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

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Psych 100A Winter 2009 Lecture 7: Normal Distribution and Statistical Inference Statistical Inference 1. Statistical inference: techniques oint stimates 2. Point estimates 3. Interval estimates rocedures Procedures •e x a m p l e s 1
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. Finding areas under the normal curve 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 . lood Blood Pressures 1 121 133.5 x mmHg z = (x-121)/12.5 01 (no measured units) 2
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a. What percent of the population has blood ressure between 121 33 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 From Table A .1: % 13 . 34 3413 . 0 ) 1 , 0 ( 3
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b. What percent of the population has blood pressure between 103-133.5 mmHg? Blood Pressures 121 133.5 x mmHg 103 01 -1.44 Solution. Standardize, use Table A.1 and symmetry   ) 1 , 44 . 1 ( , ) 5 . 133 , 103 ( 5 . 12 121 5 . 133 5 . 12 121 103 A ) 1 , 0 ( ) 0 , 44 . 1 ( ) 1 , 44 . 1 ( rom able A 413 251 4 ) 1 , 0 ( ) 44 . 1 , 0 ( From Table A : % 64 . 76 7664 . 3413 . 0 4251 . 0 ) 1 , 0 ( ) 44 . 1 , 0 ( or 4
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c. What percent have blood pressures greater than 150 mmHg? Blood Pressures 121 150 x mmHg 0 2.32 Solution. Standardize, use Table A.1 and symmetry   ) 32 . 2 ( ) 150 ( 5 . 12 121 150 A ) 32 . 2 , 0 ( ) , 0 ( ) 32 . 2 ( rom able A 898 2 ) 32 . 2 , 0 ( 5 . 0 From Table A : % 1 ~ 0102 . 4898 . 0 5 . 0 ) 32 . 2 , 0 ( 5 . 0 or 5
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d. Find the 90 percentile for the population of blood pressures. N(0,1) 90% z 0 0 z 50% 40% 90 Solution. Work backwards from Table A to z to x . 121 x = + z   90 , 9 . 0 % 90 z A ) , 0 ( ) 0 , ( 4 . 0 5 . 0 0  ) , 0 ( 4 . 0 90 From Table A : 8 90 mmHg x Thus 137 28 . 1 5 . 12 121 , . 28 . 1 90 90 6
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e. Find the median and IQR for this BP population. (0 1) N(0,1) z 0 z 25 z 75 Solution: By symmetry, median = mean = 121 mmHg . ) , ( 25 . 0 5  z A ) , 0 ( 5 . ) , ( 75 . 0 5 5  IQR = x 75 –x 25 . From Table A : 68 . 0 ) 0 , ( 25 . 0 25 25 25 5 . 112 ) 5 . 12 ( 121 25 25 x 68 . 0 ) , 0 ( 25 . 0 75 75 75 75 5 . 129 ) 5 . 12 ( 121 75 75 IQR = x 75 25 = 129.5 – 112.5 = 17 mmHG .
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This note was uploaded on 03/15/2010 for the course PSYCHOLOGY 100B taught by Professor Firstenberg,i. during the Winter '10 term at UCLA.

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

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