Unformatted text preview: 1 Ïƒ 3 âˆš 2 Ï€ e( xÎ¼ ) 2 2 Ïƒ 2 ( xÎ¼ ) and f 00 ( x ) =1 Ïƒ 3 âˆš 2 Ï€ â€¢ e( xÎ¼ ) 2 2 Ïƒ 2 Â· 1 + ( xÎ¼ ) e( xÎ¼ ) 2 2 Ïƒ 2 Â±2( xÎ¼ ) 2 Ïƒ 2 Â¶â€š . Factoring e( xÎ¼ ) 2 2 Ïƒ 2 from the expression above and setting the result equal to 0 yields ( xÎ¼ ) 2 = Ïƒ 2 so that xÎ¼ = Â± Ïƒ or x = Î¼ Â± Ïƒ are the inï¬‚ection points. 3.60: If X âˆ¼ N (1 , 4) then Î¼ = 1 and Ïƒ = 2 so that P (1 < X 2 < 9) = P (3 < X <1) âˆª P (1 < X < 3) = P Â±31 2 < X1 2 <11 2 Â¶ + P Â± 11 2 < X1 2 < 31 2 Â¶ = Î¦(1)Î¦(2) + Î¦(1)Î¦(0) = 0 . 477 using the Z distribution tables in the back of the text. 1...
View
Full Document
 Fall '07
 Song
 Normal Distribution, Probability, Probability theory, MATHEMATICAL PROBABILITY ASSIGNMENT

Click to edit the document details