# 8.5 - 8.5(a Let X be the car weight in kips X ~ N(3.33 1.04...

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8.5 (a) Let X be the car weight in kips; X ~ N(3.33, 1.04). Hence P(X > 4.5) = P( 04 . 1 33 . 3 5 . 4 > σ µ X ) = P(Z > 1.125) = 1 – Φ (1.125) = 1 - 0.8697 0.130 (b) Let Y be the gasoline mileage. For linear regression, we assume E(Y | X = x) = α + β x and seek the best (in a least squares sense) estimates of α and β in the model. Based on the following data, x i (kips) y i (mpg) x i y i x i 2 i i x y β α ˆ ˆ ' + = (y i – y i ’) 2 2.5 25 6.25 62.5 24.33641 0.440358 4.2 17 17.64 71.4 16.94624 0.002891 3.6 20 12.96 72 19.55453 0.198442 3.0 21 9.00 63 22.16283 1.352165 sum 13.3 83.0 45.9 268.9 1.993856 average 3.325 20.75 2 1 2 1 x n x y x n y x n i i n i i i = = = ± = (45.9 - 4 × 3.325 × 20.75) / (268.9 – 4 × 3.325 2 ) = 1.468 / 6.268 - 4.35, and x y ˆ = ± = 20.75 – 4.347158218 × 3.325 35.20 Also, by Eq. 8.6a, using the results in the preceding table, we estimate the variance of Y (which is assumed constant) = = n i i i x Y y y n S 1 2 ' 2 | ) ( 2 1 = 1.993856 / (4 – 2) 0.997 Hence when the car weighs 2.3 kips, i.e. x = 2.3, Y has the estimated parameters

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8.5 - 8.5(a Let X be the car weight in kips X ~ N(3.33 1.04...

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