STAT177 Assignment #1 solutions
1)
|
Class |
Frequency |
Percent |
Cum. % |
|
Less
than 40 |
1 |
2.50% |
2.50% |
|
40
to less than 80 |
3 |
7.50% |
10.00% |
|
80
to less than 120 |
4 |
10.00% |
20.00% |
|
120
to less than 160 |
4 |
10.00% |
30.00% |
|
160
to less than 200 |
18 |
45.00% |
75.00% |
|
200
to less than 240 |
7 |
17.50% |
92.50% |
|
240
to less than 280 |
1 |
2.50% |
95.00% |
|
280
to less than 320 |
2 |
5.00% |
100.00% |
2) P(less than 200) = 75% from the table
3) P(at least 120) = 100% - P(less than 120) = 100% - 20% = 80%
4) P(at least 80 but less than 160) = 10% + 10% = 20%
5)
|
|
with
309.69 |
without
309.69 |
difference |
|
Mean |
168.9115 |
165.3018 |
3.6097 |
|
Median |
173.6 |
170.91 |
2.6900 |
For the next set of questions, we create a crosstab:
|
|
Own |
Not own |
Total |
|
LOC |
4.6 |
7.4 |
12 |
|
No LOC |
63.4 |
24.6 |
88 |
|
Total |
68 |
32 |
100 |
6) P(home or LOC) = 68% + 12% - 4.6% = 75.4%
7) P(LOC | home) = 4.6/68 = 6.76%
8) P(not own | no LOC) = 24.6/88 = 27.95%
9) P(LOC and not own) = 7.4% from crosstab
10) P(support) = 225/500 = 45%
11) P(support | employed) = (92 + 95)/(234 + 205) = 187/439 = 42.6%
12) P(not employed | oppose) = 12/186 = 6.45%
13) P(employed | not support) = (98 + 76 + 44 + 34)/(186 + 89) = 252/275 = 91.64%
14) The technical answer is that there are no crosstab values of zero. The intuitive answer is that each position is represented in each employment category.