Question

For the data set: 0.09, 0.10, 0.11, 0.13, 0.09, 0.11, 0.10, 0.07

a) To obtain information of the precision of the data set the standard deviation would be?
b) Which variance would describe the precision of the data?

Answer 1

Sol:

X	Xbar	X-xbar	(X-xbar)^2
0.09	0.1	-0.01	0.0001
0.10	0.1	0.00	0
0.11	0.1	0.01	1E-04
0.13	0.1	0.03	0.0009
0.09	0.1	-0.01	0.0001
0.11	0.1	0.01	1E-04
0.10	0.1	0.00	0
0.07	0.1	-0.03	0.0009
		total	0.0022
xbar=sumof x values/n
xbar=0.80/8
xbar=0.1

sample  standard deviation=s=sqrt(0.0022/8-1)

=0.01772811

et the standard deviation would be=0.01772811

b) Which variance would describe the precision of the data?

sample variance

sample variance=s^2=0.01772811*0.01772811=0.0003142859