HW 3 Solutions

HW 3 Solutions - d. We need the top .0005 and the bottom...

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STATW1211.005 Prof. Zhang TR 02:40P-03:55P Homework 3 – Solutions Textbook: Probability and statistics for engineering and the sciences (7 th edition) by Jay L. Devore Chapter 4: Question 10 a. θ b. c. P(X b) = d. P(a X b) = Question 24 a. E(X) = b. E(X) = c. E(X 2 ) = , so Var(X) = d. Var(X) = , since E(X 2 ) = . e. E( X n ) = , which will be finite if n – ( k +1) < –1, i.e. if n < k . Question 44 a. P( μ – 1.5 σ X μ + 1.5 σ ) = P(–1.5 Z 1.5) = Φ (1.50) – Φ (–1.50) = .8664 b. P( X < μ – 2.5 σ or X > μ + 2.5 σ ) = 1 – P( μ – 2.5 σ X μ + 2.5 σ ) = 1 – P(–2.5 Z 2.5) = 1 – .9876 = .0124 c. P( μ – 2 σ X μ σ or μ + σ X μ + 2 σ ) = P(within 2 sd’s) – P(within 1 sd) = P( μ – 2 σ X μ + 2 σ ) – P( μ σ X μ + σ ) = .9544 – .6826 = .2718 Question 49 2. X N(3432, 482) a. ; b. c. We will use the conversion 1 lb = 454 g, then 7 lbs = 3178 grams, and we wish to find
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Unformatted text preview: d. We need the top .0005 and the bottom .0005 of the distribution. Using the Z table, both .9995 and . 0005 have multiple z values, so we will use a middle value, 3.295. Then 3432(482)3.295 = 1844 and 5020, or the most extreme .1% of all birth weights are less than 1844 g and more than 5020 g. e. Converting to lbs yields mean 7.5595 and sd 1.0608. Then This yields the same answer as in part c . STATW1211.005 Prof. Zhang TR 02:40P-03:55P Question 62 a. Clearly E(X) = 0 by symmetry, so V(X) = E(X 2 ) = = = . Solving = (40.9) 2 yields = 0.034577 b. P(|X 0| 40.9) = = = 1 e 40.9 = .75688 Question 66 a. = 20, 2 = 80 = 20, 2 = 80 = , = 5 b. P(X 24) = = F(6;5) = .715 c. P(20 X 40) = F(10;5) F(5;5) = .411...
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This note was uploaded on 01/31/2012 for the course STAT 1211 taught by Professor Hernandez during the Spring '08 term at Columbia.

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HW 3 Solutions - d. We need the top .0005 and the bottom...

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