Homework5key

# Homework5key - Homework 5(12 points due Feb 17(1.2 pt...

This preview shows pages 1–3. Sign up to view the full content.

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 𝑃 4 ????? = 23,468 112,356 = 0.209 b) exactly four rooms, given that it has at least two rooms. P(4 rooms|there are at least 2 rooms) N(at least 2 rooms) = N(all rooms) - N(1 room) = 112,356 - 471 = 111,885 𝑃 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 𝑃 ?? ???? 4 ????? ?? ????? 2 ????? = 35,653 111,885 = 0.328 (1.6 pt.) 4.8a-e. The following is a joint probability distribution for living arrangement and age of U.S. citizens 15 years of age and older. Age 15-24 A 1 25-44 A 2 45-64 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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document