Lect13_L6_L7_handout2

Lect13_L6_L7_handout2 - lecture 13, Continuous Random...

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Unformatted text preview: 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) 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 : 5 10 15 20 0.00 0.10 0.20 Distribution of Bin(20,0.2) x p(x) 5 10 15 20 0.000.050.100.15 Distribution of Bin(20,0.5) x p(x) 5 10 15 20 0.00 0.10 0.20 Distribution of Bin(20,0.8) x p(x) 20 40 60 80 100 0.00 0.04 0.08 Distribution of Bin(100,0.2) x p(x) 20 40 60 80 100 0.000.020.040.060.08 Distribution of Bin(100,0.5) x p(x) 20 40 60 80 100 0.00 0.04 0.08 Distribution of Bin(100,0.8) x p(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 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 = . 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 * . 5 = 25 σ = p np ( 1- p ) = √ 50 * . 5 * . 5 ≈ 3 . 5 • Q: Shall we use the probability of the normal rv taking value 23 to approximate P ( X = 23 ) ? • A: No! The probability of the normal rv taking value 23 is 0! 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 interval “(22.5,23.5)" (and similarly for other integer values) • With continuity correction, to find the probability P ( 22 . 5 ≤ Y ≤ 23 . 5 ) where Y is a normal...
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Lect13_L6_L7_handout2 - lecture 13, Continuous Random...

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