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Theoretical Probability Distributions Three major types: Binomial, Poisson, Normal Quick look at binomial and then concentrate on Normal Distribution

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Binomial Distribution Based on events for which there are only 2 alternative possibilities: Heads or tails Girl or boy Pregnant or not

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Multiple “Attempts” The binomial distribution has the form: For one toss of the coin, what is the prob’y of a head For two tosses, what is the prob’y that both will be heads For three tosses, what is the prob’y that 3 will be heads For three tosses, what is the prob’y that 2 will be heads, etc.

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What does “attempt” mean? Can mean individuals, patients, etc. Pregnant or not pregnant Out of the next 10 patients (10 attempts), what is the probability that 2 will be pregnant.

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Exposed to a disease Get an infection or don’t get it If I have an accidental needle stab, what is the probability that I will contract Hepatitis B…….Let’s say it is 0.1 What if I have 2 needle stabs, what is the probability that I will contract Hepatitis B.

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Separate Distributions For one toss For two tosses For three tosses For four tosses (or one needle stab, or two patients, or rain on any of 3 days)

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Ultra-Simple Distribution For one toss P(0 heads) = 0.5 P(1 head) = 0.5

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Two Tosses What’s the probability that one will be a head? One out of two??? That would be 0.5 That’s correct in this case but have to be very careful. Really need some additional steps to show this.

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Combinations All Possible Events for Two Tosses P(h, t) P(h, h) P(t, t) P(t, h) Add up the P(1 head) = P(1 is a head and the other is a tail) = 2/4 = 0.5

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Combinations cont’d Look at another question. What’s the probability that neither will be a head? That’s the same as saying: P(0) = P(both are tails) = 0.5 X 0.5 = 0.25 What’s the probability of at least one head P = 0.75 (Means the same as 1 – P(0).)

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But… Before, we said that the probability of one head is 0.5 What’s the difference? Ah…. Here we said, at least one head So, we mean probability of one head or two heads

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Binomial Distribution Can be simple. You can put the data we just did into a table. Tail, tail = 0.25 Head, tail = 0.25 Tail, head = 0.25 Tail, tail = 0.25 1.00

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For two tosses P(0 Heads) = 0.25 P(1, only 1) = 0.50 P(2) = 0.25

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Probability Distribution HeadsProbability 00.25 10.50 20.25 What’s the probability of one or more heads: That is either one head or two so add the 2 possibilities = 0.75

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More Complicated More than 2 attempts Or the probability of the single event might not be 0.5 Can always be done by the combinations Or we use a formula

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Another Factor in B.D., p In the first example, the probability that any one toss will be a head is taken to be 0.5 Look at smoker or non- smoker. If we just consider undergraduate students at the current time, the probability of a smoker might be 0.3. What happens to the binomial distribution?

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For two students The same combinations Non, non Non, smoker Smoker, non Smoker, smoker

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But what are the probabilities? The same combinations Non, non0.49 Non, smoker0.7 X 0.3 = 0.21 Smoker, non0.21 Smoker, smoker0.09 1.00

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Factors in B.D. Number of “Attempts” Probability of any individual “success”

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Binomial aka Bernoulli Distribution Just Make a Table of Combinations for the number of “attempts” And then pick out the probabilities Either add or multiply or subtract from 1, as needed.

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