7.0 (a 50 Number of Observato 00 50 00 50 0 3 4 5 6 7 8 9 0 Acceptance Gap G, (sec (b The mddle pont of each range s used to calculated the sample mean and sample varance as follows: No. of G ng observaton( n ( G G n( G G 0 0 7.54 0.000 6 8.046 08.73 3 34 0 0.550 358.683 4 3 58 5.054 667.063 5 79 895.558 78.793 6 8 308 0.06 3.408 7 83 8 0.566 03.487 8 46 68 3.070 448.48 9 69 6 7.574 5.57 0 30 300 4.078 4.35 3 33.58 67.745 0 0 33.086 0.000 Σ 000 648 990.496 G = 6.48 s =.993 G Perform the Ch-square test for normal dstrbuton Interval Observed frequency n Theoretcal frequency e ( n e ( n e e 0.5-.5 0.586 6.688.586.5-.5 6.3 37.368 3.085.5-3.5 34 40.966 48.50.84
3.5-4.5 3 00.063 09.948 0.93 4.5-5.5 79 76.577 5.870 0.033 5.5-6.5 8 5.50 5.6 0.7 6.5-7.5 83 07.45 597.887.88 7.5-8.5 46 38. 6.074 0.449 8.5-9.5 69 66.443 6.537 0.098 9.5-0.5 30 3.088 47.774.069 0.5-.5 3 5.794 7.804.347.5-.5 0.050.0.050 000 5.04 Perform the Ch-square test for lognormal dstrbuton Interval Observed frequency n Theoretcal frequency e ( n e ( n e e 0.5-.5 0 0.000 0.000 0.000.5-.5 6 0.60 9.047 47.58.5-3.5 34.354 35.6 6.067 3.5-4.5 3 9.06 68.59.47 4.5-5.5 79 7.509 353.093 0.343 5.5-6.5 8 4.300 54.883.50 6.5-7.5 83 79.69.434 0.064 7.5-8.5 46 07.48 50.746 4.003 8.5-9.5 69 55.6 78.97 3.8 9.5-0.5 30 6.330 3.47 0.5 0.5-.5 3.745 76.484 6.5.5-.5 0 5.044 5.44 5.044 000 97.009 Where the lognormal dstrbuton parameter s calculated as follows: λ = ln µ ζ = ln(6.48 0.5.993 =.795 σ.993 ζ = ln( + = ln( + = 0.7 µ 6.48 As 5.04<97.009, the normal dstrbuton s more sutable for ths problem. (c Acceptance gap sze (secs G No. of Observatons n G n (G -µ G (G -µ G n / Σn (x0 n -3 0 0 7.56 0 6 0.0 8.065 08.375 3 34 0.04 0.565 359.5 4 3 0.58 5.065 668.50 5 79 0.895.565 79.688
µ G = 6.5 sec Var(G =.99 σ G =.79 6 8.308 0.065 3.65 7 83.8 0.565 0.938 8 46.68 3.065 447.5 9 69 0.6 7.565 5.83 0 30 0.300 4.065 4.875 3 0.033.565 67.688 0 0 33.065 0 Σ 000 6.5 990.50x0-3
7. Rearrange the gven data n ncreasng order, we obtan the followng table: m Observed Ranfall Intensty X, n m/n+ 39.9 0.03 -.83 40.78 0.07 -.50 3 4.3 0.0 -.8 4 4.96 0.3 -. 5 43. 0.7-0.97 6 43.30 0.0-0.84 7 43.93 0.3-0.73 8 44.67 0.7-0.6 9 45.05 0.30-0.5 0 45.93 0.33-0.43 46.77 0.37-0.34 47.38 0.40-0.5 3 48. 0.43-0.7 4 48.6 0.47-0.08 5 49.57 0.50 0.00 6 50.37 0.53 0.08 7 50.5 0.57 0.7 8 5.8 0.60 0.5 9 53.0 0.63 0.34 0 53.9 0.67 0.43 54.49 0.70 0.5 54.9 0.73 0.6 3 55.77 0.77 0.73 4 58.7 0.80 0.84 5 58.83 0.83 0.97 6 59. 0.87. 7 63.5 0.90.8 8 67.59 0.93.50 9 67.7 0.97.83 (a Plot the data ponts n a lognormal probablty paper
00.0 y = 50.6e 0.59x Observed Ranfall Intensty, n 0.0.0-3.00 -.00 -.00 0.00.00.00 3.00 tandard Normal Varate, (b From the data n probablty paper x m = 50.6 λ = ln(x m = 3.9 x 0.84 = 58.8 ζ = ln(x 0.84 /x m = 0.59 Perform Ch-square test for log-normal dstrbuton Interval (n Observed frequency n Theoretcal frequency e (n -e ( n e e <40.8.50 0.590 40-45 7 4.784 4.93.07 45-50 7 7.08 0.000 0.000 50-55 7 6.69 0.37 0.0 55-60 4 4.495 0.45 0.054 >60 3 3.956 0.95 0.3 Σ 9 9.9 The degree of freedom for the lognormal dstrbuton s f=6--=3. On the bass of the observaton data ˆ µ = 50.354 ˆ σ = 6.79, thus ˆ ν = 0.83 k ˆ = 4.445 Perform Ch-square test for Gamma dstrbuton Interval (n Observed frequency n Theoretcal frequency e ( n e ( n e e <40.454.6 0.86 40-45 7 4.947 4.5 0.85
45-50 7 7.73 0.030 0.004 50-55 7 6.74 0.067 0.00 55-60 4 4.4 0.78 0.040 >60 3 3.63 0.069 0.0 9 9.790 The degree of freedom for the Gamma dstrbuton s f=6--=3. ( n For a sgnfcance level α = 5%, c.95,3 = 7.8. Comparng wth the e e calculated, both the Gamma dstrbuton and Lognormal appear to be vald model for the ranfall ntensty ( n e at the sgnfcance level of α = 5%. As the n Gamma dstrbuton s less than e that of the lognormal dstrbuton, the Gamma dstrbuton s superor to the lognormal dstrbuton n ths problem.
8. (a Plot of stoppng dstance vs speed of travel toppng Dstance, n m 50 45 40 35 30 5 0 5 0 5 0 0 0 40 60 80 00 0 40 peed, n kph (b Vehcle No. peed toppng Dstance (kph (m X Y X Y XY Y'=a+bX (Y-Y' 40 5 600 5 600 3.43.46 9 8 4 8.46 0.88 3 00 40 0000 600 4000 36.59.598 4 50 5 500 5 750 7.9 5.5 5 5 4 5 6 60 3.78 0.048 6 65 5 45 65 65 3.08 3.676 7 5 5 65 5 5 7.64 6.97 8 60 5 3600 65 500.5 4.804 9 95 30 905 900 850 34.66.755 0 65 4 45 576 560 3.08 0.84 30 8 900 64 40 9.57.467
5 45 565 05 565 46.5.55 Total: 679 38 563 690 8953 7.76 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X = 679/ = 56.6 kph, Y = 38/ = 9.8 m and correspondng sample varances, x = (563 56.6 y = 89.84 = (690 9.8 = 00.50, From Eq. 8.4 & 8.3, we also obtan, 8953 56.6 9.8 β = = 0.388, α = 9.8 0.388 56.6 = -.6 563 56.6 From Eq. 8.6a, the condtonal varance s, n ' Y X = ( y y n = = 7.76 / ( = 7.7 and the correspondng condtonal standard devaton s Y x =.678 m From Eq. 8.9, the correlaton coeffcent s, ρ = n n = ˆ x s y x s nx y y = 8953 56.6 9.8 = 0.98 89.84 00.50 (c To determne the 90% confdence nterval, let us use the followng selected values of X = 9, 30, 60 and 5, and wth t 0.95,0 =.8 from Table A.3, we obtan, At X = 9;
< µ Y X At X = 30; < µ Y X At X = 60; < µ Y X At X = 5; (9 56.6 > 0.90 =.33±.8.679 + = (.063 (563 56.6 3.73 m (30 56.6 > 0.90 = 9.48 ±.8.679 + = (7.708.5 m (563 56.6 (60 56.6 > 0.90 =. ±.8.679 + = (9.7 (563 56.6.58 m < µ Y X (5 56.6 > 0.90 = 46.34 ±.8.679 + = (43.0 (563 56.6 49.460 m
8.4 (a Plot of per capta energy consumpton vs per capta GNP Per Capta Energy Consumpton 5000 4500 4000 3500 3000 500 000 500 000 500 Y Y' lower upper 0 0 000 4000 6000 8000 0000 000 Per Capta GNP (b Country # Per Capta GNP Per Capta Energy Consumpton X Y X Y XY Y'=a+bX (Y-Y' 600 000 360000 000000 600000 950.37 463.69 700 700 790000 490000 890000 535.64 69886.734 3 900 400 840000 960000 4060000 59.38 3664.478 4 400 000 7640000 4000000 8400000 953.68 45.38 5 300 500 960000 650000 7750000 647. 774.944 6 5400 700 960000 790000 4580000 88. 69643.906 7 8600 500 73960000 650000 500000 379.96 4634.3
8 0300 4000 06090000 6000000 400000 3653.74 9893.03 Total: 37800 6800 550000 4340000 99980000 880.796 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X = 37800/8 = 475, Y = 6800/8 = 00 and correspondng sample varances, x = (550000 8 475 = 055985.7, 7 y = (4340000 8 00 = 374.86 7 From Eq. 8.4 & 8.3, we also obtan, 99980000 8 475 00 β = = 0.79, α = 00 0.79 475 = 78.75 550000 8 475 (c From Eq. 8.9, the correlaton coeffcent s, ρ = n n = ˆ x s y x s nx y y = 7 99980000 8 475 00 055985.7 374.86 = 0.85 (d From Eq. 8.6a, the condtonal varance s, n ' Y X = ( y y n = = 887.55 / 6 = 36980.799 Y X and the correspondng condtonal standard devaton s = 608.3 (e To determne the 95% confdence nterval, let us use the followng selected values of X = 600, 5400 and 0300, and wth t 0.975,6 =.447 from Table A.3, we obtan, At X = 600;
< µ Y X (600 475 > 0.95 = 950.37 ±.447 608.3 + = (6.63 835.997 8 (550000 8 475 At X = 5400; < µ Y X (5400 475 > 0.95 = 88. ±.447 608.3 + = (749.407 87.53 8 (550000 8 475 At X = 0300; < µ Y X (0300 475 > 0.95 = 3653.74 ±.447 608.3 + = (556.39 4754.469 8 (550000 8 475 (f mlarly, β =. 588, α = -709.673 and = 853.080 for predctng the per capta X Y GNP on the bass of the per capta energy consumpton.
8.8 Cost vs floor area 50 00 thousand $ 50 00 50 0 0 3 4 000 sq ft (a The standard devaton of Y s 0. 005 = 0.05. When X = 0.35, the mean of Y s E(Y X = 0.35 =. 0.35 + 0.05 = 0.44, hence P(Y > 0.3 X = 0.35 = P(Y 0.3 X = 0.35 0.3 0.44 = Φ( = 0.9977 0.05 Also, as preparaton for part (b, f X = 0.4 E(Y X = 0.4 =. 0.4 + 0.05 = 0.498 P(Y > 0.3 X = 0.4 = P(Y 0.3 X = 0.4 0.3 0.498 = Φ( 0.05 = 0.999963 (b (c Theorem of total probablty gves P(Y > 0.3 = P(Y > 0.3 X = 0.35P(X = 0.35 + P(Y > 0.3 X = 0.4P(X = 0.4 = 0.9977 (/5 + 0.999963 (4/5 0.9995 Let Y A and Y B be the respectve actual strengths at A and B. nce these are both normal, Y A ~ N(0.44, 0.05; Y B ~ N(0.498, 0.05, Hence the dfference D = Y A Y B ~ N(0.44 0.498, 0.05 + 0.05,.e. D ~ N( 0.056, 005. 0 Hence P(Y A > Y B = P(Y A Y B > 0 = P(D > 0
= Φ[ 0 ( 0.056 ] 0.005 = Φ(0.79959595 = 0.785807948 0.4