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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 ...
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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.
 Spring '08
 Staff
 Probability

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