WEBVTT 1 00:00:03.090 --> 00:00:03.923 Hello students. 2 00:00:03.923 --> 00:00:06.360 So this is 1F, problem 1F, 3 00:00:06.360 --> 00:00:09.240 which is the last problem from this set. 4 00:00:09.240 --> 00:00:13.170 And in this one, what we are being asked is 5 00:00:13.170 --> 00:00:14.283 which measure, 6 00:00:18.501 --> 00:00:19.334 the standard deviation 7 00:00:19.334 --> 00:00:21.690 or the interquartile range is a better measure 8 00:00:21.690 --> 00:00:24.633 of dispersion and justify our answer. 9 00:00:26.135 --> 00:00:28.440 So because there are outliers, the best measure 10 00:00:28.440 --> 00:00:31.800 of dispersion is going to be our standard deviation. 11 00:00:31.800 --> 00:00:36.800 So because no outliers exist 12 00:00:45.420 --> 00:00:46.710 in our dataset, 13 00:00:46.710 --> 00:00:48.363 which we just calculated, 14 00:00:56.490 --> 00:00:58.920 best measure of dispersion 15 00:00:58.920 --> 00:01:00.933 is going to be standard deviation. 16 00:01:18.000 --> 00:01:19.860 And we have already calculated 17 00:01:19.860 --> 00:01:22.503 that before and that is what? 29. 18 00:01:25.350 --> 00:01:30.240 So now I think in your mind, hopefully the next question is 19 00:01:30.240 --> 00:01:32.610 so what if we do have outliers? 20 00:01:32.610 --> 00:01:34.260 What will we do then? 21 00:01:34.260 --> 00:01:37.780 Okay, in a little while, I'm going to answer that question 22 00:01:41.970 --> 00:01:45.243 and you will learn what to do when we do have outliers.