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MS-Sol-EE-C15

# MS-Sol-EE-C15 - CHAPTER 15 SOME SPECIAL DISCRETE...

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CHAPTER 15 SOME SPECIAL DISCRETE DISTRIBUTIONS EXERCISE 15.1 Section 15.1 The Bernoulli distribution (page 332) 1. (a) Bernoulli, Q two possible outcomes: Male, Female (b) Bernoulli, Q two possible outcomes: Fail, Pass (c) not Bernoulli, Q more than two possible outcomes such as Christian, Catholic, Buddhist (d) Bernoulli, Q two possible outcomes: Hit, Miss 2. (a) Bernoulli, two possible outcomes: Valid, Not valid (b) not Bernoulli, there can be many makes, not just 2 (c) Bernoulli, two possible outcomes: Work, Fail (d) not Bernoulli, more than two possible outcomes such as Chinese, English, German 3. An appropriate random variable for each of (a) (d) is X such that (a) = not work does chip the if 0, works chip the if , 1 X (b) = male is e interviewe the if 0, female is e interviewe the if , 1 X (c) = negative is test the if 0, positive is test the if , 1 X (d) = otherwise 0, standard specified a meets fitness physical the if , 1 X 4. An appropriate random variable for each of (a) (d) is X such that (a) = flowering not is plant the if 0, flowering is plant the if , 1 X (b) = car a own not does it if 0, car a owns household the if , 1 X (c) = smoker - non a is student the if 0, smoker a is student the if , 1 X (d) = infertile is horse the if 0, fertile is horse the if , 1 X 199

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C HAPTER 15 S OME S PECIAL D ISCRETE D ISTRIBUTIONS 5. Mean μ = = 1 0 ) ( x x f x = 0 × 0.8 + 1 × 0.2 = 0.2 Variance 2 σ = ) ( ) ( 1 0 2 x f x x - = μ = (0 – 0.2) 2 × 0.8 + (1 – 0.2) 2 × 0.2 = 0.16 6. x 0 1 f ( x ) 5 7 2 7 mean = E( X ) = ) ( x f x = 7 2 1 + 7 5 0 × × = 2 7 variance = - ) ( ) ( 2 x f x μ = - - 7 2 7 2 1 + 7 5 7 2 0 2 2 = 10 49 7. (a) p + 0.38 = 1 p = 0.62 (b) x 0 1 f ( x ) 0.38 0.62 mean = ) ( x f x = 0 × 0.38 + 1 × 0.62 = 0.62 variance = - ) ( ) ( 2 x f x μ = (0 - 0.62) 2 × 0.38 + (1 - 0.62) 2 × 0.62 = 0.235 6 standard deviation = 0 235 . 6 = 0.485 4 8. (a) = - otherwise , 0 1 , 0 , 3 1 3 2 = ) ( 1 t t f t t 200
S ECTION 15.1 T HE B ERNOULLI D ISTRIBUTION (b) mean = p = 2 3 variance = 9 2 = 3 1 3 2 = ) 1 ( × - p p (c) E(3 T + 2) = 3E( T ) + 2 = 3 2 3 × + 2 = 4 Var (3 T + 2) = 3 2 Var ( T ) = 9 2 9 × = 2 9. (a) - otherwise , 0 1 0, = , (0.27) (0.63) = ) ( 1 w w f w w (b) mean = p = 0.63 variance = p (1 - p ) = 0.63 × 0.27 = 0.170 1 standard deviation = 0.170 1 = 0.412 4 (c) E(5 - 2 W ) = 5 - 2E( W ) = 5 - 2 × 0.63 = 3.74 Var (5 - 2 W ) = ( - 2) 2 Var ( W ) = 4 × 0.170 1 standard deviation of (5 - 2 W ) = 2 × 0.170 1 = 0.824 9 10. (a) - otherwise , 0 1 0, = , (0.08) (0.92) = ) ( 1 x x f x x (b) mean = 0.92 variance = 0.92 × 0.08 = 0.073 6 11. (a) The probability distribution of X is shown below. x 0 1 P( X = x ) 6 5 6 1 (b) Mean of X = 0 × 6 5 + 1 × 6 1 = 6 1 = = = i x x X x μ 0 ) ( P Variance of X = 2 2 2 6 1 6 1 1 6 5 0 - × + × = - = = i x μ x X x σ 0 2 2 2 ) ( P = 36 5 (c) The mean of X is the proportion of times a white ball is obtained. The variance of X indicates the variability of the actual outcome.

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