Review Problems Solutions

Xx 10 5 0 5 10 15 20 p x x 018 008 007 010 021 011

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Unformatted text preview: s A and B break down but not machine C or D. Solution: P (A ∩ B ∩ C c ∩ Dc ) = P (A)P (B )P (C c )P (Dc ) = (.002)(.003)(1 − .001)(1 − .002) = .000005982 (E) Machine A breaks down but not machines B or C or D. Solution: P (A ∩ B c ∩ C c ∩ Dc ) = P (A)P (B c )P (C c )P (Dc ) = (.002)(1 − .003)(1 − .001)(1 − .002) = .001988 7 Problem 10 The probability distribution of the random variable X is shown in the accompanying table. X=x -10 -5 0 5 10 15 20 P (X = x) 0.18 0.08 0.07 0.10 0.21 0.11 0.25 (A) Find P (X = −10). Solution: P (X = −10) = .18 (B) Find P (X ≥ 5). Solution: P (X ≥ 5) = .10 + .21 + .11 + .25 = .56 (C) Find P (0 ≤ X ≤ 15). Solution: P (0 ≤ X ≤ 15) = .07 + .10 + .21 + .11 = .49 (D) Find E (X ). Solution: E (X ) = −10(.18) + (−5)(.08) + 0(.07) + 5(.10) + 10(.21) + 15(.11) + 20(.25) = 7.05 Problem 11 Two cards are drawn from a well-shuffled deck of 52 playing cards. Let X denote the number of aces drawn. (A) What are all the possible values X can take? Solution: X can take the values 0, 1, or 2. (B) Give the probability distribution of X in the form of a table. X=x Solution: 0 1 2 P ( X = x) C (4,0)C (48,2) C (52,2) C (4,1)C (48,1) C (52,2) C (4,2)C (48,0) C (52,2) = .8507 = .1448 = .0045 (C) What is P (X = 1)? Solution: P (X = 1) = .1448 (D) What is E (X )? Solution: E (X ) = 0(.8507) + 1(.1448) + 2(.0045) = .1538 8 Problem 12 After the private screening of a new television pilot, audience members were asked to rate the new show on a scale of 1 to 10 (10 being the highest rating). From a group of 140 people, the following responses were obtained. Rating 1234 5 6 7 8 9 10 Frequency 2 3 4 12 22 17 21 22 12 25 Let the random variable X denote the rating given to the show by a randomly chosen audience member. (A) Find the probability distribution of the random variable X . Solution: X=x P ( X = x) 1 2/140 2 3/140 3 4/140 4 12/140 5 22/140 6 17/140 7 21/140 8 22/140 9 12/140 (B) What is P (X = 5)? Solution: P (X = 5) = 22 140 (C) What is P (X ≥ 8)? Solution: P (X ≥ 8) = 22 140 + 12 140 + 25 140 = 59 140 ≈ .4214 (D) What rating can a viewer be expected to give this show? 2 3 4 12 22 17 21 22 12 25 E (X ) = 1( 140 ) + 2( 140 ) + 3( 140 ) + 4( 140 ) + 5( 140 ) + 6( 140 ) + 7( 140 ) + 8( 140 ) + 9( 140 ) + 10( 140 ) ≈ 6.86 9 10 25/140...
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This document was uploaded on 03/17/2014 for the course MATH 1331 at Texas Tech.

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