Chapter 6

# Chapter 6 - KVANLI PAVUR KEELING Chapter6...

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Click to edit Master subtitle style   Chapter 6 Discrete  Probability  Distributions KVANLI PAVUR KEELING

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3/10/11 Chapter Objectives At the completion of this chapter, you should be able to answer the following questions: ∙ What is meant by the term " probability distribution ?" ∙ What is the formula for the mean and variance
3/10/11 Example Flip a coin 2 times Suppose we’re interested in the number of heads in the 2 flips Chapter 5 way: Define A: H on 1st flip and H on 2nd flip B: H on 1st flip and T on 2nd flip C: T on 1st flip and H on 2nd flip

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3/10/11 Example Chances of getting exactly one H in the 2 flips is P(B or C) = P(B) + P(C) since B and C are mutually exclusive P(B) = P(H on 1st flip and T on 2nd flip) = P(H on 1st flip) · P(T on 2nd
3/10/11 An Easier Way Define X to be the number of H’s in the 2 flips 0 with probability X = 1 with probability 2 with probability X=0 How can this occur? T on 1st flip and T on 2nd flip For a fair coin, this is ½ · ½ since coin flips are ¼

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3/10/11 Defining the Random Variable 0 with probability X = 1 with probability 2 with probability X=1 How can this occur? H on 1st flip and T on 2nd flip - OR - T on 1st flip and H on 2nd flip P(X = 1) is ¼ + ¼ = ½ ¼ As before, this is ½ · ½ = ¼ And this is ½ · ½ = ¼ ½
3/10/11 Defining the Random Variable 0 with probability X = 1 with probability 2 with probability X=2 How can this occur? H on 1st flip and H on 2nd flip P(X = 2) is ½ · ½ = ¼ ¼ Once again, this is ½ · ½ = ¼ ½ ¼

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3/10/11 A Discrete Random Variable X = number of H’s in 2 coin flips is called a discrete random variable Why discrete? Suppose you observe X You would get something like {1, 1, 0, 2, 1, 2, 0, 1, …,1} These are discrete data This is a sample
3/10/11 Continuous Random Variable The other type of random variable is a continuous random variable Here, X = (measuring something) Examples: Height, weight, length, length of time Consider a sample of heights: {5.82’, 6.45’,…., 5.41’} These are continuous data

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3/10/11 Another Discrete Random Variable Container with 30 poker chips 10 of these 10 of these 10 of these 1 2 3 Select 2 chips with replacement Let X = total of the 2 chips
3/10/11 X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=2 How can this occur? on the 1st draw and on the 2nd draw The probability of selecting a on each draw is 1/3 since 10 9 1 1/9 Remember: We replace the 1st chip before drawing the 2nd chip

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3/10/11 X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=3 How can this occur? on the 1st draw and on the 2nd draw -- OR -- on the 1st draw and 1/9 2/9
3/10/11 X = Total of the 2 Chips 2 with probability 3 with probability X = 4 with probability 5 with probability 6 with probability X=4 How can this occur? on the 1st draw and

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## This note was uploaded on 03/09/2011 for the course DSCI 2710 taught by Professor Hossain during the Fall '08 term at North Texas.

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Chapter 6 - KVANLI PAVUR KEELING Chapter6...

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