mean = 41.21median = 42.5s = 7.59 x - 1s – x + 1s – assignment #1

Download Mean = 41.21Median = 42.5s = 7.59 x - 1s – x + 1s – Assignment #1

If you can't read please download the document

Post on 12-Jan-2016




2 download

Embed Size (px)


  • Mean = 41.21Median = 42.5s = 7.59x - 1sx + 1sAssignment #1

  • Course Schedule

  • Probabilities in GeographyThe analyses of many problems (daily or geographic) are often based on probabilities, such as:What are the chances of having rain over the weekend?What is the likelihood that the 100-year flood will occur within the next ten years?How likely is it that a pixel on a satellite image is correctly classified or misclassified?

  • Probability & Probability DistributionWe summarize a sample statistically and want to make some inferences about the population (e.g., what proportion of the population has values within a given range) The concept of probability is the key to making statistical inferences by sampling a populationWhat we are doing is trying to ascertain the probability of an event having a given outcome This requires us to be able to specify the distribution of a variable before we can make inferences

  • Probability & Probability DistributionsPreviously, we looked at some proportions of area under the normal curve:Source: Earickson, RJ, and Harlin, JM. 1994. Geographic Measurement and Quantitative Analysis. USA: Macmillan College Publishing Co., p. 100.

  • Probability & Probability DistributionsBUT before we could use the normal curve, we have to find out if this is the right distribution for our variable While many natural phenomena are normally distributed, there are other phenomena that are best described using other distributionsBackground on probabilities (terminology & rules), and a few useful distributions:Discrete distributions: Binomial and PoissonContinuous distributions: Normal and its relatives

  • Probability-Related ConceptsAn event Any phenomenon you can observe that can have more than one outcome (e.g., flipping a coin)An outcome Any unique condition that can be the result of an event (e.g., flipping a coin: heads or tails), a.k.a simple event or sample pointsSample space The set of all possible outcomes associated with an evente.g., flip a coin heads (H) and tails (T)e.g., flip a coin twice HH, HT, TH, TT

  • Probability-Related ConceptsAssociated with each possible outcome in a sample space is a probabilityProbability is a measure of the likelihood of each possible outcomeProbability measures the degree of uncertaintyEach of the probabilities is greater than or equal to zero, and less than or equal to oneThe sum of probabilities over the sample space is equal to one

  • Probability ExamplesExample I Flip a coinTwo possible outcomes: heads, tailsEach outcome is equally likelyheads and tails have the same probability (0.5)The sum of probabilities over the sample space is one# of heads and # of tails will be nearly equal

  • Probability ExamplesExample II Flip a coin twiceFour outcomes are equally likelyTosses of the coin are independentEach outcome has probability 1/4 The probability of a head on Flip 1 and a head on Flip 2 is 1/2 * 1/2 = 1/4

  • How To Assign Probabilities to Experimental Outcomes?There are numerous ways to assign probabilities to the elements of sample spacesClassical method assigns probabilities based on the assumption of equally likely outcomesRelative frequency method assigns probabilities based on experimentation or historical dataSubjective method assigns probabilities based on the assignors judgment or belief

  • Classical MethodThis approach assumes that each outcome is equally likelyIf an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome.It is an appropriate way to assign probabilities to the outcomes in special kinds of experiments

  • Classical MethodExample I: Rolling a dieSample Space: S = {1, 2, 3, 4, 5, 6}Probabilities: Each sample point has a 1/6 chance of occurring.

  • Classical MethodExample II Flip four coinsLet 0 represent heads and 1 represents tailsFor each toss, the probability of heads or tails is Assuming that outcomes of the four tosses are independent from one anotherSixteen possible outcomes Probability of each outcome: * * * = 1/16 = 0.0625

  • Relative Frequency MethodThe second way is to assign them on the basis of relative frequenciesExampleGiven a weather pattern, a meteorologist may note that in 65 out of the last 100 times that such a pattern prevailed there was measurable precipitation the next dayIf there were such a weather pattern today, what would the probability of having rain tomorrow be?The possible outcomes rain or no rain tomorrow are assigned probabilities of 0.65 and 0.35, respectively

  • Subjective MethodWhen extreme weather conditions occur it might be inappropriate to assign probabilities based solely on historical dataWe can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occurThe best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimates.

  • Probability RulesRules for combining multiple probabilitiesA useful aid is the Venn diagram - depicts multiple probabilities and their relations using a graphical depiction of sets The rectangle that forms the area of the Venn Diagram represents the sample (or probability) space, which we have defined above Figures that appear within the sample space are sets that represent events in the probability context, & their area is proportional to their probability (full sample space = 1)

  • Probability RulesWe can use a Venn diagram to describe the relationships between two sets or events, and the corresponding probabilitiesThe union of sets A and B (written symbolically is A B) is represented by the areas enclosed by set A and B together, and can be expressed by OR (i.e. the union of the two sets includes any location in A or B) The intersection of sets A and B (written symbolically as A B) is the area that is overlapped by both the A and B sets, and can be expressed by AND (i.e. the intersection of the two sets includes locations in A AND B)

  • Addition RuleIf sets A and B do not overlap in the Venn diagram, the sets are disjoint, and this represents a case of two independent, mutually exclusive eventsThe union of sets A and B here uses the addition rule, whereP(AB) = P(A) + P(B)You can think of this in terms of areas of the events, where the union in this case is simply the sum of the areasThe intersection of sets A and B here results in the empty set (symbolized by ), because at no point do the circles overlapP(AB) = P(A) + P(B)P(AB) =

  • For example, suppose set A represents a roll of 1 or 2 on a 6-sided die, so P(A)=2/6, and set B represents a roll of 3 or 4, so P(B)=2/6Probability Rules The union of sets A and B here uses the addition rule, whereP(AB) = P(A) + P(B)P(AB) = 2/6 + 2/6P(AB) = 4/6 = 2/3 = 0.67

    The outcomes represented here are mutually exclusive, thus there is no intersection between sets A and B, thus P(AB) =

  • Probability Rules General Addition RuleIf sets A and B do overlap in the Venn diagram, the sets are independent but not mutually exclusiveThe union of sets A and B here isP(AB) = P(A) + P(B) - P(AB)because we do not wish to count the intersection area twice, thus we need to subtract it from the sum of the areas of A and B when taking the union of a pair of overlapping sets

    The intersection of sets A and B here is calculated by taking the product of the two probabilities, a.k.a. the multiplication rule:

  • General Addition RuleConsider set A to give the chance of precipitation at P(A)=0.4 and set B to give the chance of below freezing temperatures at P(B)=0.7The intersection of sets A and B here is P(AB) = P(A) * P(B)P(AB) = 0.4 * 0.7 = 0.28This expresses the chance of snow at P(AB) = 0.28The union of sets A and B here isP(AB) = P(A) + P(B) - P(AB)P(AB) = 0.4 + 0.7 0.28 = 0.82This expresses the chance of below freezing temperatures or precipitation occurring at P(AB) = 0.82

  • ComplementConsider set A to give the chance of precipitation at P(A)=0.4 and set B to give the chance of below freezing temperatures at P(B)=0.7The complement of set A isP(A) = 1 - P(A)P(A) = 1 0.4 = 0.6This expresses the chance of it not raining or snowing at P(A) = 0.6The complement of the union of sets A and B isP(AB) = 1 [P(A) + P(B) - P(AB)]P(AB) = 1 [0.4 + 0.7 0.28] = 0.18This expresses chance of it neither raining nor being below freezing at P(AB) = 0.18P(A) = 1 - P(A)P(AB) = 1 [P(A) + P(B) - P(AB)]

  • Probability RulesWe can also encounter the situation where set A is fully contained within set B, which is equivalent to saying that set A is a subset of set B:

    For example, set A might represent precipitation events with >= 5 inches, whereas set B denotes any events with >= 1 inch A is contained with B because anytime A occurs, B occurs as well In probability terms, this situation occurs when outcome B is a necessary precondition for outcome A to occur, although not vice-versa (in which case set B would be contained in set A instead)

  • Probability ExampleExample # of malls within cities City # of Malls A1 B4 C4 D4 E2 F3

    We might wonder if we randomly pick one of these six cities, what is the probability (chance) that it will have n malls?Each count of the # of malls in a city is an event

  • Random VariablesWhat we have here is a random variable defined as a function that associates a unique numerical value with every outcome of an experimentTo put this another way, a random variable is a function defined on the sample space this means that we are interested in all


View more >