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516hw7

# 516hw7 - Solution to Homework 7 Section 4.1 1 Treat the...

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Solution to Homework 7 Section 4.1 1. Treat the density function as constant over those small intervals. a) (0.001)( 1 2 π )=0.000399 b) (0.001)( 1 2 π )( 0.5 e - )=0.00025 2. a) 4 1 1 cx dx c - = c = 3 b) mean = 3/2 c) variance = ¾ 3. a) 1 0 (1 ) 1 cx x dx - = c = 6 b) By symmetry, ( 1/ 2) ( 1/ 2) 1/ 2 P X P X d = = c) 1/3 0 ( 1/3) 6 (1 ) P X x x dx c = - c =7/27 d) 1/2-7/27=13/54 e) Mean = ½ Variance =1/20 5. a)

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b) 2 2 1 1 1 ( 1 2) 7 /12 2 (1 | |) P X dx x - - < < = = + c) 2 1 1 (| | 1) 2 ( 1) 1/ 2 (1 ) P X P X dx x c = = = + d) Expectation doesn’t exist because | | ( ) x f x dx c - = . 6. a) ( 1) 2/3 ( 1) 1/3 P X P X == = , also ( 0) 1/3 P X c = , by symmetry we have 0.5 μ = . Since 0 0.5 1 ( 0) ( ) ( ) 1/3 2 P X P Z P Z σ σ - = = - = , we have: From table 1 0.4303 2 σ - = - , so 1.162 σ = b) 1 1 ( 1) 3/ 4 ( ) 0.6742 P X P Z μ μ σ σ - - = = = , ( 0) 1/3 ( ) 0.4303 P X P Z μ μ σ σ - - = = = - so we have 0.3896, 0.904 μ σ = = 8. a) 11.8 12 12.2 12 2 2 (11.8 12.2) ( ) ( ) 1.1 1.1 11 11 P X P Z P Z - - < < = < < = - < < =0.1443 b) 11.8 12 12.2 12 0.2 0.2 (11.8 12.2) ( ) ( ) 11 11 1.1/ 100 1.1/ 100 P X P Z P Z - - < < = < < = - < < =0.9307. The measurements need not follow normal distribution individually; as the CLT says their sum/average would be approximately normal as long as they are ‘iid’ [independent and identically distributed] with finite variance.
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