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Unformatted text preview: 1 Chapter 6: Continuous Distributions Spring 2010 Sada Soorapanth 2 Discrete vs Continuous variable Discrete random variable Continuous random variable Takes on finite or countable numbers X= 0,1,2,… Takes on any value over a given interval 0 ≤ X ≤ 1.2 Probability of the random variable assuming particular value P(X=1) = 0.5 Probability of the random variable assuming a value within some given interval P(0.5 ≤ X ≤ 1) =0.7 Probability distribution is characterized probability mass function, P(X) Probability distribution is characterized by probability density function or p.d.f or f(X) 3 Requirements for a Probability Density Function Suppose that a continuous random variable x has the value between a and b 1. f(x) ≥ 0 for all x between a and b . 2. P( a ≤ x ≤b) = 1 or the total area under the graph of f(x) is 1. a b f x ( ) x Area = 1 4 Area as a Measure of Probability Area under the graph of f(x) over the in the interval represents the probability of the interval. X = volume of soft drink per bottle (min=0, max=1.2L) The probability of each individual value is zero. P(X=0.9)=0 P(0.9≤X ≤1)= P(0.9<X <1) 1.2 f x ( ) x 0.9 1.0 P(0.9≤x ≤1.0) 5 Some Special Distributions Discrete binomial Continuous uniform normal exponential 6 Uniform Random Variable A uniform random variable can assume any value in the interval between a and b with equal probability . X = Flight time from Chicago to New York, where x is equally likely between 100 and 140 mins x ~ Uniform (100, 140) Typical question: What is the probability that the flight time is between 110 and 130 mins? P(110 ≤X ≤130)=? 7 Uniform Probability Distribution ≤ ≤ = es other valu all 1 ) ( for b x a for a b x f Area = 1 f x ( ) x 1 b a a b The Uniform distribution is described by the function 8 Calculate the Probability of range of values P X b a x x x x ( ) 1 2 2 1 ≤ ≤ = ? ) 130 ( ? ) 125 ( ? ) 125 ( 5 . 40 20 100 140 110 130 ) 130 110 ( = ≤ = ≥ = ≤ = = = ≤ ≤ X P X P X P X P 110 130 f x ( ) x 100 140 5 . 100 140 110 130 = Area = 0.5 9 Uniform Distribution Mean and Standard Deviation M e a n = + μ a b 2 Standard Deviation σ = b a 12 10 Example: Gasoline Sales The amount of gasoline sold daily at a service station is uniformly distributed with a minimum of 2,000 gallons and maximum of 5,000 gallons a) Find the probability that daily sales will fall between 2,500 and 3,000 gallons....
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This note was uploaded on 09/10/2011 for the course DS 212 taught by Professor Saltzman during the Spring '08 term at S.F. State.
 Spring '08
 Saltzman

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