Problem 7.54: Fair Coin? P = 0.5 binomialcdf(25,0.5,7) 1 - binomialcdf(25,0.5,17) P(x 18) = 0.043

Problem 7.54: Fair Coin? P = 0.5

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bability

Distribution PlotBinomial, n=25, p=0.5

binomialcd

f(25,0.5,7)

1 - binomialcdf(25,0.5,17)

P(x < 7 or x > 18) = 0.043

Problem 7.54: Fair Coin? P = 0.9

26252423222120191817161514131211109876543210

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bability

Distribution PlotBinomial, n=25, p=0.9

binomialcdf(2

5,0.5,7)

1 - binomialcdf(25,0.5,17)

P(x < 7 or x > 18) = 0.998

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bability

Distribution PlotBinomial, n=25, p=0.1

Problem 7.54: Fair Coin? P = 0.1

binomialcdf(2

5,0.5,7)

1 - binomialcdf(25,0.5,17)

P(x < 7 or x > 18) = 0.998

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bability

Distribution PlotBinomial, n=25, p=0.6

Problem 7.54: Fair Coin? P = 0.6

binomialcdf(2

5,0.5,7) 1 - binomialcdf(25,0.5,17)

P(x < 7 or x > 18) = 0.155

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bability

Distribution PlotBinomial, n=25, p=0.4

Problem 7.54: Fair Coin? P = 0.4

binomialcdf(2

5,0.5,7)

1 - binomialcdf(25,0.5,17)

P(x < 7 or x > 18) = 0.155

Error: Type 1 or Type 11

• What do you think of our decision rule?– If not bias, wrong 4% of the time.– If bias at 60% or 40%, then we are wrong 85% of the

time.– If bias at 90% or 10%, then we are wrong less than

1% of the time.

– Moral of the story, you can’t have it both ways (can’t have your cake and eat it too) and often times we have to decide what we will accept and work with that.

Problem 7.56: Exit Polling

• A. 1 - binomialcdf(25, .9, 20) = 0.902• B. 1 - binomialcdf(25, .9, 20) = 0.902• C. μ = (25)(0.9) = 22.5, σ = =1.5• D. If fewer than 20 favor the ban (0.098), is this

at odds with the assertion that (at least) 90% of the populace favors the ban?

)9.1)(9)(.25(

The Geometric Distribution

• Suppose that a sequence of trials we are interested in the number of the trials on which the first success occurs.

• Geometric distribution

,3,2,1)1();( 1 xforpppxg x

Mean and variance

• Mean of geometric distribution

• Variance of geometric distribution

p

1

22

1 p

p

Waiting for Reggie JacksonThe Geometric Distribution

Scenario

Children’s cereals sometimes contain small prizes. For example, not too long ago, boxes of Kellogg’s Frosted Flakes contained one of three posters: Ken Griffey, Jr., Nolan Ryan, or Reggie Jackson. A young boy wanted a Reggie Jackson poster and had to buy eight boxes until getting his poster.

Questions

Should this young boy consider himself especially unlucky?

On average, how many boxes would a person have to buy to get the Reggie Jackson poster?

What assumptions would you have to make to answer this question?

Objective

In this activity you will become familiar with the GEOMETRIC DISTRIBUTION, or WAIT TIME DISTRIBUTION, including the shape of the distribution and how to find its mean.

Activity

SIMULATION:

1 = Reggie Jackson

2 = Ken Griffey, Jr.

3 = Nolan Ryan

Count the number of times you enter until you get a “1”.

Repeat several times.

Using your calculator, simulate finding a poster by entering:

randInt(1,3) enter

Exercises

Make a histogram of the number of boxes everyone in the class requires to get their first Reggie Jackson Poster.

Describe its shape and distribution. What was the average number of “boxes”

purchased to get a Reggie Jackson poster?

Exercises

Estimate the chance that the boy would have to buy eight or more boxes to get his poster.

What assumptions are made in the simulation about the distribution of the prizes?

Do you think the assumptions are reasonable ones?