Random variable RV a variable that assumes numerical v - PDF document

1 Random variable (RV)--a variable that assumes numerical values associated with the random outcomes of an experiment, where only one numerical value is assigned to each sample point---random variables that can assume a countable number of valuesContinuous RVs---random variables that can assume values corresponding to any of the points contained in one or more intervals statistics textbook3. Number of commercials duringyour favorite TV show4. The length of the first commercialshown during your favorite TV5. The number of registered voterswho vote in a national election is the mean value of the variable in the sample space, or population of possible outcomes. can be interpreted as the mean value that would be obtained from an infinite number of observations of the random variable. A complete description of a discrete requires specification of•The possible values that the can assume•The probability associated with each valueThe probability distribution of a discrete can be represented by a graph, table, or formula that specifies the probabilities associated with each possible value of 1. for any value of Probability Distributions for Discrete RV 01 PX all 1 x PX Consider the following probability distribution function (pdf): .113.2.2.3.2P(X=x)14121110x 10 11 12 13 14 0 0.1 0.20.3 P(X= x)x Cumulative Distribution Function (cdf) .8131.0.7.5.214121110x () PX .113.2.2.3.2P(X=x)14121110x 2 mean , or expected value , of a discrete RV is determined by its probability distribution of a discrete RV isCalculator formula: all ii x EXxPx 222 all ii x VXEXPx 222 all Px The standard deviation of a discrete RV is equal to the square root of the variance:This quantifies how spread out the possible values of a discrete RV might be, weighted by how likely each value is to occur. 2 Example from Mathematical Statistics with Applications, Mendenhall et al, 1981, p. 99Given the probability distributionfind the mean, variance and SD forY Expected Values for Discrete RVs .253.3752.251.1250P(Y = y)y 41 01251252375325 175 EYyPy .... . Example from Mathematical Statistics with Applications, Mendenhall et al, 1981, p. 99 (continued) 01751175 5 21753175 5 09375 EYPy . 2 09375097 YY .. If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct. •Range rule of thumb:•Use probabilitiesor more) 0.05or fewer) 0.05 2 Based on past results found Information Please , there is a 0.1818 probability that a baseball World Series contest will last four games, a 0.2121 probability that it will last five games, a 0.2323 probability that it will last six games, and a 0.3737 probability that it will last seven games. Is it unusual for a team to “sweep”by winning in four games? .37377.23236.21215.18184P(Y = y)y number of in a world 3 Characteristics of a binomial experiment1.The experiment consists of 2.The trials are3.The experiment results in a dichotomousresponse; i.e., there are only two possible outcomes on each trial. One outcome is denoted by (success) and the other by (failure)4.The probability of , denoted as remains the same from trial to trial. The probability of , denoted as , is equal to binomial random variable, is the number of ’s in trials 14Example Record the sequence of heads and tails in 3 tosses of an unfair coin where the (H) = .6 and the (T) = .4. We are interested in the distribution of the number of tails.What is , the number of trials?Are the trials identical?Are the trials independent?What is What is How many possible outcomes are there?Number of outcomes = = 8Possible outcomes are: 3 2 P(X)X 0.6 x 0.6 x 0.6 = 0.2160.6 x 0.6 x 0.4 = 0.1440.6 x 0.4 x 0.4 = 0.0960.4 x 0.4 x 0.4 = 0.064P(THH) =P(HTT) = P(THT) =P(TTH) =P(TTT) =0.2160.432 where 012xnx PXpq,,,,n   probability of success on a single tria l number of trials number of successes in trials p -pxnx  Mean, Variance and SD for a Binomial RVVariance:The Binomial Probability Distribution np 2 npq npq For the tossing three coins example we can calculate various quantities:The Binomial Probability Distribution 32 .41.43.4.6.288 PX  3(.4)1.2 X np 3(.4)(.6)0.72; 0.720.85 X VXnpq 4 Generic Example: 60 1.411.6.0467 PX  61 .41.46.4.6.1866 PX  62 .41.415.4.6.3110 PX  Can also calculation cumulative probabilities: 2012 0467 PXPXPXPX . 212 1 4557 PXPX .   20 Table A-1, in Triola TI 83/84 binompdf() and binomcdf( )Press 2[DISTR]Press ALPHA for binompdforALPHA for binomcdforscroll through the list and press enter= 6 and = 0.4. = 3) use binompdf(6, .4, 3) To find individual probabilities for more than one value of at a time use binompdf(6, .4, {3, 4}) {.27648 .13824}3) use binomcdf(6, .4, 3) )ebinomcdf(6, .4, 3) usebinomcdf(6, .4, 3) –binomcdf(6, .4, ).774144 What is the probability that is in the interval ?What is ? , , and  15 Px 2 2 Consider the discrete probability distribution: Calculate Calculate Example a)Show that is a binomial RV.b)Use the information in the FTC study toestimate for the binomial experiment.c)What is the probability that exactly one ofthe five items is priced incorrectly by thescanner?d)What is the probability that at least one ofthe five items is priced incorrectly by thescanner?e)What is the probability that interval ?Problem 4.47 from McClaveand Sincich, 9th edition, pg. 200 A Federal Trade Commission (FTC) study of the pricing accuracy of electronic checkout scanners at stores found that one of every 30 items is priced incorrectly. Suppose the FTC randomly selects five items at a retail store and checks the accuracy of the scanner price of each. Let represent the number of the five items that is priced incorrectly. 2