WEBVTT 1 00:00:01.470 --> 00:00:02.303 <v Instructor>Hello, students.</v> 2 00:00:02.303 --> 00:00:06.150 Now we are in part C, which is slightly different. 3 00:00:06.150 --> 00:00:08.790 So first I'm going to read the problem again. 4 00:00:08.790 --> 00:00:10.680 So, it's problem three. 5 00:00:10.680 --> 00:00:13.860 Total cholesterol in children age 10 to 15 years of age 6 00:00:13.860 --> 00:00:16.050 is assumed to follow a normal distribution 7 00:00:16.050 --> 00:00:21.050 with a mean of 191 and a standard deviation of 22.4. 8 00:00:21.450 --> 00:00:24.540 So in part C, the question is if a sample 9 00:00:24.540 --> 00:00:27.840 of 20 children is selected, what is the probability 10 00:00:27.840 --> 00:00:30.660 that the mean cholesterol level in the sample 11 00:00:30.660 --> 00:00:32.763 will exceed 200? 12 00:00:34.170 --> 00:00:36.580 So what is going on here is that 13 00:00:37.440 --> 00:00:41.120 now instead of looking at the entire population, 14 00:00:43.140 --> 00:00:46.260 we are looking at a sample of 20 children. 15 00:00:46.260 --> 00:00:48.900 So we will draw the normal distribution, 16 00:00:48.900 --> 00:00:51.120 just like we did before, 17 00:00:51.120 --> 00:00:53.880 but now instead of the total population, 18 00:00:53.880 --> 00:00:57.213 this will be our sampling distribution of means. 19 00:00:59.160 --> 00:01:01.350 And as per the central limit theorem, 20 00:01:01.350 --> 00:01:03.780 the mean of our sampling distribution 21 00:01:03.780 --> 00:01:07.830 is going to be equal to our mu, our population mean, 22 00:01:07.830 --> 00:01:10.980 and the standard error of that distribution 23 00:01:10.980 --> 00:01:12.840 will be equal to the sigma, 24 00:01:12.840 --> 00:01:14.670 or population standard deviation, 25 00:01:14.670 --> 00:01:19.170 divided by the square root of our sample size, 26 00:01:19.170 --> 00:01:20.520 which is here, 20. 27 00:01:20.520 --> 00:01:23.650 So, without further delay I'm going to start 28 00:01:25.170 --> 00:01:27.150 writing some of this information. 29 00:01:27.150 --> 00:01:30.213 So here X-bar is equal to mu, 30 00:01:31.770 --> 00:01:35.550 and SE, which is the abbreviation for standard error, 31 00:01:35.550 --> 00:01:40.020 is equal to sigma divided by square root of 20, 32 00:01:40.020 --> 00:01:41.850 which is our end. 33 00:01:41.850 --> 00:01:44.820 So now I'm going to draw our normal distribution, 34 00:01:44.820 --> 00:01:46.860 because nothing changes there. 35 00:01:46.860 --> 00:01:48.160 We are going to draw that. 36 00:01:51.300 --> 00:01:55.800 So here now is 191, 37 00:01:55.800 --> 00:01:58.023 but this now will be an X-bar. 38 00:02:00.240 --> 00:02:01.073 Okay? 39 00:02:04.230 --> 00:02:07.350 Now, we will do exactly the same thing as we did before 40 00:02:07.350 --> 00:02:10.320 because looking at the probability 41 00:02:10.320 --> 00:02:15.030 that we are trying to find out the value greater than 200. 42 00:02:15.030 --> 00:02:18.780 So just like before, we will have to transform it 43 00:02:18.780 --> 00:02:21.600 to the standard normal distribution 44 00:02:21.600 --> 00:02:24.810 so that we can use our table in the back of our book 45 00:02:24.810 --> 00:02:26.550 to find the probability. 46 00:02:26.550 --> 00:02:29.280 So now, again, I'm going to draw the area 47 00:02:29.280 --> 00:02:33.483 we are looking for under the curve, 48 00:02:34.740 --> 00:02:36.693 which is, again, that 200. 49 00:02:37.680 --> 00:02:39.273 Kind of seen this before. 50 00:02:41.610 --> 00:02:42.960 Okay, this is the area. 51 00:02:44.250 --> 00:02:46.750 And now we have transformed this to 52 00:02:49.380 --> 00:02:51.783 the distribution, and this is zero. 53 00:02:53.038 --> 00:02:57.300 Okay, so we have our Z-axis here with a mean of zero, 54 00:02:57.300 --> 00:02:59.610 and we need to now calculate the values. 55 00:02:59.610 --> 00:03:01.380 Again, this time the formula 56 00:03:01.380 --> 00:03:03.150 is going to be a little different. 57 00:03:03.150 --> 00:03:05.290 So it's going to be Z equal to 58 00:03:06.690 --> 00:03:10.350 X-bar minus mu 59 00:03:10.350 --> 00:03:14.913 divided by sigma and square root of N. 60 00:03:17.910 --> 00:03:20.510 So, again, the probability we are looking for is the 61 00:03:23.760 --> 00:03:26.700 value above 200, and we have to go through 62 00:03:26.700 --> 00:03:29.070 all the transformation as before. 63 00:03:29.070 --> 00:03:32.973 So now it is Z greater than. 64 00:03:35.730 --> 00:03:38.433 So we are going to use the formula here. 65 00:03:44.700 --> 00:03:45.843 Square root of 20. 66 00:03:48.420 --> 00:03:52.990 Alright, so once we do the algebra, what we find is 67 00:03:56.850 --> 00:03:59.820 this will be 1.80. 68 00:03:59.820 --> 00:04:03.990 Now, again, we have to go to the back of our text, 69 00:04:03.990 --> 00:04:07.140 and what our text will provide us, as before, 70 00:04:07.140 --> 00:04:09.453 is the value less than. 71 00:04:10.530 --> 00:04:13.560 So this is what we are getting 72 00:04:13.560 --> 00:04:16.203 from the back of our textbook. 73 00:04:20.580 --> 00:04:24.180 So, again, going back to our normal distribution 74 00:04:24.180 --> 00:04:28.080 and using a different color, 75 00:04:28.080 --> 00:04:32.460 this is what we have right now. 76 00:04:32.460 --> 00:04:35.490 Oops, color didn't work well. 77 00:04:35.490 --> 00:04:36.903 Let's see, yes. 78 00:04:39.180 --> 00:04:43.890 So this is the area for which we have the value, 79 00:04:43.890 --> 00:04:45.810 and we need to find the value 80 00:04:45.810 --> 00:04:49.230 for the area that's kind of highlighted in red. 81 00:04:49.230 --> 00:04:51.030 So what do we do here, again? 82 00:04:51.030 --> 00:04:52.620 Thankfully we know 83 00:04:52.620 --> 00:04:57.240 that the entire area under the curve is one. 84 00:04:57.240 --> 00:04:59.970 So basically we have to deduct it from one. 85 00:04:59.970 --> 00:05:02.913 So let's do that without any further delay. 86 00:05:05.370 --> 00:05:10.320 So P, probability 87 00:05:10.320 --> 00:05:14.010 that the Z is greater than 1.80 88 00:05:14.010 --> 00:05:19.010 is basically one minus the probability 89 00:05:19.590 --> 00:05:23.133 that the Z less than 1.80. 90 00:05:24.930 --> 00:05:28.443 So the what's gonna then be one minus point- 91 00:05:33.120 --> 00:05:34.858 Oops, sorry. 92 00:05:34.858 --> 00:05:36.690 It should be 0.9641, 93 00:05:41.502 --> 00:05:43.540 and this will be 94 00:05:45.360 --> 00:05:50.040 0.0359. 95 00:05:50.040 --> 00:05:51.730 And that is basically 96 00:05:53.580 --> 00:05:55.043 3.59%. 97 00:05:58.380 --> 00:05:59.520 So what does that mean? 98 00:05:59.520 --> 00:06:01.560 So, again, given the values, 99 00:06:01.560 --> 00:06:05.730 if we take a sample of 20 children from this population, 100 00:06:05.730 --> 00:06:10.730 then there is a 3.59% probability 101 00:06:11.070 --> 00:06:14.130 that the mean cholesterol of those 20 children 102 00:06:14.130 --> 00:06:15.667 will be greater than 200.