{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture7 - Psych 100A Winter 2009 Lecture 7 Normal...

This preview shows pages 1–9. Sign up to view the full content.

L t 7 N l Di t ib ti d Psych 100A Winter 2009 Lecture 7: Normal Distribution and Statistical Inference Statistical Inference 1. Statistical inference: techniques 2 Point estimates 2. 3. Interval estimates Procedures examples 1

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

View Full Document
4. Finding areas under the normal curve Example: Suppose a population of blood pressures is normally distributed with mean =121 mmHg and SD =12.5 mmHg . Blood Pressures 1 H 121 133.5 x mmHg z = (x-121)/12.5 (n m d nit ) 0 1 (no measured units) 2
a. What percent of the population has blood pressure between 121 133 5 mmHg? pressure between 121-133.5 mmHg? Solution. The area 1 SD above and below the mean = 68% Since 133 5 mmHg is 1 SD ( =12 5 ) = 68%. Since 133.5 mmHg is 1 SD ( =12.5 above the mean ( =121 ), the area = ½(68%) = 34% by symmetry. Alternatively , standardize and use Table A.1 to find the area (A) ) 1 , 0 ( , ) 5 . 133 , 121 ( 5 . 12 121 5 . 133 5 . 12 121 121 A A A From Table A .1: % 13 . 34 3413 . 0 ) 1 , 0 ( A 3

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

View Full Document