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Set #3: Discrete Probability Functions

Define: Random Variable – numerical measure of the outcome of a probability experimentValue determined by chanceDenoted using letters, such as X

Recall and extend: differences between “discrete random variable” and “continuous random variable”

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Discrete Probability Functions

Probability Distribution may take form asTableGraphHistogramMathematical Formula

Probability Functions will also haveMeanExpected ValueVarianceStandard Deviation

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Rules for Discrete Probability Function

Let P(x) denote the probability that a random variable X equals x; then

1. Σ P(x) = 1

i.e.: Sum of all probabilities of x = 1

2. 0 ≤ P(x) ≤ 1 i.e. P(x) is between 0 and 1 inclusive

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Distributions of Random Variables

Define: Probability Histogram – a histogram where

x-axis corresponds to value of random variabley-axis represents the probability of each value of

the discrete random variable

Construct Probability Histogram the same way as a relative frequency histogram, except vertical axis is probability

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Distributions of Random Variables

Define: Mean of discrete random variable μx= sum of (value of random variable times the probability of observing the random variable x) μx = Σ [x*P(x)]

OR: mean of a discrete random variable is the average outcome if experiment is repeated many, many times

that is: – μx approaches zero

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Distributions of Random Variables

Define: Expected Value E(x) is the sum of (value of random variable times probability of observing that random variable x) E(X) = μx = Σ [x*P(x)]

OR: the expected value of a discrete random variable is the mean of that discrete random variable

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Distributions of Random Variables

Define: Variance of a discrete random variable - is the weighted average of the squared deviations where the weights are the probabilities

σ2x = Σ [(x - μx)2 *P(x)]

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Distributions of Random Variables

Define: Standard Deviation of a discrete random variable is the square root of the variance

σx = √σ2x

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Distributions of Random Variables

Define: trial - each repetition of an experiment.

Define: disjoint outcome - two mutually exclusive outcomes

Typically the disjoint outcomes are called “success” and “failure”

Probability of success = One minus probability of failure

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Distributions of Random Variables

Distributions of Discrete Random Variables• Thus far have had:

• Mean• Expected Value• Variance• Standard Deviation

• Will have next:• Binomial Probability Function• Poisson Probability Function • And . . .

Distributions of Continuous Random Variables

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Distributions of Random Variables

Binomial Probability Experiment Criteria1. Experiment performed fixed number of

times (i.e. trials discrete)2. Trials are independent3. For each trial there are two disjoint

outcomes: success & failure4. The probability of success is the same for

each trial of the experiment

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Distributions of Random Variables

Binomial Probability Experiment Notations:

n is the number of independent trials

p is the probability of success

x is number of successes in n independent trials

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Distributions of Random Variables

Computing the Probability of Binomial Experiments

P(x) = nCx px (1-p)n-x

For x = 0, 1, 2, . . . N

WoW this is a very powerful result

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Distributions of Random Variables

Keep on going along this trail

Construct Binomial Probability Histogram

Use mean, standard deviation, and empirical rule to check for unusual results

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Distributions ofRandom Variables

1) Computing the Probability of Binomial ExperimentsP(x) = nCx px (1-p)n-x For x = 0, 1, 2, . . . N

2) Mean (or Expected Value) = μx= n*p

3) Std deviation σx = √ n*p*(1-p)

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More Distributions of Random Variables

Geometric Probability Distribution

Hypergeometric Probability Distribution

Negative Binomial Probability Distribution

Poisson Probability Distribution

Note: These distributions will haveMeanStandard Deviation

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Distributions of Random Variables

Geometric Probability DistributionNumber of trials until successSame requirements as Binomial Distr.See page 357, problem # 58

Negative Binomial Probability DistributionTo compute the number of trials necessary

to observe “r” successesSame requirements as Binomial Distr.See page 357, problem # 59

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Distributions of Random Variables

Poisson Process Conditions:1. Probability of two or more successes in

any sufficiently small subinterval is zero

2. Probability of success is the same of any two intervals of equal length

3. Number of successes in any interval is independent of number of successes in any other interval

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Distributions of Random Variables

Poisson Probability Distribution Function: for X number of successes in an interval of fixed length t, and average number of occurrences λ (lamba)

P(x) = [(λt)x /x!] e-λt

Expected value equals mean µx = λtStandard deviation = sq root (mean)e = base of the natural log system ~ 2.718281828

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Distributions of Random Variables

Poisson Probability Distribution Example #6.3.13 Insect Fragments legally in Peanut Butter

1. Calculate the Expected Value = mean µx = λt = rate * interval = 0.3 fragments * 5 gram sample = 1.5 fragments per sample = λt

2. Probability of exactly 2 fragments3. Probability of fewer than 2 frags4. Probability of at least 2 fragments

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Notes of the Day

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