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

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