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# Lecture 4 notes STAT 960

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This 5 page Class Notes was uploaded by Vendi on Thursday November 26, 2015. The Class Notes belongs to STAT 960 at Rutgers taught by in Fall 2015. Since its upload, it has received 14 views.

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Date Created: 11/26/15

Sample point D most basic outcome of an experiment Sample space S D collection of all possible outcomes depends on experimenter A standard deck of cards has 52 cards in total 13 of each suit including jack queen and king What is probability 1 Numerical measure of the likelihood that event will occur rounded to three decimal places 2 All sample point probabilities must lie between 0 and 1 3 The probabilities of all sample points within a sample space must sum to 1 Calculating the probability of an event 1 Define the experiment 2 List the sample space 3 Determine the frequency count of each sample point and assign a probability to each 4 Determine the set of events 5 Sum the probabilities of the events in the set Visualizing Probability Writing Sample Space Venn Diagram Tree Diagram Toss 2 coins D H D H HH D T HT D T D H TH E T TT 1 Coin 2 Coin Total Head Tail Head HH HH HT HT Tail TH TH quotIT TT Total HH TH HT quotIT S Contingency Table Counting Rules 1 If any one of k different mutually exclusive and collectively exhaustive events can occur on each of n trials the number of possible outcomes is equal to k 2 If there are k1 events on the first trial k2 events on the second trial and kn events on the nth trial the number of possible outcomes is k1k2 kn 3 The number of ways that n items can be arranged in order is n nn 1 1 4 We do care about repetitions so we do include repetitions The number of ways that n I items can be arranged in order in x different ways with repetitions is an 5 We do not care about repetitions so we do not include repetitions The number of ways n that n items can be arranged in order in x different ways w1thout repetitions 1s an nX X Ex choosing people for a committee where their order does not matter 0 1 Union OR statement D U Intersection AND statement D upsidedown U Mutually exclusive D when one outcome cannot occur because another outcome has already happened Collectively exhaustive D when we account for all possible outcomes in an experiment Complementary Events The event that A does not occur All events not in A Denote complement of A by AC PA PAC 1 Applying the Addition Rule PA or B if A and B are mutually exclusive PA PB PA or B if A and B are not mutually exclusive meaning that they ARE independent PA PB PA and B If PA and B 0 then events A and B are mutually exclusive Type Color Total Red Black Ace 2 2 4 NonAce 24 24 48 Total 2 6 2 6 5 2 PAce or Black PAce PBlack PAce and Black 452 2652 252 2852 Conditional Probability conditional new sample space PAB PA and BPB PAceBlack 226 Events A and B are independent when 1 PAB PA 2 PBA PB 3 PA and B PA X PB Bayes s Rule 1 Permits revising old probabilities based on new information 2 Application of conditional probability 3 Mutually exclusive events Bayes s Rule Example A company manufactures MP3 players at two factories Factory I produces 60 of the MP3 players and Factory II produces 40 Two percent of the MP3 players produced at Factory I are defective while 1 of Factory 11 s are defective An MP3 player is selected at random and found to be defective What is the probability that it came from Factory I Answer 06 D Factory I D 002 D Defective D 098 D Good 04 D Factory II D 001 D Defective D 099 D Good PIPDv1 ooeo02 PID PIPDIPIIPDVII ooeo02o4o01 Random Variable 1 Discrete random variable U need whole number histogram x of counts and y probability Experiment Toss 2 coins Let x of heads p1 X Value 0 1 2 4 possible outcomes TTx O or 1p1p TIIx 1 or p1p HT X 1 or 10 HH X 2 or 1010 Probability of Outcomes 025 1 050 2 025 1 Distributions Binomial Bernoulli one trial of a binomial distribution with fx pX 1p139X A fixed number of observations n Two mutually exclusive and collectively exhaustive categories Generally called success or failure p probability of success 1 p q probability of failure Constant probability for each observation Observations are independent I of Outcomes n zm Probability for outcome px 1pnx I PX n 10X 39 11Cgt 39X of outcomes probability for n x x one outcome Pxgr binomcdfnpr and Pxr binompdfnpr Expected value p EX Z xpx np expected number b Variance U2 fh pLH5 EH22U 0 2 Poisson Unlike the binomial distribution the Poisson distribution deals with limited time space or physical entity interval that there will be no hits during manager is monitoring the website What is the probability that The random variable x is the number of occurences of an event over some The occurences must be random The occurences must be independent of each other The occurences must be uniformly distributed over the interval being used The mean is p y expected number of events PX e V X Ex A webpage receives an average of 60 hitshour What is the probability a 30 second interval in which the system there will be more than 2 hits during the 30 second interval Let x of hits in 30 seconds a y 05 05 O P0 e3905061 b y 505050 PXgt21 PX21 P0 P1 m 1T e 050 51 e 050 52 1 2 1 e3951 12 18 0014 EX Customers arrive at a rate of 72hour What is the probability of 4 customers arriving in 3 minutes y rt 123 36 36364 PM 4 2 Continuous random variable U do not need a whole number normal curve Distributions 1 Normal Bellshaped Symmetrical Mean median and mode are equal Location is determined by the mean u Spread is determined by the standard deviation 0 0 The chart gives you the probability to the LEFT of a zscore EX What is the probability that z is between 068 and 182 Answer 09656 02483 07173 7173 2 oc the probability of getting a zscore or higher Finding zscore given probability D find the closest possible probability value z in the table if it is exactly in the middle then take the average zscore Finding X given probability D Find zscore corresponding with the Xvalue described and substitute into X z o p EX Given p 08 if n 500 20yearolds find the probability that a between 375 and 425 of them inclusive will be alive at age 65 np 50008 400 210 1p 50002 100 210 p 50008 400 xnp1 p 7500o8o2 894 D II x np 4255 400 Z W 7 4ooo2 09978 3755 400 Z W 7400mm 00022 P PZ4255 P23755 09978 00022 09956 f 05 for a correction factor np gt 10 and n1p gt10 to assume that the binomial is normally distributed Fl Z X3Hn1 nfpgt lFOR BINOMIALS THAT ARE NORMAL Sample distribution from population distribution Central Limit Theorem D Even if the population is not normal the distribution will be approximately normal as long as the sample size is 3 30 Xbar the mean of more than one things from a sample I l Xbar I1 G C a Xbar W a Pomt estimate 1 Cr1t1ca1va1ueStandard error E Xbar i z of gc 2 90 90 alpha of 005 1645 or 165 95 95 alpha of 0025 196 usually rounded to 2 98 98 alpha of 001 233 99 99 alpha of 0005 258 Confidence Intervals a o known xbari b proportion D Xbar i Z M phat is called the point estimate Tl Hypothesis Testing 1 Ho null hypothesis Ha alternative hypothesis type of test 2 Define the decision method 3 Gather data pValue comes from calculated zvalue 4 Make a decision Miscellaneous P on 0th estimate is Esimated standard error is HI 1 2 5 x 45 2 Sui 3392 15 455 455 3 33 1 where u fix FEW l jd 34095 1 FEW 4E020H0303 c4095 E 2 FEW EEUEUJEEUEU 31536 3 FEW llz jgli jl 30256 4 HM imam c0015 lT tal 1

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