1. Compare
the variability of the following two sets of data by using both the population
standard deviation and the population coefficient of variation.
|
|
Data Set 1 |
Data Set 2 |
|
|
|
49 |
159 |
|
|
|
82 |
121 |
|
|
|
77 |
138 |
|
|
|
54 |
152 |
|
|
σ2 = |
809 |
845 |
|
|
σ = |
28.44293 |
29.06888 |
|
|
CV = |
0.434243 = 3.4% |
0.203992 = 20.4% |
2. A
random sample of voters in Nashville, Tennessee is classified by age group as
shown by the following data.
|
Age Group |
Frequency |
|
18-under 24 |
17 |
|
24-under 30 |
22 |
|
30-under 36 |
26 |
|
36-under 42 |
35 |
|
42-under 48 |
33 |
|
48-under 54 |
30 |
|
54-under 60 |
32 |
|
60-under 66 |
21 |
|
66-under 72 |
15 |
- Calculate the mean of the data
- Calculate the mode
- Calculate the variance
- Calculate the standard deviation
|
|
f |
Midpoint (M) |
fM |
fM2 |
variance |
std dev. |
|
18-under 24 |
17 |
(24+18)/2 = 21 |
357 |
7497 |
186.3788 |
13.65206 |
|
24-under 30 |
22 |
(24+30)/2 = 27 |
594 |
16038 |
|
|
|
30-under 36 |
26 |
(30+36)/2 = 33 |
858 |
28314 |
|
|
|
36-under 42 |
35 |
(36+42)/2 = 39 |
1365 |
53235 |
|
|
|
42-under 48 |
33 |
(42+48)/2 = 45 |
1485 |
66825 |
||
|
48-under 54 |
30 |
(48+54)/2 = 51 |
1530 |
78030 |
||
|
54-under 60 |
32 |
(54+60)/2 = 57 |
1824 |
103968 |
||
|
60-under 66 |
21 |
(60+66)/2 = 63 |
1323 |
83349 |
|
|
|
66-under 72 |
15 |
(66+72)/2 = 69 |
1035 |
71415 |
|
|
|
|
|
|
10371 |
508671 |
|
|
