pset_03_solutions_2012

pset_03_solutions_2012 - Statistics 100 — Spring 2012 —...

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Unformatted text preview: Statistics 100 — Spring, 2012 — Problem Set 3 Solutions 1. a) We could define ”most people” to be the middle 95% of people. Then an interval containing most people’s body temperature would be ( t low ,t high ) where P ( t low < T < t high ) = 0 . 95 By symmetry P ( T < t high ) = 0 . 975 Converting to the standard Normal scale this becomes P Z < t high- 98 . 2 . 7 = 0 . 975 Looking in the tables we can then deduce t high- 98 . 2 . 7 = 1 . 96 Thus t high = 99 . 57 and by symmetry t low = 98 . 2- (99 . 57- 98 . 2) = 96 . 83 So the interval is (96.83, 99.57). If we had defined ”most people” to be the middle 90% then the interval would have been (97.05, 99.35) and if we had defined it to be the middle 68% then it would have been (97.5, 98.9). b) 1 P ( T > 98 . 6) = P Z > 98 . 6- 98 . 2 . 7 = 1- P ( Z < . 571) = 0 . 28 So about 7/25 (or 28%) of people have a body temperature higher than 98.6F. c) P ( T < t cool ) = 0 . 2 Converting to the standard Normal scale we have P Z < t cool- 98 . 2 . 7 = 0 . 2 From the tables we can deduce that t cool- 98 . 2 . 7 =- . 8416 Thus t cool = 97 . 61 2 So the coolest 20% of people have a body temperature of below 97.61F. 2. a) The correlation between the pairs of observations ( x 1 ,y 1 ) = (2 , 10) , ( x 2 ,y 2 ) = (4 , 9) , ( x 3 ,y 3 ) = (1 , 0) is given by r = 1 n- 1 n X i =1 ( x i- ¯ x ) s x ( y i- ¯ y ) s y First we compute some summary statistics: ¯ x = 1 3 (2 + 4 + 1) = 7 3 = 2 . 33 ¯ y = 1 3 (10 + 9 + 0) = 19 3 = 6 . 33 s x = r 1 2 { (2- 2 . 33) 2 + (4- 2 . 33) 2 + (1- 2 . 33) 2 } = 1 . 528 s y = r 1 2 { (10...
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pset_03_solutions_2012 - Statistics 100 — Spring 2012 —...

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