hw6-solution

hw6-solution - According to our denition of N , we know N...

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

View Full Document Right Arrow Icon
solution to hw6 1 , X U ( a,b ), that is, its pdf is: f ( x ) = ( 1 b - a , if a 6 x 6 b ; 0, otherwise . Hence, E [ X ] = Z b a x 1 b - a d x = b + a 2 ; and E [ X 2 ] = Z b a x 2 1 b - a d x = b 3 - a 3 3 1 b - a = b 2 + ab + a 2 3 . Therefore, V ar [ X ] = b 2 + ab + a 2 3 - ( b + a 2 ) 2 = ( b - a ) 2 12 . 5.11 , Define, X := to be the length of the left part (i.e, it represents the distance from the starting point to the cutting point.) R := to be the ratio of the short to the long. So we have the can express R in terms of X like following: R = ( X L - X , if X 6 L 2 ; L - X X , if X L 2 . We want to compute P ( R 6 1 4 ) = P ( R 6 1 4 ,X 6 L 2 ) + P ( R 6 1 4 ,X > L 2 ) = P ( X L - X 6 1 4 ,X 6 L 2 ) + P ( X L - X 6 1 4 ,X > L 2 ) = P ( X 6 L 5 ) + P ( X > 4 L 5 ) = Z L 5 0 1 L d x + Z L 4 L 5 1 L d x = 2 5 5.13 , Define X := to be the time you will have to wait for the bus; it is uniformly distributed over [0 , 30]. P ( X > 10) = Z 30 10 1 30 d x = 2 3 ; 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
the question want us to compute conditional probability P ( X > 15 + 10 | X > 15) = P ( X > 25) P ( X > 15) = 5 / 30 15 / 30 = 1 / 3 . 5.16 , Define X := to be the annual rainfall in the region; N := to be the year it takes until a year occurs having a rainfall of over 50 inches. Assumption: the annual rainfall is independent every year. p = P ( X > 50) = P ( Z > 50 - 40 4 ) = P ( Z > 2 . 5) = 0 . 0062 which is the probability that the rainfall is over 50 inches for each year.
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: According to our denition of N , we know N Geo ( p ) P ( N > 10) = X n =11 p (1-p ) n-1 = p * (1-p ) 10 1-(1-p ) = (1-p ) 10 = 0 . 9396101 . 5.19 , It is a problem of nding a z score given a probability. P ( X > c ) = P ( Z > c-12 4 ) = 0 . 1 , and by checking the normal table, we can nd the z score corresponding 0 . 1 is -1.28. So by solving equation c-12 2 =-1 . 28 , we can compute c = 9 . 44. 5.21 , Dene X := the height of a man; we know that X N (71 , 6 . 25). P ( X > 74) = P ( Z > 3 2 . 5 ) = 0 . 1150697; next question is of computing a conditional probability P ( X > 77 | X > 72) = P ( X > 77) P ( X > 72) = . 008197536 . 3445783 = 0 . 02379006 . 2...
View Full Document

This note was uploaded on 06/05/2010 for the course STAT PStat 120a taught by Professor Rohinikumar during the Spring '10 term at UCSB.

Page1 / 2

hw6-solution - According to our denition of N , we know N...

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