DSC330_hw1_solutions

# DSC330_hw1_solutions - DSC 330(Spring 2011 HW1 SOLUTIONS 1...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: DSC 330 (Spring 2011) HW1 SOLUTIONS 1. Suppose a random variable, u1D44C 1 is from the population u1D441 (1 , 9). Calculate the proba- bility u1D443u1D45F ( u1D44C 1 1) and u1D443u1D45F ( u1D44C 1 4). Solutions : Since u1D44C 1 is symmetric about 1, u1D443u1D45F ( u1D44C 1 1) = 0 . 5. u1D707 = 1 and u1D70E = 3, so u1D443u1D45F ( u1D44C 1 4) = u1D443u1D45F ( u1D44C 1 1 3 4 1 3 ) = u1D443u1D45F ( u1D44D 1) = 0 . 5 . 3413 = . 1587 2. Textbook 1.38(a) (This question asks you to calculate the probability u1D443u1D45F ( u1D465 6) ) Solutions : For this problem, u1D707 u1D465 = u1D707 = 4 . 29 and u1D70E u1D465 = u1D70E u1D45B = 2 . 95 50 = . 4172 by the Central Limit Theorem. a. u1D443u1D45F ( u1D465 6) = u1D443u1D45F ( u1D465 4 . 29 . 4172 6 4 . 29 . 4172 ) = u1D443u1D45F ( u1D44D 4 . 10) = 0 . 5 . 5 = 0. Since the probability of observing a sample mean CAHS score of 6 or higher is so small ( u1D45D is essentially 0), we would not expect to see a sample mean of 6 or higher.essentially 0), we would not expect to see a sample mean of 6 or higher....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

DSC330_hw1_solutions - DSC 330(Spring 2011 HW1 SOLUTIONS 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online