ksi a b 8.1 (a) B = stress = y / 0.1; A = strain = x / 10 Let, E(B A = a) = + a a i Nos. a i b i a i b i

Transcription

1 8. (a) B = stress = / 0.; A = stran = / 0 Let, E(B A = a) = + a Nos. a b a b a b a 35.50, 6 0 b 35 6 a 3 b nab a na = k/n 3 4 = b a o Young s modulus = = 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 ks

2 8.4 (a) Plot of per capta energ consumpton vs per capta GNP Per Capta Energ Consumpton Y Y' lower upper Per Capta GNP (b) Countr # Per Capta GNP Per Capta Energ Consumpton X Y X Y XY Y'=a+bX (Y-Y')

3 Total: 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, ( ) , 7 ( ) From Eq. 8.4 & 8.3, we also obtan, , = = (c) From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n (d) From Eq. 8.6a, the condtonal varance s, n ' Y X ( ) n = / 6 = Y X and the correspondng condtonal standard devaton s = (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;

4 Y X ( ) ( ) 8 ( ) At X = 5400; Y X ( ) ( ) 8 ( ) At X = 0300; Y X ( ) ( ) 8 ( ) (f) mlarl,. 588, = and = for predctng the per capta X Y GNP on the bass of the per capta energ consumpton.

5 8.8 Cost vs floor area thousand $ sq ft (a) The standard devaton of Y s = When X = 0.35, the mean of Y s E(Y X = 0.35) = = 0.44, hence P(Y > 0.3 X = 0.35) = P(Y 0.3 X = 0.35) = ( ) = Also, as preparaton for part (b), f X = 0.4 E(Y X = 0.4) = = P(Y > 0.3 X = 0.4) = P(Y 0.3 X = 0.4) = ( ) 0.05 = (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) = (/5) (4/5) 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( , ),.e. D ~ N( 0.056, ) Hence P(Y A > Y B ) = P(Y A Y B > 0) = P(D > 0)

6 = [ 0 ( 0.056) ] = ( ) =

7 8.0 (a) Car # Rated Mleage (mpg) Actual Mleage (mpg) X Y X Y XY Y'=a+bX (Y-Y') Total: 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, ( ) 34.7, ( ) From Eq. 8.4 & 8.3, we also obtan, , = = (b) From Eq. 8.9, the correlaton coeffcent s, n ˆ n s s n From Eq. 8.6a, the condtonal varance s,

8 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 ) = = 7.03 mlarl E(Y R ) = = 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) ( ) = ( 0.809) 0..3

9 (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.) ( ) mpg 6 ( ) At X =0; Y X At X = 30; (0 4.) ( ) mpg 6 ( ) Y X (30 4.) (.33 6 ( ) 4.769) mpg

10 8.3 (a) Gven I = 6, E(D) = = 00.5 ( ) Hence P(D>50) = (.65) (b) Gven I = 7, E(D) = = 5.5 Gven I = 8, E(D) = = 30.5 Hence E(D) = E(D6)0.6 + P(D7)0.3 + E(D7)0. = = 08 mllon $

11 8.4 (a) Load Deflecton tons cm X Y X Y XY Y'=a+bX (Y-Y') Total: 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, ( ) 6.96, ( ) From Eq. 8.4 & 8.3, we also obtan, , = = 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 = cm.

12 (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) ( ( ) 3.07) cm Y X (0. 7.3) (4.4 4 ( ) 6.65) cm (c) Under a load of 8 tons, E(Y) = = 4.0 cm Let 7 5 be the 75-percentle reflecton P(Y < 75 ) = 0.75 = Hence (0.75) and cm

13 8.5 (a) Deformaton Brnell Hardness mm kg/mm X Y X Y XY Y'=a+bX (Y-Y') Total: 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, ( ).97, ( ) From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n (b) From Eq. 8.4 & 8.3, we also obtan,

14 , = = From Eq. 8.6a, the condtonal varance s, n ' Y X ( ) n = 9.4 / 4 = and the correspondng condtonal standard devaton s =.7 kg/mm. Y X (c) E(YX=0) = = 49.7 Hence, P(40<Y 50) = (0.) ( 3.58)

15 8.6 (a) Car # Travel peed mph toppng Dstance ft X Y X Y XY Y'=a+bX (Y-Y') Total: 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, ( ) 33.6, ( ) From Eq. 8.9, the correlaton coeffcent s, n n ˆ s s n

16 (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 Travel speed, n mph (c) E(Y X) = a + b + c ; let = o E(Y X) = a + b + c 95, ' 55, 49 ( ) 8.5, ( ' ') ( )( ' ') , ( )( ) 564 ( ' ')( )

17 9.5, ' 5.5, c = 564 and b c = det b 8.5 det det c = = = o E(Y X) = = ( ) = s ; s ft Y X Y X 0 (d) At a speed of 50 mph, the epected stoppng dstance s E(Y) = = 90. Let 90 be the dstance allowed P(Y < 90 ) = ( ) Hence, (0.9).8 and ft