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Unformatted text preview: x. . âˆ« âˆ« âˆ« âˆ« âˆž âˆžâˆžâˆž + = + + = = 3 / 1 3 3 / 1 ) 3 ( ) ( xdx dx x xdx dx x xf Î¼ (3x 2 /2) = 3 (1/9)/ 2 = 1/6 _ 0.90____ 2. P(Z< 1.28) = Î¦ (1.28). From the Table 3: Î¦ (1.28)= 0.90 _ 0.01 ____3. P( Z > 2.33) =____ P(Z > 2.33) = 1 P(Z< 2.33) = 1 Î¦ (2.33) = 1 0.99 = 0.01 (from Table 3: Î¦ (2.33)=0.99) __ 1.28 ___ 4. z 0.1 = i.e., Find c such that P(Z> c) = 0.1. P(Z > z 0.1 ) = 0.1. So P(Z< z 0.1 ) = Î¦ (z 0.1 )= 0.9 ( Z has standard normal c.d.f.) From Table 3(or from Quiz question 2): Î¦ ( 1.28 )= 0.90. So z 0.1 =1.28 2...
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 Fall '03
 Mendell
 Normal Distribution, Standard Deviation, mean value, normal distribution function

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