Asst7-solved

Asst7-solved - Statistics 427: Homework #7 Solutions to...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Statistics 427: Homework #7 Solutions to Assigned Problems 4.11 (a) The cdf is F ( x ) = x < x 2 4 x < 2 1 2 x To find P ( X 1) we compute F (1) = 1 / 4 = 0 . 25. (b) P (0 . 5 X 1) = F (1)- F (0 . 5) = 1 / 4- 1 / 16 = 3 / 16 = 0 . 1875 . (c) P ( X > . 5) = 1- P ( X . 5) = 1- 1 / 16 = 15 / 16 = 0 . 9375 . (d) The median is the solution to 0 . 5 = 2 / 4 which is = 2. (e) The density function is f ( x ) = d dx F ( x ) = d dx x 2 4 = x 2 . To be complete, the pdf is therefore f ( x ) = x 2 x < 2 o.w. (f) E [ X ] = Z 2 wf ( w ) dw = Z 2 w w 2 dw = Z 2 w 2 2 dw = w 3 6 2 = 8 6 = 4 3 . 1 (g) To compute the variance of X we will first compute E X 2 : E X 2 = Z 2 w 2 f ( w ) dw = Z 2 w 3 2 dw = w 4 8 2 = 16 8 = 2 . The variance is therefore Var [ X ] = E X 2- E [ X ] 2 = 2- 16 9 = 2 9 . The standard deviation is = p Var [ X ] = 2 / 3 . (h) This was done above: E X 2 = $2. 4.13 (a) The pdf is f ( x ) = k x 4 x > 1 x 1 ....
View Full Document

Page1 / 5

Asst7-solved - Statistics 427: Homework #7 Solutions to...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online