The following data represent the concentration of dissolved organic carbon (mg/L) collected from 20 samples of organic soil. Assume that the population is normally distributed. Complete parts (a) through (c) on the right.
5.20
8.81
30.91
19.80
29.80
22.49
14.86
14.86
27.10
20.46
17.50
8.09
16.51
14.90
15.35
15.72
11.90
33.67
9.72
18.30
(a) Find the sample mean.
The sample mean is
17.8017.80.
b) Find the sample standard deviation.
The sample standard deviation is
(Round to two decimal places as needed.)
the necessary calculation table:-
x | x2 |
5.2 | 27.0400 |
8.81 | 77.6161 |
30.91 | 955.4281 |
19.8 | 392.0400 |
29.8 | 888.0400 |
22.49 | 505.8001 |
14.86 | 220.8196 |
14.86 | 220.8196 |
27.1 | 734.4100 |
20.46 | 418.6116 |
17.5 | 306.2500 |
8.09 | 65.4481 |
16.51 | 272.5801 |
14.9 | 222.0100 |
15.35 | 235.6225 |
15.72 | 247.1184 |
11.9 | 141.6100 |
33.67 | 1133.6689 |
9.72 | 94.4784 |
18.3 | 334.8900 |
sum=355.95 | sum=7494.3015 |
a).the mean be:-
b).the sample standard deviation be:-
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