**Unformatted text preview: **x. . ∫ ∫ ∫ ∫ ∞ ∞-∞-∞ + = + + = = 6 6 6 ) 6 / 1 ( ) ( xdx dx x xdx dx x xf μ (x 2 /12) = 36/ 12 = 3.0 _ 0.05____ 2. P(Z< -1.65) = from the Table 3: Φ (-1.65)= 0.05 _ 0.005 ____3. P( Z > 2.58) =____ P(Z > 2.58) = 1- P(Z< 2.58) = 1- Φ (2.58) = 1- 0.995 = 0.005 (from Table 3: Φ (2.58)=F(2.58)=0.995) __ 1.96 ___ 4. z 0.025 = i.e., Find c such that P(Z> c) = 0.025. P(Z > z 0.025 ) = 0.025. So P(Z< z 0.025 ) = 0.975 and Φ (z 0.025 )=0.975. From Table 3: Φ (1.96)= 0.975 . So z 0.025 =1.96...

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- Fall '03
- Mendell
- Normal Distribution, Standard Deviation, mean value