In Examples - Statistics Project Example

of the of the of the Stat 2300 Exam 1. An applicant has applied for positions at Company A and at Company B. The probability of getting an offer from Company A is 0.4, and the probability of getting an offer from Company B is 0.3. Assuming that the two job offers are independent of each other, what is the probability that a. The applicant gets an offer from both companies? The probability of getting the offer from A is 0.4 and that of getting the offer from B is 0.3. Hence, the probability of getting an offer from both the companies = 0.4*0.3 = 0.12. b. The applicant will get atleast one offer? The probability of the applicant receiving at least one offer = offer from A + offer from B + offer from both = p(A)*p(B) + p(A)*p(B) + p(A)*p(B) = 0.4*0.7 + 0.6*0.3 + 0.4*0.3 = 0.58. c. The applicant will not be given an offer from either of the companies? The probability of the applicant not being given an offer from either of the companies = 1-0.58 = 0.42. 2. A press produces parts used in the manufacture of large-screen plasma televisions. If the press is correctly adjusted, it produces parts with a scrap rate of 5%. If it is not adjusted correctly, it produces scrap at 50% rate. From company records, the machine is known to be correctly adjusted 90% of the time. A quality control inspector randomly selects one part from those recently produced by the press and discovers it is defective. What is the probability that the machine is incorrectly adjusted? Let the events be Event E1 : The machine is correctly adjusted Event E2 : The machine is incorrectly adjusted Event A : The machine produces defective parts The respective probabilities are P(E1) = 0.9 P(E2) = 0.1 Probability of machine producing defective parts after correctly adjusted P(A| E1) = 0.05, and Probability of machine producing defective parts after incorrectly adjusted P(A| E2) = 0.5 From Bayes’ theorem The required probability that the machine is incorrectly adjusted after it produced the defective part P(E2|A) =  =  =  =  = 0.5263. 3. Thirty-five percent of the students who enroll in a statistics course go to the statistics laboratory on a regular basis. Past data indicates that 20.5% of students make a grade of B or better. Additionally, 6.5% of students who do not go to the lab on a regular basis make a grade of B or better. Let L be the event that a student attended the lab on a regular basis. Let B be the event that a student earned a grade of B or better. a. Show the joint probability table for events L,Lc,B and Bc. The joint probability table for events L,Lc,B and Bc is B Bc Total L 0.14 0.21 0.35 Lc 0.065 0.585 0.65 Total 0.205 0.795 1 b. If a student is selected at random, what is the probability that the student both attended the lab and earned a grade of B or better? From the above depicted joint probability table, the required probability P(L ^B) = 0.14 c. Given that student regularly attended the lab, what is the probability that he or she earned a grade of B or better? The required probability P(B|L) = P (B ^L) / P(L) = 0.14 / 0.35 = 0.4. d. If a particular student made an A, determine the probability that she or he used the lab on a regular basis. The required probability P(L|B) = P(L^B)/P(B) = 0.14/0.205 = 0.6829 e. Are L and B mutually exclusive events? Explain using probabilities. For mutually exclusive events, the probability of an event occurring after one of them has already taken place is equal to zero. Hence if L, B are mutually exclusive then P(L|B)=P(B|L)=0. Let us calculate P(L|B) from the joint probability table P(L|B) = P(L ∩ B)/P(B) = 0.14 / 0.205 = 0.6829 Since this not equal to zero, the two events can take place in conjunction. Hence L,B are not mutually exclusive events. 4. A poll by the Gallup Organization sponsored by Philadelphia-based CIGNA Integrated Care found that about 40% of the employees have missed work dude to a musculoskeletal (back) injury of some kind. A random sample of 10 workers is to be drawn from a particular manufacturing plant. a. Find the probability that none of the workers have missed work due to back injuries, i.e find P(x=0). Let x be the no of workers who missed work due to back injuries n be the number of workers in sample = 10, Therefore, p the probability of employees missing work due to back injury = 0.4 q = 1-p = 0.6 From the binomial distribution we have P(x=0) =  × 0 × (0.6)10 = 0.0060. b. Find the probability that more than 7 of the workers have missed work due to back injuries, i.e find P(x>7) The required probability P(x>7) = P(x=8) + P(x=9) +P(x=10) =  × 8 × (0.6)2+× 9 × (0.6)1 + × 10 × (0.6)0 = 45 × 0.00065 ×0.36 + 10 × 0.00026×0.6 + 0.00010 = 0.0122. c. Suppose another random sample of 10 workers is selected the following year and it is found that 8 of the workers missed work due to back injuries. Based on your calculation in part C, do you believe the rate of missed work has gone up, or could it just be due to chance? If 8 people missed work in selected random sample of 10, it can be assumed that the rate of missed work has gone up. 5. Assume that the number of network errors experienced in a day in a local area network (LAN) is distributed as a passion random variable. The mean number of network errors experienced in day is 2.4. a. What is the probability that in any given day zero network errors will occur? P(x=0) = e-2.4 × 2.40 / 0! = 0.0907. b. What is the probability that in any given day at least 2 network errors will occur? Required probability P(x>1) = 1- P(x Read More