Midterm_II_Practice_Problem_Solutions

# Midterm_II_Practice_Problem_Solutions - STA 100: Midterm II...

This preview shows pages 1–3. 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: STA 100: Midterm II Practice Questions For all questions you must show your working. This enables us to understand your thought process, give partial credit and prevent crude cheating. Notes About These Practice Questions: These practice questions are not a practice midterm! i.e., you would not be expected to complete all of these questions in the the time given for the midterm. Solutions to these questions will be posted after you have had sufficient time to practice them. 1. HDL Cholestrol, otherwise known as ‘Good’ Cholestrol, has received much attention in re- cent years. Higher levels of HDL are considered to help reduce the risk of heart disease. HDL Cholestrol levels are measured in miligrams per deciliter of blood (mg/dl). There are variations in HDL levels between men and women. HDL levels in adult men in the general population are normally distributed with mean 52 mg/dl, and standard deviation 8 mg/dl. For women, HDL levels are normally distributed with mean 50 mg/dl and standard deviation 6 mg/dl. HDL levels for men and women are assumed to be independent. (Information from the American Heart Association). (a) Among adult males, an HDL level below 40 mg/dl provides an increased risk of heart disease. What is the proportion of men in the general population will be at increased risk of heart disease? Let X 1 =HDL levels in men in the general population. Then X 1 ∼ N (52 , 8 2 ) . P ( X 1 < 40) = P X 1- 52 8 < 40- 52 8 = P ( Z <- 1 . 5) = 0 . 067 So 6.7% of men are at increased risk of heart disease. (b) Among adult females, an HDL level below 40 mg/dl provides an increased risk of heart disease. What is the proportion of women in the general population will be at increased risk of heart disease? Let X 2 =HDL levels in women in the general population. Then X 2 ∼ N (50 , 6 2 ) . P ( X 2 < 40) = P X 2- 50 6 < 40- 50 6 = P ( Z <- 1 . 67) = 0 . 048 So 4.8% of women are at increased risk of heart disease. (c) Higher HDL levels are considered to be preventive of heart disease. An HDL level of above 60 mg/dl, for both males and females, reduces your risk of heart disease. Among the general population, are there a higher proportion of men or women that have reduced risk of heart disease? Show your working. Lets just do the math: P ( X 1 > 60) = P X 1- 52 8 > 60- 52 8 = P ( Z > 1 . 0) = 1- P ( Z < 1 . 0) = 0 . 159 P ( X 2 > 60) = P X 2- 50 6 > 60- 50 6 = P ( Z > 1 . 67) = 1- P ( Z < 1 . 67) = 0 . 048 So, a higher proportion of men (15.9%) than women (4.8%) have reduced risk of heart disease. (d) You have two friends, Bob and Jane. What is the probability that Bob has a higher level of HDL cholestrol than Jane? We are interested in P ( X 1- X 2 > 0) . Lets call Y = X 1- X 2 . Then, just like in the homework, we compare to the general formula: Y = c 1 X 1 + c 2 X 2 and see that c 1 = 1 , c 2 =- 1 . Since X 1 and X 2 are independent: E [ Y ] = (1) · E [ X 1 ] + (- 1) · E [ X 2 ] = 52- 50 = 2 , Var( Y ) = (1) 2 · Var( X 1 ) + (- 1) 2 · Var(...
View Full Document

## This note was uploaded on 01/17/2012 for the course STAT 100 taught by Professor drake during the Fall '10 term at UC Davis.

### Page1 / 11

Midterm_II_Practice_Problem_Solutions - STA 100: Midterm II...

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

View Full Document
Ask a homework question - tutors are online