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Introduction to Mathematical Statistics

by: Dr. Filomena Hegmann

Introduction to Mathematical Statistics APPM 5520

Marketplace > University of Colorado at Boulder > Applied Math > APPM 5520 > Introduction to Mathematical Statistics
Dr. Filomena Hegmann

GPA 3.76


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This 2 page Study Guide was uploaded by Dr. Filomena Hegmann on Thursday October 29, 2015. The Study Guide belongs to APPM 5520 at University of Colorado at Boulder taught by Staff in Fall. Since its upload, it has received 40 views. For similar materials see /class/231873/appm-5520-university-of-colorado-at-boulder in Applied Math at University of Colorado at Boulder.


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Date Created: 10/29/15
APPM 45520 Solutions for Review Problems for Exam II Sections 7374 H The denominator factors into w 7 3w 7 4 A partial fractions decomposition gives 2w1 7 77 9 wzi7w12iwi3 will So 2w1 9 7 d 7d 7 d 91 74771 73 C w277w12 7 w74 7 w73 7 nlw I nlw H 7 This can also be written as In 0 2 32 9 is not reducible so a partial fractions decomposition is y481 i ByC DyE 332 9V T 3 32 9 32 92 Solving this gives A 1 B 0 C 0 D 718 and E 0 So 34 81 1 183 WP 9V T 32 92 Integrating the rst one gives ln The second one is a simple u substitution u 32 9 The nal answer is 4 3 81 9 d ln C y32 9V y lyl 32 9 This is an improper fractioni the degree of the numerator is larger than the degree of the denominator Before we do anything we must perform the lonf division 9 2 4 2 W 23 2 W That last term can be rewritten as 2 partirac 7 y21gty1 371 y21 So 234 d721d 1 d y d 1 d y3iy2y71 37 y y 3713 y21y y21y 1 y22y1ny1 1Hy2llitan 1ygt0 4 Let x sin 6 Then dw cos Ode and 1 7 32 1 7 sin2 6 cos2 6 and the integral becomes 01 cos 63 sin6 6 cos2 32 4 6 cos d d sin 6 6 sin cos theta d6 cot4 6 csc2 GdO Since the derivative of cotangent is negative cosecant squared7 let u cot 6 then du 7 csc2 6 and the integral becomes 5 1 7 2 7u4du71u507 cot5 C7 0 5 w This is a separable equation dw 562474 1 d 33 We could have left the 3 on the right side lntegrating both sides 1d 7 div 3 y 702 4 1 1 71 31 7 itan C gives 3 y i tan 1w2 C different C Now we use the intital condition ya gtan 122 C it 0 3 7r 37r 77 C 0 C 77 gt 2 4 gt 8 So7 the nal solution is 3 3 y ita1r1gc2 7


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