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Thermo & Stat Mech - Spring 2006 Class 16 More Discussion of the Binomial Distribution: Comments & Examples jl
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The Binomial Distribution applies ONLY to cases where there are only 2 possible outcomes: heads or tails, success or failure, defective or good item, etc. The requirements justifying the use of the Binomial Distribution are: 1. The experiment must consist of n identical trials. 2. Each trial must result in only one of 2 possible outcomes. 3. The outcomes of the trials must be statistically independent. 4. All trials must have the same probability for a particular outcome.
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Binomial Distribution The Probability of n Successes out of N Attempts is: p = Probability of a Success q = Probability of a Failure q = 1 – p (p + q) N = 1
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Thermo & Stat Mech - Spring 2006 Class 16 Mean of the Binomial Distribution l
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5 Standard Deviation ( ) of the Binomial Distribution 1 2 3
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For The Binomial Distribution
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Common Notation for the Binomial Distribution r items of one type & (n – r) of a second type can be arranged in n C r ways. Here: n C r is called the binomial coefficient In this notation, the probability distribution can be written: W n (r) = n C r p r (1-p) n-r ≡ probability of finding r items of one type & n – r items of the other type. p = probability of a given item being of one type. ≡
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Binomial Distribution: Example Problem: A sample of n = 11 electric bulbs is drawn every day from those manufactured at a plant. The probabilities of getting defective bulbs are random and independent of previous results. The probability that a given bulb is defective is p = 0.04. 1. What is the probability of finding exactly three defective bulbs in a sample? (Probability that r = 3?) 2. What is the probability of finding three or more defective bulbs in a sample? (Probability that r ≥ 3?)
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Thermo & Stat Mech - Spring 2006 Class 16 9 Binomial Distribution, n = 11 Number of Defective Bulbs, r Probability 11 C r p r (1-p) n-r p = 0.04 0 11 C 0 (0.04) 0 (0.96) 11 = 0.6382 1 11 C 1 (0.04) 1 (0.96) 10 = 0.2925 2 11 C 2 (0.04) 2 (0.96) 9 = 0.0609 3 11 C 3 (0.04) 3 (0.96) 8 = 0.0076 l
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Thermo & Stat Mech - Spring 2006 Class 16 10 Question 1: Probability of finding exactly three defective bulbs in a sample? P(r = 3 defective bulbs) = W 11 (r = 3) = 0.0076 Question 2: Probability of finding three or more defective bulbs in a sample? P(r ≥ 3 defective bulbs) = 1- W 11 (r = 0) – W 11 (r = 1) – W 11 (r = 2) = 1 – 0.6382 - 0.2925 – 0.0609 = 0.0084 l
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Thermo & Stat Mech - Spring 2006 Class 16 11 Binomial Distribution, Same Problem, Larger r Number of Defective Bulbs, r Probability 11 C r p r (1-p) n-r 0 11 C 0 (0.04) 0 (0.96) 11 = 0.638239 1 11 C 1 (0.04) 1 (0.96) 10 = 0.292526 2 11 C 2 (0.04) 2 (0.96) 9 = 0.060943 3 11 C 3 (0.04) 3 (0.96) 8 = 0.007618 4 11 C 4 (0.04) 4 (0.96) 7 = 0.000635 5 11 C 5 (0.04) 5 (0.96) 6 = 0.000037 l
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Thermo & Stat Mech - Spring 2006 Class 16 12 Binomial Distribution n = 11, p = 0.04 l Distribution of Defective Items Distribution of Good Items
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Thermo & Stat Mech - Spring 2006 Class 16 13 Consider a perfect coin. There are only 2 sides, so the probability associated with coin flipping is The Binomial Distribution. Problem: 6 perfect coins are flipped. What is the probability that they land with n heads & 1 – n tails ? Of course, this only makes sense if 0 ≤ n ≤ 6! For this case, the Binomial Distribution has the form: l The Coin Flipping Problem
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Thermo & Stat Mech - Spring 2006 Class 16 14 Binomial Distribution for Flipping 1000 Coins l Note: The distribution peaks around n = 500 successes (heads), as we would expect ( = 500)
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Thermo & Stat Mech - Spring 2006 Class 16 15 Binomial Distribution for Selected Values of n & p l n = 20, p = 0.5 n = 10, p = 0.1 & n =10, p = 0.9
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Thermo & Stat Mech - Spring 2006 Class 16 Binomial Distribution for Selected Values of n & p l n = 5, p = 0.1 n = 5, p = 0.5 n = 10, p = 0.5
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Thermo & Stat Mech - Spring 2006 Class 16 Binomial Distribution for Selected Values of n & p l n = 5, p = 0.5 n = 20, p = 0.5 n = 100, p = 0.5
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Binomial Distribution for Selected Values of n & p
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