#### Transcript LectureNotes(Chapter5) - University of South Alabama

Chapter 5 Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama Discrete Variables A random variable is a variable whose value is determined by the outcome of a random experiment. A random variable assigns a numerical value to an event in S Example: In tossing a coin S = {H,T} Define random variable as follows: X = 1 when H occurs X = 0 when T occurs. Here X is a discrete random variable. Discrete Random Variable In some cases the outcomes of the experiment are themselves numerical values, in such a case we may not have to define a random variable separately. Example: rolling a die S ={1,2,3,4,5,6}. All outcomes are numerical thus we do not have to define the random variable separately. Define X = # dots showed up X takes values 1,2,3,4,5,6 Discrete Random variables Toss three coins and note the number of heads showed up. Let X = # heads occurred. Then x can take the values 0,1,2,3 Number of vehicles owned by a family Number of dependents in a family Number of goals scored by a player Continuous Random Variable • When outcomes of an experiment are numerical values in an interval then X is continuous • Example: Recording the GPA of students • X = GPA of a student • Then X [0,4] and X can take any value in this interval. • X= highest temperature on a given day • X= Time taken by a runner to complete a race. Probability Distribution Probability distribution of a discrete random variable is the set of all values of X and the set of corresponding probabilities P(X=x). Example: toss a single coin and observe the number of heads occurred. Define X = # heads Then X can take two values 0 or 1 X P(X=x) 0 .5 1 .5 These two columns together are called probability distribution of X Probability distribution Toss three coins and define X = # heads then X=0,1,2,3 S = {TTT,TTH,THT,HTT,HHT,HTH,THH,HHH} The probability distribution x is : X P(X=x) 0 1/8 1 3/8 2 3/8 3 1/8 Properties of distribution P(X=x) is denoted by f(X=x) or just f(x) . f(x) is called probability mass function. Two properties 1. f(x) 0 2. f(x) = 1 X P(X=x) = f(x) 0 .5 1 .5 X P(X=x)=f(x) 0 1/8 1 3/8 2 3/8 3 1/8 Exercise 5.8 x P(x) x P(x) 0 .10 2 .35 1 .05 3 .28 2 .45 4 .20 3 .40 5 .14 ΣP(x) = 1.00 x P(x) 7 -.25 8 .85 9 .40 ΣP(x) = 1.00 ΣP(x) = .97 Exercise 5.10 x 0 1 2 3 4 5 6 P(x) .11 .19 .28 .15 .12 .09 .06 a. P(x=3) = .15 b. P(x ≤ 2) = P(x=0)+P(x=1)+P(x=2) = c. P(x ≥ 4) = P(x=4)+P(x=5)+P(x=6) = d. P(1 ≤x ≤4) = P(x=1)+P(x=2)+P(x=3)+P(x=4) e. P(x<4) = 1- P(x ≥ 4) f. P(x>2) = 1- P(x ≤ 2) g. P(2 ≤x ≤5) = P(x=2)+P(x=3)+P(x=4)+p(x=50) Exercise 5.14 X = # TV sets owned by a family Size of the dataset is 2500 (This is a population) Use classical approach to find probability distribution of x. # TV sets owned (x) # families (f) P(x) 0 120 .048 1 970 .388 a. The two highlighted columns together form the probability distribution of X b. These probabilities are exact because this is a 3 410 .164 population data and we 4 270 .108 are using classical approach to compute sum 2500 1.00 probabilties. C. P(x=1) = .388, P(x≥ 3) = P(x=3)+P(x=4) = .064+.108 2 730 .292 P(2 ≤ x ≤ 4) = P(x=2)+P(x=3)+P(x=4) = .292+.064+.108 P(x<4) = 1-P(x≥4) = 1-P(x=4) = 1- .108 Mean of a discrete random variable Mean of a discrete random variable x is the value that is expected to occur per repetition of the experiment. xP(x) # TV sets owned (x) P(x) xP(x) 0 .048 0 1 .388 .388 2 .292 .584 3 .164 .492 4 .108 .432 sum 1.00 ΣxP(x)=1.896 Mean = 1.896 TV sets Interpretation : If we repeat this experiment number of times then on the average a family owns 1.896 TV sets. Variance of discrete random variable 2 x 2 P( x) 2 x P( x) 2 2 # TV sets owned (x) P(x) xP(x) x2 x2P(x) 0 .048 0 0 0 1 .388 .388 1 .388 2 .292 .584 4 1.168 3 .164 .492 9 1.476 4 .108 .432 16 1.728 sum 1.00 ΣxP(x)=1.896 σ2 = 4.76 – (1.896)2 = 1.165184 σ = 1.0794 4.76 Exercise 5.28 Machines sold/day (x) P(x) xP(x) x2 x2P(x) 4 .08 0.32 16 1.28 5 .11 0.55 25 2.75 6 .14 0.84 36 5.04 7 .19 1.33 49 9.31 8 .20 1.6 64 12.8 9 .16 1.44 81 12.96 10 .12 1.2 100 12 7.28 Mean of x = 7.28 56.14 σ (Standard deviation) = 7.4927 Interpretation: if experiment of collecting data on machines sales is collected for a large number of days then the on the average 7.28 machines would be sold per day. Counting revisited Useful links: http://www.unc.edu/~knhighto/econ70/lec4 /lec4.htm http://pavlov.psyc.queensu.ca/~flanagan/202 _1999/lecture9/lecture9.html Factorials Example: three students with names A B C and three chairs red blue and yellow. Question: in how many ways students can be allocated to the chairs? A B C A C B B A C B C A C A B C B A In general n distinct objects can be arranged in n places in n! (factorial n) ways n! = n*(n-1)*…*1 3! = 3*2*1 = 6 ways Permutations Example: four students A B C D and two chairs Question: How many ways a team of two students can occupy the chairs? A B A C A D B C B D C D B A C A D A C B D B D C # ways we can arrange n items in r places Is called permutation nPr = n*(n-1)*…(n-r+1) 4P2 = 4*3 = 12 Combinations Example: four students A B C D and two identical chairs. Question: In how many ways we can allocate students to the chairs? A B A C A D B C B D C D Here order does not matter (that is it does not matter which student occupies which chair because chairs are identical) # ways choosing r items out of n distinct items is n! n Cr r!(n r )! 4C 2 = 6 Table III on page C7 lists the values of combinations Bernoulli trials An experiment which has only two possible outcomes is a Bernoulli trial with probability of one outcome p and that of the other as 1-p. Example: toss a coin : has only two outcomes H and T if P(H) =.5 then P(T) = 1-.5 = .5 If the coin is not fair and P(H) = .7 then P(T) = 1-.7 = .3 Bernoulli distribution More examples of Bernoulli trials: Inspecting a car at an assembly plant and declaring it as lemon or good. If P(L) = .001 then P(G) = .999 People entering into a football stadium. Classifying them into one of the genders. M or F if P(M) = .62 then P(F) = .38 Binomial Experiment A binomial experiment consists of n independent Bernoulli trials. That is repeating the Bernoulli experiment n times. Each repetition is independent of the other. Example: Toss three fair coins. n=3 Each coin is a Bernoulli trial. Outcome of one toss does not affect that of the others i.e. all tosses are independent. And p=.5 for all the three tosses. Binomial experiment Definition: 1. The experiment consists of n identical trials. 2. Each trial has only two possible outcomes 3. The probability of outcomes remain constant at each trial. 4. The trials are independent. Binomial probability distribution In a Binomial experiment define a random variable x as X =# heads occurred in n tosses or X = # successes in n trials. Then x takes values 0, 1,2 …,n Such an X is called Binomial random variable with n and p specified. If we repeat the binomial experiment a large number of times then we are interested probabilities of x assuming different values i.e in tossing n coins what is P(X=0) or P(X=1) and so on In other words we are interested in probability distribution of x. Binomial distribution Consider a binomial experiment with n trials and p as probability of success in a single trial then X = # successes in n independent trials is a binomial variable and its probability distribution is called Binomial distribution with parameters n and p P(x=k) = nCk pk (1-p)n-k for x = 0,1, …n Alternatively replace 1-p =q P(x=k) = nCk pk qn-k for x = 0,1, …n Exercise 5.51 a. b. c. Rolling a die many times and observing # spots is not a binomial experiment because there are more than two possible outcomes of a roll Rolling a die many times and observing whether the number is even or odd is a binomial experiment because at each roll there are only two possible outcomes: even or odd. Besides all the rolls are independent. Yes this a binomial experiment because we are selecting a few people, each person has only two possible answers: in favor or not in favor. The probability of an individual being in favor is known i.e. .54 and all persons answers are independent. Exercise 5.53 a. x= binomial vairate with n=8 and p=.7 P(x=5) = 8C5 *p5 (1-p)8-5 = 8C5 *p5 (1-p)3 = 8C5 *(.7)5 (.3)3 = .2441 (from table iv) b. Given n=4, p= .40 P(x=3) = 4C3 *p3 (1-p)4-5 = 4C3 *(.4)3 (.6) = Mean and Variance of binomial Let x be a binomial variable with parameters n and p. then mean of x µ = Σx. nCk pk (1-p)n-k = np µ = np Variance σ2 = npq Exercise 5.58 Question asked: Do you eat home cooked food three or more times a week? Yes: 85% No: 15% P(Y) = p = .85 P(N) = 1-p =q = .15 a. n=12 (a random sample of 12 Americans) is selected X = # Americans from the sample of 12 who say Yes Then X has binomial distribution with parameters n=12 and p= .85 X may take values between 0 to 12 That is there may be o Americans who say yes or may be 2 .. 0r at the most 12 will say yes. b. What is the probability that 10 out of 12 Americans say yes to the posed question? P(X=10) = 12C10 *p10(1-p)12-10 = 12C10 *p10(1-p)2