The middle point of each range is used to calculated the sample mean and sample variance as follows:

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Ernest Hood
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1 7.0 (a 50 Number of Observato 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 Σ 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

2 Perform the Ch-square test for lognormal dstrbuton Interval Observed frequency n Theoretcal frequency e ( n e ( n e e Where the lognormal dstrbuton parameter s calculated as follows: λ = ln µ ζ = ln( =.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 µ G = 6.5 sec Var(G =.99 σ G = Σ x0-3

4 7. Rearrange the gven data n ncreasng order, we obtan the followng table: m Observed Ranfall Intensty X, n m/n (a Plot the data ponts n a lognormal probablty paper

5 00.0 y = 50.6e 0.59x Observed Ranfall Intensty, n 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 < > Σ The degree of freedom for the lognormal dstrbuton s f=6--=3. On the bass of the observaton data ˆ µ = ˆ σ = 6.79, thus ˆ ν = 0.83 k ˆ = Perform Ch-square test for Gamma dstrbuton Interval (n Observed frequency n Theoretcal frequency e ( n e ( n e e <

6 > 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.

7 8. (a Plot of stoppng dstance vs speed of travel toppng Dstance, n m peed, n kph (b Vehcle No. peed toppng Dstance (kph (m X Y X Y XY Y'=a+bX (Y-Y'

8 Total: 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 = ( y = = ( = 00.50, From Eq. 8.4 & 8.3, we also obtan, β = = 0.388, α = = 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 = = (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;

9 < µ Y X At X = 30; < µ Y X At X = 60; < µ Y X At X = 5; ( > 0.90 =.33± = (.063 ( m ( > 0.90 = 9.48 ± = ( m ( ( > 0.90 =. ± = (9.7 ( m < µ Y X ( > 0.90 = ± = (43.0 ( m

10 8.4 (a Plot of per capta energy consumpton vs per capta GNP Per Capta Energy Consumpton Y Y' lower upper Per Capta GNP (b Country # Per Capta GNP Per Capta Energy Consumpton X Y X Y XY Y'=a+bX (Y-Y'

11 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, x = ( = , 7 y = ( = From Eq. 8.4 & 8.3, we also obtan, β = = 0.79, α = = (c From Eq. 8.9, the correlaton coeffcent s, ρ = n n = ˆ x s y x s nx y y = = 0.85 (d From Eq. 8.6a, the condtonal varance s, n ' Y X = ( y y 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;

12 < µ Y X ( > 0.95 = ± = ( ( At X = 5400; < µ Y X ( > 0.95 = 88. ± = ( ( At X = 0300; < µ Y X ( > 0.95 = ± = ( ( (f mlarly, β =. 588, α = and = for predctng the per capta X Y GNP on the bass of the per capta energy consumpton.

13 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 = = Φ( = 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 = = Φ( 0.05 = (b (c Theorem of total probablty gves P(Y > 0.3 = P(Y > 0.3 X = 0.35P(X = P(Y > 0.3 X = 0.4P(X = 0.4 = (/ (4/ 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

14 = Φ[ 0 ( ] = Φ( =

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

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 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