 Now, in some cases, you'll be the one collecting all of the data, but in other cases, you'll rely on data that's collected by others, and rather than dropping a whole mess of numbers into your lab, typically this data will be summarized in tabular form, and so it'll be important to be able to find measures of center from tables. So for example, let's suppose that they have a large class of students taking a five-point quiz, and their tabulated scores look something like this. And from this table, we want to find the maximum, minimum, mid-range, mode, mean, and median. So we can see from the table the highest score represented is a five, while the lowest score is a zero, which gives us our maximum and minimum. And once we have the maximum and minimum, we can find the mid-range. Remember, the mode is the most common value, and so according to our table, we see that the most common value is four, which appears 35 times, which is more than anything else. How about the median? Remember the table is a compact way of listing all the scores. What it really indicates is that our scores are 0, 3 times, 1, 5 times, 2, 18 times, where we won't list all 18 twos, but we'll use the ellipsis to indicate we've dropped out a whole bunch of scores that are still there, but just aren't going to be written down. And after that we have a bunch of threes, a bunch of fours, and a bunch of fives. Now remember, the median is going to be the middle score when all of the scores are put in order. So now all of our scores have been put in order. We just have to figure out where the middle is. So first, we can compute how many scores we actually have. That would be 3, plus 5, plus 18, plus 25, plus 35, plus 21. That's 107 scores altogether. So now I want the middle number. So since there's 107 scores, which is an odd number, if I subtract 1 I get 106, and I want half of those scores to be above the median, and half of those scores to be below the median. So that means there's going to be 53 scores above the median, and 53 scores below the median, and the median itself will be the 54th score from either end. So there are 35 plus 21, 56 scores of 4 or 5, and so that median is going to be listed somewhere in this set of fours. And so that means the median will be one of the fours. How about the mean? Since we have all the scores, we can find the mean by adding all the scores together and then dividing by the number of scores. There are three zeros, five ones, 18 twos, and so I can start to add them up. Three zeros, five ones, that's a lot of twos. Maybe there's an easier way of doing this. And definitions are the whole of mathematics. All else is commentary, and here the one definition that you'll want to remember from previous courses is that multiplication is repeated addition. So let's take those three zeros we're supposed to add together. If I add three zeros together, it's the same as computing three times zero, and so that sum is going to be zero. There are five ones, so I should add those five ones together, but that's the same as computing five times one, which will be five. Now the real time saver comes when we try to add together those 18 twos. So we don't actually have to add them. We can say that's the same as 18 times two, or 36. The 25 threes added together will give us a sum of 25 times three, or 75. If I add together 35 fours, I get 140, and if I add together 21 fives, I get 105. So I can find the sum of all scores. That'll be three zeros, five ones, 18 twos, 25 threes, 35 fours, and 21 fives, which gives me a total of 361 points, and there are three plus five plus eight plus 25 plus 35 plus 21, that's 107 scores, and so the mean will be the total 361 over the number 107, which will be about 3.4. One problem arises if our bins are larger than a single unit. So here maybe we surveyed 50 graduates of a college, and they reported their annual income, and we might get the following data. We'll still try to find the median and mean income of the graduates. So first of all, we notice that there are 50 graduates, so that means if we put the income of the graduates in order, that median will be midway between the incomes of the 25th and 26th graduates. Now the 25th and 26th graduates will be someplace in this second bin, between 20,000 and 29,999, and that's because the nine lowest incomes are the first bin, the next 17 incomes are the second bin, and the 25th and 26th will be among the next 17. The problem is we don't actually know what the income of the 25th and 26th graduates are, so how can we find the median? Since a bin includes all data values that fall within its range, we don't know what values are actually present, but we might use the middle of the bin as our value, and so we might take the median 20,000 plus 29,999 over 2 is 24,999.5. Now we can't use this number because that would commit us to all six significant figures, so we might say the median is 25,000. How about the mean? We could do the same thing as we did before, which is to use the middle of the bin as our representative value. So for the first bin, the income values are between zero and 1999, so we use the midrange, and it's important to keep in mind that we have to use this value without rounding because we're using it in a computation. Likewise, the income values of the second bin can all be treated as having a value equal to the midrange, 24,999,50, and for the third bin, fourth bin, and fifth bin. So our sum of values, well, there's nine at this 99,999,17 at 24,999,50, 12 at 39,999,50, 8 at 74,999,50, and 4 at 149,999,50, and that gives us a total income of, and there's 50 incomes altogether, so our mean is going to be, and we could say that the mean income of the graduates is $43,899.50. And if we wanted to lie, we could say that, but we can't guarantee this accuracy. Since we're basing our conclusion on a table where we only know the incomes to within one significant figure, we should only include one significant figure in our answers, so the correct way of stating our conclusion is to say that the mean income of the graduates is $40,000.