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What is Probability?Measurement of uncertaintyTheory of choice and chanceAllows intelligence guess about futureHelps to quantify riskPredicts outcomes
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PROBABILITY DEFINITION-OBJECTIVEFrequency ConceptBased on empirical observationsNumber of times an event occurs in along series of trials
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PROBABILITY DEFINITION -SUBJECTIVEMerely expresses degree of beliefBased on personal experience
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Basic TerminologiesExperiment(Process of conductingtrials)Trial (Act of an experiment.)Outcome ( Result of a Particulartrial)Event (Particular outcome orsingle result of an experiment)
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Classical ProbabilityCount number favorable to event E = aCount number unfavorable to event E = bTotal favorable and unfavorable = a+bAssume E can occur in n possible waysAssume occurrence of events equally likelyTotal number of possible ways =a+b = n
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Probability of an event-EProbability of E = a = Pr(E)a+bNumber of times event favorable dividedby number of all possible ways.
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Probability Thermometer. 1.0 - sure to occur– - 0.50- cannot occur0>Pr (E) < 1
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Type of EventsSimple eventsCompound eventsMutually exclusive eventsIndependent events
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Simple eventsEvents with single outcomestossing a fair coin
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Compound eventsCompound events is the combination oftwo or more than two simple events.Suppose two coins are tossedsimultaneously
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Probability RulesAddition ruleMultiplication rule
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Addition ruleSingle 6-sided die is rolled.What is the probability of rolling a 2 or a5?P(2) = 1/6P(5) = 1/6P(2 or 5) = P(2) + P(5) = 1/6 + 1/6=2/6
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Mutually Exclusive EventsPr (A or B) = Pr (A) + Pr (B)IF not mutually exclusivePr (A or B) = Pr (A) + Pr (B) - Pr (A and B)
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QUESTION ON MUTUALLYEXLUSIVE EVENTSFrom the records at an STC, 4 girlshad HIV, 4 other girls had gonorrheawhile 2 girls have both gonorrhea andHIV.What is the probability that any girlselected will havei. HIV onlyii. HIV or Gonorrhea.
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SOLUTIONProb. Of HIV only =4/10– Prob. of HIV or Gonorrhea = Pr(HIV) +Prob.(Gonorrhea) - Pr(HIV andGonorrhea) = 4/10 + 4/10 - 2/ 10
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Independent eventsChoosing a marble from a jar AND landingon heads after tossing a coin.Choosing a 3 from a deck of cards,replacing it, AND then choosing an ace asthe second card.Rolling a 4 on a single 6-sided die, ANDthen rolling a 1 on a second roll of the die.
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Multiplication RuleWhen two events, A and B, areindependent, the probability of bothoccurring is:P(A and B) = P(A) · P(B)
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A coin is tossed and a single 6-sided die isrolled.Find the probability of landing on the headside of the coin and rolling a 3 on the die.P(head) = 1/2P(3) = 1/6P(head & 3) = P(head) · P(3)=1/2 x 1/6 = 1/12
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Conditional ProbabilityIn probability theory, a conditionalprobability is the probability that an eventwill occur, when another event is known tooccur or to have occurred.
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Conditional ProbabilityEvents not independentPr (A given B) = Pr (A and B)Pr (B)
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On the “Information for the Patient” label ofa certain antidepressant, it is claimed thatbased on some clinical trials, there is a14% chance of experiencing sleepingproblems known as insomnia (denote thisevent by I),26% chance of experiencing headache(denote this event by H), and there is a 5%chance of experiencing both side effects (Iand H).
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Suppose that the patient experiencesinsomnia; what is the probability that thepatient will also experience headache?Since we know (or it is given) that thepatient experienced insomnia, we arelooking for P(H | I). According to thedefinition of conditional probability:P(H | I) = P(H and I) / P(I) = 0.05/0.14 =0.357.
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Random VariablesA real valued function, defined over thesample space of a random experiment iscalled the random variable, associated tothat random experiment.That is the values of the random variablecorrespond to the outcomes of the randomexperiment.
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Random VariablesTake specified values with specifiedprobabilitiesDiscrete Random variable–E.g. no of children in a family, no ofpatients in a doctors surgeryContinuous Random variable
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The probability distribution of adiscrete random variable is a list ofprobabilities associated with each of itspossible values.It is also sometimes called theprobability function or the probabilitymass function
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Continuous random variables are usuallymeasurements.Examples include height, weight, theamount of sugar in an orange, the timerequired to run a mile, etc
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BINOMIAL DISTRIBUTIONSuccessive trials are independentOnly two outcomes are possible ineach trial or observationChance of success in each trial is knownSame chance of success from trial to trial
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BINOMIAL FORMULAPr (r out of n events) = n ! pr qn-rr! (n-r) !where n ! =n(n-1)(n-2)(n-3)….2.1e.g. 3! =3x2x1
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BINOMINAL TERMSN = Number of trialsr = Number of successesp = Probability of success in each trialq = 1-p = Probability of failure in eachtrial! = Factorial sign
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EXAMPLE BINOMINALIt is known that 10% of patientsdiagnosed to have a condition survivefollowing surgical treatment. What isthe chance of 2 people surviving outof 5 diagnosed with the condition andtreated surgically.
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The Shape of a DistributionSymmetrical– can be divided at the center so that each halfis a mirror image of the otherAsymmetrical
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Skewness– If a distribution is asymmetric it is eitherpositively skewed or negatively skewed.– A distribution is said to be positively skewed ifthe values tend to cluster toward the lowerend of the scale (that is, the smaller numbers)with increasingly fewer values at the upperend of the scale (that is, the larger numbers).
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With a negatively skeweddistribution, most of the values tend tooccur toward the upper end of the scalewhile increasingly fewer values occurtoward the lower end.
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Properties of Normal CurveBell shaped and symmetric about centreCompletely determined by its mean and standarddeviationMean, median and mode have same valueTotal area under curve is 1 (100%).68% of all observations lie within one standarddeviations of the mean.95% of observations lie within 1.96 standarddeviations of the mean valueGives probability of falling within interval if data hasnormal distribution.
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Importance of Normal DistributionFits many practical distributions of variables inmedicineIf variables are not normallydistributed, transformation techniques to makethem normal exist.Sampling distributions of means and proportions areknown to have normal distributionsIt is the cornerstone of all parametric tests ofstatistical significance.
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Presentation of Normal Distribution.As a mathematical equationGraphTable
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-1. Mathematical Equation- 1/2 (x - )2y = 1___ e2IIII and e are constantsis arithmetic meanis standard deviation
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The Standardized normaldistributionAll normal distributions have same overallshapePeak and spread may be differentHowever markers of 68th and 95Thpercentiles will still be located at 1 and 2SDThis attribute allows for standardization ofany normal distribution
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Can define distance along x axis in termsof SD from the mean instead of the truedata pointCondenses all normal distributions intoone through a mathematical equationZ= x- μσ
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Each data point is converted into astandardized value, and its new value iscalled a Z score
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Z ScoreStandardizing data on one scale so that acomparison can be madeStandard score or Z score is:– The number of standard deviations fromthe meanconvert a value to a Standard Score:– first subtract the mean,– then divide by the Standard Deviation
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Z ScoreThe z-score is associated with the normaldistribution and it is a number that may beused to:– tell you where a score lies compared with therest of the data, above/below mean.– compare scores from different normaldistributions
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Table of AreaAreas under a standard normal curveGives probability of falling within aninterval.Standard normal curve has a mean = 0and standard deviation = 1Need to transform data to standard normalcurve to use this table.
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1. Transformation to standard Normal Curve.- Use Z = (x - )Z is standardized normal deviate or normalscore.- Read corresponding area from table.- Z is in the Ist column in the table.- Area in the heart of the table.
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The IQs of a group of students arenormally distributed with a mean of 100and a standard deviation of 12. Whatpercentage of students will have an IQ of110 or more?
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Z= x- μ/ σZ= (110-100)/12Z=0.83, this corresponds to 0.2033 fromthe table20% of students will have an IQ of 100 ormoreWhat % of students will have IQ between100 and 110?
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If the heights of a population of men areapproximately normally distributed withmean of 172m and standard deviation of6.7cm. What proportion of men wouldhave heights above 180cm.
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solutionZ= x- µσ180-172 = 1.196.7In the table of normal distribution, theprobability of obtaining a standardisednormal deviate greater than 1.19 is0.117(11.7%)
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Therefore around 12% of the populationwould have heights above 180cm.
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In summaryNormal distribution as a predictor ofeventsDirect applications in statisticsTesting for significanceBackbone of inferential statistics
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