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# slide_smpling - problem:The 2 distribution with 4 degrees...

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2 / 2 problem:- The distribution with 4 degrees of freedom is given by 0 0 ( ) 1 ( ) 0 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 0 1 1 4 0.2873 x xe dx - = - =

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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 E e E e E e μ σ μ σ μ σ = = = = - + - = - = = = = = = 2 2 2 2 2 2 2 2 1 2 1 2 ... ] ] [ . ..... ] [ ]. [ ] ..... [ ] k k k Z tZ tZ tZ tZ tZ tZ E e e e E e E e E e + + = =
2 2 2 2 2 r / 2 1 ( ) 2 Since all Z are independent.

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