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Chapter 5

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

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Chapter 5

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  1. Chapter 5 Nutan S. Mishra Department of Mathematics and Statistics University of South Alabama

  2. 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.

  3. 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

  4. 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

  5. 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.

  6. 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 These two columns together are called probability distribution of X

  7. 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 :

  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 • f(x)  0 • f(x) = 1

  9. Exercise 5.8

  10. Exercise 5.10 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)

  11. 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. • The two highlighted columns together form the probability distribution of X • These probabilities are exact because this is a population data and we are using classical approach to compute probabilties. • P(x=1) = .388, P(x≥ 3) = P(x=3)+P(x=4) = .064+.108 • 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

  12. 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. 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.

  13. Variance of discrete random variable σ2 = 4.76 – (1.896)2 = 1.165184 σ = 1.0794

  14. Exercise 5.28 Mean of x = 7.28 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.

  15. Counting revisited Useful links: http://www.unc.edu/~knhighto/econ70/lec4/lec4.htm http://pavlov.psyc.queensu.ca/~flanagan/202_1999/lecture9/lecture9.html

  16. C A B C B A C A B C A B B A C C A 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? 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 B

  17. A D B D A B D B C C A C C A B D D C A A C B B D Permutations Example: four students A B C D and two chairs Question: How many ways a team of two students can occupy the chairs? # ways we can arrange n items in r places Is called permutation nPr = n*(n-1)*…(n-r+1) 4P2 = 4*3 = 12

  18. A A A B B C B C D C D D Combinations Example: four students A B C D and two identical chairs. Question: In how many ways we can allocate students to the chairs? 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 4C2 = 6 Table III on page C7 lists the values of combinations

  19. 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

  20. 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

  21. 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.

  22. Binomial experiment Definition: • The experiment consists of n identical trials. • Each trial has only two possible outcomes • The probability of outcomes remain constant at each trial. • The trials are independent.

  23. 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.

  24. 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

  25. Exercise 5.51 • 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.

  26. 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) =

  27. 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

  28. 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 • 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

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