AF Sections 6.1-6.3 handouts

AF Sections 6.1-6.3 handouts - STAT 2000 From Probability...

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1 STAT 2000 Sections 6.1-6.3 Probability Distributions Agresti-Franklin From Probability to Probability Distributions Toss two coins: S = { HH, HT, TH, TT} Assign probabilities and count the number of heads for each outcome. prob: ¼ ¼ ¼ ¼ S= { HH, HT, TH, TT} # heads 2110 Another way to display this information is in a table similar to the frequency tables from Chapter 2. We call this a probability distribution. Probability distribution for # heads in two tosses: #heads Probability xP (x) This distribution represents all possible outcomes of two coin tosses, and is considered a population. We call the number of heads, X, a random variable . A random variable is a variable with numerical values that is associated with outcomes in the sample space. In this example, X is a discrete random variable. (Recall discrete vs. continuous data.) The probability distribution of a discrete random variable is a listing of the values of the random variable, together with the associated probability. (Recall frequency and relative frequency tables/distributions.) Requirements for a Probability Distribution Similar to the rules for probability, there are two requirements for a probability distribution. (1) Every probability must be between 0 and 1. (2) The sum of all the probabilities must equal 1. x P(x) There are many practical applications of probability distributions, especially in the gambling and insurance industries. Suppose you have been given the opportunity to select one of 500 unmarked envelopes. One of these envelopes contains $1000. Ten of these envelopes contain $100. Twenty of these envelopes contain $10. The remaining envelopes contain nothing. What is the probability distribution for X, the amount you could win? Winning Amount: Probability: $1000 $100 $10 $0
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2 Suppose an insurance company sells you a $50,000 policy. They may have long term data to indicate that 70% of the time you will have no losses, 20% of the time you will have a $25,000 loss, and 10% of the time you will have a $50,000 loss. (This is a VERY simple example!) P b bilit di t ib ti f th t f l Probability distribution for the amount of loss: Amount of Loss Probability $ 0 .70 $25,000 .20 $50,000 .10 Insurance companies focus on how much they will pay out ‘on average’ as opposed to being interested in an individual payout. This is called the expected value, or the amount they ‘expect’ to have to pay. Just as we found the mean for a frequency distribution, we can find the mean of a probability distribution. It is a weighted average, with the probabilities serving as the weights. (Your grade calculation is a weighted average.) How much does the insurance company expect to have to pay out on a $50,000 policy? Amount of Loss Probability $ 0 .70 $25,000 .20 $50,000 .10 The mean of a probability distribution for a discrete random variable is given by: μ = Σ xP(x) Note that the parameter μ is used because a probability distribution either represents a population or is used as a model for a population.
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This note was uploaded on 03/22/2009 for the course STAT 2000 taught by Professor Smith during the Spring '08 term at University of Georgia Athens.

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AF Sections 6.1-6.3 handouts - STAT 2000 From Probability...

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