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MATH 2565 (19)
Lecture

# PPS7_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 339 PPS7 (Solution) Problem 17.27: Tennis, anyone? The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0.70. Finally, we are assuming that each serve is independent of the others. Since she is serving 6 times, use Binom(6,0.70). Let X  the number of successful serves in n = 6 first serves. a) P(all six servesin ) P(X 6)   (0.70) (0.30) 0 6  0.118 b) P(exactlyfour servesin)  P(X  4)   (0.70) (0.30) 2 4 0.324 c) P(at least four servesin) P(X  4) P(X  5)  P(X  6)   4 2   5 1   6 0  40.70) (0.30)  5 .70) (0.30)  6 .70) (0.30)  0.744 d) P(no more thanfourservesin)  P(X  0) P(X  1) P(X  2) P(X  3) P(X  4)   (0.70) (0.30) 6 (0.70) (0.30) 5  (0.70) (0.30) 4 0 1 2   (0.70) (0.30) 3  (0.70) (0.30) 2 3 4 0.580 Problem 17.29: And more tennis. The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0.70. Finally, we are assuming that each serve is independent of the others. Since she is serving 80 times, use Binom(80,0.70). Let X  the number of successful serves in n = 80 first serves. a) E(X) np 80(0.70) 56 first serves in. 340 Part 4 Randomness and Probability SD(X)  npq  80(0.70)(0.30)  4.10 first serves in. b) Since np = 56 and nq = 24 are both greater than 10, approximated by the Normal model, N(56, 4.10). c) According to the Normal model, in matches with 80 serves, she is expected to make between 51.9 and 60.1 first serves approximately 68% of the time, between 47.8 and 64.2 first serves approximately 95% of the time, and between 43.7 and 68.3 first serves approximately 99.7% of the time. d) Using Binom(80, 0.70): P(at least 65 firstserves)  P(X  65)  P(X  65) P(X  66)  (0.70) (0.30) 65  0.0161 According to the Binomial model, the probability that she makes at least 65 first serves out of 80 is approximately 0.0161. Using N(56, 4.10): x  z  6556 z 4.10 z 2.195 P(X 65)  P(z 2.195) 0.0141 Problem17.31: Apples. a) A binomial model and a normal model are both appropriate for modeling the number of cider apples that may come from the tree. Let X = the number of cider apples found in the n = 300 apples from the tree. The quality of the apples may be considered Bernoulli trials. There are only two possible outcomes, cider apple or not a cider apple. The probability that an apple must be used for a cider apple is constant, given as p = 0.06. The trials are not independent, since the population of apples is finite, but the apples on the tree are undoubtedly less than 10% of all the apples that the farmer has ever produced, so model with Binom(300, 0.06). E(X) np  300(0.06) 18 cider apples. SD(X)  npq  300(0.06)(0.94)  4.11 cider apples. Since np = 18 and nq = 282 are both greater than 10, approximated by the Normal model, N(18, 4.11). b) Using Binom(300, 0.06): P(at most 12 cider apples)  P(X  12)  P(X  0) 30 0  00.06) (0.94)  0.085 According to the Binomial model, the probability that no more than 12 cider apples come from the tree is approximately 0.085. Using N(18, 4.11): x 
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