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mont4e_sm_ch03_sec01

# mont4e_sm_ch03_sec01 - CHAPTER 3 Section 3-1 3-1 3-2 3-3...

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Unformatted text preview: CHAPTER 3 Section 3-1 3-1. 3-2. 3-3. 3-4. 3-5. The range of X is {0,1,2,...,1000} The range of X is {0,12,...,50} , The range of X is {0,12,...,99999} , The range of X is {0,12,3,4,5} , The range of X is 1,2,...,491 . Because 490 parts are conforming, a nonconforming part must be selected in 491 selections. The range of X is {0,12,...,100} . Although the range actually obtained from lots typically might not , exceed 10%. { } 3-6. 3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is , {0,12,...} The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is , {0,12,...} The range of X is 3-8. 3-9. 3-10. {0,1,2,...,15} The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8. ⎧1 3 1 5 6⎫ Therefore the range of X is ⎨ , , , , ⎬ ⎩4 8 2 8 8 ⎭ The range of X is 3-11. {0,1,2, … ,10000} 3-12.The range of X is {100, 101, …, 150} 3-13.The range of X is {0,1,2,…, 40000) Section 3-2 3-14. f X (0) = P ( X = 0) = 1 / 6 + 1 / 6 = 1 / 3 f X (1.5) = P ( X = 1.5) = 1 / 3 f X (2) = 1 / 6 f X (3) = 1 / 6 a) P(X = 1.5) = 1/3 b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2 c) P(X > 3) = 0 d) P ( 0 ≤ X < 2) = P ( X = 0) + P ( X = 1.5) = 1/ 3 + 1/ 3 = e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2 2/3 3-15. All probabilities are greater than or equal to zero and sum to one. 3-1 ...
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