WEBVTT

1
00:00:00.000 --> 00:00:06.536
Here is the problem: Sample standard deviation of 1, 2, 3, 4, 5.

2
00:00:06.986 --> 00:00:16.866
Treat the data as a sample from a larger population (Bessel prime s correction, divide by n minus 1). The variance is the square of the standard deviation.

3
00:00:17.666 --> 00:00:18.525
Find the mean.

4
00:00:18.975 --> 00:00:21.320
Square each deviation from the mean.

5
00:00:21.770 --> 00:00:23.662
Add the squared deviations.

6
00:00:24.112 --> 00:00:28.048
Divide by n minus 1 to get the sample variance.

7
00:00:28.848 --> 00:00:30.055
Take the square root.

8
00:00:30.855 --> 00:00:34.419
So the answer is s squared equals 2.5.

9
00:00:34.869 --> 00:00:42.625
Standard deviation measures how spread out the values are around the mean. A small value means the data are clustered near the mean.

10
00:00:43.424 --> 00:00:51.644
Variance equals 2.5, and 1.58113883 squared equals 2.5.

11
00:00:52.094 --> 00:00:58.561
Population SD for comparison: 1.414213562.
