24장

24장 - Homework 1 24.1 1. (a)...

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Homework 1 § 24 . 1 1. (a) Represent the data 403 399 398 401 400 401 401 by a stem-and-leaf plot, a histogram, and a boxplot. (b) In(a), find the mean and compare it with the median. Find the standard devi- ation and compare it with the interquartile range. Sol. (sorted) 398 399400 401 401401 403 a stem-and-leaf plot Leaf unit= 1 . 0 2 39 | 89 7 40 | 01113 median : q M = 401, upper quartile : q U = 401, lower quartile : q L = 399 We omit a histogram, and a boxplot. (b) mean : 403+399+398+401+400+401+401 7 = 400 3 7 400 . 43, median : q M = 401 variance : σ 2 = [(403 - 400 . 43) 2 + (399 - 400 . 43) 2 + (398 - 400 . 43) 2 + (401 - 400 . 43) 2 + (400 - 400 . 43) 2 + (401 - 400 . 43) 2 + (401 - 400 . 43) 2 ] / 7 = 2 . 24 standard deviation : σ = 2 . 24 = 1 . 497, interquartile range : IQR= q U - q L = 401 - 399 = 2 2. Prove that ¯ x must be always lie between the smallest and largest data values. Sol. Consider x 1 , x 2 , ··· , x n and assume that x 1 is the smallest value and x n is the largest value. ¯ x - x 1 = x 1 + x 2 + ··· + x n n - x 1 = ( x 1 - x 1 ) + ( x 2 - x 1 ) + ··· + ( x n - x 1 ) n 0 , ¯ x - x n = x 1 + x 2 + ··· + x n n - x n = ( x 1 - x n ) + ( x 2 - x n ) + ··· + ( x n - x n ) n 0 . Thus ¯ x must be always lie between the smallest and largest data values. (CSE) C. K. Ko
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Homework 2 § 24 . 3 1. Three screws are drawn at random from a lot of 100 screws, 10 of which are de- fective. Find the probability that the screws drawn will be nondefective in drawing (a) with replacement, (b) without replacement. Sol. (a) 90 100 · 90 100 · 90 100 = 729 1000 , (b) 90 100 · 89 99 · 88 98 2. If you need a right-handed screw from a box containing 20 right-handed and 5 left-handed screws, what is the probability that you get at least one right-handed screw in drawing 2 screws with replacement? Sol. probability that you get all two right-handed screw in drawing 2 screws with replacement : 5 25 · 5 25 = 1 25 . Thus, probability that you get at least one right-handed screw in drawing 2 screws with replacement : 1 - 1 25 = 24 25 . 3. A motor drives an electric generator. During a 30 - day period, the motor needs repair with probability 8% and the generator needs repair with the probability 4%. What is the probability that during a given period, the entire apparatus(consisting of a motor and a generator) will need repair? Sol. 1 - 92 100 · 96 100 = 0 . 1168 4. Let A,B and C be events in a sample space S . (a) Show that if B is a subset of A , then P ( B ) P ( A ). (b) Let P ( A ) 6 = 0 , P ( B ) 6 = 0 , P ( A B ) 6 = 0. Show that P ( A B C ) = P ( A ) P ( B | A ) P ( C | A B ) . Sol. (a) Since A = B ( A - B ) and 0 P ( A - B ) 1, P ( A ) = P ( B ) + P ( A - B ) P ( B ) .
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24장 - Homework 1 24.1 1. (a)...

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