Tales by Dots and Lines

Every day numbers arrive in bunches. The runs a batter scored in her last ten matches. The marks of forty students in a unit test. The price of one kilogram of onions on the first day of every month. The rainfall of a district in June, July and August. A bunch of numbers collected about one thing is called data, and each single number inside it is called an observation.

Collected raw, data is just a heap. Here are the marks out of 5 that twenty students scored in a class test, written down in the order the answer sheets came in:

3, 4, 5, 2, 3, 4, 4, 5, 1, 3, 4, 2, 5, 3, 4, 4, 2, 4, 5, 3

You can stare at that line as long as you like and it will tell you almost nothing. Now count how many times each mark appears and set it out in a frequency table. The frequency of a value is simply the number of times that value turns up.

  Marks (out of 5)    Number of students
         1                     1
         2                     3
         3                     5
         4                     7
         5                     4
      Total                   20

The heap has become five tidy rows. Notice the check built into the last line: the frequencies must add up to the number of students, and 1 + 3 + 5 + 7 + 4 = 20, so no answer sheet has been lost or counted twice. Always do that addition.

Straight away the table shows you things the list hid. The most common mark is 4, because it has the biggest frequency, 7. A value that occurs most often is called the mode. The marks run from a lowest of 1 to a highest of 5, so the range is 5 − 1 = 4, which tells you how widely spread the class is.

But a table is still a list, and two questions come up again and again with data. This chapter answers both.

The first question is: what is a typical value? If a parent asks how the class did, you do not want to read out twenty numbers; you want one number that stands in for all of them. Such a number is called a representative value. The two you will use here are the mean, which is the ordinary average, and the median, which is the value standing in the middle.

The second question is: how is the quantity behaving? Is it climbing, falling, steady, or jumping about? Is everybody crowded together with one odd exception? A column of figures will not tell you, but a picture will. This chapter uses two pictures: the dot plot, which puts one dot for every single observation, and the line graph, which plots a quantity against time and joins the points. Dots and lines really are the tools that let a page of figures tell its tale.

One warning before you start. A representative value throws information away on purpose. The moment you say the class average is 3.5, you have deliberately forgotten who scored 1 and who scored 5. That is exactly what a summary is for, but it is also its danger, and a good part of this chapter is about knowing which summary deserves your trust.

The mean is the number you get when the total is shared out equally among all the observations. That one sentence is the whole idea; the formula only writes it down.

Mean = sum of all the observations ÷ number of observations

Worked example. A batter scores 42, 18, 65, 30 and 25 runs in five innings. Find her mean score.

• Add them: 42 + 18 = 60; 60 + 65 = 125; 125 + 30 = 155; 155 + 25 = 180.
• Count them: there are 5 innings.
• Mean = 180 ÷ 5 = 36 runs.

So 36 is what she would have scored in every single innings if those 180 runs had been spread out evenly.

Worked example. Six friends get daily pocket money of ₹20, ₹35, ₹15, ₹40, ₹25 and ₹45. Find the mean.

• Total = 20 + 35 + 15 + 40 + 25 + 45 = ₹180.
• There are 6 friends, so the mean = 180 ÷ 6 = ₹30.

Notice that the unit travels with the answer. The mean of six amounts of money is money, the mean of five scores is runs, and the mean of seven heights is centimetres. An answer written as a bare number has lost half its meaning.

Mean from a frequency table. Go back to the twenty test marks. You could add all twenty numbers one at a time, but the table does it faster. The mark 4 occurred 7 times, so instead of adding 4 seven times you multiply: 4 × 7 = 28.

• 1 × 1 = 1
• 2 × 3 = 6
• 3 × 5 = 15
• 4 × 7 = 28
• 5 × 4 = 20

Total of all the marks = 1 + 6 + 15 + 28 + 20 = 70, and the number of students is 20, so the mean = 70 ÷ 20 = 3.5 marks.

Three facts about the mean are worth carrying around with you.

The mean need not be one of the observations. Not a single student scored 3.5, yet 3.5 is the correct mean. The mean of 2 and 3 is 2.5, which is neither 2 nor 3. If a survey reports a mean of 2.4 children per family, that is not nonsense; it is a summary, not a child.

The mean never falls outside the data. It always sits somewhere from the smallest observation to the largest one, and it can equal one of those two ends only when every observation in the set is the same number. The batter scored between 18 and 65, and her mean, 36, sits comfortably inside that range. If you ever calculate a mean that is bigger than every number you started with, or smaller than every one of them, you have made a slip somewhere. Go back and check the addition.

The gaps above and below the mean cancel out exactly. Subtract the mean from each of the batter’s scores: 42 − 36 = +6, 18 − 36 = −18, 65 − 36 = +29, 30 − 36 = −6, and 25 − 36 = −11. Now add those differences: 6 − 18 + 29 − 6 − 11 = 0. This happens for every data set, which makes it a quick and satisfying way to check a mean you have just worked out. It is also why the mean is sometimes described as the balance point of the data.

Finally, the formula can be turned around, and this is the single most useful move in the whole chapter: sum = mean × number of observations. If a cricket team of 11 players has a mean age of 26 years, then the total of all their ages must be 26 × 11 = 286 years. The next section uses this in every single question.

Questions about the mean are rarely as simple as add and divide. Usually something changes. One more observation arrives, one is taken away, or one turns out to have been copied down wrongly, and you are asked what the mean becomes. Every one of these is solved by the same first move: turn the mean back into a total, using sum = mean × number.

Adding one more observation. The batter’s mean over 5 innings was 36, so her total was 36 × 5 = 180 runs. In her sixth innings she scores 48.

• New total = 180 + 48 = 228.
• New number of innings = 6.
• New mean = 228 ÷ 6 = 38 runs.

The mean climbed from 36 to 38, and you could have predicted the direction before dividing anything, because the new score, 48, is bigger than the old mean of 36, so it pulls the mean up. Try the other two possibilities with the same total of 180.

• If the sixth score is 36, exactly the old mean, the new mean = (180 + 36) ÷ 6 = 216 ÷ 6 = 36, completely unchanged.
• If the sixth score is 18, below the old mean, the new mean = (180 + 18) ÷ 6 = 198 ÷ 6 = 33, which is lower.

So the rule is easy to remember: a new observation above the mean raises it, one below the mean lowers it, and one equal to the mean leaves it exactly where it was.

Removing an observation. Seven students have heights of 138, 140, 142, 144, 146, 148 and 150 cm.

• Total = 138 + 140 + 142 + 144 + 146 + 148 + 150 = 1008 cm.
• Mean = 1008 ÷ 7 = 144 cm.

Now the tallest student, 150 cm, moves to another school.

• New total = 1008 − 150 = 858 cm.
• New mean = 858 ÷ 6 = 143 cm.

Taking away someone above the mean pulls the mean down. Remove the shortest student instead and the new total is 1008 − 138 = 870, so the new mean is 870 ÷ 6 = 145 cm, and the mean goes up. Removing a value works the opposite way round to adding one, with one exception: if the value taken out is exactly equal to the mean, the mean does not move at all, just as it would not have moved had you put that value in.

Finding a missing value to hit a target average. This is the mean formula worked backwards. A student scored 72, 85, 64 and 79 in four unit tests. What must she score in the fifth test so that her mean over all five tests is 78?

• Total she needs = target mean × number of tests = 78 × 5 = 390.
• Total she already has = 72 + 85 + 64 + 79 = 300.
• Required fifth score = 390 − 300 = 90.

Check it: (72 + 85 + 64 + 79 + 90) ÷ 5 = 390 ÷ 5 = 78. It works. One extra habit is worth building here. If the required score had come out as 115 when each test is out of 100, the honest answer is that the target is impossible, not that the answer is 115.

The same idea appears with different wording. Suppose the mean of six numbers is 25 and five of them are 20, 22, 27, 31 and 24. Find the sixth. The total must be 25 × 6 = 150, the five given numbers add up to 20 + 22 + 27 + 31 + 24 = 124, so the sixth number is 150 − 124 = 26.

Correcting a wrong observation. The mean of 10 observations was found to be 42, but later it turned out that one observation, which was really 52, had been copied down as 25. There is no need to start all over again.

• Wrong total = 42 × 10 = 420.
• Correct total = 420 − 25 + 52 = 447. Take out what should not be there, put in what should.
• Correct mean = 447 ÷ 10 = 44.7.

Combining two groups. One class has 20 students with a mean score of 62, and another has 30 students with a mean of 72. What is the mean of all 50 students?

• Total of the first group = 62 × 20 = 1240.
• Total of the second group = 72 × 30 = 2160.
• Total of all 50 = 1240 + 2160 = 3400, so the combined mean = 3400 ÷ 50 = 68.

It is very tempting to average 62 and 72 and answer 67. That is wrong. The second group has more students in it, so it pulls the combined mean towards its own 72. Averaging the averages is only safe when the two groups are exactly the same size.

Changing every observation at once. If every number in a set is increased by 4, the total rises by 4 for each observation, so the mean also rises by exactly 4. If every number is doubled, the total doubles and so does the mean. Take eight numbers with a mean of 25, so a total of 200. Adding 4 to each gives a total of 200 + 32 = 232 and a mean of 232 ÷ 8 = 29, which is 25 + 4. Doubling each gives a total of 400 and a mean of 400 ÷ 8 = 50, which is 25 × 2.

The mean uses every observation, which is usually a strength and occasionally a disaster. The median works quite differently. It takes no notice of how large the largest values are or how small the smallest ones are; what matters is the order the values come in. The median is the value standing in the middle once the data has been arranged from smallest to largest, and when two values share the middle it is the number halfway between those two.

When the number of observations is odd. Take 12, 15, 9, 20, 18, 11 and 14.

• Arrange them in ascending order: 9, 11, 12, 14, 15, 18, 20.
• Count them: n = 7.
• The middle position is (n + 1) ÷ 2 = (7 + 1) ÷ 2 = 4, so the median is the 4th value, which is 14.

Three values lie below it and three above it, which is what standing in the middle means.

The easiest mistake to make here is skipping the sorting. If you take the 4th number of the original unsorted list you get 20, which happens to be the largest value in the whole set and is obviously not a middle at all. Arrange first, every time, without exception.

When the number of observations is even there is no single middle value; there are two of them, and the median is their mean. The daily wages of eight workers, in rupees, are 250, 300, 275, 400, 320, 260, 350 and 290.

• In ascending order: 250, 260, 275, 290, 300, 320, 350, 400.
• Here n = 8, so the two middle positions are the 8 ÷ 2 = 4th and the one after it, the 5th.
• The 4th value is 290 and the 5th is 300, so the median = (290 + 300) ÷ 2 = 590 ÷ 2 = ₹295.

Notice that ₹295 is not one of the wages at all. A median can miss the data just as a mean can.

What happens to the median when the data changes. Start with 4, 7, 9, 12, 15, which is already in order. Here n = 5, so the median is the 3rd value, 9.

• Add 20 to the set, giving 4, 7, 9, 12, 15, 20. Now n = 6, so the median is the mean of the 3rd and 4th values: (9 + 12) ÷ 2 = 10.5. Adding a value at the top shifted the middle half a step to the right.
• Instead remove 15 from the original set, leaving 4, 7, 9, 12. Now n = 4 and the median = (7 + 9) ÷ 2 = 8. Taking a value off the top shifted the middle to the left.

Notice how gentle those shifts are. Adding or taking away one value moves the middle position only half a step along the sorted list, so the median steps across to a neighbouring value instead of leaping about. That steadiness is the whole point of the median.

Why the median exists at all. Change the 15 in 4, 7, 9, 12, 15 into a monstrous 150 and watch what each summary does.

• The mean was (4 + 7 + 9 + 12 + 15) ÷ 5 = 47 ÷ 5 = 9.4. It is now (4 + 7 + 9 + 12 + 150) ÷ 5 = 182 ÷ 5 = 36.4.
• The median was 9, and it is still 9.

One unusual value, called an outlier, dragged the mean from 9.4 all the way to 36.4, which is now bigger than four of the five numbers in the set. The median did not budge a millimetre, because 9 is still the third value along.

A real example. Seven people who work in a small shop earn, per month, ₹9,000, ₹9,500, ₹10,000, ₹10,500, ₹11,000, ₹11,500 and ₹61,000, the last being the owner.

• Total = ₹1,22,500, so the mean = 122500 ÷ 7 = ₹17,500.
• The values are already in order and n = 7, so the median is the 4th value, ₹10,500.

Six of the seven people earn less than the mean. A notice saying the average monthly earning here is ₹17,500 would be arithmetically true and thoroughly misleading, because nobody except the owner earns anything like that. The median of ₹10,500 describes what a typical worker actually takes home. So: use the mean when the values sit reasonably close together, and use the median when a few very large or very small values would otherwise take over the answer.

A dot plot is the simplest picture of data there is. Draw a number line covering the values in your data. Then, for every observation, put one dot above the value it equals, stacking the dots neatly whenever a value repeats. That is the whole method: one dot, one observation.

Twenty students were asked how many brothers and sisters they have. The answers, in the order they were given, were:

1, 2, 0, 1, 3, 1, 2, 1, 0, 4, 1, 2, 3, 1, 0, 2, 1, 3, 2, 1

Here is the dot plot of that data.

Number of brothers and sisters (20 students)

        •
        •
        •
        •   •
        •   •
    •   •   •   •
    •   •   •   •
    •   •   •   •   •
  --+---+---+---+---+--
    0   1   2   3   4

Now read it. Every question you might ask is answered by counting dots.

• How many students altogether? Count every dot: 3 + 8 + 5 + 3 + 1 = 20. If the dots do not add up to the number of people surveyed, a dot has gone missing.
• Mode. The tallest stack sits above 1 and holds 8 dots, so having exactly one brother or sister is the commonest situation in this class.
• Range. The dots stretch from 0 to 4, so the range is 4 − 0 = 4.
• Mean. Multiply each value by the number of dots above it: (0 × 3) + (1 × 8) + (2 × 5) + (3 × 3) + (4 × 1) = 0 + 8 + 10 + 9 + 4 = 31. Now divide by the 20 dots: mean = 31 ÷ 20 = 1.55.
• Median. With n = 20 the median is the average of the 10th and the 11th values. Count dots from the left: the 3 dots above 0 are values 1 to 3, and the 8 dots above 1 are values 4 to 11. So both the 10th and the 11th values are 1, and the median is 1.

A dot plot also shows things that no single number can. The dots crowd together at 1 and 2, and a crowd like that is called a cluster. One lonely dot sits far out at 4, and a value standing apart like that is an outlier. You can even see the shape of the data: this plot leans to the left, with a thin tail trailing away to the right.

A second example, where the outlier really bites. Twelve students were asked how many minutes they take to reach school. Arranged in order, the answers were 5, 5, 10, 10, 10, 15, 15, 20, 20, 25, 30 and 60. On a dot plot the dots crowd between 5 and 25, then comes a wide stretch of empty number line, and finally one lonely dot at 60.

• Mean = (5 + 5 + 10 + 10 + 10 + 15 + 15 + 20 + 20 + 25 + 30 + 60) ÷ 12 = 225 ÷ 12 = 18.75 minutes.
• Median: n = 12, so it is the average of the 6th and 7th values, which are 15 and 15, giving 15 minutes.

The single student who travels a full hour pushes the mean 3.75 minutes above the median. The empty gap on the dot plot is the visual warning that this is happening, which is a warning you would never get from the bare list of numbers.

Dot plot or bar graph? A bar graph draws one bar whose height is the frequency, so the individual observations disappear into a solid block. A dot plot keeps them separate, so you can still count them one by one. Dot plots suit small sets of data whose values are whole numbers and do not spread too far. If you had the heights of 200 people measured to the nearest millimetre, a dot plot would be hopeless, and you would group the data into intervals instead.

Some data is not about how many; it is about how things change. The temperature of a patient taken every two hours, the price of tomatoes on the first of each month, the number of cycles a shop sells month by month. For data like this the natural picture is a line graph: plot each value as a point against its time, then join consecutive points with straight segments.

How to draw one.

1. Put time along the horizontal axis, always. The quantity you measured goes up the vertical axis.
2. Choose a scale that fits your largest value. If prices run up to ₹60 and you have twelve squares of height to spare, let 1 square stand for ₹5, because 60 ÷ 5 = 12.
3. Label both axes with what they show and in which unit. A graph without units cannot be read.
4. Plot each point carefully, then join consecutive points with a ruled straight line.
5. Give the graph a title so a stranger knows what it is about.

Worked example. The price of one kilogram of tomatoes on the first day of six months was: January ₹20, February ₹30, March ₹25, April ₹45, May ₹60, June ₹42. Plot the six points, join them up, and the graph will answer a whole set of questions.

• Highest and lowest. The peak is May at ₹60 and the lowest point is January at ₹20, a difference of 60 − 20 = ₹40.
• Rises and falls. Work out each step in turn: January to February +₹10; February to March −₹5; March to April +₹20; April to May +₹15; May to June −₹18.
• Steepest segment. The biggest single jump is March to April, ₹20 in one month, so that is the steepest climb on the graph. The steepest fall is May to June, ₹18.
• Mean price. Total = 20 + 30 + 25 + 45 + 60 + 42 = ₹222, so the mean = 222 ÷ 6 = ₹37.
• Median price. In ascending order: 20, 25, 30, 42, 45, 60. With n = 6 the median is the average of the 3rd and 4th values, (30 + 42) ÷ 2 = ₹36.
• Net change. The price ended at ₹42 having started at ₹20, so across the whole graph there was a net rise of 42 − 20 = ₹22, even though the price fell twice along the way.

Reading a value off a line graph. To find the April price, go up the vertical line through April until you meet the graph, then move straight across to the price axis and read off the number. To answer the reverse kind of question, such as in which months was the price above ₹40, draw a light horizontal line at 40 and see which points lie above it. Here they are April, May and June.

What the slope tells you. A segment going up means the quantity increased over that period; a segment going down means it decreased; a flat, horizontal segment means the reading at the far end was the same as the reading at the near end. And the steeper the segment, the faster the change, because a steep line covers more units of quantity in the same one month. Comparing steepness is often the quickest way to answer a question about when something rose or fell fastest, without doing a single subtraction.

One honest warning. The joining line is a guide for the eye, not a measurement. The tomato price was recorded only on the first of each month. The line drawn between April and May passes through the height ₹52.50 halfway along, but nobody actually measured that, and the real price may well have shot up to ₹70 and come back down in between. When the quantity genuinely changes smoothly, such as the temperature of a cooling cup of tea or the height of a growing plant, the in-between points are meaningful. When the data comes only as separate snapshots, treat the line as a connector and not as evidence.

Newspapers, weather bulletins, shop notices and school reports are full of graphs, and reading them well is a skill of its own. Before believing the shape of any graph, run through five questions.

1. What quantity is on each axis, and in what unit: rupees, millimetres, degrees, numbers of people?
2. What is the scale, that is, how much does one square or one step stand for?
3. Where does the vertical axis start? At zero, or somewhere higher up?
4. What stretch of time does the graph cover?
5. If there is more than one line, which is which? Look for the key.

A rainfall chart. The rainfall recorded at a place during one monsoon season was: June 120 mm, July 300 mm, August 280 mm, September 160 mm, October 60 mm.

• Total. 120 + 300 + 280 + 160 + 60 = 920 mm across the five months.
• Mean. 920 ÷ 5 = 184 mm per month.
• Median. In ascending order: 60, 120, 160, 280, 300. With n = 5 the median is the 3rd value, 160 mm.
• Wettest and driest. July is the wettest at 300 mm and October the driest at 60 mm, a difference of 300 − 60 = 240 mm. July received 300 ÷ 60 = 5 times as much rain as October.
• How concentrated it was. July and August together gave 300 + 280 = 580 mm, which is 580 out of 920, or about 63% of the whole season’s rain in just two months.

That last fact matters more than the mean does, because the rain was nowhere near evenly spread. Quoting 184 mm per month describes no month that actually happened. This is the mean hiding the story again, and the graph is what reveals it.

Two lines on one grid. Suppose the tomato prices from before, ₹20, ₹30, ₹25, ₹45, ₹60 and ₹42, are drawn along with the onion prices for the same six months, which were ₹20, ₹24, ₹30, ₹30, ₹25 and ₹42. Drawn on one grid with a key, this double line graph answers comparison questions instantly.

• Where the two lines meet at a plotted point, the prices really were equal. That happened in January, when both were ₹20, and again in June, when both were ₹42. The lines also cross twice in between, once between February and March and once between March and April, because tomatoes were dearer in February, onions were dearer in March and tomatoes were dearer again in April. Those two crossings sit on the joining lines, not on any recorded price, so all they tell you is that the two vegetables swapped places somewhere in that month.
• The vertical gap between the lines is the difference on that date. In May the gap is 60 − 25 = ₹35, the widest of the six months, so May is when tomatoes were dearest compared with onions.
• In March the onion line rises above the tomato line, ₹30 against ₹25, and that is the only month in which onions cost more.

How a graph can mislead without telling a single lie. Suppose a company shows monthly sales of 102, 104 and 106 units on a graph whose vertical axis starts at 100 and stops at 107. The line will climb steeply right across the page and look like an explosion of growth. But the real rise is 106 − 102 = 4 units on a starting figure of 102, which is (4 ÷ 102) × 100, or about 4%. Redraw the same three points with the axis starting at zero and the line looks nearly flat. Nothing in the first graph is false. The scale simply does the exaggerating.

Two more tricks are worth spotting. One is time labels that are not equally spaced, so a gap of one month is drawn the same width as a gap of one year and the line lies about how fast things moved. The other is a vertical scale stretched or squashed to make a change look dramatic or trivial. The defence is always the same. Read the numbers printed on the axes, and where you can, work out the actual change and the percentage change for yourself. A graph is an argument, and you are allowed to check the arithmetic behind it.