Warm-up Day 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 Review 1) # of 6Combined 2 1/ / /36 a. 10/ 36 b. 1 3) # of H Prob. 3 1/8 2 3/8 1 3/8 0 1/8 Mean: 1.5 Standard Deviation: 0.866

#4, Labeled as #5 a. 0.3b c.

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# terms and formulas 2pts each for 20 pts 1) Probability Review 2) 6.1 (Day 1) 3) 6.1 (Day 2) 4) 6.2 5) notes with warm-ups (16 pts each) = 80 pts

Formulas not A.P. Statistics Formula sheet Geometric Distribution Linear Transformation Combining Data

Answers to Textbook Review

Or 1 – geometcdf (.1, 3) = 0.729

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

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

b. (geomcdf 0.2, 4) = c. 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 be 1/(2/5)= 2.5 boxes + 1/.2 for the second box so it will take 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 the distribution how you would normally would. This means writing out the probability distribution and finding the mean and std. dev. The second way was describe a simulation. You had to state how you would carry it out. Stating assumptions and your model, and how you would carry out the repetitions. If you got less than a 90%, you can come in during 7 th block today or tomorrow to make corrections and I will give you partial credit back, BUT only if you do them in this class during 7 th block.

Warm-up Day of Ch. 6 Practice Test 4% of people have AB blood. What the probability that there is a Type AB donor among the first five people checked? I understand why you use the geometcdf(0.04,5) to calculate the probability of finding success on or before the 5 th person. But what I don’t understand is why you can’t also use binompdf (5,0.04,1) to find the probability of one success in five people. This was a question posted on A.P. Statistics teacher mailing list.

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

Remaining Practice Test Answers b. 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

Rules for Transformations and addition and subtraction of Data