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York University (33,691)
MATH 2565 (19)
Lecture

# PPS6_Solution_.doc

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School
York University
Department
Mathematics and Statistics
Course
MATH 2565
Professor
Jagmohan Chawla
Semester
Winter

Description
Chapter 17 Probability Models 83 PPS6 (Solution) Problem 17.17: Coins and intuition. a) Intuitively, we expect 50 heads. b) E(heads)= np = 100(0.5) = 50heads. Problem17.19: Lefties. These may be considered Bernoulli trials. There are only two possible outcomes, left-handed and not left-handed. Since people are selected at random, the probability of being left-handed is constant at about 13%. The trials are not independent, since the population is finite, but a sample of 5 people is certainly fewer than 10% of all people. Let X = the number of people checked until the first lefty is discovered. Y = Let the number of lefties among n = 5. a) Use om (0.13). P(firstleftyisthefifthperson )= P(X = 5)= (0.87 Bm (5,0.13) b) Use . P(someleftiesamongthe5people ) = 1− P(noleftiesamongthefirst5people = 1− P(Y = 0)   = 1− 0 ≈ 0.502 c) Use Geom (0.13 ). P(firstleftyis second orthirdperson ) = P(X = 2)+ P(X = 3) = (0.87)(0.13 d) Use Binom (5,0.13). P(exactly3leftiesin thegroup ) = P(Y = 3) = Binom (5,0.13) e) Use . P(atleast3leftiesin thegroup ) = P(Y = 3)+ P(Y = 4) + P(Y = 5)   3 = 3(0.13) (0.87 ≈ 0.0179 f) Use Binom (5,0.13). P(atmost3leftiesin thegroup ) = P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3)   0 = 0(0.13) (0.87   + 2(0.13 ≈ 0.9987 € € € 84 Part 4 Randomness and Probability Problem17.23: Still more lefties. a) In a previous exercise, we determined that the selection of lefties (and also righties) could be considered Bernoulli trials. Since our group consists of 12 people, and now we are considering the righties, use Binom(12 ,0.87). Let Y = the number of righties among n = 12. E(Y) = np = 12(0.87)= 10.44 righties SD(Y)= npq = 12 (0.87)(0.1) ≈ 1.16righties b) P(notallrighties ) = 1− P(allrightie) = 1− P(Y = 12 )   12 0 = 1− 12 (0.87) (0.13) ≈ 0.812 c) P(nomorethan10righties) = P(Y ≤ 10 ) = P(Y = 0)+ P(Y = 1)+ P(Y = 2)+K + P(Y = 10 ) =  (0.87) (0.13) +  (0.87) (0.13) +K +  (0.87) (0.13)2 0 1 10 ≈ 0.475 d) P(exactly sixofeach)= P(Y = 6) =  (0.87) (0.1)6  6 ≈ 0.00193 e) P(majorityrighties) = P(Y ≥ 7) = P(Y = 7)+ P(Y = 8)+ P(Y = 9)+K + P(Y = 12 ) =  (0.87) (0.13) +  (0.87) (0.13) +K +   (0.87) (0.13)0 7 8 12 ≈ 0.998 Problem17.25 : Vision. The vision tests can be considered Bernoulli trials. There are only two possible outcomes, nearsighted or not. The probability of any child being nearsighted is given as p = 0.12. Finally, since the population of children is finite, the trials are not independent. However, 169 is certainly less than 10% of all children, and we will assume that the children in this district are representative of all children in relation to nearsightedness. Use Binom(169,0.12). μ = E(nearsighted)= np = 169(0.12) = 20.28children. Chapter 17 Probability Models 85 σ = SD(nearsighted)= npq = 169(0.12)(0.88) ≈ 4.22children. Problem17.27: Ten
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