The Binomial Distribution

1 / 20

# The Binomial Distribution - PowerPoint PPT Presentation

The Binomial Distribution. For the very common case of “Either-Or” experiments with only two possible outcomes. Recognize Binomial Situations. Only two possible outcomes in each trial. Probability for one of the outcomes. Probability for the other outcome.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'The Binomial Distribution' - alayna

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### The Binomial Distribution

For the very common case of “Either-Or” experiments with only two possible outcomes

Recognize Binomial Situations
• Only two possible outcomes in each trial.
• Probability for one of the outcomes.
• Probability for the other outcome.
• Some definite number of trials, .
• They’re independent trials. don’t change.
• We’re interested in , probability of a certain count of how many times event happens in those trials.
A special kind of probability distribution
• It’s the familiar probability distribution
• But only two rows for the two outcomes.
• Note that
• Because probabilities must always sum to 1.00000
• And this leads to .
The Binomial Probability Formula
• Question: If we have trials, what is the probability of occurrences of the “success” event (the one with probability )
Practice with the Formula
• Experiment: Roll two dice
• Event of interest: “I rolled a 7 or an 11”
• Probability of success: (from
• Probability of failure:
• Number of trials
Practice with the Formula
• Find P(2) successes in the seven/eleven game
• Probability of at least three wins in five trials
• P(X≥3) = P(X=3) + P(X=4) + P(X=5) add them up!
• Probability of more than three wins
• P(X>3) = P(X=4) + P(X=5)
• Probability of at most three wins
• P(X≤3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
• Probability of fewer than three wins
• P(X<3) = P(X=0) + P(X=1) + P(X=2)
Use the Complement to save time
• Example:
• “Probability of at least 3 wins”
• P(X≥3) = P(X=3) + P(X=4) + … + P(X=49) + P(X=50)
• This means 48 calculations and sum results.
• EASIER: The complement is “fewer than 3”
• Take 1 – [ P(X=0) + P(X=1) + P(X=2) ]
TI-84 Computations
• binompdf(n, p, X) = probability of X successes in n trials
• Recompute the table and make sure we get the same results as the by-hand calculations.
• The “pdf” in “binompdf” stands for “probability distribution function”
TI-84 Computations
• binomcdf(n, p, x) = P(X=0) + P(X=1) + … P(X=x) successes in n trials
• binomcdf(n, p, x) does lots of little binompdf() for you for x = 0, x = 1, etc. up to the x you told it, and it adds up the results
• The “cdf” in “binomcdf” stands for “cumulative distribution function”
binomcdf() and complements
• Sevens or elevens, n = 50 trials again
• P(no more than 10 successes)
• binomcdf(50, 8/36, 10)
• P(fewer than 10 successes)
• binomcdf(50, 8/36, 9)
• P(more than 10 successes) – use complement!
• 1 minus binomcdf(50, 8/36, 10)
• P(at least 10 successes) – use complement!
• 1 minus binomcdf(50, 8/36, 9)
Mean, Variance, and Standard Deviation
• We had formulas and methods for probability distributions in general.
• The special case of the Binomial Probability Distribution has special shortcut formulas
• Mean =
• Variance =
• Standard deviation =
Mean, Variance, and Standard Deviation
• Compute these for the seven-eleven experiment with n = 5 trials
• Mean =
• Variance =
• Standard deviation =
Mean, Variance, and Standard Deviation
• Compute these for the seven-eleven experiment with n = 50 trials
• Mean =
• Variance =
• Standard deviation =
Mean, Variance, and Standard Deviation
• Compute these for the seven-eleven experiment with n = 100 trials
• Mean = and Standard deviation =
• “Expected value” – in 100 tosses of two dice, how many seven-elevens are expected?
• Remember, the mean of a probability distribution is also called the “expected value”
Standard Deviation
• What happens to the standard deviation in the seven-eleven experiment as the number of trials, n, increases?