Homework_1_Solution

# Homework_1_Solution - conclude that Alaska Airlines has the...

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Homework 1 Solution 1.1-2 (5 points) a. S = {ggg, ggb, gbg, gbb, bgg, bgb, bbg, bbb} b. S = {female, male} c. S = {100, 101, 102,…, 999} No point is subtracted if you list S = {000, 001, 002,…, 999} 1.1-4 (10 points) a. Clutch Size Count Relative Frequency 4 3 0.025641026 5 5 0.042735043 6 7 0.05982906 7 27 0.230769231 8 26 0.222222222 9 37 0.316239316 10 8 0.068376068 11 2 0.017094017 12 0 0 13 1 0.008547009 14 1 0.008547009 Total 117 1 b. c. The mode is 9

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1.1-13 (10 points) a. On 5 individual cities, Alaska Airlines has the better on time performance. b. America West has the better on time performance. c. This example illustrates the Simpson’s paradox. It should be careful when drawing statistical inference from the aggregate (combined) data. In this problem, if we are only provided with the combined data, we may conclude that America West has the better on time performance. However, when taking into account the individual cities’ data, we can
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Unformatted text preview: conclude that Alaska Airlines has the better on time performance. 1.2-2 (10 points) a. S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} b. A = {HHHH, HHHT, HHTH, HTHH, THHH} B = {HHTT, HTHT, HTTH, HTTT, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} C = {HHHH, HHHT, HTHH, HTHT, THHH, THHT, TTHH, TTHT} D = {HTTT, THTT, TTHT, TTTH} i. P(A) = 5/16 ii. P(A ⋂ B) = 0 iii. P(B) = 11/16 iv. P(A ⋂ C) = 4/16 = 1/4 v. P(D) = 4/16 = 1/4 vi. P(A ⋃ C) = P(A) + P(C) – P(A ⋂ C) = 5/16 + 8/16 – 4/16 = 9/16 vii. P(B ⋂ D) = P(D) = 1/4 (D is a subset of B) 1.2-6 (5 points) a. P(A ⋃ B) = P(A) + P(B) – P(A ⋂ B) = .4 + .5 - .3 = .6 b. P(A ⋂ B’) = P(A) – P(A ⋂ B) = .4 - .3 = .1 c. P(A’ ⋃ B’) = 1 – P(A ⋂ B) = 1 - .3 = .7 (use De Morgan’s law)...
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Homework_1_Solution - conclude that Alaska Airlines has the...

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