Warm-upDay of Ch. 6 Textbook Review

1)2)3)4)

#5 and Free Response will be the warm-up next block

G is for Googol

Binomial and Geometric Distribution Review1) # of 6 Combined 2 1/36 1 10/36

0 25/36a. 10/ 36 b. 13) # of H Prob. 3 1/8 2 3/8 1 3/8 0 1/8Mean: 1.5Standard Deviation: 0.866

#4, Labeled as #5

a. 0.3 b. 0.1029 c. 3.3

Textbook Review

pg 401 E#44 to #49 (Skip E#45) and E#54 (Skip part e.)Please explain the full simulation for 54 a. pg 406 AP#1-4

10 terms and formulas 2pts each for 20 pts1) Probability Review 2) 6.1 (Day 1) 3) 6.1 (Day 2) 4) 6.25) 6.3 5 notes with warm-ups (16 pts each) = 80 pts

Formulas not A.P. Statistics Formula sheetGeometric Distribution

Linear Transformation Combining Data

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Answers to Textbook Review

Or 1 – geometcdf (.1, 3) = 0.729

E49. a. 1- binomcdf (20, 0.80, 9) = 0.9994b. 1 – binomcdf (20, 0.65, 9) = 0.9468This is assuming the region has the same statistics reported for the rest of the country.c. For both girls and boys (0.8 + 0.65)/2 = 0.725So combined 1 – binomcdf(40,0.725, 19) = 0.99928

E54. a. Assumptions: First I have to assume that all 5 athletesare equally likely to be found in each of the cereal boxes. Model: I am assigning each of the following digits to the 5athletes. I am going to pretend that the athlete I want isAthlete 1.

b. (geomcdf 0.2, 4) = 0.5904c. Expected a # is 1/0.2 = 5. It will be on average 5 boxes on average to get the desired poster.d. For the first of one of the two athletes it would be1/(2/5)= 2.5 boxes + 1/.2 for the second box so it willtake 7.5 boxes.AP 1) B 2) B 3) C 4) D

Feedback on H.W.6.1 D7 I gave you specific directions to solve for thedistribution how you would normally would. This meanswriting out the probability distribution and finding the meanand std. dev. The second way was describe a simulation. Youhad to state how you would carry it out. Stating assumptionsand your model, and how you would carry out therepetitions.If you got less than a 90%, you can come in during 7th

block today or tomorrow to make corrections and Iwill give you partial credit back, BUT only if you do them inthis class during 7th block.

Warm-upDay of Ch. 6 Practice Test

4% of people have AB blood. What the probability thatthere is a Type AB donor among the first five people checked?

I understand why you use the geometcdf(0.04,5) tocalculate the probability of finding success on or before the5th person.

But what I don’t understand is why you can’t also usebinompdf (5,0.04,1) to find the probability of one successin five people.

This was a question postedon A.P. Statistics teacher mailing list.

Ch. 6 Practice Test Answers1) B (pg 371 and 372 show the rules)2) E (Law of Larger Numbers)3) C mean 42 (times 1.5 + 7 ) = 70 new mean variance 9 (1.52) then square root for S.D.4)C = 4 (1/6) + 0.5 (1/6) - 1 (4/6) = 0.085)C = -1000(0.13) + 1000(0.24) + 2000(0.35) + 3000 (0.13)Free Response1. a. 0.651 1 – binomcdf(10,.1,0) b. 0.549 1 – binomcdf (100, .1,9) c. 0.04 1 – binomcdf (100,.1, 15) d. 0.7292. a. 24 outcomes

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Remaining Practice Test Answersb. Sample space: {(1, 1) (1, 2) (1, 3) (1,4) (2, 1) … (6, 4)}c. 6/24 = ¼ d.2/24 = 1/12 e. Not disjoint b/c (2,2) is a double and a sum of 4. Not independent P(4 l Doubles) = P(4)

A. P. Statistics: 5 multiple choice questions and 3 free response

Statistics : 5 multiple choice questions and 2 free response

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Rules for Transformationsand addition and subtraction of Data

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gSubtractinandAdding

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222

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