8. (a) B = stress = / 0.; A = stran = / 0 Let, E(B A = a) = + a Nos. a b a b a b 90-4 0 90-3 80-8 00 00-4 0 400-3 4000-8 400 3 80-4 30 840-3 7840-8 900 4 40-4 40 640-3 680-8 600 5 50-4 50 600-3 7040-8 500 6 630-4 60 3780-3 3969 0-8 3600 30-4 0 9350-3 9690-8 900 4 30 4 a 35.50, 6 0 b 35 6 a 3 b nab 9350 63535.50 8 a na 9690 635.5 0 = 9. 0 3 k/n 3 4 = b a 35 9.0 35.50 35 3.7. 3 o Young s modulus = = 9. 0 3 k/n 4 8 (b) Assume E(B A = a) = a 6 ( b ) a Now, 6 ( b a )( a ) 0 or, a b 3 3 a 9350 9690 8 9.70 ks
8.4 (a) Plot of per capta energ consumpton vs per capta GNP Per Capta Energ 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) Countr # Per Capta GNP Per Capta Energ 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, (550000 8 475 ) 055985.7, 7 (4340000 8 00 ) 374.86 7 From Eq. 8.4 & 8.3, we also obtan, 99980000 8 475 00 0.79, = 00 0.79475 = 78.75 550000 8 475 (c) From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n 7 99980000 8 475 00 055985.7 374.86 0.85 (d) From Eq. 8.6a, the condtonal varance s, n ' Y X ( ) 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;
Y X (600 475) 0.95 950.37.447608.3 (6.63 835.997) 8 (550000 8 475 ) At X = 5400; Y X (5400 475) 0.95 88..447608.3 (749.407 87.53) 8 (550000 8 475 ) At X = 0300; Y X (0300 475) 0.95 3653.74.447608.3 (556.39 4754.469) 8 (550000 8 475 ) (f) mlarl,. 588, = -709.673 and = 853.080 for predctng the per capta X Y GNP on the bass of the per capta energ 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 = ( ) = 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 = ( ) 0.05 = 0.999963 (b) (c) Theorem of total probablt gves P(Y > 0.3) = P(Y > 0.3 X = 0.35)P(X = 0.35) + P(Y > 0.3 X = 0.4)P(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)
= [ 0 ( 0.056) ] 0.005 = (0.79959595) = 0.785807948 0.4
8.0 (a) Car # Rated Mleage (mpg) Actual Mleage (mpg) X Y X Y XY Y'=a+bX (Y-Y') 0 6 400 56 30 5.54 0.5 5 9 65 36 475 9.9 0.086 3 30 5 900 65 750 3.05 3.807 4 30 900 484 660 3.05.00 5 5 8 65 34 450 9.9.67 6 5 5 44 80.78 0.048 Total: 45 3675 94 835 6.97 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X =45/6 = 4., Y = /6 = 8.7 and correspondng sample varances, (3675 6 4. ) 34.7, (94 6 8.7 ) 0. 67 5 5 From Eq. 8.4 & 8.3, we also obtan, 835 6 4.8.7 0.75, = 8.7 0.754. = 0.5 3675 6 4. (b) From Eq. 8.9, the correlaton coeffcent s, n ˆ n s s n 835 6 4.8.7 0.97 5 34.7 0.67 From Eq. 8.6a, the condtonal varance s,
n ' Y X ( ) n = 6.97 / 4 =.73 and the correspondng condtonal standard devaton s Y X =.36 mpg. (c) Y Q = actual mlage of model Q car Y R = actual mlage of model R car X Q = rated mlage of model Q car = mpg E(Y Q ) = 0.5 + 0.75 = 7.03 mlarl E(Y R ) = 0.5 + 0.754 = 8.54 Hence Y Q = N(7.03,.3); Y R = (8.54,.3) Assume Y Q, Y R to be statstcall ndependent, P(Y Q > Y R ) = P(Y R - Y Q < 0) (8.54 7.03) = ( 0.809) 0..3
(d) To determne the 90% confdence nterval, let us use the followng selected values of X = 5, 0 and 30, and wth t 0.95,4 =.3 from Table A.3, we obtan, At X = 5; Y X (5 4.) 0.90.78.3.36 (9.446 4.4) mpg 6 (3675 6 4. ) At X =0; Y X At X = 30; (0 4.) 0.90 5.54.3.36 (4.066 7.04) mpg 6 (3675 6 4. ) Y X (30 4.) 0.90 3.05.3.36 (.33 6 (3675 6 4. ) 4.769) mpg
8.3 (a) Gven I = 6, E(D) = 0.5 + 56 = 00.5 (50 00.5) Hence P(D>50) = (.65) 0.95 0. 05 30 (b) Gven I = 7, E(D) = 0.5 + 57 = 5.5 Gven I = 8, E(D) = 0.5 + 58 = 30.5 Hence E(D) = E(D6)0.6 + P(D7)0.3 + E(D7)0. = 00.50.6 + 5.50.3 +30.50. = 08 mllon $
8.4 (a) Load Deflecton tons cm X Y X Y XY Y'=a+bX (Y-Y') 8.4 4.8 70.56 3.04 40.3 4.44 0.6 6.7.9 44.89 8.4 9.43 3.43 0.76 3 4.0.0 6 4 8.8 0.037 4 0. 5.5 04.04 30.5 56. 5.5 0.00 Total: 9.3 5. 35.49 65.70 3.85 0.440 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X =9.3/4 = 7.3, Y = 5./4 = 3.8 and correspondng sample varances, (35.49 4 7.3 ) 6.96, (65.70 43.8 ). 65 3 3 From Eq. 8.4 & 8.3, we also obtan, 3.85 47.33.8 0.599, = 3.8 0.5997.3 = -0.59 35.49 47.3 From Eq. 8.6a, the condtonal varance s, n ' Y X ( ) n = 0.44 / = 0.0 and the correspondng condtonal standard devaton s Y X = 0.469 cm.
(b) To determne the 90% confdence nterval, let us use the followng selected values of X = 4 and 0., and wth t 0.95, =.9 from Table A.3, we obtan, At X = 4; Y X At X =0.; (4 7.3) 0.90.8.90.469 (0.597 4 (35.49 47.3 ) 3.07) cm Y X (0. 7.3) 0.90 5.5.90.469 (4.4 4 (35.49 47.3 ) 6.65) cm (c) Under a load of 8 tons, E(Y) = -0.59 + 0.5998 = 4.0 cm Let 7 5 be the 75-percentle reflecton P(Y < 75 ) = 0.75 = 75 4.0 0.469 Hence 75 4.0 0.469 (0.75) 0.675 and 75 4.5cm
8.5 (a) Deformaton Brnell Hardness mm kg/mm X Y X Y XY Y'=a+bX (Y-Y') 6 68 36 464 408 68.65 0.49 65 45 75 6.88 9.70 3 3 59 69 348 767 59.8 0.03 4 44 484 936 968 47.00 9.000 5 8 37 784 369 036 38.88 3.543 6 35 3 5 04 0 9.4 6.699 Total: 5 305 89 6659 504 9.4 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X =5/6 = 9., Y = 305/6 = 50.8 and correspondng sample varances, (89 6 9. ).97, (6659 650.8 ) 30. 97 5 5 From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n 504 69.50.8 0.99 5.97 30.97 (b) From Eq. 8.4 & 8.3, we also obtan,
504 69.50.8.353, = 50.8 +.3539. = 76.765 89 69. From Eq. 8.6a, the condtonal varance s, n ' Y X ( ) n = 9.4 / 4 = 7.353 and the correspondng condtonal standard devaton s =.7 kg/mm. Y X (c) E(YX=0) = 76.765.3530 = 49.7 Hence, P(40<Y 50) = 50 49.7 40 49.7 (0.) ( 3.58) 0.544 0.000 0.544.7.7
8.6 (a) Car # Travel peed mph toppng Dstance ft X Y X Y XY Y'=a+bX (Y-Y') 5 46 65 6 50 4.48.384 5 6 5 36 30 3.9 7.370 3 60 0 3600 00 6600.07.5 4 30 46 900 6 380 5.8 39.437 5 0 6 00 56 60 3.08 8.50 6 45 75 05 565 3375 8.68 44.578 7 5 6 5 56 40.88 47.377 8 40 76 600 5776 3040 7.88 6.993 9 45 90 05 800 4050 8.68 69.77 0 0 3 400 04 640 3.68 0.465 Total: 95 53 55 37405 0665 47.536 On the bass of calculatons n the above table we obtan the respectve sample means of X and Y as, X =95/0 = 9.5, Y = 53/0 = 5.3 and correspondng sample varances, (55 0 9.5 ) 33.6, (37405 05.3 ) 3. 0 9 9 From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n 0665 0 9.55.3 0.99 9 33.6 3.0
(b) We can sa that there s a reasonable lnear relatonshp between the stoppng dstance and the speed of travel. Plot of stoppng dstance vs travel speed 0 00 toppng Dstance, n ft 80 60 40 0 0 0 0 0 30 40 50 60 70 Travel speed, n mph (c) E(Y X) = a + b + c ; let = o E(Y X) = a + b + c 95, ' 55, 49 ( ) 8.5, ( ' ') 86.06 0 4 ( )( ' ') 77387.5, ( )( ) 564 ( ' ')( ) 3589.5
9.5, ' 5.5, 49. 68.5 + c77387.5 = 564 and b77387.5 + c86.060 4 = 35895 564 det 35895 b 8.5 det 77387.5 77387.5 86.060 77387.5 4 86.060 4 3.348590.000 8.5 564 det 77387.5 35895.0644750 c 9.000.000 7 9 9 9.666 0.0053 = 49..6669.5 0.00535.5 = 49. 49.47 6.085 = -6.055 o E(Y X) = -6.055 +.666 + 0.0053 = ( ) = 376.55 376.55 s 53.793 ; s 7. 334 ft Y X Y X 0 (d) At a speed of 50 mph, the epected stoppng dstance s E(Y) = -6.055 +.6650 + 0.05350 = 90. Let 90 be the dstance allowed 90 90. P(Y < 90 ) = ( ) 0. 9 7.334 Hence, 90 90. 7.334 (0.9).8 and 90 7.334.8 90. 99.6 ft