231hw5solutionsF11

# 231hw5solutionsF11 - Will Landau September 26, 2011 STAT...

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Will Landau September 26, 2011 STAT 231 Homework 5 Solutions Exercise 5.1 (Devore 4.11) . f. E ( X ) = R -∞ xf ( x ) dx = R -∞ xF 0 ( x ) dx = R 2 0 x · x 2 dx = (1 / 2) R 2 0 x 2 dx = h x 3 6 i 2 0 = 8 6 1.333 g. E ( X 2 ) = R -∞ x 2 f ( x ) dx = R -∞ x 2 F 0 ( x ) dx = R 2 0 x 2 · x 2 dx = (1 / 2) R 2 0 x 3 dx = h x 4 8 i 2 0 = 2. Hence V ( X ) = E ( X 2 ) - E 2 ( X ) = 2 - 8 6 2 = 8 36 . 222 . Also, σ X = p V ( X ) = .471 h. E ( h ( X )) = E ( X 2 ) = 2 from part g. Exercise 5.2 (Devore 4.13) . First, we should get part a: a. cdf: for x > 1 , F X ( x ) = R x -∞ f ( t ) dt = R x 1 kt - 4 dt = ± - k 3 t - 3 ² x 1 = - k 3 x 3 + k 3 . We need lim x →∞ F ( x ) = k 3 = 1, so k=3 and F ( X ) = 1 - x - 3 if x > 1 and 0 if x 1. d. E ( X ) = R 1 x · k x 4 dx = R 1 k x 3 dx = ± - k 2 x - 2 ² 1 = 0 + k/ 2 = k/2=1.5 . E ( X 2 ) = R 1 x 2 · k x 4 dx = R 1 k x 2 dx = ± - kx - 1 ² 1 = 0 + k = k=3 V ( X ) = E ( X 2 ) - E 2 ( X ) = k - k 2 4 , σ X = V = q k - k 2 4 = 0 . 866 e. P ( X ( E ( X ) - σ X ,E ( X ) + σ X )) = F ( E ( X ) + σ X ) - F ( E ( X ) - σ X ) = F (1 . 5 + . 866) - F (1 . 5 - . 866) = F (2 . 366) - F ( . 634) = (1 - 2 . 366 - 3 ) - 0 (part a) = .9245 Exercise 5.3 (Devore 4.28) . a. Φ(2 . 17) - Φ(0) = .4850 b. Φ(1) - Φ(0) = .3413 c. Φ(0) - Φ( - 2 . 5) = .4938 d. Φ(2 . 5) - Φ( - 2 . 5) = .9876 1

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e. Φ(1 . 37) = .9147 f. 1 - Φ - 1 . 75 = .9599 g. Φ(2) - Φ( - 1 . 5) = .9104 h. Φ(2 . 5) - Φ(1 . 37) = .0791 i. 1 - Φ(1 . 5) = .0668 j. P ( | Z | ≤ 2 . 5) = P ( - 2 . 5 Z 2 . 5) = Φ(2 . 5) - Φ( - 2 . 5) = .9876 Exercise 5.4 (Devore 4.29) . a. You can ﬁnd .9838 is found in the 2.1 row and the .04 column of the standard normal table (Table A-3 page A-7) so c = 2.14 . b. Φ( c ) - Φ(0) = . 291 Φ( c ) = . 791 c =.81 c. P ( c Z ) = . 121 1 - P ( clZ ) = P ( Z < c ) = Φ( c ) = 1 - . 121 = . 8790 c=1.17 d. Φ( c ) - Φ( - c ) = Φ( c ) - (1 - Φ( c )) = 2Φ( c ) - 1 Φ( c ) = . 9920 c=.97 . e. . 016 = P ( c ≤ | Z | ) 1 - . 016 = 1 - P ( c ≥ | Z | ) = P ( - c Z c ) = Φ( c ) - Φ( - c ) = 2Φ( c ) - 1 Φ( c ) = . 9920 c=2.41 Exercise 5.5 (Devore 4.30) . a. Φ( c ) = . 9100 c 1.34 (Use the table. 0.9099 is the entry in the 1.3 row, 0.04 column) b. Since the normal distribution is symmetric with mean 0, 9th percentile = -91st percentile = -1.34 c. Φ( c ) = . 7500 c .675 since .7486 and .7517 are in the .67 and .68 entries of the table, respectively. d. 25th percentile = -75th percentile =
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## This note was uploaded on 02/11/2012 for the course STAT 231 taught by Professor Staff during the Fall '08 term at Iowa State.

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231hw5solutionsF11 - Will Landau September 26, 2011 STAT...

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