Pam210_PS2_answer

# Pam210_PS2_answer - PAM 210 Spring 2008 Professor Owens...

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

PAM 210 Spring 2008 Professor Owens Problem Set 2 The Suggested Solution 1) breathing rates X ~ N(12, 2.3) a) P(9.7<=X<=14.3)=P((9.7-12)/2.3<=Z<=(14.3-12)/2.3)=P(- 1<=Z<=1)=0.8413-0.1587=0.6826 (or, based on the empirical rule of a normal distribution, 68% of the observations fall between one standard deviation of the mean). b) P(7.4<=X<=16.6)=P((7.4-12)/2.3<=Z<=(16.6-12)/2.3)=P(- 2<=Z<=2)=0.9772-0.0228=0.9544 (or, based on the empirical rule of a normal distribution, 95% of the observations fall between two standard deviations of the mean). c) P(9.7<=X<=16.6)=P((9.7-12)/2.3<=Z<=(16.6-12)/2.3)=P(- 1<=Z<=2)=0.9772-0.1587=0.8185. d) P(X<=5.1 or X>=18.9)=P(X<=5.1)+P(X>=18.9)=P(Z<=(5.1- 12)/2.3)+P(Z>=(18.9-12)/2.3)=P(Z<=-3)+P(Z>=3)=0.0013*2=0.0026 2) The mean of the amounts of chloroform present in water sources is 34 per liter, and the standard deviation is 53 per liter. Think about the shape of a normal distribution. It is bell-shaped, unimodal, and symmetric density curve. If the chloroform amounts has a normal distribution, its density curve should also be

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.

## This note was uploaded on 02/10/2009 for the course PAM 210 taught by Professor Abdus,s. during the Spring '08 term at Cornell.

### Page1 / 3

Pam210_PS2_answer - PAM 210 Spring 2008 Professor Owens...

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

View Full Document
Ask a homework question - tutors are online