MATH 1324 Homework 12 Answers

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MATH 1324 Homework 12 Answers - Page 1 preview image

1MATH 1324 - Finite Math with ApplicationsHomework 12 (Untimed,1Attempt)DUE November 25, 2022 11:59:00 PM!QI. The choicesforproblem number 8 partb,from Section 7.3 inthe bookare given below.*A. 6.9676*B. 2.6396*C. 4.5727*D. 4.8776*E. 8.0278*F. None of the above.*x___________I- 3 - 1 1 3 5P(X = xj|0.170.280.150.270.13*Part B: Findthevariance.*EX = x1p.+x2p2 +x2p2= - 3 : 0 . 1 7-11'0.281 0.15+3 '0.27'+5' 0.131 = 0.82 = p*Vor A’=p. xx —| Z +p2lx,-p2+p2= 0.17(-3 - 0.82)2+ 0.28(-l - 0.82)2+ 0.15(1 - 0.82) + 0.27(3 - 0.82)2+ 0.13(5 - 0.82)2= 6.9676Q2. The choices for problem number10panb,from Section 7.3 in diebookare given belowT.*A.312.4215*B. 284.8275*C. 248.9658*D. 298.9746*E. 315.9872*F. None of the above.*x|015304560P(A' = x)|0.240.460.130.110.06*Part B: Find die variance.*EA" =x1p+x2p2 +x2pt= 0(0.24) + 15(0.46) + 30(0.13) + 45(0.11) + 60(0.06) = 19.35 = pVor A =p.xx—p)2-pj.x2- p'+p;x3“ p ?= 0.24(0 - 19.35)2+ 0.46(15 - 19.35)2+ 0.13(30 - 19.35)2+ 0.11(45 - 19.35)2+ 0.06(60 - 19.35)2= 284.8275

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MATH 1324 Homework 12 Answers - Page 2 preview image

2Q3. The choices for problem number 12 part c, from Section 7 3 in the book are given below.*A. 5.4866'B. 7.4532*C. 4.5727'D. 5.0197*E. 6.8147*F. None of the above.*xI-6-1___08P(X = x)|0.210.250.260.28*Part C: Find die standard deviation.*Standard deviation = o = VVar (X J*E(X) = -6(0.21) + (-1X0.25) + 0(0.26) + 8(0.28) = 0.73 = p*Vai(X) = 0.21(-6 - 0.73)2- 0.25( 1 - 0.73)2+ 0.26(0 - 0.73)2+ 0.28(8 - 0.73)3= 25.1971*o = <25.1971 «5.0197*5.0197Q4. The choices for problem number 20 part c, from Section 7.3 in the book are given below.*A. 0.9897*B. 0.7940*C. 1.0135*D. 0.9236*E. 0.8688*F. None of the above.•*The table below gives the number of days it takes for packages to be delivered from an online retailerand the probability associated with each number of days.*Number of Days|34567Probability|0.010.140.460.250.14*Part C: Find the standard deviation of the random variable.*Standard deviation = o = VVor (x )*E(X) = 3(0.01) + 4(0.14) + 5(0.46) + 6(0.25) + 7(0.14) = 5.37 = p'Var(X) = 0.01(3 - 5.37)2+ 0.14(4 - 5.37)2+ 0.46(5 - 5.37)2+ 0.25(6 - 5.37)2+ 0.14(7 - 5.37)2= 0.8531*cr = <0.8531a0.9236*0.9236

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MATH 1324 Homework 12 Answers - Page 3 preview image

3Qz>. The choices for problem number 26 pan a, from Section 7.3 in the book are given below.*A. 6*B. 0.9722*C. 0.8333*D. 0.5556*E. 0.4444*F. None of the above.A probability distribution has a mean of 12 and a standard deviation of. Use Chebyshev's inequalityto approximate the probability that an outcome of the experiment lies between:p = 12, o = 7Pan A: P(10 < x < 14)solve for kp - ko = small|p - ko = Big12 “k1;=10 |12+kj=14~11/k= -2133k = 6|k = 6111I ’T-1= 1-—= 1 -RJ0.9722k26s0.9722Q6. The choices for problem number 26 pan b, from Section 7.3 in the book are given below.*A. 12*B. 0.9167*C. 0.4375*D. 0.9931*E. 0.5625*F. None of the above.*A probability distribution has a mean of 12 and a standard deviation of. Use Chebyshev's inequalityto approximate the probability that an outcome of the experiment lies between:1*p = 1 2 , o = 3Pan B: P(8 < x < 16)solve for k*p - ko = small|p - ko = Big12— k ’ i - SI12+k 1 1 = 1 6313|l k = 4k =12|k = 12=1-lb= 1--fhRs0.9931144-----*0.9931

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