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This is the Past Exam of Advanced Quantitative Methods which includes Sample Space, Interviewed Agrees, Response Combination, Furniture Store, Furnish, Models, Technicians, Salaries, Problems etc. Key important points are: Sample Space, Interviewed Agrees, Response Combination, Furniture Store, Furnish, Models, Technicians, Salaries, Problems, Four Digit Number
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Show all steps if possible Marks for each question are included in [ ]on the right, [total marks/80] Give answers to 2 decimal places unless otherwise noted
a. Describe the sample space; that is, list all possible response combinations to the two statements. b. Assuming each response combination in the sample space is equally likely, what is the probability the person being interviewed agrees with at least one of the two policy statements? c. Assuming each response combination in the sample space is equally likely, what is the probability the person being interviewed agrees with exactly one of the two political policy statements? d. Assuming each response combination in the sample space is equally likely, what is the probability the person being interviewed agrees with the two political policy statements?
Smoking Habits Gender Smoker (S) Nonsmoker (N) Total Male (M) 2 24 26
Female (F) 6 8 14
Total^8 32
a. What is the probability the person does not smoke? b. What is the probability the person is female?
c. What is the probability the person is female and does not smoke? d. If the person was female, what is the probability she does not smoke. e. What is the probability the person is female or does not smoke?
a. Let W= x 1 + x 2 be a random variable representing the total time to examine and repair the computer. Calculate the mean , variance and standard deviation for the random variable W b. There is a flat rate of $1.50 per minute to examine the computer, and if no repairs are ordered, there is also an additional $50 service charge. Let L=1.5 x 1 +50. Calculate the mean , variance and standard deviation for the random variable L
a. Find the expected average weight gain in pounds per month for a calf? b. Find the variance of the weight gain. c. What is P( x 5)?
a. How many questions should the student expect to answer correctly? b. What is the standard deviation of the number of questions answered correctly? Give answer to three decimal places
x p( x ) 0 0. 5 0. 10 0. 15 0.
Clerk 1 2 3 4 5 6 Rate in old system 110 100 97 85 117 101 Rate in new system 120 112 115 83 125 109
a. State the null and the alternate hypotheses. b. What is the value of the sample test statistic? c. Find (or estimate) the P -value. d. State your conclusions in the context of the application.
a. State the null and the alternate hypotheses. b. What is the value of the sample test statistic? c. Find (or estimate) the P -value. d. State your conclusions in the context of the application.
Type of fish Percentage stocked Number of fishes sampled after seven years Bass 30% 150 Carp 25%^180 Perch 5% 30 Trout 40% 300
a. State the null and the alternate hypotheses. b. What is the value of the sample test statistic? c. Find (or estimate) the P -value. d. State your conclusions in the context of the application.
Answers:
a) The sample space is S = {AA, AN, AD, NA, NN, ND, DA, DN, DD}. b) 5/9 0.556; c) 4/9 0.444; d) 1/9 0.
a) Number of choices the decorator has = C 2^6 C^82 C 24 =(15)(28)(6)=2520; b) P 220 =
a) C 1012 C 08 ; b) C 1020 ; c) 20
10
8 0
12 10 C
a) P(N) = 32 / 40 = 0.80; b) P(F) = 14 / 40 = 0.35; c)P(F N) = 8 / 40 = 0.20; d) P(N / F) = P(F N) / P(F) = 0.2 / 0.35 = 0.5714 ; e) P ( F N ). 35 . 8 . 2 . 95
a) Define the following events: M : Lily chooses McDonald, B : Lily chooses Burger King, and F : Lily orders French Fries. Then, P( M ) = 0.25, P( B ) = 0.75, P( F / M ) = 0.10, and P( F / B ) =0.8. Therefore, P(M F) = P( M ).P( F / M ) = (0.25)(0.1) = 0.025. b) P( B / F ) = P( B F ) / P( F ) = P( B ). P( F / B ) / P( F ) = (0.75)(0.8)/0.625=0.
c) P( x 5) = P( x = 5) +P(x=10)+P(x=15)= 0.
a) μ = np = 1; b) σ = npq = 0.894; c) P( x ≥ 3) = 0.
a) P(x > 450) = 0.1788; (b) P( (^) x > 450) = 0.
a) P ( x ≥ 49.5) = P ( z ≥ – 2.02) = 0.9783;(b) P (49.5 < x < 65.5) = P (–2.02 < z < 1.97) = 0.
(a) H0: μ = 10.7 oz; H1: μ > 10.7 oz ; (b) t = 1.7166 ; (c) P value is between 0.05 and 0.075.
(d) Do not reject H0. We cannot conclude that the mean weight of all packages mailed this week is greater than 10.7 oz.
(d) Reject H0. Fewer than 55% of all voters favor the project.
(c) P value between 0.005 and 0.010; (d) Reject H0. The mean work rate is higher with the new work- flow system.
(d) Reject H0. The systolic blood pressure for the second group is higher.
(b) ^2 = 18. (c) p-value < 0. (d) Reject H0. The distribution of fish has changed.