Hw4sol - HW4 Solution Section 2.5 8(a 5 5 1= y =0 5 x=0 A(20 − x − 2y)dx dy 5 =A x2 20x − =A − 2xy 2 y =0 5 y =0 x=0(87.5 − 10y)dy =

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: HW4 Solution Section 2.5 8. (a) 5 5 1= y =0 5 x=0 A(20 − x − 2y )dx dy 5 =A x2 20x − =A − 2xy 2 y =0 5 y =0 x=0 (87.5 − 10y )dy = 312.5A dy Since 312.5A = 1, that is A = 1/312.5 = 0.0032. (b) P (1 ≤ X ≤ 2, 2 ≤ Y ≤ 3) = 0.0432. (c) fX (x) = 0.016(15 − x), fY (y ) = 0.008(35 − 4y ), 0≤x≤5 0≤y≤5 (d) Since f (x, y ) = fX (x)fY (y ), X, Y are NOT indepenent. (e) 15 37 7 . EX = , EX 2 = , V arX = 3 2 18 (f) 20 71 13 . EY = , EY 2 = , V arY = 6 3 36 (g) 17 − 2y , 0 ≤ y ≤ 5. fY |X =3 (y ) = 60 (h) 1 EXY = 5, Cov (X, Y ) = − . 18 (i) Corr (X, Y ) = −0.0276 9. (Check solution of Textbook.) Section 2.6 1. (Check solution of Textbook.) 3. (Check solution of Textbook.) 6. µ = 1.12, σ = 0.03 2 2 ¯ ¯ (a) n = 25, E X = µ = 1.12, V ar (X ) = σ = .03 = 3.6 × 10−5 n 25 ¯ (b) To have standard deviation no more than 0.005, that is V ar ( X ) ≤ 0.0052 ⇒ 3 9. (a) A = 4 . σ2 ≤ 0.0052 n ⇒n≥ 0.032 = 36 .0052 1 (b) V = 4 πR3 , find the cdf of V first 3 4 FV (v ) = P (V ≤ v ) = P ( πR3 ≤ v ) 3 1/3 3v =P R≤ 4π = FR Where FR (r ) is the cdf for R, r 3v 4π 1/3 FR (r ) = Therefore, 0 1 3 3 (1 − (y − 1)2 )dy = − r 3 + r 2 4 4 4 3 3v + 16π 4 3v 4π 2/3 FV (v ) = − The pdf of V is fV (v ) = The range of v is: 0 < v < Therefore, the pdf of V is d 3 1 FV (v ) = − + dv 16π 2 3 4π 2/3 v −1/3 32 3 π. fV (v ) = − 1 3 + 16π 2 3 4π 2/3 v −1/3 , 0<v< 32 π. 3 (c) The expected value of V can be computed by two methods: Method 1. 32π/3 32 vfV (v )dv = · · · = EV = π. 15 0 Method 2. 4 EV = E πR3 = 3 2 0 4 32 3 (1 − (r − 1)2 ) πr 3 dr = · · · = π. 4 3 15 14. µ = 77, σ = 9. Variance of Y is 102 , that is b2 92 = 102 ⇒ b = ± 10 . 9 When b = 10 , a = 914.44. 9 When b = − 10 , a = 1085.56. 9 15. (Check solution in Textbook). 2 ...
View Full Document

This note was uploaded on 04/16/2010 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

Page1 / 2

Hw4sol - HW4 Solution Section 2.5 8(a 5 5 1= y =0 5 x=0 A(20 − x − 2y)dx dy 5 =A x2 20x − =A − 2xy 2 y =0 5 y =0 x=0(87.5 − 10y)dy =

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online