Name of state |
Data value |
|
State 1 |
Alabama |
2.2 |
State 2 |
Arkansas |
0.7 |
State 3 |
Connecticut |
0.9 |
State 4 |
Florida |
6.6 |
State 5 |
Idaho |
0.6 |
State 6 |
Iowa |
1.2 |
State 7 |
Louisiana |
2.9 |
State 8 |
Massachusetts |
1.9 |
State 9 |
Mississippi |
1.3 |
State 10 |
Nebraska |
0.5 |
State 11 |
New Jersey |
2.2 |
State 12 |
North Carolina |
4.6 |
State 13 |
Oklahoma |
1.4 |
State 14 |
Rhode Island |
0.4 |
State 15 |
Tennessee |
3.4 |
Analysis:
Compute the mean and standard deviation of the sample data:
Mean =
Standard deviation =
Sample size =
Sample Size = 15
Sample Mean () isgiven by:
x | x - | (x - )2 |
2.2 | 0.1467 | 0.0215 |
0.7 | -1.3533 | 1.8315 |
0.9 | -1.1533 | 1.3302 |
6.6 | 4.5467 | 20.6722 |
0.6 | -1.4533 | 2.1122 |
1.2 | -0.8533 | 0.7282 |
2.9 | 0.8467 | 0.7168 |
1.9 | -0.1533 | 0.0235 |
1.3 | -0.7533 | 0.5675 |
0.5 | -1.5533 | 2.4128 |
2.2 | 0.1467 | 0.0215 |
4.6 | 2.5467 | 0.4855 |
1.4 | -0.6533 | 0.4268 |
0.4 | -1.6533 | 2.7335 |
3.4 | 1.3467 | 1.8135 |
Total = | 41.8973 |
Sample Standard Deviation (s) is given by:
Answers are:
Mean = 2.0533
Standard Deviation = 1.7299
Sample Size = 15
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