hw2_sol_cont - . 5 z Z P 0.25 [0.501,0.545] 0.3...

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Question 3 a) We have,   2 / 1 X E , since X in uniformly distributed over [0,1],   2 / 1 ) 1 ( 1 ) 1 ( 1 ) 1 /( 1 1 0 1 1 0 2 2 y dy y Y E Then,         . 4 1 ) 1 /( 1 ) 1 /( 2 2 Y E X E Y X E Z E We also have,   3 / 1 3 1 1 0 3 1 0 2 2 x dx x X E ,     24 / 7 ) 1 ( 3 1 1 1 1 ) 1 /( 1 1 0 3 1 0 4 4 y dy y Y E Then                 144 5 4 1 24 7 3 1 ) 1 /( 1 ) 1 /( 1 ) 1 /( ) 1 /( ) ) 1 /( ( ) ( 2 2 2 4 2 2 2 4 2 2 Y E X E Y E X E Y X E Y X E Y X Var Z Var . b)   . 4 / 1 } { Z E Z E n     . 144 5 1 n Z Var n Z Var n
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c) We fix the sample size at 2000. Then, for z =0.25, 0.3, 0.35, 0.4 the accompanying spreadsheet computes a 95% confidence interval for   z Z P 5 .The value of z is contained in cell U1. The results are: (You will see slightly different results as the calculations are redone.) Z 95% confidence interval for
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Unformatted text preview: . 5 z Z P 0.25 [0.501,0.545] 0.3 [0.728,0.766] 0.35 [0.854,0.883] 0.4 [0.952,0.969] d) We have z P{N<z} 0.25 0.5 0.3 0.725747 0.35 0.88493 0.4 0.96407 See the accompanying spreadsheet for the calculations. These probabilities are close to the probabilities estimated in Part c. n Z is the average of n i.i.d. random variables. Therefore, by the central limit theorem, the distribution of n Z converges to the normal distribution as n approaches infinity. For a finite and large n , the distribution of n Z is approximately normal. Interestingly, for the central limit theorem to take effect, n =5 is large enough....
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hw2_sol_cont - . 5 z Z P 0.25 [0.501,0.545] 0.3...

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