1 / 9

The Binomial Distribution

The Binomial Distribution. Permutations: . How many different pairs of two items are possible from these four letters: L, M. N, P. L,M L,N L,P M,L M,N M,P N,L N,M N,P P,L P,M P,N. P = # of Permutations N = # of items r = how many taken at a time. Combinations.

Antony
Download Presentation

The Binomial Distribution

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Binomial Distribution Permutations: How many different pairs of two items are possible from these four letters: L, M. N, P. L,M L,N L,P M,L M,N M,P N,L N,M N,P P,L P,M P,N P = # of Permutations N = # of items r = how many taken at a time

  2. Combinations (order is not important) C = # of Combinations N = # of items r = how many taken at a time L,M L,N L,P M,N M,P N,P

  3. If we flip a single coin, there are two possible outcomes: head or tail (H or T). P(one head) = .5 P(no head) = .5 Together the probabilities of these possible outcomes sum to 1.0. If we flip two coins there are four possible outcomes. H,H H,T T,H T,T P(two heads) = .25 P(one head) = .50 P(no heads) = .25 Total = 1.0 Can be calculated by… P = P(H) and Q = P(T) Probability of 2 heads Probability of 1head and 1 tail Or Probability of 2 tails * Think about the frequency distribution.

  4. Flip 3 Coins: 8 possible outcomes P(3 heads) = 1/8 or .13 P(2 heads) = 3/8 or .37 P(1 head) = 3/8 or .37 P(0 heads) = 1/8 or .37 Total = 1.0 H,H,H H,H,T H,T,H T,H,H T,T,H T,H,T H,T,T T,T,T Can be calculated by….. or

  5. Binomial Distribution A binomial distribution is produced when each of a number of Independent trials results in one of two Mutually Exclusive outcomes. A single flip of a coin is a Bernoulli trial. They reflect a discrete variable. P(X) = Probability of X number of an outcome N = Number of trials P = Probability of success on any given trial Q = (1 – P) The number of combinations of N things taken X at a time

  6. Example: Guess if I am thinking of the colour black or white. Because I randomly choose one of the two colours, the probability that I am thinking of black is .5 on any given trial. There will be 10 trials. What is the probability of guessing 8 of them correctly? = 45(.009765) = .0439425 This is the probability of * OR MORE correct, but of exactly 8 out of 10 correct.

  7. Binomial Distribution & Testing a Hypothesis Let us repeat the previous example, but now the question is what is the probability of guessing 8 or more of the trials correctly. We know that P(8) = .44 That is less than .05, But that is P(8) against 10, including P(9) and P(10). We need to test the probability of 8 or more correct. P(8) = .044 P(9) = .010 P(10) = .001 Total P = .055 The probability of guessing 8 or more correctly out of 10 turns out to have a probability greater than .05, thus we fail to reject the null hypothesis. Which was what?

  8. Binomial Distribution We can calculate the probability of 0 to 10 correct. # Correct Probability • 0 .001 • .010 • .044 • .117 • .205 • .246 • .205 • .117 • .044 • .010 • .001 Mathematical Sampling Distribution

  9. As N gets large, all binomial distributions approach normality, regardless of P. Thus, Mean = NP For example, 10(.5) = 5 That is, if you examine the data on the previous slide, you will see that the mean of the distribution is 5 (the number of correct trials). Variance = NPQ For example, 10(.5)(.05) = 2.25 Standard Deviation = = 1.58

More Related