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Unformatted text preview: 2 /2 problem: The distribution with 4 degrees of freedom is given by 0 ( ) 1 ( ) 4 Find the probability that the variance of a random sampl x x f x xe x χ ≤ = 2 2 2 2 2 e of size 5 from a normal population with 12 will exceed 180. solution: ( 180) [( 1) / 4(180)/144] ( 5) 1 ( 5) P S P n S P P σ σ χ χ = =  = =  ≤ 5 /2 1 1 4 0.2873 x xe dx =  = ∫ 2 2 1 lemma 1: If the random variables ( 1,2,..., ) are normally and independently distributed with means and variance , then follows distribution with degrees of freedom. Pr r k r r r r r r X r k X k μ μ σ χ σ = =  ∑ 2 1 2 2 1 2 1 2 2 1 1 [ oof: Let , then has standard normal distribution. Let then, the m.g.f of U is given by ( ) ( ) [ ] [ ] [ k r r r r k r r r r r r r k k r r r r r r X t tu u t Z t Z Z X Z Z X U Z mg t E e Ee Ee Ee μ σ μ σ μ σ = = = =  + =  = = ∑ = = ∑ = = ∑ ∑ 2 2 2 2 2 2 2 2 1 2 1...
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This note was uploaded on 02/23/2011 for the course MATH 112 taught by Professor Ritadubey during the Spring '11 term at Amity University.
 Spring '11
 ritadubey
 Degrees Of Freedom, Probability, Variance

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