STAT 400 Lecture AL1 Homework #1 (due Friday, January 30, by 3:00 p.m.) Spring 2015 Dalpiaz Please include your name ( with your last name underlined ), your NetID, and your discussion section number at the top of the first page. 1 – 4. Do NOT use a computer. You may only use +, –, ×, ÷, and on a calculator. Show all work. 1. Evaluate the following integrals: a) ∫∞− 0 32dxxxe; b) ()∫π 02 sin dxxx; c) ()∫∞+14 1 dxxx. a) ∫∞− 0 32dxxxe = 0 333232 323231 ∞−−−−−−⋅⋅⋅⋅⋅xxxeexex = 272. b) ( )∫π 02 sin dxxx = ( )( )∫π 2 21 0 2sin dxxx = ( )0π 2 2 cos−x = 1. c) ( )∫∞+14 1 dxxx = ( )∫∞−24 1 duuu = ∫∫∞−∞2423 1 1 duuduu = 24181− = 121.2. Let c > 0. Consider f ( x, y ) = x + c y and A = { ( x, y ) : 0 < y < x < 1 }. a) Sketch A. That is, sketch { ( x, y ) : 0 < y < x < 1 }. b) Set up the double integral(s) of f ( x, y ) over A with the outside integral w.r.t. x and the inside integral w.r.t. y. ( )∫ ∫+10 0 dxdyycxx c) Set up the double integral(s) of f ( x, y ) over A with the outside integral w.r.t. y and the inside integral w.r.t. x. ( )∫ ∫+10 1 dydxycxy d) Find the value of c such that ( )∫∫A , dydxyxf = 1. ( )∫ ∫+10 0 dxdyycxx = ∫+10 2 21 dxxc = +21 31c = 631c+ = 1. ⇒ c = 4.3. Consider f ( x, y ) = 3 yx and A = { ( x, y ) : 0 < y < 1, y < x < 2 }. a) Sketch A. That is, sketch { ( x, y ) : 0 < y < 1, y < x < 2 }. b) Set up the double integral(s) of f ( x, y ) over A with the outside integral w.r.t. x and the inside integral w.r.t. y. dxdyyxx 10 0 3 ∫ ∫ + dxdyyx 2110 3 ∫ ∫ c) Set up the double integral(s) of f ( x, y ) over A with the outside integral w.r.t. y and the inside integral w.r.t. x. dydxyxy 102 3 ∫ ∫ d) Use either (b) or (c) to evaluate ( )∫∫A , dydxyxf. dydxyxy 102 3 ∫ ∫ = dyyxyxx 2 10 3 2 21==∫ = dyyy 105 3 212∫− = 01 6 4 12121==−yyyy = 12121− = 125.OR dxdyyxx 10 0 3 ∫ ∫ + dxdyyx 2110 3 ∫ ∫ = dxx 105 4 ∫ + ∫21 4dxx = −+8121241 = 125.4. Evaluate the following sums ( do NOT use a calculator ): a) ∑∞=3 52 kk; b) ∑∞=+112 6.0 kk; c) ∑∞=5 ! 2 kkk. a) ∑∞=3 52 kk = ...525252526543 ++++ Geometric series. Base = 51. ∑∞=3 52 kk = basetermfirst−1 = 511 523 − = 54 1252 = 501 = 0.02. b) ∑∞=+112 6.0 kk = 0.6 3 + 0.6 5 + 0.6 7 + 0.6 9 + … Geometric series. Base = 0.6 2 = 0.36. ∑∞=+112 6.0 kk = basetermfirst−1 = 36.01 6.03 − = 64.0 0.216 = 8027 = 0.3375. c) ∑∞=5 ! 2 kkk = !!!!!! 4 3 2 1 00 42322212022 −−−−−∞∑=kkk = 32342212 −−−−−e = e 2 – 7.5. Suppose P ( A ) = 0.70, P ( B ) = 0.50, P ( A ∪ B' ) = 0.80. a) Find P ( A ∩ B ). b) Find P ( A ∪ B ). c) Find P ( A | B ). P ( A ) = + = 0.70, P ( B ) = + = 0.50, P ( A ∪ B' ) = + + = 0.80. ⇒ = 0.10. P ( B' ) = + = 1 – P ( B ) = 0.50, ⇒ = 0.40. ⇒ = 0.30. ⇒ = 0.20. a) P ( A ∩ B ) = 0.30. b) P ( A ∪ B ) = 0.90. c) P ( A | B ) = 50.030.0 = 0.60.6. Suppose P ( A ) = 0.40, P ( B ) = 0.34, P ( C ) = 0.55, P ( A ∩ B ) = 0.19, P ( A ∩ C ) = 0.25, P ( B ∩ C ) = 0.17, P ( A ∩ B ∩ C ) = 0.07. a) Find P ( A ∪ B ∪ C ). P ( A ∪ B ∪ C ) = 0.75.b) Find P ( ( A ∩ B ) ∪ C ). P ( ( A ∩ B ) ∪ C ) = 0.67. c) Find P ( A ∩ ( B ∪ C ) ). P ( A ∩ ( B ∪ C ) ) = 0.37.7. Let a > 1. Suppose S = { 2, 3, 4, 5, 6, … } and P ( k ) = kac , k = 2, 3, 4, 5, 6, … . a) Find the value of c that makes this is a valid probability distribution. Must have ( )∑xx all P = 1. ∑∞=2 kkac = ...5432 ++++acacacac = basetermfirst−1 = 11 2 aac− = () 1 −aac = 1. ⇒ c = a ( a – 1 ). b) Find P ( outcome is odd ). P ( outcome is odd ) = P ( 3 ) + P ( 5 ) + P ( 7 ) + P ( 9 ) + … = ( )( ) ( ) ( )...11119 7 5 3 +−+−+−+−aaaaaaaaaaaa = basetermfirst−1 = 11 122 aaa−− = 112 −−aa = 11+a. OR P ( outcome is odd ) = P ( 3 ) + P ( 5 ) + P ( 7 ) + P ( 9 ) + … = ( ) ( ) ( ) ( )...11119 7 5 3 +−+−+−+−aaaaaaaaaaaaP ( outcome is even ) = P ( 2 ) + P ( 4 ) + P ( 6 ) + P ( 8 ) + … = ( )( )( )()...11118 6 4 2 +−+−+−+−aaaaaaaaaaaa = a P ( outcome is odd ). P ( outcome is odd ) + P ( outcome is even ) = 1. ⇒ P ( outcome is odd ) + a P ( outcome is odd ) = 1. ⇒ P ( outcome is odd ) = 11+a. c) Find P ( outcome is less than or equal to 5 ). P ( outcome is less than or equal to 5 ) = P ( 2 ) + P ( 3 ) + P ( 4 ) + P ( 5 ) = ( ) ( ) ( ) ()5 4 3 2 1111aaaaaaaaaaaa −+−+−+− = ( )+++− 1 1234aaaaa = 44 1aa− = 4 11a−. OR P ( outcome is less than or equal to 5 ) = 1 – [ P ( 6 ) + P …
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