Solutions hw1

Solutions hw1 - W1211 Homework 1 February 2008 Section 1 1...

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Unformatted text preview: W1211: Homework 1 February 2008 Section 1 1. Crash test. a. The sample proportion of successes is 6/10 or 60%. b. ¯ x = 0 . 6 or 60% . This is identical to part (a). c. Suppose we include 15 more cars. Denote by x the number of successes within these additional 15 cars. If . 8 = (6 + x ) / (10 + 15) → x = 20- 6 = 14 . 2. Oxygen consumption. a. The range is [23 . 5 , 49 . 3] . b. ¯ x = 1 n ∑ x i = 30 . 89 . Yielding deviations d = ( x- 30 . 89) = (-1.39 18.41 -0.79 -2.69 -2.89 -4.79 3.01-1.59 -7.39 0.11). Then s 2 = 1 10 ∑ d 2 i = 44 . 62 . c. ∑ x 2 = 9988 . 15 , ∑ x = 308 . 9 . s 2 = 1 9 (9988 . 15- 1 10 308 . 9 2 ) ≈ 49 . 58 . d. The sample standard deviation is √ s 2 = √ 49 . 58 ≈ 7 . 04 . 3. For d i = x i- ¯ x , notice ∑ d i = ∑ x i- n ¯ x = 0 . Thus, . 3 + . 9 + 1 + 1 + d 5 = 0 and d 5 =- 3 . 2 . There is no unique sample which might’ve generated these deviations. However, reaction times are most likely positive (though not necessarily, consider a logarithmic scale). Supposing nothing exotic, (4.5, 5.1, 5.2, 5.2, 1.0) are possible samples. 4. Sodas A, B & C. Barring experimental malfeasance and interesting psychological models, all permutations of A, B & C are equally likely. a. There are 3!=6 arrangements of A,B,C. In only two of them, CAB & CBA, does C occur first. Thus the probability is 2 / 6 = 1 / 3 ≈ 33% ....
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Solutions hw1 - W1211 Homework 1 February 2008 Section 1 1...

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