WEBVTT

1
00:00:00.000 --> 00:00:05.480
Here is the problem: Sample standard deviation of 1, 2, 3, 4.

2
00:00:05.930 --> 00:00:12.931
Treat the data as a sample from a larger population (Bessel prime s correction, divide by n minus 1).

3
00:00:13.731 --> 00:00:14.579
Find the mean.

4
00:00:15.029 --> 00:00:17.293
Square each deviation from the mean.

5
00:00:17.743 --> 00:00:19.612
Add the squared deviations.

6
00:00:20.062 --> 00:00:23.742
Divide by n minus 1 to get the sample variance.

7
00:00:24.542 --> 00:00:25.796
Take the square root.

8
00:00:26.596 --> 00:00:32.494
So the answer is s equals 1.290994449.

9
00:00:32.944 --> 00:00:40.514
Standard deviation measures how spread out the values are around the mean. A small value means the data are clustered near the mean.

10
00:00:41.314 --> 00:00:54.886
Variance equals 1.666666667, and 1.290994449 squared equals 1.666666667.

11
00:00:55.336 --> 00:01:01.698
Population SD for comparison: 1.118033989.