What Does an Average Hide?

The middle hardly moves

The mean, the average of Grade 8, adds up everyone’s earnings and divides by how many people there are. Every rupee the newcomer earns goes into the total, so the mean follows him: each Rs 1,000 more he earns adds Rs 1,000 ÷ 10 = Rs 100 to the mean. Give him Rs 1,000,000 (ten lakh) a day and the mean is over Rs 100,000.

The median is the middle person when everyone is put in order. With the nine regulars it is the fifth, the carpenter on Rs 1,100: four earn less, four earn more. With ten people there are two in the middle, the fifth and the sixth, and the median is halfway between them.

Three numbers

Smaller cases make the pattern plain. Take three numbers, 2, 3 and 10. Their mean is (2 + 3 + 10) ÷ 3 = 5 and their median, the middle one, is 3.

Now change the 10 into 100, then into 1,000:

Numbers	Mean	Median
2, 3, 4	3	3
2, 3, 10	5	3
2, 3, 100	35	3
2, 3, 1,000	335	3

The median doesn’t move at all, because the order doesn’t change: 3 is still in the middle. The mean grows by a third of whatever the biggest number gains, since the gain is shared among three. A value far away from all the others, like the 1,000 or the businessman, is called an outlier.

The mean shares everything out

The mean answers one question exactly: if everything were put together and shared out equally, how much would each get? If all ten people at the tea shop put their day’s earnings in one pile, Rs 110,800, and shared it, each would get Rs 11,080. That is true about the pile, not about any one person.

So the mean is the right average whenever the total is what matters. A school buying rice for its midday meal needs the mean a child eats, times the number of children. A bus company wants the mean number of passengers a trip, because the total pays for the diesel.

The mean is a balance point

Imagine each value as a coin placed along a ruler. The ruler balances on a finger at exactly one place: the mean. Why? To share equally, the people above the mean must give away what they have above it, and that must be exactly what the people below need to come up to it. So the distances below the mean add up to the same as the distances above it.

Try it with 2, 3 and 10, mean 5. Below it: 2 is 3 short and 3 is 2 short, 5 in all. Above it: 10 is 5 over. It balances: 5 and 5.

In every position, the blue and the amber totals come out the same. That is the balance, and it is why the mean is sensitive: an outlier far out on one side has a long distance, and the balance point must slide towards it to even things up. The median feels none of that, because it counts people, not distances. A measure that a few extreme values can’t push around is called resistant: the median is resistant, and the mean is not.

Which average should you use?

They answer different questions:

• The mean: what each would get if the total were shared equally. Use it when the total matters, and when there are no big outliers.
• The median: what the person in the middle has. Use it for “typical” whenever a few values are far from the rest, as with earnings, house prices or the sizes of towns.
• The mode: the most common value. Use it for choices and categories (shoe sizes to stock, the favorite subject in a class).

When the mean and the median are far apart, that is news in itself: it says that a few values are pulling the mean away from the middle. Report both, and the reader sees it at once.

Two classes take the same test, marked out of 20. Both have a mean of 12 and a median of 12. Are they doing equally well?

Not at all (Figure 28.1). Class A is bunched near 12; Class B is spread from 3 to 19, and its three students below the pass mark would be invisible to a teacher who looked only at the averages.

So an average needs a second number beside it: how spread out the values are.

• The range is the biggest value minus the smallest: 15 − 9 = 6 for Class A, 19 − 3 = 16 for Class B. It is easy, but it depends only on the two most extreme values, so one outlier ruins it, like the mean: at the tea shop the range jumps from Rs 1,300 to Rs 99,300 when the businessman sits down.
• The interquartile range (IQR) is resistant, like the median. Put the values in order and find the quartiles: the first quartile, Q₁, has a quarter of the values below it, and the third, Q₃, three quarters. (Grade 10 finds them as the $\frac{N + 1}{4}$th and $\frac{3(N + 1)}{4}$th values.) The IQR is Q₃ − Q₁, the width of the middle half. Class A’s is 13 − 11 = 2 marks; Class B’s is 17 − 7 = 10.

A spread built on the mean is the mean absolute deviation, the mean distance from the mean (in the momo widget, the two readouts added and divided by 7); its cousin the standard deviation comes in later grades.

When is a value an outlier?

“Far from the rest” needs a rule. A common one was proposed by the American statistician John Tukey in 1977: build a fence one and a half IQRs beyond each quartile, and call anything outside the fences an outlier.

With the businessman counted, Q₁ = Rs 875 and Q₃ = Rs 1,775 (the 2.75th and 8.25th values, by the Grade 10 rule), so the IQR is Rs 900 and the upper fence is Rs 1,775 + 1.5 × Rs 900 = Rs 3,125 (Figure 28.2). The biggest of the regulars’ earnings, Rs 2,000, is well inside; Rs 100,000 is an outlier by any rule.

An outlier is a reason to look harder, not to throw the value away. It might be a mistake (one zero too many), to be corrected. But the businessman is real: leaving him out would hide something true, so report the median, or both averages.

Adding to everyone, doubling everyone

The missing mark

Sunita’s first five tests have a mean of 72. She wants her mean to be 75 after the sixth. What must she score?

The trick is to think in totals, not averages. Five tests with a mean of 72 have a total of 5 × 72 = 360. Six tests with a mean of 75 need a total of 6 × 75 = 450. So the sixth test must be 450 − 360 = 90. (A mean of 80 would need 480 − 360 = 120, in a test out of 100.)

Bhaskara’s gold

In the Lilavati, written in the 12th century, Bhaskara II asks a goldsmith’s question. In Colebrooke’s 1817 English translation:

Parcels of gold weighing severally ten, four, two and four mashas, and of the fineness of thirteen, twelve, eleven and ten respectively, being melted together, tell me quickly, merchant, who art conversant with the computation of gold, what is the fineness of the mass?

A masha (Sanskrit माष) is a small weight, and the fineness, or touch, says how pure the gold is, judged from the color it leaves on a touchstone. You might add the four finenesses and divide by four: (13 + 12 + 11 + 10) ÷ 4 = 11.5. But that treats a 2-masha lump as if it mattered as much as a 10-masha one. Bhaskara’s rule multiplies each fineness by its weight, adds, and divides by the total weight:

$$\frac{10 \times 13 + 4 \times 12 + 2 \times 11 + 4 \times 10}{10 + 4 + 2 + 4} = \frac{130 + 48 + 22 + 40}{20} = \frac{240}{20} = 12$$

The melted gold is of fineness 12, not 11.5: the big lump of 13 counts most. This is a weighted mean, each value counted as many times as its weight, and it is exactly the mean you get if you think of the gold as 20 one-masha pieces. Bhaskara also asks it backwards, like Sunita’s test (Question c).

The average of averages

A school has two Grade 7 sections. Section A has 20 students with a mean mark of 60, and Section B has 30 students with a mean of 70. Is the mean for the whole grade 65?

No. Like the gold, the bigger section counts more. The total marks are 20 × 60 + 30 × 70 = 1,200 + 2,100 = 3,300, shared among 50 students: a mean of 66. The average of two averages is right only when the groups are the same size.

The twist: one move, and both averages go up

A puzzle that seems to break arithmetic. Two tea shops, side by side:

• Shop A’s three regulars earn Rs 600, 800 and 1,000 a day: mean Rs 800.
• Shop B’s four regulars earn Rs 1,200, 1,500, 1,800 and 2,100: mean Rs 1,650.

One morning the person earning Rs 1,200 changes shops, from B to A. Now:

• Shop A: Rs 600, 800, 1,000 and 1,200: mean Rs 900, up Rs 100.
• Shop B: Rs 1,500, 1,800 and 2,100: mean Rs 1,800, up Rs 150.

Both averages went up, and nobody earns a rupee more. The person who moved earns less than Shop B’s average, so B’s rises without him, and more than Shop A’s, so A’s rises with him. The mean of all seven people is the same before and after, Rs 9,000 ÷ 7 ≈ Rs 1,286. Statisticians call this the Will Rogers phenomenon, after a joke attributed to the American comedian Will Rogers: when people from Oklahoma moved to California, they raised the average intelligence of both states. It is a real trap in medicine: when a better scan moves patients from the “mild” group to the “serious” one, the average survival of both groups can rise without anyone living a day longer.

• Can a Graph Lie? The same numbers drawn to look like a crisis or like nothing at all.
• How Much Rice Does Nepal Eat in a Day? Estimating a total you can’t count, from averages you can.
• Is This Game Fair? The mean of a game’s results over a long run: expected value.
• Can a Treatment Win in Every Group but Lose Overall? (Grades 9–12) The average of averages at its most surprising: Simpson’s paradox.
• Can 1,000 People Speak for 3 Crore? (Grades 9–12) How a sample’s average can stand for a whole country.
• How Do You Count Tigers You Can’t See? (Computing, Going Further) Estimating from samples. (Coming soon.)
• Grade 10 Math: Statistics The mean, median, mode and quartiles of grouped data, for the SEE.