M3 Problem Set

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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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M3: Problem Set Due No due date Points 5 Questions 11 Time Limit None Attempt History Attempt Time Score LATEST Attempt 1 149 minutes 5 out of 5 Score for this quiz: 5 out of 5 Submitted Dec 3 at 7:04pm This attempt took 149 minutes. Question 1 0/0pts According to the National Oceanic and Atmospheric Administration (NOAA), between 1851 and 2013 there were 290 hurricanes that hit the U.S. Coast. Of these, 117 were Category 1 hurricanes, 76 were Category 2 hurricanes, 76 were Category 3 hurricanes, 18 were Category 4 hurricanes, and 3 were Category 5 hurricanes. Make a probability distribution for this data. If a hurricane hits the U.S. coast, what is the probability that the hurricane will be a Category 1 hurricane. Your Answer: Category|Probability 1 117/290 = 0.403 76/290 = 0.262 76/290 = 0.262 18/290 = 0.062 5/290= 0.011 o~ ODN There is a .403 probability that it will be a category 1 hurricane.

Solution: Category Probability 117/290 = 403 76/290 = 262 76/290 = 262 18/290 = .062 3/290 = 011 Q| B QO N — We can see from the probability distribution, that if a hurricane hits the U.S. coast, there is a .403 probability that it is a category 1 hurricane. Question 2 0/0 pts Find the answer to each of the following by first reducing the fractions as much as possible: a) P(17,8)= b) C(19,15)= Your Answer: a) P (n,r)=n!/ (n-r)! P(17, 18) = 171/ (17-8)! = 171/9! = 17(16)(15)(14)(13)(12)(11)(10) = 9 b) C(n, r) = n!/ rl(n-r)! C(19, 15) = 19! / 151(19-15) = 19! / 15141 = 19(18)(17)(916) / 4(3)(2)(1) = 3876

Solution. a) For permutations, we will use: n! P(n, r) - m o oSy 5 = 980,179,200 P(17,8) = ar—e)l_ o 17 (16)(15)(14)(13)(12)(11)(10) = ,179, b) For combinations, we will use: n! i i ey B 19! _ 190 19 (18)(17)(16) _ €(19.15) = 15! (19 —15)! 15!4! 4(3)(2)(1) 3876 Question 3 0/0 pts Suppose you are going to make a password that consists of 5 characters chosen from {2,3,6,7,a,c,k,n,p}. How many different passwords can you make if you cannot use any character more than once in each password? Your Answer: Ordering matters P(n, r) = N!/ (n-r)! 91/ (9 - 5)! = 9! / 41 = 9(8)(7)(6)(5) = 15,120

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Solution. For passwords, order matters. For example, the password 63pn2 is different than the password p2n36 even though both passwords use the exact same characters. Therefore, we want to count the total number of ways that we can choose and order the characters. That means that we should use permutations. We have a total number of nine characters to choose from, so n=9. We will be selecting five characters, so r=5. We will use: n! P(n,r) = =) ! 9! P(9,5) = @35 = Fm = 9(8)(7)(6)(5) = 15,120. Question 4 0/0 pts Suppose that 10 countries submit bids for the summer Olympics. The Olympic committee will select 4 finalists out of these 10 countries. In how many ways can the Olympic committee pick these 10 finalists? Your Answer: Order does not matter C(n, r) = n!'/ ri(n-r)! 10! / 41(10-4)! = 10! / 416! = 10(9)(8)(7) / 4(3)(2)(1) = 210

Solution. Here order does not matter. So, we will use combinations. There are ten countries to choose from, so n=10. Four countries will be chosen, so r=4. We will use Gl rl (n - r)! 10t 10t 10(9)(8)(7) €104) = 41 (10—-4)! 416! 4(3)(2)(1) = Question 5 0/0 pts Suppose A and B are two events with probabilities: P(A) =.40,P(B) =.45,P(AN B) =.25. Find the following: a) P(A U B). b) P(A°). c) P(B°). Your Answer: P(A) = .40, P(B) = .45, P(B) - 25 a) PAlU[B) = P(a) + P(B) - P(AINB) 40+.45-.25=.6 b) P(A)=1-P(Ac) converted to (Ac)=1-P(A) 1-.40=.60

C) P(B)=1-P(Bc) converted to (Bc) =1-P(B) P(Bc)=1-.45-55 Solution. We are given: P(A) = .40, P(B) = .45, P(AnB) =.25. a) For P(AUB). Use P(AUB)=P(A)+P(B)—P(ANB). P(AUB)=40+ 45-.25= 6. b) For P(A°). Use P(A) = 1 — P(A®) which may be rearranged to (4°) =1 - P(4) . P(A®) = 1 - .40 = .60. c) For P(B¢). Use P(B) = 1 — P(B®) which may be rearranged to (B°) = 1 — P(B) . P(B€) =1 - .45 = .55, Question 6 0/0 pts Suppose A and B are two events with probabilities: P(A°) = .40, P(B) = .25, P(AUB) =.75. Find the following: a) P(ANn B). b) P(A). c) P(B®). d) P((A n B)®). Your Answer:

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) P(B)=P(A)+P(B)-P(AB) converted to P(B)=P(A)+P(B)-P(A o P(A)=1-P(Ac) so P(A)=1-.40=.60 P(B)=.60+.25-.75=.1 b) Found above P(A)=.60 C) P(B)=1-P(Bc) converted to (Bc)=1-P(B) P(Bc)=1-25=.75 d) P((Bc)‘c)=1-P(B) 1-.1=.9 Solution. We are given: P(A°) = 40, P(B) = .25, P(AUB) =.75 a)ForP(AnB). Use P(AUB)=P(A)+ P(B)—- P(An B) and rearrange to P(AnB)=P(A) + P(B) — P(A U B). But for this equation, we need P(A4) which we can find by using P(4) = 1 — P(A%). So, P(A) = 1 — .40 = .60. P(AnB)=.60+.25-.75=.1 b) P(A) was found above as .60. c) For P(B¢). Use P(B) = 1 — P(B®) which may be rearranged to (B°) = 1 — P(B) . P(B°) =1-25=.75. d) For P((ANnB)®).Use P((ANB)°)=1-P(ANB). P((ANB)Y)=1-.1=.9.

Question 7 0/0 pts Suppose A and B are two events with probabilities: P(A) = .50, P(B) = .40, P(ANB) = .25. a) What is (A|B) ? b) What is (B|A4) ? Your Answer: ) P(A|B)=P(B) /P(B) = .25/.40 = 625 b) P(B|A)=P(B) IP(A)=.25/.50=.50 Solution. We are given: P(A) = .50, P(B) = .40, P(An B) =.25. a) For P(A|B) use P(A|B) = 5%;2 P(A|B) = f—i o £IE b) For P(Bl4A)use P(B|A) = it P(B|A) = =2 50. P(A) 50 Question 8 0/0 pts Suppose A and B are two events with probabilities:

P(A°) = .30, P(B|4) = .40. Whatis(ANB)? Your Answer: P(A)=1-P( A® )=P(A)=1-.30=.70 P(B|A)=P(AB) IP(A) converted to P(B) =P(BIA)P(A) P(B) =(.40)(.70)=.28 Solution. We are given: P(A%) = .30, P(B]A) = .40. We must first find P(A). By using P(A) = 1 - P(A°). So, P(4) =1-.30=.70. In order to find P(A N B), we can use P(ANB PlA) P(BlA) = ! and rearrange to get P(An B) = P(B|A)P(A). So, P(AnB) =(.40)(.70) = .28. Question 9 0/0 pts Suppose that in a certain type of washing machine that the probability that a belt will fail is .03 and the probability that the agitator will fail is .01. The probability that the belt and the agitator will fail simultaneously is .008. Suppose that you discover that the belt in a washing machine failed, what is the probability that the agitator has also failed? Your Answer:

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A = event agitator will fail, B= event belt fails P(A)=.01, P(B)=.03, P(B) =.008 P(A|B)=P(AB) /P(B) =.008/.03=.267 Solution Suppose that in a certain type of washing machine that the probability that a belt will fail is .03 and the probability that the agitator will fail is .01. The probability that the belt and the agitator will fail simultaneously is .008. Suppose that you discover that the belt in a washing machine failed, what is the probability that the agitator has also failed? Let’s call: B the event that the belt fails. A the event that the agitator will fail. Then P(A) = .01, P(B) = .03, P(A N B) =.008 So, the probability that the agitator has failed, given that the belt has failed is expressed as: P(AnB) _ .008 P(AlB) = P(B) 03 = .267. Question 10 0/0 pts

In a manufacturing plant, three machines A,B, and C produce 40 %, 35 %, and 25 %, respectively, of the total production. The company's quality control department determined that 1 % of the items produced by machine A, 1.5 % of the items produced by machine B, and 2 % of the items produced by machine C are defective. If an item is selected at random and found to be defective, what is the probability that it was produced by machine B? Your Answer: "D" to designate "Defective" A part produced by machine "A" has a defect probability of: P(D|A)=.01 A part produced by machine "B" has a defect probability of: P(D|B)=.015 A part produced by machine "C" has a defect probability of: P(D|C)=.02 We are told the probability that a part was made by machine "A" "B", "C" are: P(A) = 0.40, P(B) = 0.35, P(C) = 0.25 We need to find P(B|D) so we use: P(B|D)=(.35)(.015) / (.40)(.01)+(.35)(.015)+(.25)(.02) = .368

Solution In a manufacturing plant, three machines A, B, and C produce 40 %, 35 %, and 25 %, respectively, of the total parts production. The company's quality control department determined that 1 % of the parts produced by machine A, 1.5 % of the parts produced by machine B, and 2 % of the parts produced by machine C are defective. If a part is selected at random and found to be defective, what is the probability that it was produced by machine B? If we use Def to designate “defective”. We are told that given that a part was produced by machine A, the probability that it has a defect is: P(Def |A) = .01. We are told that given that a part was produced by machine B, the probability that it has a defect is: P(Def |B) =.015. We are told that given that a part was produced by machine C, the probability that it has a defect is: P(Def |C) =.02. Furthermore, we are told that the probability that a part was produced by machine A, B, and C, are respectively: P(A) = .40, P(B) = .35, P (C) = .25. We want to find P ( B |Def), so use: P(B) = P(Def|B) P(B|Def) = (P(A) ™ P(DefIA)) + (P(B) * P(Dele)) + P(C) = P(DefIC). .35+.015 = .368. P(BIDef) = o+ 0T +.35+.015+ 25+.02

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Question 11 5/5 pts As a reminder, the questions in this review quiz are a requirement of the course and the best way to prepare for the module exam. Did you complete all questions in their entirety and show your work? Your Answer: Yes Quiz Score: 5 out of 5