By Leonard J. Kazmier
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Extra resources for Business Statistics
5 5! 3! a. Because order is important, we’ll use P8,3 = 8(7)(6) = 336. b. Order doesn’t matter here, so we use C30,6 = 593,775. c. From each group we choose 2: d. The numerator comes from part c and the denominator from part b: e. We use the same denominator as in part d. We can have all zinfandel, all merlot, or all cabernet, so P(all same) = P(all z) + P(all m) + P(all c) = 30. 002 30 593,775 6 31. a. (n 1 )(n 2 ) = (9)(27) = 243 b. (n 1 )(n 2 )(n 3 ) = (9)(27)(15) = 3645, so such a policy could be carried out for 3645 successive nights, or approximately 10 years, without repeating exactly the same program.
The median is the most resistant to outliers because it uses only the middle value (or values) in its computation. The trimmed mean is somewhat resistant to outliers. The larger the trimming percentage, the more resistant the trimmed mean becomes. The midhinge, which uses the quartiles, is reasonably resistant to outliers because both quartiles are resistant to outliers. 77. a. 5 stem: ones leaf: tenths 41 Chapter 1: Overview and Descriptive Statistics b. Interval Frequency Rel. Freq. 00 0 2 4 6 10 20 30 Repair Time 78.
B. 02 c. 03 from the Venn Diagram d. 24 59 Chapter 2: Probability 26. 27. 28. a. 88 b. 06 c. 05 d. 99 Outcomes: a. 1 b. P(at least one C) = c. 7 There are 27 equally likely outcomes. a. P(all the same) = P[(1,1,1) or (2,2,2) or (3,3,3)] = 3 27 = 1 9 b. P(at most 2 are assigned to the same station) = 1 – P(all 3 are the same) 3 24 = 1 − 27 = 27 = 89 c. 3 29. a. (5)(4) = 20 (5 choices for president, 4 remain for vice president) b. (5)(4)(3) = 60 c. 5 5! 3! a. Because order is important, we’ll use P8,3 = 8(7)(6) = 336.