http://www.ck12.org Chapter 15. Concepts of Statistics
59.8, 57.1, 58.2, 58.6, 57.8, 57.9, 58.0, 57.3
Solution:x=^18 ( 59. 8 + 57. 1 + 58. 2 + 58. 6 + 57. 8 + 57. 9 + 58. 0 + 57. 3 ) = 58. 0875
This is asample, so you should use the sample variance formula.
s^2 = 8 −^11 ·[(μ− 59. 8 )^2 +(μ− 57. 1 )^2 +(μ− 58. 2 )^2 +(μ− 58. 6 )^2 +(μ− 57. 8 )^2
+(μ− 57. 9 )^2 +(μ− 58. 0 )^2 +(μ− 57. 3 )^2 ]
=^17 [(− 1. 7125 )^2 + 0. 98752 +(− 0. 1125 )^2 +(− 0. 5125 )^2 + 0. 28752 + 0. 18752 + 0. 08752 + 0. 78752 ]
≈^17 [ 2. 9327 + 0. 9751 + 0. 0126 + 0. 2626 + 0. 0826 + 0. 0351 + 0. 0076 + 0. 6201 ]
≈^17 [ 4. 9288 ]
≈ 0. 7041Example C
Use a calculator to calculate the variance from Example B.
Solution:To calculate variance on your calculator, enter the data in a list, choose 1-Var Stats and run the 1-Var Stats
on the list you entered the data.
The two outputs that are important for you to interpret are:
Sx= 0. 839110924
σx= 0. 7848163968
Since the calculator does not know whether the data is a population or a sample, it produces both. Since this problem
is about a sample, the number of interest isSx. This number does not match the variance from Example B because
it is the sample standard deviation which means it is the square root of the sample variance. The calculator produces
standard deviation. You need to square that number to produce the appropriate variance.
0. 83912 ≈ 0. 7041