4.3-4.4 - Randomvariables(4.3,4.4) Discrete random...

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Random variables (4.3, 4.4) Discrete random variables Continuous random variables Probability distributions Normal probability distributions Mean of a random variable Variance of a random variable Law of large numbers
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Discrete random variables A random variable is a variable whose value is a numerical outcome of a random phenomenon. A fair coin is tossed three times. We define X as the number of times the head appears. Then X is a discrete random variable. X can only take the values 0, 1, 2, or 3.
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The probability distribution of a random variable X lists the values and their probabilities: The probabilities p i must add up to 1. A fair coin is tossed three times. X is the number of times the head appears Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 H - H HMM HHM MHM HMH MMM MMH MHH HHH
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Value of X 0 1 2 3 Probability 1/8 3/8 3/8 1/8 What is the probability that H appears at least twice (“at least two” means “two or more”)? P ( X ≥2) = P ( X =2) + P ( X =3) = 3/8 + 1/8 = 1/2 What is the probability that H appears fewer than three times? P ( X <3) = P ( X =0) + P ( X =1) + P ( X =2) = 1/8 + 3/8 + 3/8 = 7/8 or P ( X <3) = 1 – P ( X =3) = 1 – 1/8 = 7/8
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A fair coin is tossed 4 times.  X = the # of times H is observed The probability distribution of X is  P(0)=p(4)=0.0625, p(1)=p(3)=0.25, p(2)=0.375 Probability histogram
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Continuous random variables Example: Let X be a random number drawn from the interval [0,1]
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Continuous random variables X Example: Let X be a random number drawn from the interval [0,1] How do we assign probabilities to events in an infinite sample space? We use density curves and represent the probability of an event as the area under the density curve for the values of X that make up the event
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The probability that X falls between 0.3 and 0.7 is 0.4, which is equal to the area under the density curve for that interval: P (0.3 ≤ X ≤ 0.7) = (0.7 – 0.3)*1 = 0.4 X The density function in this case is f(x) = 1 for 0 < x < 1, and 0 outside the interval Uniform distribution over (0,1)
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P ( X < 0.5 or X > 0.8) = P ( X < 0.5) + P ( X > 0.8) = 1 – P (0.5 < X < 0.8) = 0.7 For continuous random variables, the probability of a single value is zero, and the probability of an interval is the same whether the boundary values are included or excluded. Height = 1 X P (0 ≤ X ≤ 0.5) = (0.5 – 0)*1 = 0.5 P (0 < X < 0.5) = (0.5 – 0)*1 = 0.5 P (0 ≤ X < 0.5) = (0.5 – 0)*1 = 0.5
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Y to be their sum. Y can take any value between 0 and 2. It can be shown that the density curve for Y is: 0 1 2 Height = 1. We know this because the base = 2, and the area under the curve has to equal 1 by definition. The area of a triangle is
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This note was uploaded on 09/04/2010 for the course STAT 131 taught by Professor Isber during the Spring '08 term at University of California, Berkeley.

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4.3-4.4 - Randomvariables(4.3,4.4) Discrete random...

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