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Chapter 6: Binomial Probability DistributionsPowerPoint Presentation

Chapter 6: Binomial Probability Distributions

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Chapter 6: Binomial Probability Distributions

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Chapter 6: Binomial Probability Distributions

In Chapter 6:

6.1 Binomial Random Variables

6.2 Calculating Binomial Probabilities

6.3 Cumulative Probabilities

6.4 Probability Calculators

6.5 Expected Value and Variance ..

6.6 Using the Binomial Distribution to Help Make Judgments

- Bernoulli trial≡ a random event with two possible outcomes (“success” or “failure”)
- Binomial random variable ≡ the random number of successes in n independent Bernoulli trials, each trial with the same probability of success
- Binomials have two parameters:n number of trialsp probability of success of each trial

- Only two outcomes are possible (success and failure)
- The outcome of each trial does not depend on the previous trial (independence)
- The probability for success p is the same for each trial
- Trials are repeated a specified number of times n

Formula:

where

nCx≡ the binomial coefficient (next slide)

p≡ probability of success for a single trial

q≡ probability of failure for single trial = 1 – p

Formula for the binomial coefficient:

where ! represents the factorial function:

x!= x (x – 1) (x – 2) … 1

For example, 4! = 4 3 2 1 = 24

By definition 1! = 1 and 0! = 1

For example:

The binomial coefficient tells you the number of ways you could choose x items out of n

nCx the number of ways to x items out of n

For example, 4C2 = 6 Therefore, there are 6 ways to choose 2 items out of 4.

“Four patients example”: X ~ b(4,.75).

Note q = 1 −.75 = .25.

What is the probability of 0 successes?

Pr(X = 1) = 4C1· 0.751 · 0.254–1

= 4 · 0.75 · 0.0156

= 0.0469

Pr(X = 2) = 4C2· 0.752 · 0.254–2

= 6 · 0.5625 · 0.0625

= 0.2106

Pr(X = 3) = 4C3· 0.753 · 0.254–3

= 4 · 0.4219 · 0.25

= 0.4219

Pr(X = 4) = 4C4· 0.754 · 0.254–4

= 1 · 0.3164 · 1

= 0.3164

Pr(X = 2)

=.2109 × 1.0

This figure illustrates Pr(X 2) on X ~b(4,.75)

Pr(X =0) + Pr(X = 1)

Pr(X =0) + Pr(X = 1) + Pr(X = 2)

Pr(X =0) + Pr(X = 1) + … + Pr(X = 3)

Pr(X =0) + Pr(X = 1) + … + Pr(X = 4)

Cumulative probability function (cdf) = cumulative probabilities for all outcome

Example: cdf for X~b(4, 0.75)

Pr(X 0) = 0.0039

Pr(X 1) = 0.0508

Pr(X 2) = 0.2617

Pr(X 3) = 0.6836

Pr(X 4) = 1.0000

StaTable is a free computer program that calculates probabilitiesfor many types of random variables, including binomials

Number of successes x

Binomial parameter p

Binomial parameter n

Calculates Pr(X = x)

Calculates Pr(X≤ x)

StaTable

Exact and cumulative probability of “2” for X~b(n = 4, p = .75)

x = 2

p = .75

n = 4

Pr(X = 2) = .2109

Pr(X≤ 2) = .2617

- Expected value μ
- Variance σ2
- Shortcut formulas:

For X~b(4,.75)

μ = n∙p= (4)(.75) = 3

σ2 = n∙p∙q= (4)(.75)(.25) = 0.75

- Suppose we observe 2 successes in a “Four patients” experiment?
- Assume X~b(4, .75)
- 3 success are expected
- Does the observation of 2 successes cast doubt on p = 0.75?

Pr(X 2) = 0.2617. What does this infer?