To accompany Hawkes lesson 5.2 Original content by D.R.S

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The Binomial Distribution For the very common case of “Either-Or” experiments with only two possible outcomes. To accompany Hawkes lesson 5.2 Original content by D.R.S. Recognize Binomial Situations. Only two possible outcomes in each trial. Probability for one of the outcomes. - PowerPoint PPT Presentation

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The Binomial DistributionFor the very common case of

“Either-Or” experiments with only two possible outcomes

To accompany Hawkes lesson 5.2Original content by D.R.S.

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 .

Outcomes Probabilities

One of the events

The other event

Total Exactly 1

The Binomial Probability Formula

• Question: If we have trials, what is the probability of occurrences of the “success” event (the one with probability )

• Answer:

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

• How about 3 successes?

Compute them all

Summary of the 7-11 experiment

X successes P(X successes)

0 times (no sevens or elevens)

1 time

2 times

3 times

4 times

5 times (all sevens and elevens)

Total (must equal 1.000000 !!)

Sometimes you add probabilities

• 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”

Try and verify binomcdf(n,p,x)

X successes P(X successes)Using binompdf

P(0 thru X) successesUsing binomcdf

0 times (no sevens or elevens)1 time

2 times

3 times

4 times

5 times (all sevens and elevens)Total (must equal 1.000000 !!)

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?

trials Standard deviation

5

50

100

Advanced TI-84 Exercise

• Y1=binompdf(20,8/36,X)

• seq(X,X,0,20) STO> L1

• seq(Y1(X),X,0,20) STO> L2

• STAT PLOT for these two lists, histogram• WINDOW

Xmin=0, Xmax=20, Ymin=-0.1,Ymax=0.6• ZOOM 9:ZoomStat

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