Thermo & Stat Mech - Spring 2006 Class 16 More Discussion of the Binomial Distribution: Comments & Examples jl

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.

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

Thermo & Stat Mech - Spring 2006 Class 16 Mean of the Binomial Distribution l

5 Standard Deviation (  ) of the Binomial Distribution  1 2 3

For The Binomial Distribution

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. ≡

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 = 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?)

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 = C 0 (0.04) 0 (0.96) 11 = C 1 (0.04) 1 (0.96) 10 = C 2 (0.04) 2 (0.96) 9 = C 3 (0.04) 3 (0.96) 8 = l

Thermo & Stat Mech - Spring 2006 Class Question 1: Probability of finding exactly three defective bulbs in a sample? P(r = 3 defective bulbs) = W 11 (r = 3) = 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 – – = l

Thermo & Stat Mech - Spring 2006 Class 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 = C 1 (0.04) 1 (0.96) 10 = C 2 (0.04) 2 (0.96) 9 = C 3 (0.04) 3 (0.96) 8 = C 4 (0.04) 4 (0.96) 7 = C 5 (0.04) 5 (0.96) 6 = l

Thermo & Stat Mech - Spring 2006 Class Binomial Distribution n = 11, p = 0.04 l Distribution of Defective Items Distribution of Good Items

Thermo & Stat Mech - Spring 2006 Class 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

Thermo & Stat Mech - Spring 2006 Class Binomial Distribution for Flipping 1000 Coins l Note: The distribution peaks around n = 500 successes (heads), as we would expect ( = 500)

Thermo & Stat Mech - Spring 2006 Class Binomial Distribution for Selected Values of n & p l n = 20, p = 0.5 n = 10, p = 0.1 & n =10, p = 0.9

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

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

Binomial Distribution for Selected Values of n & p