1 00:00:08,370 --> 00:00:13,950 Today, inshallah, we'll start chapter six. Chapter 2 00:00:13,950 --> 00:00:20,350 six talks about the normal distribution. In this 3 00:00:20,350 --> 00:00:24,810 chapter, there are mainly two objectives. The 4 00:00:24,810 --> 00:00:30,470 first objective is to compute probabilities from 5 00:00:30,470 --> 00:00:34,530 normal distribution. And mainly we'll focus on 6 00:00:34,530 --> 00:00:37,270 objective number one. So we are going to use 7 00:00:37,270 --> 00:00:40,290 normal distribution in this chapter. And we'll 8 00:00:40,290 --> 00:00:43,830 know how can we compute probabilities if the data 9 00:00:43,830 --> 00:00:46,810 set is normally distributed. You know many times 10 00:00:46,810 --> 00:00:50,690 you talked about extreme points or outliers. So 11 00:00:50,690 --> 00:00:54,490 that means if the data has outliers, that is the 12 00:00:54,490 --> 00:00:57,290 distribution is not normally distributed. Now in 13 00:00:57,290 --> 00:01:01,090 this case, If the distribution is normal, how can 14 00:01:01,090 --> 00:01:04,350 we compute probabilities underneath the normal 15 00:01:04,350 --> 00:01:10,030 curve? The second objective is to use the normal 16 00:01:10,030 --> 00:01:13,210 probability plot to determine whether a set of 17 00:01:13,210 --> 00:01:18,150 data is approximately normally distributed. I mean 18 00:01:18,150 --> 00:01:25,550 beside box plots we discussed before. Beside this 19 00:01:25,550 --> 00:01:30,190 score, how can we tell if the data point or 20 00:01:30,190 --> 00:01:35,350 actually the entire distribution is approximately 21 00:01:35,350 --> 00:01:39,410 normally distributed or not. Before we learn if 22 00:01:39,410 --> 00:01:44,110 the point is outlier by using backsplot and this 23 00:01:44,110 --> 00:01:46,750 score. In this chapter we'll know how can we 24 00:01:46,750 --> 00:01:51,630 determine if the entire distribution is 25 00:01:51,630 --> 00:01:54,770 approximately normal distributed. So there are two 26 00:01:54,770 --> 00:01:56,710 objectives. One is to compute probabilities 27 00:01:56,710 --> 00:01:59,370 underneath the normal curve. The other, how can we 28 00:01:59,370 --> 00:02:05,310 tell if the data set is out or not? If you 29 00:02:05,310 --> 00:02:09,330 remember, first class, we mentioned something 30 00:02:09,330 --> 00:02:13,130 about data types. And we said data has mainly two 31 00:02:13,130 --> 00:02:17,930 types. Numerical data, I mean quantitative data. 32 00:02:18,690 --> 00:02:22,630 and categorical data, qualitative. For numerical 33 00:02:22,630 --> 00:02:26,190 data also it has two types, continuous and 34 00:02:26,190 --> 00:02:30,430 discrete. And discrete takes only integers such as 35 00:02:30,430 --> 00:02:35,310 number of students who take this class or number 36 00:02:35,310 --> 00:02:40,190 of accidents and so on. But if you are talking 37 00:02:40,190 --> 00:02:45,320 about Age, weight, scores, temperature, and so on. 38 00:02:45,560 --> 00:02:49,260 It's continuous distribution. For this type of 39 00:02:49,260 --> 00:02:53,320 variable, I mean for continuous distribution, how 40 00:02:53,320 --> 00:02:56,300 can we compute the probabilities underneath the 41 00:02:56,300 --> 00:02:59,640 normal? So normal distribution maybe is the most 42 00:02:59,640 --> 00:03:02,380 common distribution in statistics, and it's type 43 00:03:02,380 --> 00:03:07,820 of continuous distribution. So first, let's define 44 00:03:07,820 --> 00:03:12,010 continuous random variable. maybe because for 45 00:03:12,010 --> 00:03:15,230 multiple choice problem you should know the 46 00:03:15,230 --> 00:03:19,110 definition of continuous random variable is a 47 00:03:19,110 --> 00:03:22,070 variable that can assume any value on a continuous 48 00:03:23,380 --> 00:03:27,020 it can assume any uncountable number of values. So 49 00:03:27,020 --> 00:03:31,080 it could be any number in an interval. For 50 00:03:31,080 --> 00:03:35,720 example, suppose your ages range between 18 years 51 00:03:35,720 --> 00:03:39,580 and 20 years. So maybe someone of you, their age 52 00:03:39,580 --> 00:03:44,000 is about 18 years, three months. Or maybe your 53 00:03:44,000 --> 00:03:47,580 weight is 70 kilogram point five, and so on. So 54 00:03:47,580 --> 00:03:49,780 it's continuous on the variable. Other examples 55 00:03:49,780 --> 00:03:53,140 for continuous, thickness of an item. For example, 56 00:03:53,740 --> 00:03:54,440 the thickness. 57 00:03:58,260 --> 00:04:02,490 This one is called thickness. Now, the thickness 58 00:04:02,490 --> 00:04:05,930 may be 2 centimeters or 3 centimeters and so on, 59 00:04:06,210 --> 00:04:09,730 but it might be 2.5 centimeters. For example, for 60 00:04:09,730 --> 00:04:13,030 this remote, the thickness is 2.5 centimeters or 2 61 00:04:13,030 --> 00:04:16,510 .6, not exactly 2 or 3. So it could be any value. 62 00:04:16,650 --> 00:04:19,450 Range is, for example, between 2 centimeters and 3 63 00:04:19,450 --> 00:04:23,010 centimeters. So from 2 to 3 is a big range because 64 00:04:23,010 --> 00:04:25,670 it can take anywhere from 2.1 to 2.15 and so on. 65 00:04:26,130 --> 00:04:28,810 So thickness is an example of continuous random 66 00:04:28,810 --> 00:04:31,190 variable. Another example, time required to 67 00:04:31,190 --> 00:04:36,010 complete a task. Now suppose you want to do an 68 00:04:36,010 --> 00:04:39,710 exercise. Now the time required to finish or to 69 00:04:39,710 --> 00:04:45,150 complete this task may be any value between 2 70 00:04:45,150 --> 00:04:48,730 minutes up to 3 minutes. So maybe 2 minutes 30 71 00:04:48,730 --> 00:04:52,150 seconds, 2 minutes 40 seconds and so on. So it's 72 00:04:52,150 --> 00:04:55,550 continuous random variable. Temperature of a 73 00:04:55,550 --> 00:05:00,140 solution. height, weight, ages, and so on. These 74 00:05:00,140 --> 00:05:03,720 are examples of continuous random variable. So 75 00:05:03,720 --> 00:05:08,040 these variables can potentially take on any value 76 00:05:08,040 --> 00:05:11,340 depending only on the ability to precisely and 77 00:05:11,340 --> 00:05:14,020 accurately measure. So that's the definition of 78 00:05:14,020 --> 00:05:17,320 continuous random variable. Now, if you look at 79 00:05:17,320 --> 00:05:21,810 the normal distribution, It looks like bell 80 00:05:21,810 --> 00:05:25,990 -shaped, as we discussed before. So it's bell 81 00:05:25,990 --> 00:05:31,270 -shaped, symmetrical. Symmetrical means the area 82 00:05:31,270 --> 00:05:34,390 to the right of the mean equals the area to the 83 00:05:34,390 --> 00:05:37,950 left of the mean. I mean 50% of the area above and 84 00:05:37,950 --> 00:05:41,770 50% below. So that's the meaning of symmetrical. 85 00:05:42,490 --> 00:05:46,370 The other feature of normal distribution, the 86 00:05:46,370 --> 00:05:49,510 measures of center tendency are equal or 87 00:05:49,510 --> 00:05:53,170 approximately equal. Mean, median, and mode are 88 00:05:53,170 --> 00:05:55,530 roughly equal. In reality, they are not equal, 89 00:05:55,650 --> 00:05:58,210 exactly equal, but you can say they are 90 00:05:58,210 --> 00:06:01,850 approximately equal. Now, there are two parameters 91 00:06:01,850 --> 00:06:05,750 describing the normal distribution. One is called 92 00:06:05,750 --> 00:06:10,820 the location parameter. location, or central 93 00:06:10,820 --> 00:06:13,800 tendency, as we discussed before, location is 94 00:06:13,800 --> 00:06:17,160 determined by the mean mu. So the first parameter 95 00:06:17,160 --> 00:06:20,340 for the normal distribution is the mean mu. The 96 00:06:20,340 --> 00:06:24,240 other parameter measures the spread of the data, 97 00:06:24,280 --> 00:06:27,680 or the variability of the data, and the spread is 98 00:06:27,680 --> 00:06:31,860 sigma, or the variation. So we have two 99 00:06:31,860 --> 00:06:36,770 parameters, mu and sigma. The random variable in 100 00:06:36,770 --> 00:06:39,930 this case can take any value from minus infinity 101 00:06:39,930 --> 00:06:44,270 up to infinity. So random variable in this case 102 00:06:44,270 --> 00:06:50,310 continuous ranges from minus infinity all the way 103 00:06:50,310 --> 00:06:55,100 up to infinity. I mean from this point here up to 104 00:06:55,100 --> 00:06:58,380 infinity. So the values range from minus infinity 105 00:06:58,380 --> 00:07:02,080 up to infinity. And if you look here, the mean is 106 00:07:02,080 --> 00:07:05,600 located nearly in the middle. And mean and median 107 00:07:05,600 --> 00:07:10,820 are all approximately equal. That's the features 108 00:07:10,820 --> 00:07:14,740 or the characteristics of the normal distribution. 109 00:07:16,460 --> 00:07:20,360 Now, how can we compute the probabilities under 110 00:07:20,360 --> 00:07:25,840 the normal killer? The formula that is used to 111 00:07:25,840 --> 00:07:29,220 compute the probabilities is given by this one. It 112 00:07:29,220 --> 00:07:33,560 looks complicated formula because we have to use 113 00:07:33,560 --> 00:07:36,040 calculus in order to determine the area underneath 114 00:07:36,040 --> 00:07:40,120 the cube. So we are looking for something else. So 115 00:07:40,120 --> 00:07:45,300 this formula is it seems to be complicated. It's 116 00:07:45,300 --> 00:07:49,600 not hard but it's complicated one, but we can use 117 00:07:49,600 --> 00:07:52,380 it. If we know calculus very well, we can use 118 00:07:52,380 --> 00:07:55,240 integration to create the probabilities underneath 119 00:07:55,240 --> 00:07:58,900 the curve. But for our course, we are going to 120 00:07:58,900 --> 00:08:04,460 skip this formula because this 121 00:08:04,460 --> 00:08:09,340 formula depends actually on mu and sigma. A mu can 122 00:08:09,340 --> 00:08:13,110 take any value. Sigma also can take any value. 123 00:08:13,930 --> 00:08:17,310 That means we have different normal distributions. 124 00:08:18,470 --> 00:08:23,830 Because the distribution actually depends on these 125 00:08:23,830 --> 00:08:27,610 two parameters. So by varying the parameters mu 126 00:08:27,610 --> 00:08:29,790 and sigma, we obtain different normal 127 00:08:29,790 --> 00:08:32,710 distributions. Since we have different mu and 128 00:08:32,710 --> 00:08:36,310 sigma, it means we should have different normal 129 00:08:36,310 --> 00:08:38,770 distributions. For this reason, it's very 130 00:08:38,770 --> 00:08:43,430 complicated to have tables or probability tables 131 00:08:43,430 --> 00:08:46,010 in order to determine these probabilities because 132 00:08:46,010 --> 00:08:50,130 there are infinite values of mu and sigma maybe 133 00:08:50,130 --> 00:08:57,750 your edges the mean is 19. Sigma is, for example, 134 00:08:57,910 --> 00:09:01,990 5. For weights, maybe the mean is 70 kilograms, 135 00:09:02,250 --> 00:09:04,990 the average is 10. For scores, maybe the average 136 00:09:04,990 --> 00:09:08,710 is 65, the mean is 20, sigma is 20, and so on. So 137 00:09:08,710 --> 00:09:11,090 we have different values of mu and sigma. For this 138 00:09:11,090 --> 00:09:13,650 reason, we have different normal distributions. 139 00:09:18,490 --> 00:09:25,740 Because changing mu shifts the distribution either 140 00:09:25,740 --> 00:09:29,640 left or to the right. So maybe the mean is shifted 141 00:09:29,640 --> 00:09:32,440 to the right side, or the mean maybe shifted to 142 00:09:32,440 --> 00:09:37,140 the left side. Also, changing sigma, sigma is the 143 00:09:37,140 --> 00:09:40,660 distance between the mu and the curve. The curve 144 00:09:40,660 --> 00:09:45,220 is the points, or the data values. Now this sigma 145 00:09:45,220 --> 00:09:48,380 can be increases or decreases. So if sigma 146 00:09:48,380 --> 00:09:52,860 increases, it means the spread also increases. Or 147 00:09:52,860 --> 00:09:55,780 if sigma decreases, also the spread will decrease. 148 00:09:56,200 --> 00:09:59,660 So the distribution or the normal distribution 149 00:09:59,660 --> 00:10:02,820 depends actually on these two values. For this 150 00:10:02,820 --> 00:10:05,120 reason, since we have too many values or infinite 151 00:10:05,120 --> 00:10:07,600 values of mu and sigma, then in this case we have 152 00:10:07,600 --> 00:10:14,500 different normal distributions. There is another 153 00:10:14,500 --> 00:10:16,940 distribution. It's called standardized normal. 154 00:10:20,330 --> 00:10:26,070 Now, we have normal distribution X, and how can we 155 00:10:26,070 --> 00:10:31,930 transform from normal distribution to standardized 156 00:10:31,930 --> 00:10:35,310 normal distribution? The reason is that the mean 157 00:10:35,310 --> 00:10:40,310 of Z, I mean, Z is used for standardized normal. 158 00:10:40,850 --> 00:10:44,490 The mean of Z is always zero, and sigma is one. 159 00:10:45,770 --> 00:10:48,150 Now it's a big difference. The first one has 160 00:10:48,150 --> 00:10:53,160 infinite values of Mu and Sigma. Now, for the 161 00:10:53,160 --> 00:10:56,200 standardized normal distribution, the mean is 162 00:10:56,200 --> 00:11:01,540 fixed value. The mean is zero, Sigma is one. So, 163 00:11:01,620 --> 00:11:04,340 the question is, how can we actually transform 164 00:11:04,340 --> 00:11:09,720 from X, which has normal distribution, to Z, which 165 00:11:09,720 --> 00:11:13,160 has standardized normal with mean zero and Sigma 166 00:11:13,160 --> 00:11:23,330 of one. Let's see. How can we translate x which 167 00:11:23,330 --> 00:11:27,510 has normal distribution to z that has standardized 168 00:11:27,510 --> 00:11:32,190 normal distribution? The idea is you have just to 169 00:11:32,190 --> 00:11:39,170 subtract mu of x, x minus mu, then divide this 170 00:11:39,170 --> 00:11:43,150 result by sigma. So we just subtract the mean of 171 00:11:43,150 --> 00:11:49,660 x. and dividing by its standard deviation now so 172 00:11:49,660 --> 00:11:52,360 if we have x which has normal distribution with 173 00:11:52,360 --> 00:11:55,940 mean mu and standard deviation sigma to transform 174 00:11:55,940 --> 00:12:00,960 or to convert to z score use this formula x minus 175 00:12:00,960 --> 00:12:05,220 the mean then divide by its standard deviation now 176 00:12:05,220 --> 00:12:09,090 all of the time we are going to use z for 177 00:12:09,090 --> 00:12:12,230 standardized normal distribution and always z has 178 00:12:12,230 --> 00:12:15,370 mean zero and all and sigma or standard deviation. 179 00:12:16,250 --> 00:12:20,170 So the z distribution always has mean of zero and 180 00:12:20,170 --> 00:12:25,490 sigma of one. So that's the story of standardizing 181 00:12:25,490 --> 00:12:33,070 the normal value. Now the Formula for this score 182 00:12:33,070 --> 00:12:37,570 becomes better than the first one, but still we 183 00:12:37,570 --> 00:12:40,570 have to use calculus in order to determine the 184 00:12:40,570 --> 00:12:45,710 probabilities under the standardized normal k. But 185 00:12:45,710 --> 00:12:49,470 this distribution has mean of zero and sigma of 186 00:12:49,470 --> 00:12:56,910 one. So we have a table on page 570. Look at page 187 00:12:56,910 --> 00:13:00,910 570. We have table or actually there are two 188 00:13:00,910 --> 00:13:05,010 tables. One for negative value of Z and the other 189 00:13:05,010 --> 00:13:08,830 for positive value of Z. So we have two tables for 190 00:13:08,830 --> 00:13:14,730 positive and negative values of Z on page 570 and 191 00:13:14,730 --> 00:13:15,470 571. 192 00:13:17,870 --> 00:13:22,770 Now the table on page 570 looks like this one. The 193 00:13:22,770 --> 00:13:26,610 table you have starts from minus 6, then minus 5, 194 00:13:26,750 --> 00:13:32,510 minus 4.5, and so on. Here we start from minus 3.4 195 00:13:32,510 --> 00:13:38,850 all the way down up to 0. Look here, all the way 196 00:13:38,850 --> 00:13:44,490 up to 0. So these scores here. Also we have 0.00, 197 00:13:44,610 --> 00:13:51,880 0.01, up to 0.09. Also, the other page, page 571, 198 00:13:52,140 --> 00:13:56,940 gives the area for positive z values. Here we have 199 00:13:56,940 --> 00:14:01,760 0.0, 0.1, 0.2, all the way down up to 3.4 and you 200 00:14:01,760 --> 00:14:05,920 have up to 6. Now let's see how can we use this 201 00:14:05,920 --> 00:14:11,020 table to compute the probabilities underneath the 202 00:14:11,020 --> 00:14:12,460 normal curve. 203 00:14:14,940 --> 00:14:19,190 First of all, you have to know that Z has mean 204 00:14:19,190 --> 00:14:23,750 zero, standard deviation of one. And the values 205 00:14:23,750 --> 00:14:26,610 could be positive or negative. Values above the 206 00:14:26,610 --> 00:14:32,850 mean, zero, have positive Z values. The other one, 207 00:14:32,910 --> 00:14:36,690 values below the mean, have negative Z values. So 208 00:14:36,690 --> 00:14:42,770 Z score can be negative or positive. Now this is 209 00:14:42,770 --> 00:14:46,530 the formula we have, z equals x minus mu divided 210 00:14:46,530 --> 00:14:46,990 by six. 211 00:14:52,810 --> 00:15:01,170 Now this value could be positive if x is above the 212 00:15:01,170 --> 00:15:04,810 mean, as we mentioned before. It could be a 213 00:15:04,810 --> 00:15:09,870 negative if x is smaller than the mean or zero. 214 00:15:13,120 --> 00:15:18,140 Now the table we have gives the area to the right, 215 00:15:18,420 --> 00:15:21,240 to the left, I'm sorry, to the left, for positive 216 00:15:21,240 --> 00:15:26,220 and negative values of z. Okay, so we have two 217 00:15:26,220 --> 00:15:32,160 tables actually, one for negative on page 570, and 218 00:15:32,160 --> 00:15:38,260 the other one for positive values of z. I think we 219 00:15:38,260 --> 00:15:41,060 discussed that before when we talked about these 220 00:15:41,060 --> 00:15:44,080 scores. We have the same formula. 221 00:15:47,120 --> 00:15:53,700 Now let's look at this, the next slide. Suppose x 222 00:15:53,700 --> 00:16:01,880 is distributed normally with mean of 100. So the 223 00:16:01,880 --> 00:16:06,470 mean of x is 100. and the standard deviation of 224 00:16:06,470 --> 00:16:11,110 50. So sigma is 50. Now let's see how can we 225 00:16:11,110 --> 00:16:17,750 compute the z-score for x equals 200. Again the 226 00:16:17,750 --> 00:16:22,790 formula is just x minus mu divided by sigma x 200 227 00:16:22,790 --> 00:16:28,330 minus 100 divided by 50 that will give 2. Now the 228 00:16:28,330 --> 00:16:33,910 sign of this value is positive That means x is 229 00:16:33,910 --> 00:16:37,950 greater than the mean, because x is 200. Now, 230 00:16:37,990 --> 00:16:42,270 what's the meaning of 2? What does this value tell 231 00:16:42,270 --> 00:16:42,410 you? 232 00:16:48,230 --> 00:16:55,430 Yeah, exactly. x equals 200 is two standard 233 00:16:55,430 --> 00:16:58,690 deviations above the mean. Because if you look at 234 00:16:58,690 --> 00:17:05,210 200, the x value, The mean is 100, sigma is 50. 235 00:17:05,730 --> 00:17:09,690 Now the difference between the score, which is 236 00:17:09,690 --> 00:17:16,810 200, and the mu, which is 100, is equal to 237 00:17:16,810 --> 00:17:18,690 standard deviations, because the difference is 238 00:17:18,690 --> 00:17:24,230 100. 2 times 50 is 100. So this says that x equals 239 00:17:24,230 --> 00:17:29,070 200 is 2 standard deviations above the mean. If z 240 00:17:29,070 --> 00:17:34,330 is negative, you can say that x is two standard 241 00:17:34,330 --> 00:17:38,710 deviations below them. Make sense? So that's how 242 00:17:38,710 --> 00:17:42,670 can we compute the z square. Now, when we 243 00:17:42,670 --> 00:17:45,970 transform from normal distribution to 244 00:17:45,970 --> 00:17:49,490 standardized, still we will have the same shape. I 245 00:17:49,490 --> 00:17:51,350 mean the distribution is still normally 246 00:17:51,350 --> 00:17:55,800 distributed. So note, the shape of the 247 00:17:55,800 --> 00:17:58,840 distribution is the same, only the scale has 248 00:17:58,840 --> 00:18:04,500 changed. So we can express the problem in original 249 00:18:04,500 --> 00:18:10,640 units, X, or in a standardized unit, Z. So when we 250 00:18:10,640 --> 00:18:16,620 have X, just use this equation to transform to 251 00:18:16,620 --> 00:18:17,160 this form. 252 00:18:21,360 --> 00:18:23,200 Now, for example, suppose we have normal 253 00:18:23,200 --> 00:18:26,040 distribution and we are interested in the area 254 00:18:26,040 --> 00:18:32,660 between A and B. Now, the area between A and B, it 255 00:18:32,660 --> 00:18:34,700 means the probability between them. So 256 00:18:34,700 --> 00:18:39,140 statistically speaking, area means probability. So 257 00:18:39,140 --> 00:18:42,700 probability between A and B, I mean probability of 258 00:18:42,700 --> 00:18:45,380 X greater than or equal A and less than or equal B 259 00:18:45,380 --> 00:18:49,420 is the same as X greater than A or less than B. 260 00:18:50,450 --> 00:18:57,210 that means the probability of X equals A this 261 00:18:57,210 --> 00:19:02,510 probability is zero or probability of X equals B 262 00:19:02,510 --> 00:19:06,930 is also zero so in continuous distribution the 263 00:19:06,930 --> 00:19:10,630 equal sign does not matter I mean if we have equal 264 00:19:10,630 --> 00:19:15,130 sign or we don't have these probabilities are the 265 00:19:15,130 --> 00:19:19,390 same so I mean for example if we are interested 266 00:19:20,310 --> 00:19:23,450 for probability of X smaller than or equal to E. 267 00:19:24,850 --> 00:19:30,370 This probability is the same as X smaller than E. 268 00:19:31,330 --> 00:19:33,730 Or on the other hand, if you are interested in the 269 00:19:33,730 --> 00:19:39,010 area above B greater than or equal to B, it's the 270 00:19:39,010 --> 00:19:44,770 same as X smaller than E. So don't worry about the 271 00:19:44,770 --> 00:19:48,660 equal sign. Or continuous distribution, exactly. 272 00:19:49,120 --> 00:19:53,820 But for discrete, it does matter. Now, since we 273 00:19:53,820 --> 00:19:58,200 are talking about normal distribution, and as we 274 00:19:58,200 --> 00:20:01,320 mentioned, normal distribution is symmetric around 275 00:20:01,320 --> 00:20:05,900 the mean, that means the area to the right equals 276 00:20:05,900 --> 00:20:09,340 the area to the left. Now the entire area 277 00:20:09,340 --> 00:20:12,940 underneath the normal curve equals one. I mean 278 00:20:12,940 --> 00:20:16,500 probability of X ranges from minus infinity up to 279 00:20:16,500 --> 00:20:21,500 infinity equals one. So probability of X greater 280 00:20:21,500 --> 00:20:26,920 than minus infinity up to infinity is one. The 281 00:20:26,920 --> 00:20:31,480 total area is one. So the area from minus infinity 282 00:20:31,480 --> 00:20:38,080 up to the mean mu is one-half. The same as the 283 00:20:38,080 --> 00:20:42,600 area from mu up to infinity is also one-half. That 284 00:20:42,600 --> 00:20:44,760 means the probability of X greater than minus 285 00:20:44,760 --> 00:20:48,300 infinity up to mu equals the probability from mu 286 00:20:48,300 --> 00:20:52,120 up to infinity because of symmetry. I mean you 287 00:20:52,120 --> 00:20:56,160 cannot say that for any distribution. Just for 288 00:20:56,160 --> 00:20:59,000 symmetric distribution, the area below the mean 289 00:20:59,000 --> 00:21:03,780 equals one-half, which is the same as the area to 290 00:21:03,780 --> 00:21:07,110 the right of the mean. So the entire Probability 291 00:21:07,110 --> 00:21:11,330 is one. And also you have to keep in mind that the 292 00:21:11,330 --> 00:21:17,570 probability always ranges between zero and one. So 293 00:21:17,570 --> 00:21:20,030 that means the probability couldn't be negative. 294 00:21:22,870 --> 00:21:27,730 It should be positive. It shouldn't be greater 295 00:21:27,730 --> 00:21:31,710 than one. So it's between zero and one. So always 296 00:21:31,710 --> 00:21:39,020 the probability lies between zero and one. The 297 00:21:39,020 --> 00:21:44,500 tables we have on page 570 and 571 give the area 298 00:21:44,500 --> 00:21:46,040 to the left side. 299 00:21:49,420 --> 00:21:54,660 For negative or positive z's. Now for example, 300 00:21:54,940 --> 00:22:03,060 suppose we are looking for probability of z less 301 00:22:03,060 --> 00:22:08,750 than 2. How can we find this probability by using 302 00:22:08,750 --> 00:22:12,210 the normal curve? Let's go back to this normal 303 00:22:12,210 --> 00:22:16,410 distribution. In the second page, we have positive 304 00:22:16,410 --> 00:22:17,070 z-scores. 305 00:22:23,850 --> 00:22:33,390 So we ask about the probability of z less than. So 306 00:22:33,390 --> 00:22:40,690 the second page, gives positive values of z. And 307 00:22:40,690 --> 00:22:44,590 the table gives the area below. And he asked about 308 00:22:44,590 --> 00:22:49,550 here, B of z is smaller than 2. Now 2, if you 309 00:22:49,550 --> 00:22:54,910 hear, up all the way down here, 2, 0, 0. So the 310 00:22:54,910 --> 00:23:00,530 answer is 9772. So this value, so the probability 311 00:23:00,530 --> 00:23:02,130 is 9772. 312 00:23:03,990 --> 00:23:05,390 Because it's 2. 313 00:23:09,510 --> 00:23:14,650 It's 2, 0, 0. But if you ask about what's the 314 00:23:14,650 --> 00:23:20,590 probability of Z less than 2.05? So this is 2. 315 00:23:23,810 --> 00:23:30,370 Now under 5, 9, 7, 9, 8. So the answer is 9, 7. 316 00:23:34,360 --> 00:23:38,900 Because this is two, and we need five decimal 317 00:23:38,900 --> 00:23:44,820 places. So all the way up to 9798. So this value 318 00:23:44,820 --> 00:23:54,380 is 2.05. Now it's about, it's more than 1.5, 319 00:23:55,600 --> 00:23:56,880 exactly 1.5. 320 00:24:02,140 --> 00:24:04,880 1.5. This is 1.5. 321 00:24:08,800 --> 00:24:09,720 9332. 322 00:24:12,440 --> 00:24:16,300 1.5. Exactly 1.5. So 9332. 323 00:24:18,780 --> 00:24:27,990 What's about probability less than 1.35? 1.3 all 324 00:24:27,990 --> 00:24:35,250 the way to 9.115. 9.115. 9.115. 9.115. 9.115. 9 325 00:24:35,250 --> 00:24:35,650 .115. 9.115. 326 00:24:41,170 --> 00:24:42,430 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9 327 00:24:42,430 --> 00:24:42,450 .115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9 328 00:24:42,450 --> 00:24:44,050 .115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9 329 00:24:44,050 --> 00:24:50,530 .115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9 330 00:24:50,530 --> 00:24:54,980 .115. 9. But here we are looking for the area to 331 00:24:54,980 --> 00:25:01,280 the right. One minus one. Now this area equals 332 00:25:01,280 --> 00:25:05,660 one minus because 333 00:25:05,660 --> 00:25:11,420 since suppose 334 00:25:11,420 --> 00:25:18,760 this is the 1.35 and we are interested in the area 335 00:25:18,760 --> 00:25:24,030 to the right or above 1.35. The table gives the 336 00:25:24,030 --> 00:25:28,230 area below. So the area above equals the total 337 00:25:28,230 --> 00:25:31,970 area underneath the curve is 1. So 1 minus this 338 00:25:31,970 --> 00:25:39,050 value, so equals 0.0885, 339 00:25:39,350 --> 00:25:42,250 and so on. So this is the way how can we compute 340 00:25:42,250 --> 00:25:47,850 the probabilities underneath the normal curve. if 341 00:25:47,850 --> 00:25:51,090 it's probability of z is smaller than then just 342 00:25:51,090 --> 00:25:55,910 use the table directly otherwise if we are talking 343 00:25:55,910 --> 00:26:00,390 about z greater than subtract from one to get the 344 00:26:00,390 --> 00:26:04,870 result that's how can we compute the probability 345 00:26:04,870 --> 00:26:13,750 of z less than or equal now 346 00:26:13,750 --> 00:26:18,890 let's see if we have x and x that has normal 347 00:26:18,890 --> 00:26:22,070 distribution with mean mu and standard deviation 348 00:26:22,070 --> 00:26:26,250 of sigma and let's see how can we compute the 349 00:26:26,250 --> 00:26:33,790 value of the probability mainly 350 00:26:33,790 --> 00:26:38,190 there are three steps to find the probability of x 351 00:26:38,190 --> 00:26:42,490 greater than a and less than b when x is 352 00:26:42,490 --> 00:26:47,000 distributed normally first step Draw normal curve 353 00:26:47,000 --> 00:26:54,880 for the problem in terms of x. So draw the normal 354 00:26:54,880 --> 00:26:58,140 curve first. Second, translate x values to z 355 00:26:58,140 --> 00:27:03,040 values by using the formula we have. z x minus mu 356 00:27:03,040 --> 00:27:06,440 divided by sigma. Then use the standardized normal 357 00:27:06,440 --> 00:27:15,140 table on page 570 and 571. For example, Let's see 358 00:27:15,140 --> 00:27:18,420 how can we find normal probabilities. Let's assume 359 00:27:18,420 --> 00:27:23,760 that X represents the time it takes to download an 360 00:27:23,760 --> 00:27:28,580 image from the internet. So suppose X, time 361 00:27:28,580 --> 00:27:33,760 required to download an image file from the 362 00:27:33,760 --> 00:27:38,460 internet. And suppose we know that the time is 363 00:27:38,460 --> 00:27:42,060 normally distributed for with mean of eight 364 00:27:42,060 --> 00:27:46,130 minutes. And standard deviation of five minutes. 365 00:27:46,490 --> 00:27:47,510 So we know the mean. 366 00:27:50,610 --> 00:27:59,670 Eight. Eight. And sigma of five minutes. And they 367 00:27:59,670 --> 00:28:03,410 ask about what's the probability of X smaller than 368 00:28:03,410 --> 00:28:07,990 eight one six. So first thing we have to compute, 369 00:28:08,170 --> 00:28:12,190 to draw the normal curve. The mean lies in the 370 00:28:12,190 --> 00:28:18,060 center. which is 8. He asked about probability of 371 00:28:18,060 --> 00:28:22,580 X smaller than 8.6. So we are interested in the 372 00:28:22,580 --> 00:28:27,920 area below 8.6. So it matched the table we have. 373 00:28:29,980 --> 00:28:34,900 Second step, we have to transform from normal 374 00:28:34,900 --> 00:28:37,280 distribution to standardized normal distribution 375 00:28:37,280 --> 00:28:42,120 by using this form, which is X minus mu divided by 376 00:28:42,120 --> 00:28:51,430 sigma. So x is 8.6 minus the mean, 8, divided by 377 00:28:51,430 --> 00:28:57,130 sigma, gives 0.12. So just straightforward 378 00:28:57,130 --> 00:29:02,890 calculation, 8.6 is your value of x. The mean is 379 00:29:02,890 --> 00:29:12,810 8, sigma is 5, so that gives 0.12. So now, the 380 00:29:12,810 --> 00:29:17,210 problem becomes, instead of asking x smaller than 381 00:29:17,210 --> 00:29:25,110 8.6, it's similar to z less than 0.12. Still, we 382 00:29:25,110 --> 00:29:26,310 have the same normal curve. 383 00:29:29,450 --> 00:29:32,990 8, the mean. Now, the mean of z is 0, as we 384 00:29:32,990 --> 00:29:39,230 mentioned. Instead of x, 8.6, the corresponding z 385 00:29:39,230 --> 00:29:43,000 value is 0.12. So instead of finding probability 386 00:29:43,000 --> 00:29:48,580 of X smaller than 8.6, smaller than 1.12, so they 387 00:29:48,580 --> 00:29:53,760 are equivalent. So we transform here from normal 388 00:29:53,760 --> 00:29:56,980 distribution to standardized normal distribution 389 00:29:56,980 --> 00:29:59,980 in order to compute the probability we are looking 390 00:29:59,980 --> 00:30:05,820 for. Now, this is just a portion of the table we 391 00:30:05,820 --> 00:30:06,100 have. 392 00:30:10,530 --> 00:30:18,530 So for positive z values. Now 0.1 is 0.1. Because 393 00:30:18,530 --> 00:30:25,670 here we are looking for z less than 0.1. So 0.1. 394 00:30:27,210 --> 00:30:32,950 Also, we have two. So move up to two decimal 395 00:30:32,950 --> 00:30:38,190 places, we get this value. So the answer is point. 396 00:30:42,120 --> 00:30:45,860 I think it's straightforward to compute the 397 00:30:45,860 --> 00:30:49,460 probability underneath the normal curve if X has 398 00:30:49,460 --> 00:30:53,160 normal distribution. So B of X is smaller than 8.6 399 00:30:53,160 --> 00:30:56,740 is the same as B of Z less than 0.12, which is 400 00:30:56,740 --> 00:31:02,680 around 55%. Makes sense because the area to the 401 00:31:02,680 --> 00:31:07,080 left of 0 equals 1 half. But we are looking for 402 00:31:07,080 --> 00:31:12,440 the area below 0.12. So greater than zero. So this 403 00:31:12,440 --> 00:31:16,600 area actually is greater than 0.5. So it makes 404 00:31:16,600 --> 00:31:20,440 sense that your result is greater than 0.5. 405 00:31:22,320 --> 00:31:22,960 Questions? 406 00:31:25,480 --> 00:31:30,780 Next, suppose we are interested of probability of 407 00:31:30,780 --> 00:31:35,380 X greater than. So that's how can we find normal 408 00:31:35,380 --> 00:31:41,980 upper tail probabilities. Again, the table we have 409 00:31:41,980 --> 00:31:46,580 gives the area to the left. In order to compute 410 00:31:46,580 --> 00:31:50,880 the area in the upper tail probabilities, I mean 411 00:31:50,880 --> 00:31:55,620 this area, since the normal distribution is 412 00:31:55,620 --> 00:32:00,160 symmetric and The total area underneath the curve 413 00:32:00,160 --> 00:32:04,680 is 1. So the probability of X greater than 8.6 is 414 00:32:04,680 --> 00:32:11,640 the same as 1 minus B of X less than 8.6. So first 415 00:32:11,640 --> 00:32:17,020 step, just find the probability we just have and 416 00:32:17,020 --> 00:32:21,680 subtract from 1. So B of X greater than 8.6, the 417 00:32:21,680 --> 00:32:25,930 same as B of Z greater than 0.12. which is the 418 00:32:25,930 --> 00:32:30,370 same as 1 minus B of Z less than 0.5. It's 1 minus 419 00:32:30,370 --> 00:32:36,230 the result we got from previous one. So this value 420 00:32:36,230 --> 00:32:39,410 1 minus this value gives 0.452. 421 00:32:41,610 --> 00:32:45,090 So for the other tail probability, just subtract 1 422 00:32:45,090 --> 00:32:47,690 from the lower tail probabilities. 423 00:32:51,930 --> 00:32:55,750 Now let's see how can we find Normal probability 424 00:32:55,750 --> 00:33:01,750 between two values. I mean if X, for example, for 425 00:33:01,750 --> 00:33:06,610 the same data we have, suppose X between 8 and 8 426 00:33:06,610 --> 00:33:13,360 .6. Now what's the area between these two? Here we 427 00:33:13,360 --> 00:33:17,220 have two values of x, x is 8 and x is 8.6. 428 00:33:24,280 --> 00:33:33,780 Exactly, so below 8.6 minus below 8 and below 8 is 429 00:33:33,780 --> 00:33:40,840 1 half. So the probability of x between 8 430 00:33:40,840 --> 00:33:47,340 and And 8.2 and 8.6. You can find z-score for the 431 00:33:47,340 --> 00:33:52,480 first value, which is zero. Also compute the z 432 00:33:52,480 --> 00:33:55,540 -score for the other value, which as we computed 433 00:33:55,540 --> 00:34:01,580 before, 0.12. Now this problem becomes z between 434 00:34:01,580 --> 00:34:04,540 zero and 0.5. 435 00:34:07,480 --> 00:34:15,120 So B of x. Greater than 8 and smaller than 8.6 is 436 00:34:15,120 --> 00:34:20,800 the same as z between 0 and 0.12. Now this area 437 00:34:20,800 --> 00:34:25,320 equals b of z smaller than 0.12 minus the area 438 00:34:25,320 --> 00:34:26,520 below z which is 1.5. 439 00:34:31,100 --> 00:34:37,380 So again, b of z between 0 and 1.5 equal b of z 440 00:34:37,380 --> 00:34:42,840 small. larger than 0.12 minus b of z less than 441 00:34:42,840 --> 00:34:46,520 zero. Now, b of z less than 0.12 gives this 442 00:34:46,520 --> 00:34:53,060 result, 0.5478. The probability below zero is one 443 00:34:53,060 --> 00:34:56,160 -half because we know that the area to the left is 444 00:34:56,160 --> 00:34:59,320 zero, same as to the right is one-half. So the 445 00:34:59,320 --> 00:35:04,240 answer is going to be 0.478. So that's how can we 446 00:35:04,240 --> 00:35:07,540 compute the probabilities for lower 10 directly 447 00:35:07,540 --> 00:35:12,230 from the table. upper tail is just one minus lower 448 00:35:12,230 --> 00:35:18,990 tail and between two values just subtracts the 449 00:35:18,990 --> 00:35:21,970 larger one minus smaller one because he was 450 00:35:21,970 --> 00:35:26,310 subtracted bz less than point one minus bz less 451 00:35:26,310 --> 00:35:29,430 than or equal to zero that will give the normal 452 00:35:29,430 --> 00:35:36,850 probability another example suppose we are looking 453 00:35:36,850 --> 00:35:49,350 for X between 7.4 and 8. Now, 7.4 lies below the 454 00:35:49,350 --> 00:35:55,270 mean. So here, this value, we have to compute the 455 00:35:55,270 --> 00:36:00,130 z-score for 7.4 and also the z-score for 8, which 456 00:36:00,130 --> 00:36:04,090 is zero. And that will give, again, 457 00:36:07,050 --> 00:36:13,710 7.4, if you just use this equation, minus 458 00:36:13,710 --> 00:36:17,690 the mean, divided by sigma, negative 0.6 divided 459 00:36:17,690 --> 00:36:21,150 by 5, which is negative 0.12. 460 00:36:22,730 --> 00:36:31,410 So it gives B of z between minus 0.12 and 0. And 461 00:36:31,410 --> 00:36:35,700 that again is B of z less than 0. minus P of Z 462 00:36:35,700 --> 00:36:40,140 less than negative 0.12. Is it clear? Now here we 463 00:36:40,140 --> 00:36:42,260 converted or we transformed from normal 464 00:36:42,260 --> 00:36:45,960 distribution to standardized. So instead of X 465 00:36:45,960 --> 00:36:52,100 between 7.4 and 8, we have now Z between minus 0 466 00:36:52,100 --> 00:36:57,480 .12 and 0. So this area actually is the red one, 467 00:36:57,620 --> 00:37:03,740 the red area is one-half. Total area below z is 468 00:37:03,740 --> 00:37:10,700 one-half, below zero, and minus z below minus 0 469 00:37:10,700 --> 00:37:17,820 .12. So B of z less than zero minus negative 0.12. 470 00:37:18,340 --> 00:37:21,940 That will give the area between minus 0.12 and 471 00:37:21,940 --> 00:37:28,860 zero. This is one-half. Now, B of z less than 472 00:37:28,860 --> 00:37:33,270 negative 0.12. look you go back to the normal 473 00:37:33,270 --> 00:37:37,650 curve to the normal table but for the negative 474 00:37:37,650 --> 00:37:42,310 values of z negative point one two negative point 475 00:37:42,310 --> 00:37:53,290 one two four five two two it's four five point 476 00:37:53,290 --> 00:37:56,630 five minus point four five two two will give the 477 00:37:56,630 --> 00:37:58,370 result we are looking for 478 00:38:01,570 --> 00:38:06,370 So B of Z less than 0 is 0.5. B of Z less than 479 00:38:06,370 --> 00:38:12,650 negative 0.12 equals minus 0.4522. That will give 480 00:38:12,650 --> 00:38:14,290 0 forcibility. 481 00:38:16,790 --> 00:38:23,590 Now, by symmetric, you can see that this 482 00:38:23,590 --> 00:38:28,470 probability between 483 00:38:28,470 --> 00:38:38,300 Z between minus 0.12 and 0 is the same as the 484 00:38:38,300 --> 00:38:43,340 other side from 0.12 I mean this area the red one 485 00:38:43,340 --> 00:38:46,200 is the same up to 8.6 486 00:38:55,600 --> 00:38:58,840 So the area between minus 0.12 up to 0 is the same 487 00:38:58,840 --> 00:39:04,920 as from 0 up to 0.12. Because of symmetric, since 488 00:39:04,920 --> 00:39:09,680 this area equals the same for the other part. So 489 00:39:09,680 --> 00:39:15,660 from 0 up to 0.12 is the same as minus 0.12 up to 490 00:39:15,660 --> 00:39:19,100 0. So equal, so the normal distribution is 491 00:39:19,100 --> 00:39:23,200 symmetric. So this probability is the same as B of 492 00:39:23,200 --> 00:39:27,980 Z between 0 and 0.12. Any question? 493 00:39:34,520 --> 00:39:36,620 Again, the equal sign does not matter. 494 00:39:42,120 --> 00:39:45,000 Because here we have the complement. The 495 00:39:45,000 --> 00:39:49,250 complement. If this one, I mean, complement of z 496 00:39:49,250 --> 00:39:53,350 less than, greater than 0.12, the complement is B 497 00:39:53,350 --> 00:39:56,350 of z less than or equal to minus 0.12. So we 498 00:39:56,350 --> 00:40:00,070 should have just permutation, the equality. But it 499 00:40:00,070 --> 00:40:04,830 doesn't matter. If in the problem we don't have 500 00:40:04,830 --> 00:40:07,470 equal sign in the complement, we should have equal 501 00:40:07,470 --> 00:40:11,430 sign. But it doesn't matter actually if we have 502 00:40:11,430 --> 00:40:14,510 equal sign or not. For example, if we are looking 503 00:40:14,510 --> 00:40:19,430 for B of X greater than A. Now what's the 504 00:40:19,430 --> 00:40:25,950 complement of that? 1 minus less 505 00:40:25,950 --> 00:40:32,450 than or equal to A. But if X is greater than or 506 00:40:32,450 --> 00:40:37,870 equal to A, the complement is without equal sign. 507 00:40:38,310 --> 00:40:40,970 But in continuous distribution, the equal sign 508 00:40:40,970 --> 00:40:44,990 does not matter. Any question? 509 00:40:52,190 --> 00:40:58,130 comments. Let's move to the next topic which talks 510 00:40:58,130 --> 00:41:05,510 about the empirical rule. If you remember before 511 00:41:05,510 --> 00:41:16,750 we said there is an empirical rule for 68, 95, 95, 512 00:41:17,420 --> 00:41:23,060 99.71. Now let's see the exact meaning of this 513 00:41:23,060 --> 00:41:23,320 rule. 514 00:41:37,580 --> 00:41:40,460 Now we have to apply the empirical rule not to 515 00:41:40,460 --> 00:41:43,020 Chebyshev's inequality because the distribution is 516 00:41:43,020 --> 00:41:48,670 normal. Chebyshev's is applied for skewed 517 00:41:48,670 --> 00:41:52,630 distributions. For symmetric, we have to apply the 518 00:41:52,630 --> 00:41:55,630 empirical rule. Here, we assume the distribution 519 00:41:55,630 --> 00:41:58,390 is normal. And today, we are talking about normal 520 00:41:58,390 --> 00:42:01,330 distribution. So we have to use the empirical 521 00:42:01,330 --> 00:42:02,410 rules. 522 00:42:07,910 --> 00:42:13,530 Now, the mean is the value in the middle. Suppose 523 00:42:13,530 --> 00:42:16,900 we are far away. from the mean by one standard 524 00:42:16,900 --> 00:42:22,720 deviation either below or above and we are 525 00:42:22,720 --> 00:42:27,040 interested in the area between this value which is 526 00:42:27,040 --> 00:42:33,040 mu minus sigma so we are looking for mu minus 527 00:42:33,040 --> 00:42:36,360 sigma and mu plus sigma 528 00:42:53,270 --> 00:42:59,890 Last time we said there's a rule 68% of the data 529 00:42:59,890 --> 00:43:06,790 lies one standard deviation within the mean. Now 530 00:43:06,790 --> 00:43:10,550 let's see how can we compute the exact area, area 531 00:43:10,550 --> 00:43:15,250 not just say 68%. Now X has normal distribution 532 00:43:15,250 --> 00:43:18,390 with mean mu and standard deviation sigma. So 533 00:43:18,390 --> 00:43:25,280 let's compare it from normal distribution to 534 00:43:25,280 --> 00:43:29,700 standardized. So this is the first value here. Now 535 00:43:29,700 --> 00:43:34,940 the z-score, the general formula is x minus the 536 00:43:34,940 --> 00:43:40,120 mean divided by sigma. Now the first quantity is 537 00:43:40,120 --> 00:43:45,660 mu minus sigma. So instead of x here, so first z 538 00:43:45,660 --> 00:43:49,820 is, now this x should be replaced by mu minus 539 00:43:49,820 --> 00:43:55,040 sigma. So mu minus sigma. So that's my x value, 540 00:43:55,560 --> 00:44:00,240 minus the mean of that, which is mu, divided by 541 00:44:00,240 --> 00:44:07,900 sigma. Mu minus sigma minus mu mu cancels, so plus 542 00:44:07,900 --> 00:44:13,520 one. And let's see how can we compute that area. I 543 00:44:13,520 --> 00:44:16,980 mean between minus one and plus one. In this case, 544 00:44:17,040 --> 00:44:23,180 we are interested or we are looking for the area 545 00:44:23,180 --> 00:44:28,300 between minus one and plus one this area now the 546 00:44:28,300 --> 00:44:31,360 dashed area i mean the area between minus one and 547 00:44:31,360 --> 00:44:39,460 plus one equals the area below one this area minus 548 00:44:39,460 --> 00:44:44,980 the area below minus one that will give the area 549 00:44:44,980 --> 00:44:48,200 between minus one and plus one now go back to the 550 00:44:48,200 --> 00:44:52,500 normal table you have and look at the value of one 551 00:44:52,500 --> 00:45:02,620 z and one under zero what's your answer one point 552 00:45:02,620 --> 00:45:11,520 one point now without using the table can you tell 553 00:45:11,520 --> 00:45:17,360 the area below minus one one minus this one 554 00:45:17,360 --> 00:45:17,840 because 555 00:45:23,430 --> 00:45:29,870 Now the area below, this is 1. The area below 1 is 556 00:45:29,870 --> 00:45:31,310 0.3413. 557 00:45:34,430 --> 00:45:37,590 Okay, now the area below minus 1. 558 00:45:40,770 --> 00:45:42,050 This is minus 1. 559 00:45:46,810 --> 00:45:49,550 Now, the area below minus 1 is the same as above 560 00:45:49,550 --> 00:45:50,510 1. 561 00:45:54,310 --> 00:45:58,810 These are the two areas here are equal. So the 562 00:45:58,810 --> 00:46:03,110 area below minus 1, I mean b of z less than minus 563 00:46:03,110 --> 00:46:09,130 1 is the same as b of z greater than 1. And b of z 564 00:46:09,130 --> 00:46:12,650 greater than 1 is the same as 1 minus b of z 565 00:46:12,650 --> 00:46:17,310 smaller than 1. So b of z less than 1 here. You 566 00:46:17,310 --> 00:46:19,710 shouldn't need to look again to the table. Just 567 00:46:19,710 --> 00:46:26,770 subtract 1 from this value. Make sense? Here we 568 00:46:26,770 --> 00:46:30,490 compute the value of B of Z less than 1, which is 569 00:46:30,490 --> 00:46:35,430 0.8413. We are looking for B of Z less than minus 570 00:46:35,430 --> 00:46:39,770 1, which is the same as B of Z greater than 1. 571 00:46:40,750 --> 00:46:43,850 Now, greater than means our tail. It's 1 minus the 572 00:46:43,850 --> 00:46:48,700 lower tail probability. So this is 1 minus. So the 573 00:46:48,700 --> 00:46:52,240 answer again is 1 minus 0.8413. 574 00:46:54,280 --> 00:47:00,040 So 8413 minus 0.1587. 575 00:47:11,380 --> 00:47:17,030 So 8413. minus 1.1587. 576 00:47:21,630 --> 00:47:27,570 Okay, so that gives 0.6826. 577 00:47:29,090 --> 00:47:37,550 Multiply this one by 100, we get 68.1826. 578 00:47:38,750 --> 00:47:44,010 So roughly 60-80% of the observations lie between 579 00:47:44,010 --> 00:47:50,470 one standard deviation around the mean. So this is 580 00:47:50,470 --> 00:47:53,850 the way how can we compute the area below one 581 00:47:53,850 --> 00:47:57,250 standard deviation or above one standard deviation 582 00:47:57,250 --> 00:48:03,790 of the mean. Do the same for not mu minus sigma, 583 00:48:05,230 --> 00:48:11,540 mu plus minus two sigma and mu plus two sigma. The 584 00:48:11,540 --> 00:48:14,600 only difference is that this one is going to be 585 00:48:14,600 --> 00:48:17,280 minus 2 and do the same. 586 00:48:20,620 --> 00:48:23,080 That's the empirical rule we discussed in chapter 587 00:48:23,080 --> 00:48:28,980 3. So here we can find any probability, not just 588 00:48:28,980 --> 00:48:33,660 95 or 68 or 99.7. We can use the normal table to 589 00:48:33,660 --> 00:48:36,900 give or to find or to compute any probability. 590 00:48:48,270 --> 00:48:53,090 So again, for the other one, mu plus or minus two 591 00:48:53,090 --> 00:49:00,190 sigma, it covers about 95% of the axis. For mu 592 00:49:00,190 --> 00:49:03,750 plus or minus three sigma, it covers around all 593 00:49:03,750 --> 00:49:08,450 the data, 99.7. So just do it at home, you will 594 00:49:08,450 --> 00:49:14,210 see that the exact area is 95.44 instead of 95. 595 00:49:14,840 --> 00:49:18,520 And the other one is 99.73. So that's the 596 00:49:18,520 --> 00:49:23,520 empirical rule we discussed in chapter three. I'm 597 00:49:23,520 --> 00:49:32,560 going to stop at this point, which is the x value 598 00:49:32,560 --> 00:49:38,400 for the normal probability. Now, what we discussed 599 00:49:38,400 --> 00:49:43,560 so far, we computed the probability. I mean, 600 00:49:43,740 --> 00:49:49,120 what's the probability of X smaller than E? Now, 601 00:49:49,200 --> 00:49:56,240 suppose this probability is known. How can we 602 00:49:56,240 --> 00:50:01,500 compute this value? Later, we'll talk about that. 603 00:50:06,300 --> 00:50:09,820 It's backward calculations. It's inverse or 604 00:50:09,820 --> 00:50:11,420 backward calculation. 605 00:50:13,300 --> 00:50:14,460 for next time inshallah.