Mean, Median, Mode, and Range: Free Response

The mean pools the whole season's crop and shares it out evenly, so start with the total the five plots produced.

$17 + 26 + 21 + 14 + 27 = 105$

There are five plots, so divide that total by five.

$\text{mean} = \frac{105}{5} = 21$

The mean is $21$ kilograms per plot. Read as a fair share, it says that if the five families had tipped the season's tomatoes into one heap and divided the heap evenly, each family would have carried home $21$ kilograms. Nothing is lost or invented in that redistribution: five lots of $21$ kilograms come back to the $105$ kilograms the garden actually grew.

Subtract the mean from each weight in turn, so that a plot above the mean gives a positive number and a plot below it gives a negative one.

$17 - 21 = -4, \qquad 26 - 21 = +5$

$21 - 21 = 0, \qquad 14 - 21 = -7$

$27 - 21 = +6$

Now add the five results, keeping every sign.

$-4 + 5 + 0 - 7 + 6 = 0$

The total is $0$. Splitting the five into two camps shows what is happening. The plots above the mean are $5$ and $6$ kilograms clear of it, which is $11$ kilograms of surplus, while the plots below it fall short by $4$ and $7$ kilograms, which is $11$ kilograms of shortfall. The surplus covers the shortfall exactly, and the plot that weighed $21$ kilograms contributes nothing either way.

The eleven deviations must add to $0$, and so must the deviations of any list of values whatever, however long the list is and whatever the values weigh.

The reason sits inside the definition of the mean. The mean is the total of the values divided by how many values there are, so multiplying both sides of that by the count turns it around and gives a fact worth writing down.

$\text{total of the values} = \text{count} \times \text{mean}$

Now build the deviations. Each one is a value with the mean subtracted from it, so adding all eleven deviations adds all eleven values once and subtracts the mean eleven times over.

$(\text{total of the values}) - 11 \times \text{mean}$

But the bracket is exactly $11$ times the mean, by the line above, so this is a quantity minus itself, which is $0$. Nothing about the particular weights was used anywhere in that argument, which is why the neighbouring garden's total is settled without anyone weighing a single tomato.

This is the precise sense in which the mean is a balance point. Every value above it pulls one way and every value below it pulls the other, and the two pulls always come out the same size.

The mean pools every bill and divides by the number of jobs.

$46 + 52 + 49 + 58 + 45 + 350 = 600$

$\text{mean} = \frac{600}{6} = 100$

The median is fixed by position rather than by size, so the bills have to be put in order before the middle can be found.

$45, \; 46, \; 49, \; 52, \; 58, \; 350$

Six is an even count, so no single bill sits in the middle. The third and fourth share it, and the median is their mean.

$\text{median} = \frac{49 + 52}{2} = 50.5$

The mean bill is $100$ dollars and the median bill is $50.50$ dollars. Sorting mattered only for the median: adding the six bills would have given the same total in any order at all.

Compare each bill with the mean of $100$ dollars.

$45 - 100 = -55, \qquad 46 - 100 = -54$

$49 - 100 = -51, \qquad 52 - 100 = -48$

$58 - 100 = -42, \qquad 350 - 100 = +250$

Five of the six results are negative, so five of the six jobs cost less than the mean. Adding those five shortfalls gives

$-55 - 54 - 51 - 48 - 42 = -250$

against the rebuild's $+250$. One job is $250$ dollars up on its own while five jobs are $250$ dollars down between them, which is the balance from the lesson doing its work in an extreme case: the mean has been pushed to the one place where those two totals match.

Paint the median. A customer wheeling a bicycle through the door wants to know what an ordinary repair will cost them, and the median of $50.50$ dollars sits right among the ordinary bills: three of the six jobs came within five dollars of it, and a fourth within six. The mean of $100$ dollars is higher than five of the six bills the shop actually issued, so a customer reading it would brace for roughly twice what an ordinary service costs, and would take the bicycle elsewhere for no good reason.

The comparison in part B says why the two figures part company. The rebuild stands $250$ dollars above the mean and that one job is what hauled the mean up there, because the mean answers to the size of every bill. The median never felt it. The median depends only on which bill sits in the middle of the order, so raising the largest bill from $350$ dollars to $3{,}500$ dollars would leave it exactly where it is.

The mean is not the wrong figure everywhere, though, and the shop needs it elsewhere. When the owner asks what six jobs bring in altogether, the mean is precisely the tool, because it was built to be a fair share of a total and so gives the total straight back when it is multiplied by the count.

$100 \times 6 = 600$

So the shop forecasts its takings from the mean and describes a repair to a customer with the median. Neither figure is wrong; they answer different questions, and the window is asking the customer's question.

Add the ten sizes and divide by ten.

$6 + 8 + 6 + 7 + 9 + 6 + 10 + 8 + 6 + 8 = 74$

$\text{mean} = \frac{74}{10} = 7.4$

For the median, sort the sizes first, because the median is the middle of the data in order and not the middle of the list as the morning happened to produce it.

$6, \; 6, \; 6, \; 6, \; 7, \; 8, \; 8, \; 8, \; 9, \; 10$

Ten is an even count, so the middle is shared by the fifth and sixth values, $7$ and $8$.

$\text{median} = \frac{7 + 8}{2} = 7.5$

The mean size is $7.4$ and the median size is $7.5$.

The mode is settled by counting rather than by arithmetic. Size $6$ was sold four times, size $8$ three times, and sizes $7$, $9$ and $10$ once each, so the size that came up most often is $6$.

$\text{mode} = 6$

The range subtracts the smallest size sold from the largest.

$\text{range} = 10 - 6 = 4$

The mode is size $6$, and the range is $4$ sizes wide. Notice that the range is not itself a size: it is a gap between two sizes.

The mode decides the bulk order, and for this morning that is size $6$. The manager is not asking where the middle of the morning's sizes lay. She is asking which size walked out of the door most often, and counting is what answers that question.

$\text{mean} = 7.4, \qquad \text{median} = 7.5$

$\text{mode} = 6$

Neither of the two measures of center can stand in for it. Both landed between the whole sizes, and neither $7.4$ nor $7.5$ is a size a single customer bought that morning, so neither figure names a crate she could sensibly order. That is not an accident of these particular sales either: the mean is a total shared out, and the median is the average of two middle values whenever the count is even, so both are free to land between the sizes that sold.

Even if they had come out on a whole size, they would still be answering the wrong question. Both describe the middle of the sizes sold, and the middle of the sizes sold is not the size that sold most. Four of these ten pairs were size $6$, and both measures of center sit above every one of those four. The mode is also the only one of the four summaries that would still work if the shop recorded colours instead of sizes, which is a sign of what it is for: it counts, and counting does not care what is being counted.

The range tells the manager how wide the band has to be, and nothing more than that. Every pair sold that morning lay between size $6$ and size $10$.

$10 - 6 = 4$

So the shelf has to reach across four sizes. That is a statement about spread, which is the one thing no measure of center reports: two mornings can share a mean, a median and a mode and still stretch across bands of different widths.

What the range leaves out is how the ten pairs were shared across the band, and she needs that as much as the width. Size $6$ took four of the ten pairs on its own, while sizes $9$ and $10$ took one each, so a shelf stocked to the same depth all the way across would run out at one end and gather dust at the other. The range fixes how wide the shelf must be; only the counts fix how deep each pile should be.

A group's mean and its total are two views of the same figures, because the books read add up to the number of members times the mean. That turns each group's mean into a total.

$4 \times 9 = 36, \qquad 6 \times 14 = 84$

The two groups read

$36 + 84 = 120$

books between them, and the club has $4 + 6 = 10$ members, so the club's mean comes from that pair.

$\text{mean} = \frac{120}{10} = 12$

The Tuesday group read $36$ books, the Thursday group read $84$, and the mean across all ten members is $12$ books.

Work backwards from the target, using the same relationship in the other direction. Eleven members at a mean of $13$ books would have to read

$11 \times 13 = 143$

books between them. The ten current members have already read $120$ of those, so the new member has to supply what is missing.

$143 - 120 = 23$

The eleventh member must read $23$ books. That figure sits well above the club's current mean of $12$, which is what should be expected: a mean only climbs if the value joining the set stands above it.

The $11.5$ is a real number and the arithmetic behind it is sound, but it is the mean of the two group means rather than the mean of the ten members.

$\frac{9 + 14}{2} = 11.5$

That calculation counts each group once, which hands the four Tuesday readers exactly as much say as the six Thursday readers. The club's figure has to count each member once. Thursday has half as many members again as Tuesday, so its higher mean deserves to pull the club figure further than Tuesday's pulls it back, and it does: the club mean of $12$ sits above the halfway point of $11.5$ and closer to the Thursday group's $14$.

The shortcut is not always wrong, though, and it is worth knowing when it is safe. It gives the right answer whenever the two groups hold the same number of members, because then counting each group once and counting each member once come to the same thing. (It also lands on the right figure in the one uninteresting case where the groups already share a mean, since then every way of averaging returns that same figure.) Had both groups held five members, the totals would have been $45$ and $70$, which is $115$ books over $10$ members, and the club mean would have been $11.5$ after all.

$\frac{45 + 70}{10} = \frac{115}{10} = 11.5$

Take the conditions one at a time and let each one fix part of the set. Think of the five numbers in order as a first, a second, a middle, a fourth and a fifth.

A mean of $8$ over five numbers fixes the total, because the values add up to the count times the mean.

$5 \times 8 = 40$

A median of $6$ fixes the middle number itself, since with an odd count the median is the third value in order. A range of $12$ fixes the gap between the first number and the last.

Now choose. Let the smallest be $3$, and the range forces the largest to be $15$. Those two and the middle $6$ account for $3 + 15 + 6 = 24$ of the total, which leaves $40 - 24 = 16$ to be split between the second and the fourth. The second has to land between $3$ and $6$ and the fourth between $6$ and $15$, so that the two ends stay the ends. Taking $5$ and $11$ does it.

$3, \; 5, \; 6, \; 11, \; 15$

Check all three conditions: the total is $40$, so the mean is $8$; the third value in order is $6$, so the median is $6$; and $15 - 3 = 12$, so the range is $12$. Many other sets work here, because these particular conditions leave several choices for the endpoints and the remaining pair.

Run the same construction again, this time steering a repeat into the set. The total stays at $40$, the middle value stays at $6$, and the two ends stay $12$ apart.

Let the smallest be $4$, which forces the largest to be $16$. Those two and the middle account for $4 + 16 + 6 = 26$, leaving $14$ for the second and fourth numbers, the second between $4$ and $6$ and the fourth between $6$ and $16$. Choosing $6$ and $8$ puts a repeat alongside the middle value.

$4, \; 6, \; 6, \; 8, \; 16$

The total is $40$, the third value in order is $6$, and $16 - 4 = 12$, so all three conditions still hold. The value $6$ now appears twice while every other value appears once.

$\text{mode} = 6$

The mode is not tied down by the other three summaries at all. A set can satisfy the same conditions with no repeated value whatever, as $3$, $5$, $6$, $11$, $15$ does, or with a clear mode, as this one does.

The claim fails, and a single set of five numbers is enough to break it.

Take $4$, $5$, $6$, $20$ and $25$. They total $60$, so the mean is $12$, while the third value in order is $6$, so the median is $6$. The mean is comfortably the larger, which is exactly the situation the claim is about.

$\text{mean} = \frac{60}{5} = 12, \qquad \text{median} = 6$

Now remove the single largest value, $25$, as the claim instructs. What is left is $4$, $5$, $6$ and $20$, totalling $35$ over four values, and with an even count the median is the mean of the middle pair.

$\text{mean} = \frac{35}{4} = 8.75$

$\text{median} = \frac{5 + 6}{2} = 5.5$

The mean has fallen, from $12$ to $8.75$, but it is still well above the median, so the claim is broken.

The reasoning behind the claim is worth naming, because it is half right. It assumes that one value is responsible for the whole gap between the mean and the median, and in the familiar picture of a single wild value among ordinary ones that is fair enough. Here two values sit far above the other three, so taking one away leaves the other still hauling the mean upward. The mean answers to the size of every value at once, and nothing says only one value at a time is allowed to be extreme.