M3 Exam

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Portage Learning *

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110

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Statistics

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Jan 9, 2024

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9

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M3: Exam - Requires Respondus LockDown Browser + Webcam Due No due date Points 50 Questions 7 Time Limit 90 Minutes Requires Respondus LockDown Browser Instructions You may only have the following items when taking an exam: computer, 1-2 pieces of blank scratch paper, a pen/pencil, and a calculator. You may ONLY use the equation sheets that are provided WITHIN the exam. The use of printed versions will be considered a violation of the Academic Integrity Policy. Attempt History Attempt Time Score LATEST Attempt 1 76 minutes 44 out of 50 Score for this quiz: 44 out of 50 Submitted Aug 8 at 5:58pm This attempt took 76 minutes. Question 1 5/5 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet = Find the answer to each of the following by first reducing the fractions as much as possible: a) P(490,2)=
b) C(670,665)= Your Answer: . 490! 490! _ P (490,2) = 0 = {5 = 490 (489) = 239,610 a) b) _ 670! _ 670! __ 670(669)(668)(667)(666) ‘0(670’ 665) o 665!(670—665)! ~ 670!5! 5(4)(3)(2)(1) = 1.1( < (D 4 G 4 a. 490! 490! P(490,2) = oo aeal 490(489) = 239,610. b. ! ! 70(669)(668)(667)(666) €(670,665) = 665!(:'77:-665)! - 66675(!)51 = = 62‘24)‘2:‘)3(2):17') == 1.108 x 10** Question 2 515 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https://previous.nursingabc.com/upload/images/Help_file_picture/Statistics Suppose you are going to make a password that consists of 6 characters chosen from {3,4,5,7,d,f,g.k,q,r,s,w}. How many different passwords can
you make if you cannot use any character more than once in each password? Your Answer: n' ! P(n,r) = mier = = =12(11) (10) (9) (8)(7) = 665, 280 P(12,6) = i i 12(11)(10)(9)(8)(7) = 665,280 T (az2-6) 6l T 4 G 2 Question 3 6/9 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https://previous.nursingabc.com/upload/images/Help_file picture/Statistics Suppose A and B are two events with probabilities: P(A)=.35,P(B° )=.45,P(ANB)=.25. Find the following: a) P(AuB). b) P(A®). c) P(B). 4 D 4 Your Answer:
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a.For|P(AUB)|use|P(ANB)=P(A)+P(B)— P(AUB) and rearrange to|P(AU B) = P(A) + P(B) P(AN B) P(A)=1-P(AY|=[P(4)=1— .35 = .65 P(AUB) = .65+ .45 .25 = .85 b. P(A) found above as .65 c. ForrlP(B) =1 P (B°)|rearranged to |(B°) = 1 P (B) P(B°) =1— .45 = .55
a. For P(AUB). Use P(AuB)=P(A)+P(B)-P(ANB) . But for this equation, we need P(B) which we can find by using P(B)=1- P(B®). So, P(B)=1-.45=.55. P(AUB)=.35+.55-.25=.65 b. For P(A®). Use P(A)=1-P(A® ) which may be rearranged to (A° )=1-P(A) . P(AS )=1-.35=.65. C. For P(B). Use (B)=1-P(B®) . P(B)=1-.45=.55. Question 4 3/6 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet = Suppose A and B are two events with probabilities: P(A° )=.45,P(B)=.45,P(ANB)=.20. a) What is (A| B) ?
b) What is (B| A) ? Your Answer: P(A° )=.45 P(B)=.45,P(ANB)=.20. P(ANB) a) PAIB)= —53 ‘P(A B) =2 = .444‘ P(ANB b;|P(B|A) ;iAi )| [P(B|A) = 2 = 444] a. For P(A|B) use P(ANB) 20 P(AIB) = “50 P(A|B) = 22 = 444, b. For P(Bl4) use P(B|A) =" ;‘22;’) P(B|A) = %g 3636. Question 5 10/10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https://previous.nursingabc.com/upload/images/Help_file picture/Statistics
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In a tri-state conference, 20% attendees are from California, 45% from Oregon, and 35% from Washington. As it turns out 15 % of the attendees from California, 5 % of the attendees from Oregon, and 19% of the attendees from Washington came to the conference by train. If an attendee is selected at random and found to have arrived by train, what is the probability that the person is from Oregon? Your Answer: | P(O)-P(TRAIN|O)) ‘P (O|TRAIN ))H .20+.15+.45+.05+.35+.19 189 - 4 P(Train| C)=.15.. P(Train| 0)=.05.. P(Train | W)=.19.. P(C)=.20,P(0)=.45,P(W)=.35. We want to find P(O | Train), so use: P(0) » P(Train|0) PLOITrain) = ey < PCtramiC)) + (P(0) + P(Trainl0)) + P(W) « P(TrainlW) P(0|Train) = 2 = R S 200 05+ 459.05+.350.19 Question 6 10/10 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet & (https://previous.nursingabc.com/upload/images/Help_file picture/Statistics The probability that a certain type of battery in a smoke alarm will last 2 years or more is .85. The probability that a battery will last 5 years or more
is .15. Suppose that the battery is 2 years old and is still working, what is the probability that the battery will last at least 5 years? Your Answer: E=2 years or more F= 5 years or more \P(ENF) = .15 ‘P do-_ = > ~ P(E) .8 Define E to be the event that the battery will last 2 years or more. Define F to be the event that the battery will last 5 years or more. We are given that: P(E)=.85. P(F)=.15. Note also that if a battery lasts more than 5 years, it would have had to have lasted 2 years, so P(ENF)=.15. The question is asking “given event E has occurred, what is the probability that event F will occur”? This may be found by using: __P(EnF) 15 _ P(F|E) = P .1765. Question 7 913 pts You may find the following files helpful throughout the exam: Statistics_Equation_Sheet &
Suppose that 7 out of 17 people are to be chosen to go on a mission trip. In how many ways can these 7 be chosen if the order in which they are chosen is not important. Your Answer: Since we do not want to count all of the possible orderings, we use combinations. 170 170 (17)(16)(15)(14)(13)(12)(11) 70Q17-7 " 71100 7! C(7.17) = = 19448. Quiz Score: 44 out of 50
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