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Lect13_L6_L7_handout1

# Lect13_L6_L7_handout1 - lecture 13 Continuous Random...

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lecture 13, Continuous Random Variables IV Normal Ap- proximation Continuity Correction More Examples (Review) lecture 13, Continuous Random Variables IV Outline 1 Normal Approximation Continuity Correction 2 More Examples (Review)

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lecture 13, Continuous Random Variables IV Normal Ap- proximation Continuity Correction More Examples (Review) lecture 13, Continuous Random Variables IV Normal Approximation Binomials with different n and p Recall, distributions of binomial ( n , p ) for different n and p : 0 5 10 15 20 0.00 0.10 0.20 Distribution of Bin(20,0.2) x p(x) 0 5 10 15 20 0.05 0.15 Distribution of Bin(20,0.5) x 0 5 10 15 20 Distribution of Bin(20,0.8) x 0 20 40 60 80 100 0.04 0.08 Distribution of Bin(100,0.2) x 0 20 40 60 80 100 0.000.020.040.060.08 Distribution of Bin(100,0.5) x 0 20 40 60 80 100 Distribution of Bin(100,0.8) x As n gets larger, the graph of the binomial distribution tends to have the symmetric, bell-shaped form of the normal curve
lecture 13, Continuous Random Variables IV Normal Ap- proximation Continuity Correction More Examples (Review) lecture 13, Continuous Random Variables IV Normal Approximation Normal Approximation to the Binomial Suppose X is a binomial(n,p) random variable, where ± n is the number of trials ± Each trial has a probability of success p ± The probability of failure is q = 1 - p If n and p are such that np 10 and n ( 1 - p ) 10 1 , then X is approximately normally distributed with μ = np , and σ = npq = p np ( 1 - p ) . 1 the textbook uses 5 as a reference number, which is okay, too. We are being conservative here

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lecture 13, Continuous Random Variables IV Normal Ap- proximation Continuity Correction More Examples (Review) lecture 13, Continuous Random Variables IV Normal Approximation Example: Normal Approximation to a Binomial Suppose X is a binomial rv with n = 50 and p = 0 . 5 Want the probability of X = 23, i.e., 23 successes in the n = 50 trails Want P ( X = 23 ) np 10 and n ( 1 - p ) 10 approximation applies. Approximating using the normal curve with μ = np = 50 * 0 . 5 = 25 σ = p np ( 1 - p ) = 50 * 0 . 5 * 0 . 5 3 . 5 Q: Shall we use the probability of the normal rv taking value 23 to approximate P ( X = 23 ) ? A:
lecture 13, Continuous Random Variables IV Normal Ap- proximation Continuity Correction More Examples (Review) lecture 13, Continuous Random Variables IV Normal Approximation Continuity Correction Continuity Correction Continuity correction : Because the binomial distribution can only take integer values, while the normal distribution can take any real numbers, a normal approximation to the binomial should identify the binomial event “23" with the

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Lect13_L6_L7_handout1 - lecture 13 Continuous Random...

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