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Chapter 6 Notes (1)

Chapter 6 Notes (1) - *4 ~14 6 Probability Distributions...

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Unformatted text preview: *4 ~14 6 Probability Distributions 3.1 Summarizing Possible Outcomes and Their Probabilities Random Variable A random variable is a numerical measurement of the outcome of a random phenomenon. Often, the randomness results from the use of random sampling or a randomized Experiment to gather the data. Exllz Number of touchdowns for Red Raiders, Duration of a call Probability Distribution The probability distribution of a random variable specifies its possible vnluai and their probabilities. Ex]: Probability distribution of number of home runs in a game for Boston Red Sox. The table lists the possible values for the number of home runs and the corresponding probabilities. Number ol' Home Runs Probability 0 0.23 1 0.38 2 0.22 3 0.13 4 0.03 5 0.01 6 or more T Probability dimrilmtions can be defined for both discm random variables and continuous random variables. We will first look at discrete random WTlelkS, Probability Distribntlon of 3 Discrete Random Variable A discrete random tunable X takes a set of separate values (such as 0.1.1....) Its probability distribution assigns a probability P(r) to each possible value at z. . For each z, the probability 13(1) ialls between a and i. I The sum of the prubnbilities {or all [he pmzible x valum equals 1. Ema: Refer to Exz. . for each 1. the probability huh between 0 ml 1. (all the vnlnm 0.23, 0.33, 0.22, 0.13, 0.03, ml 0.01 are between 0 ml 1) . sum or the probabilities for all the possible a: value equals 1. (0.23 + 0.33 + 0.22 + 0.13 + 0.03 + 0.01:1) Therefore, it is a probability distribution. Probability of at least 3 humernns is P(X 3 3) = P(3) + Pu) + 11(5) + [’(6) = 0.13 + (Loo +0.01Jr 0.00 = 0.17 an m n a n u an n' a 'l 1 . a I : l 1 -. Figure 1; Probability Distribution for Ex.2 ExA: F‘mm six marbles numbered as l,1,1,1,2,2. lwu marblm will be drawn at. mt (loin without replacement. Let X denote the sunle of the numbers on the selected marbles. List the possible values of X and determine the probability distribution. Outcome Pmbl bility 3/5 ®© Nola/5) = 61:5 l‘lfillflfil = #115 (lellllsl - 1/15 lzléluls) . 1/15 ...
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