STAT213
Tutorial sheet #2 Solutions
1) P(paper or plastic) = 78% + 60% - 52% = 86%
2) P(plastic | paper) = 52/78 = 66.67%. P(plastic | not paper) = 8/22 = 36.36%. Since the first probability is higher, those who recycle paper are more likely to recycle plastic. Since P(plastic) = 60% and neither conditional probability equals this, we can conclude that recycling one thing depends on recycling the other.
3) P(not paper | not plastic) = 8/22 = 36.36%
4) Since P(paper and plastic) = 52% which is not zero, a household recycling one thing is not mutually exclusive of recycling the other.
5) P(balance < $1000) = (806 + 2690)/10000 = 0.3496 = 34.96%
6) P(balance < $1000 | income between $35K and $50K) = (508 + 1377)/3029 = 0.6223 = 62.23%
7) P(balance < $1000 | income < $75K) = (107 + 508 + 191 + 169 + 1377 + 826)/(276 + 3029 + 3241) = = 0.4855 = 48.55%
8) P(income at least $50K | balance at least $1000) = (1419 + 911 + 254 + 805 + 1229 + 742)/(3474 + 3030) = 0.8241 = 82.41%
9) For each income group, we need to compute the percentage that has a balance of $1000 or more:
|
<
35K |
35K
– 50K |
50K
– 75K |
75K
– 100K |
100K
+ |
|
0/276
= 0 |
(890
+ 254)/ 3029
= 0.3777 |
(1419
+ 805)/ 3241
= 0.6862 |
(911
+ 1229)/ 2394
= 0.8939 |
(254
+ 742)/ 1060
= 0.9396 |
Thus the $100K+ group has
the largest percentage of those whose balance is $1000 or more. As we can see
from the calculations, the percentage who have a
balance of at least $1000 increases with each income group.
10)
We
first need to compute the overall market share for the under
25 age group. This output summarizes the results:
|
Brand |
% market share |
% used by under 25 age group |
Weighted average |
|
Brand A |
0.25 |
0.7 |
0.175 |
|
Brand B |
0.15 |
0.55 |
0.0825 |
|
Brand C |
0.6 |
0.4 |
0.24 |
|
|
|
Total |
0.4975 |
We see the overall market share for the under 25 age group is 49.75% and that Brand C comprises the
largest percentage at 24%. Therefore, the percentage of the under 25 age group
that uses Brand C is 24/49.75 = 48.24%.
11)
P(suburb)
= 0.762 P(transit | suburb) = 0.252 P(suburb and transit) = 0.192024
P(IC) = 0.238 P(transit | IC) = 0.128 P(IC and
transit) = 0.030464
P(transit) = 0.222488 or 22.25%
12)
P(suburb
| transit) = 0.192024/0.222488 = 0.86.31%
13)
We
first construct a crosstab:
|
|
Suburb |
IC |
Total |
|
Transit at least once per week |
0.192024 |
0.030464 |
0.222488 |
|
Transit less than once per week |
0.569976 |
0.207536 |
0.777512 |
|
Total |
0.762 |
0.238 |
1.0 |
P(IC
| transit less than once per week) = 0.207536/0.777512 = 26.69%