Sample Problem 1. Find the average (also called mean) of \(a=3+2\), \(b=3-4\), \(c=3+6\), and \(d=3-4\). How can we use this result?
We could just compute all of the sums and take the average of those numbers, but we will leave the numbers in the forms they are above to try to see if we can find any interesting patterns that we could generalize. Observe that \(\frac{3+2+3-4+3+6+3-4}{4}=\frac{3\cdot 4 +2-4+6-4}{4}=\frac{3\cdot 4}{4}+\frac{2-4+6-4}{4}=3\). We notice that because (a-3)+(b-3)+(c-3)+(d-3)=0, 3 ends up being the average. Let's try to generalize this.
Sample Problem 2. Find the average (also called mean) of \(a+d_1\), \(a+d_2\), \(a+d_3\), ..., and \(a+d_i\) where \(d_1+d_2+...+d_i=c\). How can we use this result?
First of all, we note that there are \(i\) terms. Thus, we can note that our average is \(\frac{(i)a+d_1+d_2+...+d_i}{i}=\frac{ai+c}{i}=a+\frac{c}{i}\). From this result we get a few useful realizations. First of all, note that if we let \(a\) be the average of the terms in the series, then we get \(a=a+\frac{c}{i}\implies c=0\). By thinking further about what this means, we can see that the sum of the average minus each term in the sequence is equal to \(0\). Also, note that if we are trying to quickly find the average of a sequence, we can guess a number, say n, and find the average of s-n for each s that is a term of our sequence. The average of all terms in the sequence is then just n plus the average of each s-n (this is just restating the formula we just derived).
Sample Problem 3. If the median of a set of 5 positive integers is 8 and their average is 9, what is the largest possible integer that could be in the set?
To make this problem easier to think about, we will write out the set of numbers in non-decreasing order so it looks like this: _ _ 8 _ _. Next, using what we got in the last problem, note that another way to say that we want to get the largest possible integer is to say that we want the integer that is greater than 9 by the largest possible margin. Note that 9 minus each term in our sequence adds to 0, so we want all terms but the largest term to be as small as possible. Other than the 8 term and the largest term, we need 1 term greater than or equal to 8 and two terms less than or equal to 8. Thus, the minimum values of the terms other than the largest one would be 1, 1, 8, and 8. We note that the sum of each of these integers minus 9 is -1-1-8-8=-18, so our largest possible positive integer is 9+18 =27.
Sample Problem 4. If group A had an average test score of 90, group B had an average test score of 80, and together they had an average test score of 83, find the ratio of the number of people in group A to the number of people in group B.
Using sample problem 1, we note that the total decreases from 83% equals the total increases. Thus, we will let the scores of the people in group A be \(83+a_1\), \(83+a_2\), \(83+a_3\), . . ., and \(83+a_i\). Similarly, we will let the scores of the people in group B be \(83+b_1\), \(83+b_2\), \(83+b_3\), . . ., and \(83+b_z\). Thus, \(a_1+a_2+a_3+ . . . +a_i+b_1+b_2+ . . .+b_z=0\). However, using sample problem 1, we know that \(a_1+a_2+a_3+a_4+...+a_i=-3i\) because 80-83=-3. Similarly, we know that \(b_1+b_2+b_3+b_4+...+b_z=7z\). Thus, \(-3i+7z=0\implies 3i=7z\), so the ratio of people in group A to group B is 3 to 7. Try to think about this solution to generalize this strategy.
Exercises
- If the median of set S is 8, the average is \(\frac{76}{9}\), it has n elements, the range is 1, and all of its elements are integers, find n (there could be multiple possible values of n; find them all)?
- Whenever an element with value v is removed from a set with n elements and average a, what does the new average become?
Solution 1
First of all, note because the range is 1, the median must either be just the middle term, or the two terms that average to it must both be 8. Thus, we know there is an 8 in the set. However, because the average is greater than 8, we also need numbers greater than 8, which can only be 9s. We will then break this up into cases based on if n is even or odd.
First, if n is even, we know that we have two 8s in the middle and can add and 8 and 9 pair by putting one on either side of the two middle 8s when in non-decreasing order. Thus, we always have two more 8s than 9s, so our average is \(\frac{8a+9(a-2)}{2a-2}\). Setting this equal to 8.3, we have \(\frac{8a+9(a-2)}{2a-2}=\frac{76}{9}\implies 17a-18=\frac{152}{9}a-\frac{152}{9}\implies \frac{1}{9}a=\frac{10}{9}\implies a=10\). This works because we assumed a is even, and 10 is indeed even.
Second, if n is odd, we can use similar reasoning to get \(\frac{8a+9(a-1)}{2a-1}=\frac{76}{9}\). Solving, we get a a=5. This is indeed odd, so thus we get a= 5, 10 as our solutions.
Notice that if you were to try solving the problem with an average of 8.5 you would get no solutions. The same is true for any average below 8. Try to come up with a non-algebraic, intuitive explanation for why the average must be between 8 and 8.5, including 8 and excluding 8.5. Also, note that not every value in this interval can be formed (can you find an intuitive explanation for this).
First, if n is even, we know that we have two 8s in the middle and can add and 8 and 9 pair by putting one on either side of the two middle 8s when in non-decreasing order. Thus, we always have two more 8s than 9s, so our average is \(\frac{8a+9(a-2)}{2a-2}\). Setting this equal to 8.3, we have \(\frac{8a+9(a-2)}{2a-2}=\frac{76}{9}\implies 17a-18=\frac{152}{9}a-\frac{152}{9}\implies \frac{1}{9}a=\frac{10}{9}\implies a=10\). This works because we assumed a is even, and 10 is indeed even.
Second, if n is odd, we can use similar reasoning to get \(\frac{8a+9(a-1)}{2a-1}=\frac{76}{9}\). Solving, we get a a=5. This is indeed odd, so thus we get a= 5, 10 as our solutions.
Notice that if you were to try solving the problem with an average of 8.5 you would get no solutions. The same is true for any average below 8. Try to come up with a non-algebraic, intuitive explanation for why the average must be between 8 and 8.5, including 8 and excluding 8.5. Also, note that not every value in this interval can be formed (can you find an intuitive explanation for this).
Solution 2
First, note that the sum of the elements in the set is \(a\cdot n\), where a is the average value of the elements in the set. Thus, when we remove an element of value v, the sum of the remaining elements is reduced to \(an-v\), and the average becomes \(\frac{an-v}{n-1}\). Try letting \(v=a+c\) to see what the average would become in terms of c.