lecture7note - Psych 100A Winter 2009 Lecture 7: Normal...

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1 Lecture 7: Normal Distribution and Statistical Inference Psych 100A Winter 2009 Statistical Inference 1. Statistical inference: techniques 2. Point estimates 3. Interval estimates Procedures •e x a m p l e s 1 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 121 133.5 1 x mmHg 01 z = (x-121)/12.5 (no measured units) 2 a. What percent of the population has blood 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) above the mean ( =121), the area = ½(68%) = 34% by symmetry. Alternatively standardize and use Table A 1 to 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 -1.44 b. What percent of the population has blood pressure between 103-133.5 mmHg? 121 133.5 Blood Pressures x mmHg 103 ) 1 , 44 . 1 ( , ) 5 . 133 , 103 ( 5 . 12 121 5 . 133 5 . 12 121 103 Solution. Standardize, use Table A.1 and symmetry From Table A : % 64 . 76 7664 . 3413 . 0 4251 . 0 ) 1 , 0 ( ) 44 . 1 , 0 ( or ) 1 , 0 ( ) 44 . 1 , 0 ( ) 1 , 0 ( ) 0 , 44 . 1 ( ) 1 , 44 . 1 ( 4 0 2.32 c. What percent have blood pressures greater than 150 mmHg? 121 150 Blood Pressures x mmHg ) 32 . 2 ( ) 150 ( 5 . 12 121 150 Solution. Standardize, use Table A.1 and symmetry From Table A : % 1 ~ 0102 . 4898 . 0 5 . 0 ) 32 . 2 , 0 ( 5 . 0 ) 32 . 2 , 0 ( 5 . 0 ) 32 . 2 , 0 ( ) , 0 ( ) 32 . 2 ( 5 90% z 90 d. Find the 90 percentile for the population of blood pressures. N(0,1) 121 x = + z 0 z 50% 40% 90 , 9 . 0 % 90 z Solution. Work backwards from Table A to z to x. ) , 0 ( 4 . 0 90 From Table A : mmHg x Thus 137 28 . 1 5 . 12 121 , . 28 . 1 90 90 ) , 0 ( ) 0 , ( 4 . 0 5 . 0 90  6
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2 e. Find the median and IQR for this BP population. N(0,1) z 0 z 25 z 75 Solution: By symmetry, median = mean = 121 mmHg. 68 . 0 ) 0 , ( 25 . 0 ) , ( 25 . 0 25 25 25  z A 5 . 112 ) 5 . 12 ( 121 25 25 x 68 . 0 ) , 0 ( 25 . 0 ) , 0 ( 5 . ) , ( 75 . 0 75 75 75 75  5 . 129 ) 5 . 12 ( 121 75 75 IQR = x 75 –x 25 = 129.5 – 112.5 = 17 mmHG.
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lecture7note - Psych 100A Winter 2009 Lecture 7: Normal...

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