1 / 16

Warm-up Day of Ch. 6 Textbook Review

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 6 Combined 2 1/36 1 10/36 0 25/36

tamika
Download Presentation

Warm-up Day of Ch. 6 Textbook Review

An Image/Link below is provided (as is) to download presentation 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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Warm-upDay of Ch. 6 Textbook Review 1) 2) 3) 4) #5 and Free Response will be the warm-up next block

  2. G is for Googol

  3. Binomial and Geometric Distribution Review 1) # of 6 Combined 2 1/36 1 10/36 0 25/36 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. #4, Labeled as #5 a. 0.3 b. 0.1029 c.

  5. 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 pts 1) Probability Review 2) 6.1 (Day 1) 3) 6.1 (Day 2) 4) 6.2 5) 6.3 5 notes with warm-ups (16 pts each) = 80 pts

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

  7. Answers to Textbook Review

  8. Or 1 – geometcdf (.1, 3) = 0.729

  9. E49. a. 1- binomcdf (20, 0.80, 9) = 0.9994 b. 1 – binomcdf (20, 0.65, 9) = 0.9468 This 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 = 0.725 So combined 1 – binomcdf(40,0.725, 19) = 0.99928

  10. 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.

  11. b. (geomcdf 0.2, 4) = 0.5904 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

  12. 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 7th 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 7th block.

  13. This was a question posted on A.P. Statistics teacher mailing list. Warm-upDay 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 5thperson. 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.

  14. Ch. 6 Practice Test Answers • 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) = 0.08 5)C = -1000(0.13) + 1000(0.24) + 2000(0.35) + 3000 (0.13) Free Response • 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. 0.729 2. a. 24 outcomes

  15. 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

  16. Rules for Transformationsand addition and subtraction of Data

More Related