Number Obtai Youngstown - STAT - 3717 Stat 3717 Statistical Methods(Yates) 4/16/04Project 3 (Group Project) Due Friday April 30 , 2004For your last project, you are to show what insight you can obtain from collecting and analyzing data on the proportions of individuals in a sample Youngstown - STAT - 3717 Stat 3717 Statistical Methods(Yates) 3/5/04Project 2 (Group Project) Due Friday April 2 , 2004For your first group project you are to show what insight you can get from collecting data on a single quantitative variable. You and your group membe University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer Session 1, 2007COURSE SYLLABUS Lecture: MTWRF 9:00-10:15am Saunders Hall 244 Instructor: Quincy Edwards Office hours: TBA Office location: TBA Email: quincy.edwards@hawaii.edu University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 11. THE STATISTICAL IMAGINATION 1. In a statistical relationship, the predictor variable is the: (a) constant. (b) variable that we wis University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244Instructor: Office hours:Computer lab: MW 10:30-11:45am Saunders Hall 342Quincy Edwards MW 12:00-1:00pm TRF 10:30-11:30a University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer 2007 Instructor: Office hours: Quincy Edwards MW 12:00-1:00pm TRF 10:30-11:30am Office location: Saunders Hall 226 Email: quincy.edwards@hawaii.eduLecture:MTWRF 9:00-10:15am Wake Forest - ATT - 0604 Utz-Peter Reich The Role of Money in the Measurement of Value113/ 9/051. Introduction The speed of production and consumption in an economy is customarily measured in units of currency per unit of time, dollars/year, for example. Such usage impli University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 9. HYPOTHESIS TESTING I: THE SIX STEPS OF STATISTICAL INFERENCE. 1. An adequate scientific explanation accomplishes two things: (a) it Wake Forest - ATT - 0604 (TEXTE EN FRANAIS : p.1-2) (TEXTO EN ESPAOL : p.3-4) (ENGLISH TEXT : p.5-6) Chre Madame, Cher Monsieur, J'ai le plaisir de vous faire savoir que mon dernier manuel d'conomie marxiste (2005) est disponible, la fois sous forme lectronique et sous form Wake Forest - ATT - 0604 UNPRODUCTIVE LABOR AS PROFIT RATE MAXIMIZING LABOR Grard DUMENIL and Dominique LEVY eEconomiX -CNRS and PSE -CNRSVersion: March 11, 2006. Address all mail to: PSE-CNRS, 48 bd Jourdan, 75014 Paris, France. Tel: 33 1 43 13 62 62, Fax: 33 1 43 13 6 University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 8. PARAMETER ESTIMATION USING CONFIDENCE INTERVALS. 1. A range of possible values of a parameter expressed with a specific degree of co University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 7: Using Probability Theory to Produce Sampling Distributions. Key Terms(psmaller ) (n) &gt; 5 A requirement for using the normal distribution as the sampl Wake Forest - ATT - 0604 Die Kritik der hen konomie politiscgend sterreich ie der Sozialistischen Ju AkademIMPRESSUMDie Kritik der politischen konomie Broschre der Sozialistischen Jugend sterreichAutor: Armin L. Puller Geschrieben fr die Akademie der Sozialistischen J Wake Forest - ATT - 0604 Staatsanalyse Staatskritik undgend sterreich ie der Sozialistischen Ju AkademIMPRESSUMStaatsanalyse und Staatskritik Broschre der Sozialistischen Jugend sterreichAutor: Armin L. Puller Geschrieben fr die Akademie der Sozialistischen Jugend ste Wake Forest - ATT - 0604 Philosophie und llschaftskritik Gese bei Marxgend sterreich ie der Sozialistischen Ju AkademIMPRESSUMPhilosophie und Gesellschaftskritik bei Marx Broschre der Sozialistischen Jugend sterreichAutor: Armin L. Puller Geschrieben fr die Akademie d University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 6: Probability Theory and the Normal Probability Distribution. Key Termsaddition rule States that the probability of alternative events is equal to the Wake Forest - ATT - 0604 1Marx's value theory and subjectivityRobert Albritton York University Toronto, Canada ralbritt@yorku.caIn this paper I shall argue that at the present time no single theory has greater potential for advancing the theory of subjectivity than Marx University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer 2007 Lecture: MTWRF 9:00-10:15am Saunders Hall 244 Instructor: Office hours: Quincy Edwards MW 12:00-1:00pm TRF 10:30-11:30am Office location: Saunders Hall 226 Email: quincy.edw University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 6. THE STATISTICAL IMAGINATION 1. (a) (b) (c) (d) is the analysis and understanding of chance occurrences. A normal distribution Probab University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244Instructor: Office hours:Computer lab: MW 10:30-11:45am Saunders Hall 342Quincy Edwards MW 12:00-1:00pm TRF 10:30-11:30a University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 5. THE STATISTICAL IMAGINATION 1. Statistics that describe how the scores of an interval/ratio variable are spread across its distribut University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 4: Measuring Averages. Key Termscentral tendency statistic A statistic that provides an estimate of the typical, usual, or normal score found in a distr Wake Forest - ATT - 0604 01-Glyn-Chap01.qxd11/1/0603:11 PMPage 11Challenges to CapitalWe all had the feeling it could come apart in quite a serious way. As I saw it, it was a choice between Britain remaining in the liberal nancial system of the West as opposed to Wake Forest - ATT - 0129 Lapotenzadellapovert.MarxleggeSpinozaanalyzestheconceptofpovertyin relationtotheconceptoftheSpinozianpotentia. MovingfromMarxreadingofSpinoza(HeftSpinoza,1841),theanalysisfollowstwo paths:MarxCapitalandSpinozasconceptofimaginationworkwithtimeinasimil Wake Forest - ATT - 0604 Lapotenzadellapovert.MarxleggeSpinozaanalyzestheconceptofpovertyin relationtotheconceptoftheSpinozianpotentia. MovingfromMarxreadingofSpinoza(HeftSpinoza,1841),theanalysisfollowstwo paths:MarxCapitalandSpinozasconceptofimaginationworkwithtimeinasimil University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 3: Charts and Graphs: A Picture Says a Thousand Words. Key Termsbar chart A series of vertical or horizontal bars with the length of a bar representing University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244Instructor: Office hours:Computer lab: MW 10:30-11:45am Saunders Hall 342Quincy Edwards MW 12:00-1:00pm TRF 10:30-11:30a University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 2: Statistical Analysis: Error Control and Management. Key Termsconstants Characteristics of study subjects that do not vary. codebook A concise descrip University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244 MW 10:30-11:45am Saunders Hall 342Instructor: Office hours: Office location: Email:Computer lab:Quincy Edwards MW 12:00 Minnesota - STAT - 8053 Assignment # 6, Stat 8053, Fall 2008ReadingReminder: There is no class on Wednesday, October 22.ProblemsBecause the homework problems this week are not data analysis problems, the next assigned homework team leaders will do Assignment 7, not thi Minnesota - STAT - 8801 Good GraphicsGraphsSanford Weisberg &amp; G. OehlertUniv. of MinnesotaFundamental Principal of Statistical GraphicsAbove all else show the data. Ed Tufte Graphics can be . . . all that is read in an article . . . eciently summarize a problem . . . Minnesota - STAT - 8801 The beamer classManual for version 3.06.\begin{frame} \frametitle{There Is No Largest Prime Number} \framesubtitle{The proof uses \textit{reductio ad absurdum}.} \begin{theorem} There is no largest prime number. \end{theorem} \begin{proof} \begin{ University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics LaboratoryUniversity of Hawaii at Mnoa, Summer 2007Chapter 1: The Statistical Imagination. Key Termsconstants Characteristics of study subjects that do not vary. data Systematically acquired information that is University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244 MW 10:30-11:45am Saunders Hall 342Instructor: Office hours: Office location: Email:Computer lab:Quincy Edwards MW 12:00 University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007Lecture:MTWRF 9:00-10:15am Saunders Hall 244Instructor: Office hours:Computer lab: MW 10:30-11:45am Saunders Hall 342 SINCLAIR LIBRARY RESERVE SERVICES.Quincy Edwar University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social StatisticsUniversity of Hawaii at Mnoa, Summer 2007STUDENT SELF-ASSESSMENT QUIZCHAPTER 1. THE STATISTICAL IMAGINATION 1. The field of statistics is: (a) The application of laboratory methods to the study of human beings. ( Minnesota - STAT - 8801 REVIEWOral Appliances for Snoring and Obstructive Sleep Apnea: A ReviewKathleen A. Ferguson, MD1; Rosalind Cartwright, PhD2; Robert Rogers, DMD3; Wolfgang Schmidt-Nowara, MD41Division of Respirology, University of Western Ontario, London, Ontar University of Hawaii - Hilo - UH - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer Session 2009COURSE SYLLABUS Lecture: MTWRF 9:00-10:15am Room: TBA Dates: 07/06-08/14/2009 Computer lab: MW 10:30-11:45am Room: TBA Dates: 07/06-08/14/2009 1. Classmate contact University of Hawaii - Hilo - DOCS - 476 SOCIOLOGY 476/L: Social Statistics University of Hawaii at Mnoa, Summer Session 2009COURSE SYLLABUS Lecture: MTWRF 9:00-10:15am Room: TBA Dates: 07/06-08/14/2009 Computer lab: MW 10:30-11:45am Room: TBA Dates: 07/06-08/14/2009 1. Classmate contact Minnesota - STAT - 8801 Open access, freely available onlineCorrespondenceResponse to Amir AttaranJohn W. McArthur, Jeffrey D. Sachs, Guido SchmidtTraubAmir Attarans Policy Forum [1] raises important points on the poor quality of data for some indicators used to measu University of Hawaii - Hilo - UH - 313 SOCIOLOGY 313: Survey of Sociology of WorkUniversity of Hawaii at Mnoa, Fall 2009COURSE SYLLABUS Lecture Posted: M 6:00pm Course Dates: 8/24/2009-12/18/2009 Online System: Laulima https:/laulima.hawaii.edu/ Instructor: Quincy Edwards Office hours: University of Hawaii - Hilo - UH - 313 SOC313:SurveyofSociologyofWorkMonday,8September2008Mostimportantsoc313 Dotakeownershipofyoureducationinthisclassbycompletingallreadingandwritingassignmentsontime andparticipatinginonlinediscussions. Readeachoftheassignedtextbookchaptersbefor Minnesota - STAT - 8311 Stat 8311 IGF dataTen traces were repeatedly measured for IGF-1 concentration. Presumably, we would like to see zero slopes and equal intercepts. &gt; library(nlme); data(IGF) &gt; formula(IGF) conc age | Lot &gt; postscript(&quot;IGF1.eps&quot;,horizontal=FALSE,hei Minnesota - STAT - 8311 &quot;Block&quot; &quot;Variety&quot; &quot;nitro&quot; &quot;yield&quot;&quot;I&quot; &quot;Victory&quot; 0 111&quot;I&quot; &quot;Victory&quot; 0.2 130&quot;I&quot; &quot;Victory&quot; 0.4 157&quot;I&quot; &quot;Victory&quot; 0.6 174&quot;I&quot; &quot;Golden Rain&quot; 0 117&quot;I&quot; &quot;Golden Rain&quot; 0.2 114&quot;I&quot; &quot;Golden Rain&quot; 0.4 161&quot;I&quot; &quot;Golden Rain&quot; 0.6 141&quot;I&quot; &quot;Marvellous&quot; 0 105&quot;I&quot; &quot; University of Hawaii - Hilo - UH - 313 SOC313:SurveyofSociologyofWorkMonday,15September2008Mostimportantsoc313 Dotakeownershipofyoureducationinthisclassbycompletingallreadingandwritingassignmentsontime andparticipatinginonlinediscussions. Readeachoftheassignedtextbookchaptersbefo Minnesota - STAT - 8311 deviance0.10 0.00 610 0.0 1.0 2.0620630640log(sigma^2)0.0 0.3log(Blck.(In)Density5.0 0.00 0.065.56.0 0.00 0.0346810VarietyVictorynitro2010010 0.00 0.06406080100120(Intercept)0.00 0.03VarietyMarve Minnesota - STAT - 8311 Golden Rain Marvellous Victoryq q q0.0 0.1 0.2 0.3 0.4 0.5 0.6IVIIIq q q q q q q q q q q q q q160 140 120 100 80qq q q q q q q q q q qq qqq q q q q q60yieldVI160 140 120 100 80 60q q q q q q q q q q q q q q qVIIIq Minnesota - STAT - 8311 Stat 8311 More from Chapter 4&gt; library(nlme) &gt; options(width=68,digits=5) &gt; data(Oats) &gt; m1 &lt;- lme(yield~nitro, data = Oats, random=~1|Block/Variety) &gt; m2 &lt;- lme(yield~nitro, data = Oats, + random=list(Block = pdCompSymm(~ Variety -1) &gt; m1 Linear m Minnesota - STAT - 8311 kUrGq Zeroforce travel time (nanoseconds)40 2 5 1 60 80 100cwdgfqWb`X d caxa d ta p a YRail uqbee t y t q byqb qbbqbybqbuy1b qb u Minnesota - STAT - 8311 %!PS-Adobe-3.0 %DocumentNeededResources: font Helvetica %+ font Helvetica-Bold %+ font Helvetica-Oblique %+ font Helvetica-BoldOblique %+ font Symbol %DocumentMedia: letter 612 792 0 () () %Title: R Graphics Output %Creator: R Software %Pages: (atend Minnesota - STAT - 8311 mean of y 5 10 15 20 25 30 351 3 6 Age 9 12 Water 4 8 Minnesota - STAT - 8311 mean of y 5 10 15 20 25 30 354 Water 8 Age 12 6 9 1 3 Minnesota - STAT - 8311 mean of y 5 1 3 6 9 12 Water 4 8 mean of y 5 4 Water 8 Age 12 6 9 1 3 10 15 20 25 30 35 Age 10 15 20 25 30 35 Minnesota - STAT - 8311 Two-factor anova, Stat 8311R : Copyright 2005, The R Foundation for Statistical Computing Version 2.2.0 alpha (2005-09-16 r35603) ISBN 3-900051-07-0 &gt; &gt; &gt; &gt; + &gt; &gt; + + + , # Barley sprouting data from Oehlert, p. 166 Age &lt;- factor(rep(c(1,3,6,9,12),c Minnesota - STAT - 8311 Stat 8311 Unbalanced data from SearleR : Copyright 2004, The R Foundation for Statistical Computing Version 1.9.1 (2004-06-21), ISBN 3-900051-00-3 &gt; #Table 4.1 from Searle, Unbalanced data &gt; soil &lt;- rep(c(1,2),c(7,8) &gt; var &lt;- c(rep(c(1,2,3),c(3,2,2 University of Hawaii - Hilo - UH - 313 SOC313:SurveyofSociologyofWorkMonday,29September2008Mostimportantsoc313 Dotakeownershipofyoureducationinthisclassbycompletingallreadingandwritingassignmentsontime andparticipatinginonlinediscussions. Readeachoftheassignedtextbookchaptersbefo Minnesota - STAT - 8311 Stat 8311, Fall, 2002, Binomial exampleR : Copyright 2002, The R Development Core Team Version 1.5.1 (2002-06-17) November 11, 2002 &gt; # The toxicity of a pyrethrum extract in a heavy oil was texts on the beetle &gt; # Tribolium castaneum. The experimen Minnesota - STAT - 8311 Conc Oil Meth Killed N 0.5 0.29 0 1 27 0.5 0.57 0 4 29 0.5 1.08 0 6 30 0.5 0.29 1 3 29 0.5 0.57 1 4 27 0.5 1.08 1 8 28 1 0.29 0 15 29 1 0.57 0 19 29 1 1.08 0 15 24 1 0.29 1 10 30 1 0.57 1 14 28 1 1.08 1 27 28 2 0.29 0 27 30 2 0.57 0 26 Minnesota - STAT - 8311 Chapter 7 Two-way layoutThe general normal linear model is, as usual y N(, 2I), E n The two-way layout or two-factor design, provides a particular case for the estimation space E. It illustrates many of the important issues in xed effects linea Minnesota - STAT - 8311 Stat 8311 Two random eects&gt; library(nlme) &gt; options(width=68,digits=5) &gt; data(Assay) &gt; formula(Assay) logDens ~ 1 | Block &gt; Assay$d &lt;- as.numeric(as.character(Assay\$dilut) # dilution as a vector &gt; postscript(&quot;assay1.eps&quot;,horizontal=FALSE,height=4,w