stats134-4 - Stat134, Lec 3: Midterm I Solutions Problem 1:...

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

View Full Document Right Arrow Icon
Stat134, Lec 3: Midterm I Solutions Problem 1: (a) Let X be the number of people who prefer Candidate A. P (530 X 570) = 570 X k =530 ± 1000 k ² p k (1 - p ) 1000 - k (b)Using the Normal Approximation Method, μ = 1000 p and σ = p 1000 p (1 - p ). P (530 X 570) Φ( 570 + . 5 - 1000 p p 1000 p (1 - p ) ) - Φ( 530 - . 5 - 1000 p p 1000 p (1 - p ) ) To find the maximum point, take the derivative of the above expression and let it be 0: φ ( 570 + . 5 - 1000 p p 1000 p (1 - p ) ) h 141 p - 570 . 5 2 1000( p (1 - p )) 3 / 2 i - φ ( 530 - . 5 - 1000 p p 1000 p (1 - p ) ) h 59 p - 529 . 5 2 1000( p (1 - p )) 3 / 2 i = 0 exp {- 410 10(0 . 55 - p ) p (1 - p ) } = 59 p - 529 . 5 141 p - 570 . 5 p 0 . 5501 # Although it’s computationally hard, the idea is simple. (c) Since Φ( - 2 , 2) 95%, the largest “reasonable” p is corresponding to the point such that 570+ . 5 - 1000 p 1000 p (1 - p ) = - 2 and the smallest “reasonable” p is corresponding to the point such that
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/11/2009 for the course CEE cee taught by Professor Monteiro during the Fall '05 term at University of California, Berkeley.

Page1 / 2

stats134-4 - Stat134, Lec 3: Midterm I Solutions Problem 1:...

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