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Unformatted text preview: Î¼ 1 = m 1 = â‡’ Î» = 1 n n Â± i =1 X i . Therefore, Ë† Î» , our estimator of Î» , is Ë† Î» = 1 n n Â± i =1 X i . Question 4. If X âˆ¼ Binomial(7 , p ), then E { X } = 7 p . Set m 1 = 1 n âˆ‘ n i =1 X i , Î¼ 1 = 7 p . All we need to do is to set m 1 = Î¼ 1 and solve for the unknown parameter. Î¼ 1 = m 1 = â‡’ 7 p = 1 n n Â± i =1 X i . 1 Therefore, Ë† p , our estimator of p , is Ë† p = 1 7 n n âˆ‘ i =1 X i . Question 5. Let m 1 = 1 n n âˆ‘ i =1 X i and m 2 = 1 n n âˆ‘ i =1 X 2 i . Set m 1 = Ë† n Ë† p and m 2 = m 2 1 + Ë† n Ë† p (1Ë† p ). Solving this system of equations yields: m 1 = Ë† n Ë† p = â‡’ Ë† n = m 1 Ë† p m 2 = m 2 1 + Ë† n Ë† p (1Ë† p ) = m 2 1 + m 1 Ë† p Ë† p (1Ë† p ) = â‡’ Ë† p = 1m 2m 2 1 m 1 = m 1( m 2m 2 1 ) m 1 Therefore, Ë† n = m 1 Ë† p = m 2 1 m 1( m 2m 2 1 ) (Note that whenever the sample mean is smaller than the sample variance, Ë† n and Ë† p both become negative.) 2...
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 '08
 TOPALOGLU
 Normal Distribution, Variance, Prime number, Divisor, accompanying spreadsheet

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