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Unformatted text preview: 1 Homework 5 (12 points) due Feb. 17 (1.2 pt.) 4.4abc. The following table provides a frequency distribution, with frequencies in thousands, for the number of rooms in U.S. housing units. (Note: this is the same table as was used in problem 1.4.) Rooms 1 2 3 4 5 6 7 8+ No. Units 471 1,470 11,715 23,468 24,476 21,327 13,782 15,647 Compute the following conditional probabilities directly; do not use the conditional probability rule. For a U.S. housing unit selected at random, find the probably that the unit has a) exactly four rooms. This is an unconditional probability: N( ) = 112,356 N(4 rooms) = 23,468 4 ????¡¢ = 23,468 112,356 = 0.209 b) exactly four rooms, given that it has at least two rooms. P(4 roomsthere are at least 2 rooms) N(at least 2 rooms) = N(all rooms)  N(1 room) = 112,356  471 = 111,885 4 ????¡£¤? ¥¦¤¡? 2 ????¡¢ = 23,468 111,885 = 0.210 c) at most four rooms, given that it has at least two rooms. N(at most four rooms) = N(2 rooms) + N(3 rooms) + N(4 rooms) = 1,470 + 11,715 + 23,468 = 36,653 Note: since there are at least 2 rooms, 1 room is not included in the sum ¤? ??¡? 4 ????¡£¤? ¥¦¤¡? 2 ????¡¢ = 35,653 111,885 = 0.328 (1.6 pt.) 4.8ae. The following is a joint probability distribution for living arrangement and age of U.S. citizens 15 years of age and older. Age 1524 A 1 2544 A 2 4564 A 3 Over 64 A 4 P(L j ) Living Arrangement Alone L 1 0.006 0.038 0.035 0.047 0.126 With spouse L 2 0.017 0.241 0.187 0.084 0.529 With others L 3 0.154 0.122 0.047 0.022 0.345 P(A i ) 0.177 0.401 0.269 0.153 1.000 Note: no bonus points for guessing which box that I am in 2 A U.S. citizen 15 years of age or older is selected at random. Determine the probability that the person selected a) lives with spouse. This is an unconditional probability. P(L 2 ) = 0.529 b) is older than 64. This is an unconditional probability. P(A 4 ) = 0.153 c) lives with spouse and is older than 64. This is an unconditional probability. P(L 2 ∩ A 4 ) = 0.084 d) lives with spouse, given that the person is older than 64. 2 ¡? 4 ¢ = ( 2 ∩ ? 4 ) ( ? 4 ) = 0.084 0.153 = 0.549 Note: c) and d) are NOT the same value. e) is older than 64, given that the person lives with spouse. P(A 4 L 2 ) = ( 2 ∩? 4 ) ( 2 ) = 0.084 0.529 = 0.159 Note: d) and e) are the same value because the denominators are different. That is, the condition (what is given) is different in the two cases. (1 pt.) 4.10 (6 sided die). A balanced die is tossed 12 times. Given that a 3 occurs at least once, what is the probability that it occurs four times or more? E n : the event that a 3 occurs exactly n times, n = 0, …, 12 £¤ ?...
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This note was uploaded on 02/20/2012 for the course STAT 311 taught by Professor Staff during the Spring '08 term at Purdue.
 Spring '08
 Staff

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