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 peoplechecked?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 alsouse binompdf (5,0.04,1) to find the probability of one successin five people.Why does geometcdf (0.04, 5) give a different answer thanbinompdf(5, 0.04, 1)?

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

Answers to 6.3 H.W.

2 rounds, if you round up.

Student of the day!Block 4

Student of the day!Block 5

A.P. Statistics Today

• Finish the Activity or A.P. Practice Questions.• Once you finish start the Practice test.• I will be calling up people to see their Notebooks.• When you finish work on your 3x5 card.• If you do not want to make a 3x5 card then you

have the option of using the A.P. Statistics Exam formula sheet

• I will be walking around in 20 minutes to see your progress on your activity or A.P. Practice Questions

Statistics Directions• Read and complete the quiz carefully.• When you finish turn in your quiz and work on the practice

test.• When EVERYONE is done with the quiz, then you may

discuss the practice test.• You will get to see the practice test answers 15 minutes

before the end of the class.

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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)B = 4 (1/6) + 0.5 (1/6) +0 (4/6) = 0.75, But since you pay $1.00 to play.

5)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) This is very low. The technician should

d. 0.729 1- geomcdf(0.1,3) check more to ensure getting the right amount

2. 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. 4/24 = 1/6 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) f. Not disjoint and independent. P(2T l 5R) = P(2T)Complete the Ch. 6 Summary pg. 402and A.P. Sample test for extra practiceA. P. Statistics: 5 multiple choice questions and 3 free

responseStatistics : 5 multiple choice questions and 2 free response

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