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Unformatted text preview: Chapter Five 5.1 Random variable: A variable whose value is determined by the outcome of a random experiment is called a random variable. An example of this is the income of a randomly selected family. Discrete random variable: A random variable whose values are countable is called a discrete random variable. An example of this is the number of cars in a parking lot at any particular time. Continuous random variable: A random variable that can assume any value in one or more intervals is called a continuous random variable. An example of this is the time taken by a person to travel by car from New York City to Boston. 5.3 a. a discrete random variable b. a continuous random variable c. a continuous random variable d. a discrete random variable e. a discrete random variable f. a continuous random variable 5.5 The number of cars x that stop at the Texaco station is a discrete random variable because the values of x are countable: 0, 1, 2, 3, 4, 5 and 6. 5.7 The two characteristics of the probability distribution of a discrete random variable x are: 1. The probability that x assumes any single value lies in the range 0 to 1, that is, 1 ) ( ≤ ≤ x P . 2. The sum of the probabilities of all values of x for an experiment is equal to 1, that is: 1 ) ( = ∑ x P . 5.9 a. P ( x = 1) = .17 b. P ( x ≤ 1) = P (0) + P (1) = .03 + .17 + = .20 c. P ( x ≥ 3) = P (3) + P (4) + P (5) = .31 + .15 + .12 = .58 d. P (0 ≤ x ≤ 2) = P (0) + P (1) + P (2) = .03 + .17 + .22 = .42 e. P ( x < 3) = P (0) + P (1) + P (2) = .03 + .17 + .22 = .42 f. P ( x > 3) = P (4) + P (5) = .15 + .12 = .27 g. P (2 ≤ x ≤ 4) = P (2) + P (3) + P (4) = .22 + .31 + .15 = .68 5.11 a. b. i. P (exactly 3) = P (3) = .25 ii. P (at least 4) = P ( x ≥ 4) = P (4) + P (5) + P (6) = .14 + .07 + .03 = .24 71 10 20 30 1 2 3 4 5 Number of Systems Installed P(x) iii. P (less than 3) = P ( x< 3) = P (0) + P (1) + P (2) = .10 + .18 + .23 = .51 iv. P (2 to 5) = P (2) + P (3) + P (4) + P (5) = .23 + .25 + .14 + .07 = .69 5.13 a. x P ( x ) 1 8 / 80 = .10 2 20 / 80 = .25 3 24 / 80 = .30 4 16 / 80 = .20 5 12 / 80 = .15 b. The probabilities listed in the table of part a are approximate because they are obtained from a sample of 80 days. c. i. P (x = 3) = .30 ii. P ( x ≥ 3) = P (3) + P (4) + P (5) = .30 + .20 + .15 = .65 iii. P (2 ≤ ≤ x 4) = P (2) + P (3) + P (4) = .25 + .30 + .20 = .75 iv. P ( x ≤ 4) = P (2) + P (3) + P (4) = .10 + .25 + .30 = .65 5.15 Let Y = owns a cell phone and N = does not own a cell phone. Then P ( Y ) = .64 and P ( N ) = 1 – .64 = .36 Let x be the number of adults in a sample of two who own a cell phone. The following table lists the probability distribution of x . Note that x = 0 if neither adult owns a cell phone, x = 1 if one adult owns a cell phone and the other does not, and x = 2 if both adults own a cell phone. The probabilities are written in the table using the tree diagram above. The probability that x = 1 is obtained by adding the probabilities of YN and NY...
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 Fall '07
 Guggenberger
 Poisson Distribution, Probability theory, Binomial distribution, Discrete probability distribution, µ

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