Statistics

Statistics is the branch of mathematics that collects data, presents it, sorts it and analyses it. It is very old: the rulers of Egypt, Babylon and Rome kept simple records of their people, trade and taxes. Modern statistics grew in Europe from the 1600s on, with the scientific way of measuring and testing; from 1700 on mathematicians such as Jacob Bernoulli, Thomas Bayes and Pierre-Simon Laplace worked out the mathematics of chance and error that it rests on. Today statistics is used in the natural and social sciences, in medicine, business and government, and on your phone.

Using numbers to describe a set of data (its middle, its spread, its shape) is called descriptive statistics, and showing the data in tables and pictures so that a pattern can be seen is part of it. That is this unit.

To describe how the months spread out, we first put them in order and cut the ordered list into equal parts. That is what quartiles and percentiles do.

5.1.1 Quartiles of individual and discrete data

Data comes in two forms in this unit.

• An individual series lists every value on its own: the weights of 11 students, 35 kg, 39 kg, 52 kg, …
• A discrete series gives each value with its frequency, how many times it occurs: 5 teachers aged 25, 8 aged 30, 10 aged 35, … in a table.

Here is a smaller set to try it on. The temperature in Kathmandu was read every 2 hours on a winter day:
5 ∘°C, 7 ∘°C, 3 ∘°C, 10 ∘°C, 12 ∘°C, 15 ∘°C, 9 ∘°C, 17 ∘°C, 13 ∘°C, 19 ∘°C, 8 ∘°C
In ascending order they are
3 ∘°C, 5 ∘°C, 7 ∘°C, 8 ∘°C, 9 ∘°C, 10 ∘°C, 12 ∘°C, 13 ∘°C, 15 ∘°C, 17 ∘°C, 19 ∘°C
There are 11 readings, and the 6th, 10 ∘°C, stands in the middle with 5 readings on each side: it is the median, and it cuts the data into two equal halves. To cut the data into four equal parts we need three cutting points, just as a rope needs three cuts to make four pieces.

Counting from the left, the 3rd reading, 7 ∘°C, the 6th, 10 ∘°C, and the 9th, 15 ∘°C, divide the readings into four equal parts. The values that divide ordered data into four equal parts are called quartiles:

• the first quartile or lower quartile, Q₁ (here 7 ∘°C): about 25% of the data lies below it and 75% above;
• the second quartile, Q₂, which is the median, $M_d$ (here 10 ∘°C): 50% below, 50% above;
• the third quartile or upper quartile, Q₃ (here 15 ∘°C): about 75% below, 25% above.

5.1.2 Quartiles of an individual series

With 11 values the quartiles were the 3rd, 6th and 9th. Where are they for any number of values? If an individual series has N values, in ascending order,
$$Q_1 = \left(\frac{N + 1}{4}\right)^{\text{th}} \text{ item}$$
$$Q_2 = M_d = 2\left(\frac{N + 1}{4}\right)^{\text{th}} = \left(\frac{N + 1}{2}\right)^{\text{th}} \text{ item}$$
$$Q_3 = 3\left(\frac{N + 1}{4}\right)^{\text{th}} \text{ item}$$
For the 11 temperatures, $\frac{11 + 1}{4} = 3$: the 3rd, 6th and 9th, as we found by counting.

If $\frac{N + 1}{4}$ is not a whole number, the quartile lies between two items. A place like 2.25 means “the 2nd item, plus a quarter of the way to the 3rd”:
$$\text{the } 2.25^{\text{th}} \text{ item} = \text{2nd item} + 0.25 \times (\text{3rd item} - \text{2nd item})$$

Worked examples

5.1.3 Quartiles of a discrete series

In a discrete series the values repeat, so we can’t just count along a list, but the places are found the same way. With $N = \Sigma f$, the sum of the frequencies,
$$Q_1 = \left(\frac{N + 1}{4}\right)^{\text{th}}, \quad Q_2 = 2\left(\frac{N + 1}{4}\right)^{\text{th}}, \quad Q_3 = 3\left(\frac{N + 1}{4}\right)^{\text{th}} \text{ item}$$
To find which value stands at a place, make a cumulative frequency (cf) column: the running total of the frequencies. A value’s cf says how many items there are up to and including it. The item at a place is the value whose cf is equal to the place or just greater than it.

5.1.4 Percentiles

The 99 values that divide ordered data into 100 equal parts are called percentiles, written $P_1, P_2, P_3, \ldots, P_{99}$. About n% of the data lies below $P_n$. Like the quartiles, the place of the nth percentile, in an individual or a discrete series, is
$$P_n = \left(\frac{n(N + 1)}{100}\right)^{\text{th}} \text{ item}$$

Here are the heights (cm) of 19 students:
150, 140, 132, 125, 133, 126, 123, 137, 145, 122, 149, 147, 142, 148, 136, 124, 128, 141, 144
In ascending order:
122, 123, 124, 125, 126, 128, 132, 133, 136, 137, 140, 141, 142, 144, 145, 147, 148, 149, 150
They must be in ascending order, because the percentiles grow as n grows: $P_{60}$ must be at least as big as P₅. Then, with N + 1 = 20,

• $P_5 = \left(\frac{5 \times 20}{100}\right)^{\text{th}} = 1^{\text{st}}$ item = 122 cm
• $P_{60} = \left(\frac{60 \times 20}{100}\right)^{\text{th}} = 12^{\text{th}}$ item = 141 cm
• $P_{82} = \left(\frac{82 \times 20}{100}\right)^{\text{th}} = 16.4^{\text{th}}$ item, between the 16th (147) and the 17th (148): $P_{82} = 147 + 0.4(148 - 147) = 147.4$ cm
• $P_{95} = \left(\frac{95 \times 20}{100}\right)^{\text{th}} = 19^{\text{th}}$ item = 150 cm

Notice that $P_{25} = Q_1$, $P_{50} = Q_2 = M_d$ and $P_{75} = Q_3$: quartiles are just three of the percentiles.

Question 1

Questions 2 to 5

Question 6

Question 7

Question 8

Sabita finds the three quartiles of each set of data below and says that the second quartile is the median. Find the quartiles too. Do you agree with her? Give reasons.

Questions 9 to 12

Questions 13 and 14

Here are the prices of edible oil (Rs per liter) and of a 25 kg sack of rice (Rs per sack) in a Nepali market over the last 11 months:

Month	1	2	3	4	5	6	7	8	9	10	11
Oil (Rs/liter)	180	200	230	240	245	250	260	270	300	310	315
Rice (Rs/sack)	1,600	1,700	1,850	2,040	2,145	2,250	2,260	2,300	2,350	2,400	2,450

How much the values of a set of data are scattered or spread out from their middle is called dispersion. A measure of dispersion is a single number that says how spread out the data are: small for data bunched together, large for data spread wide. The range, the largest value minus the smallest, is the simplest, but it hangs on just two values, which may be unusual. The measures in this unit use more of the data.

Each measure comes in two forms. An absolute measure (the quartile deviation, mean deviation or standard deviation itself) is in the data’s own units: rupees, kilograms, ∘°C. A relative measure, a coefficient, divides it by an average and has no units, so it can compare sets in different units or of very different sizes, such as rupees for a liter of oil and rupees for a sack of rice.

5.2.1 Quartile deviation and its coefficient

Measure the heights of all the people of a country and you will find huge differences, from toddlers to the tallest adults. Measure the heights of the players of its national football team and the differences are much smaller. The quartile deviation is one way of saying how much smaller.

At a village meeting the ages (years) of the people gathered were:
40, 46, 35, 50, 38, 57, 44, 52, 60, 48, 55, 56, 67, 70, 62
In ascending order: 35, 38, 40, 44, 46, 48, 50, 52, 55, 56, 57, 60, 62, 67, 70. With N = 15, $\frac{N + 1}{4} = 4$:

• Q₁ = the 4th item = 44 years
• Q₂ = the 8th item = 52 years
• Q₃ = the 12th item = 60 years

The middle half of the people, those between the lower and the upper quartile, are aged 44 to 60. The width of that middle half is the interquartile range:
$$\text{Interquartile range} = Q_3 - Q_1$$
Half of it is the quartile deviation (Q.D.), also called the semi-interquartile range:
$$\text{Q.D.} = \frac{Q_3 - Q_1}{2}$$
and the relative measure based on it is the coefficient of quartile deviation:
$$\text{Coefficient of Q.D.} = \frac{Q_3 - Q_1}{Q_3 + Q_1}$$
For the village ages:
$$\begin{aligned} \text{Interquartile range} &= 60 - 44 = 16 \text{ years} \\ \text{Q.D.} &= \tfrac{16}{2} = 8 \text{ years} \\ \text{Coefficient} &= \tfrac{60 - 44}{60 + 44} = \tfrac{16}{104} \approx 0.154 \end{aligned}$$
What it means: the middle half of the meeting spans 16 years of age. Half of that, 8 years, is how far, roughly, a typical person in that middle half is from the middle of the group.

The coefficient is a ratio: dividing by Q₃ + Q₁, which is twice the middle of the two quartiles, compares the spread with the size of the values. For the 11 months of prices, Q₁ and Q₃ are the 3rd and 9th prices:

• oil: $\text{Q.D.} = \frac{300 - 230}{2} =$ Rs 35, coefficient $\frac{70}{530} \approx 0.132$;
• rice: $\text{Q.D.} = \frac{2350 - 1850}{2} =$ Rs 250, coefficient $\frac{500}{4200} \approx 0.119$.

Rice’s quartile deviation is seven times oil’s, but a sack of rice costs about eight times a liter of oil. The coefficients say that, for their size, the two prices swung about equally, oil a little more.

Worked examples

Questions 1 to 3

Question 4

Questions 5 to 7

Question 8

Find the quartile deviation and its coefficient of each set of data.

Questions 9 to 13

Questions 14 and 15

The quartile deviation looks only at the two quartiles. The mean deviation uses every value: it asks how far, on average, the values are from their middle. That is often what people really want to know. A factory filling 1 kg bags of flour wants every bag close to 1 kg, and asks how far, on average, the bags are off. A weather forecast can be judged the same way, by the average size of its misses (a forecast that is 2 ∘°C too warm one day and 2 ∘°C too cold the next is off by 2 ∘°C on average, not by 0).

Two friends, Pemba and Tashi, took ten mathematics tests of 25 marks each during a year:
Pemba: 17, 19, 20, 16, 22, 23, 21, 20.5, 18, 15
Tashi: 9, 10, 3, 15, 20, 16, 5, 25, 21, 23

The average of the absolute values of the differences between each item and a measure of central tendency (the mean or the median) is called the mean deviation (M.D.), or average deviation. The absolute value |X − X| is the distance, without its sign: |3 − 9| = | − 6| = 6. Without it the minus and plus differences would cancel: the differences from the mean always add up to 0.

5.3.1 Mean deviation of an individual series

Take the eight values 3, 6, 6, 7, 8, 11, 16, 15.
First find the mean: $\overline{X} = \frac{72}{8} = 9$.
Then find each value’s distance from 9:

X	3	6	6	7	8	11	15	16
$D = \lvert X - \overline{X} \rvert$	6	3	3	2	1	2	6	7

Last, average the distances: $\text{M.D.} = \frac{30}{8} = 3.75$.
What it means: on average the values are 3.75 away from their mean, 9.

In general, for an individual series of N values:
a. Mean deviation from the mean
$$\overline{X} = \frac{\Sigma X}{N}, \qquad \text{M.D.} = \frac{\Sigma |X - \overline{X}|}{N} = \frac{\Sigma D}{N}, \quad \text{where } D = |X - \overline{X}|$$
b. Mean deviation from the median
$$M_d = \left(\frac{N + 1}{2}\right)^{\text{th}} \text{ item}, \qquad \text{M.D.} = \frac{\Sigma |X - M_d|}{N} = \frac{\Sigma D}{N}, \quad \text{where } D = |X - M_d|$$
D is always positive or 0. For the eight values, the median is the 4.5th item, $\frac{7 + 8}{2} = 7.5$, the distances from it are 4.5, 1.5, 1.5, 0.5, 0.5, 3.5, 7.5, 8.5, and the mean deviation from the median is $\frac{28}{8} = 3.5$.

The mean deviation is an absolute measure, in the data’s units. To compare sets in different units, or of different sizes, we use its coefficient:
$$\text{Coefficient of M.D. (from the mean)} = \frac{\text{M.D.}}{\overline{X}}, \qquad \text{Coefficient of M.D. (from the median)} = \frac{\text{M.D.}}{M_d}$$
For Pemba, M.D. ≈ 2.15 and the coefficient $\frac{2.15}{19.15} \approx 0.11$; for Tashi, M.D. ≈ 6.36 and $\frac{6.36}{14.7} \approx 0.43$. Tashi’s marks swing about four times as much as Pemba’s, for their size.

5.3.2 Mean deviation of a discrete series

In a discrete series each value X comes f times, so each distance counts f times. With $N = \Sigma f$:
a. From the mean
$$\overline{X} = \frac{\Sigma fX}{N}, \qquad \text{M.D.} = \frac{\Sigma f|X - \overline{X}|}{N} = \frac{\Sigma fD}{N}$$
b. From the median
$$M_d = \left(\frac{N + 1}{2}\right)^{\text{th}} \text{ item}, \qquad \text{M.D.} = \frac{\Sigma f|X - M_d|}{N} = \frac{\Sigma fD}{N}$$
The coefficients are $\frac{\text{M.D.}}{\overline{X}}$ and $\frac{\text{M.D.}}{M_d}$, as before.

Worked examples

Questions 1 to 7

Question 8

Find the mean deviation from the mean and its coefficient.

Question 9

Find the mean deviation from the median and its coefficient of the same data.

Questions 10 to 13

The mean deviation throws away the signs of the deviations to stop them canceling. There is another way to get rid of a minus sign: square it, since $(-3)^2 = 9$ is positive like $3^2$. Square each deviation from the mean, average the squares, and take the square root to get back to the data’s units. The result is the standard deviation, the most used measure of dispersion of all.

The standard deviation (S.D.) is the square root of the mean of the squares of the deviations of the items from their mean. It is written $\sigma$, the small Greek letter sigma:
$$\sigma = \sqrt{\frac{\Sigma (X - \overline{X})^2}{N}}$$
The square of the standard deviation, the mean of the squared deviations, is the variance, $\sigma^2$. The smaller the standard deviation, the more uniform the data, and the better the mean represents them.

Why is the standard deviation used more than the others? It uses every value, like the mean deviation, but squares are easy to work with in algebra (absolute values are not), so it leads on to most of the rest of statistics: the shape of the bell curve, the margins of error of surveys and the “±” after a measurement are all given in standard deviations.

5.4.1 Standard deviation of an individual series

There are three ways to work it out; all give the same answer.

a. Actual mean method. Find the mean $\overline{X} = \frac{\Sigma X}{N}$; find each deviation d = X − X; square each and add, $\Sigma d^2$; then
$$\sigma = \sqrt{\frac{\Sigma (X - \overline{X})^2}{N}} = \sqrt{\frac{\Sigma d^2}{N}}$$
This is easiest when the mean is a whole number; with a decimal mean the squares get long.

b. Direct method. No mean is subtracted: add the values, $\Sigma X$, and their squares, $\Sigma X^2$; then
$$\sigma = \sqrt{\frac{\Sigma X^2}{N} - \left(\frac{\Sigma X}{N}\right)^2}$$
In words: the mean of the squares minus the square of the mean.

c. Assumed mean method. Take a convenient value near the middle as the assumed mean A; find d = X − A for each value, then $\Sigma d$ and $\Sigma d^2$:
$$\sigma = \sqrt{\frac{\Sigma d^2}{N} - \left(\frac{\Sigma d}{N}\right)^2}, \quad \text{where } d = X - A$$
If A happens to be the actual mean, $\Sigma d = 0$ and this is method a.

5.4.2 Standard deviation of a discrete series

With frequencies, each value’s square counts f times, and $N = \Sigma f$.

a. Actual mean method.
$$\sigma = \sqrt{\frac{\Sigma f(X - \overline{X})^2}{N}} = \sqrt{\frac{\Sigma fd^2}{N}}, \quad d = X - \overline{X}$$
b. Direct method.
$$\sigma = \sqrt{\frac{\Sigma fX^2}{N} - \left(\frac{\Sigma fX}{N}\right)^2}$$
c. Assumed mean method. Choose A; find d = X − A, then fd and $fd^2$:
$$\sigma = \sqrt{\frac{\Sigma fd^2}{N} - \left(\frac{\Sigma fd}{N}\right)^2}$$

5.4.3 Coefficient of standard deviation

The standard deviation is an absolute measure, in the data’s units. The relative measure based on it is the coefficient of standard deviation:
$$\text{Coefficient of S.D.} = \frac{\sigma}{\overline{X}}$$
The smaller it is, the more uniform the data. When comparing sets, the one with the smaller coefficient is the more uniform (more consistent).

5.4.4 Coefficient of variation

The coefficient of standard deviation is often a small decimal, so it is usually multiplied by 100 and written as a percentage. That percentage is the coefficient of variation (C.V.):
$$\text{C.V.} = \frac{\sigma}{\overline{X}} \times 100\%$$
A large C.V. means more dispersion; a small one more uniformity. For Pemba and Tashi, $\sigma \approx 2.49$ and 7.28, and C.V. $\approx 13.0\%$ and 49.5%: Pemba’s marks are far more consistent.

Worked examples

Question 1

Question 2

Question 3

Find the standard deviation, its coefficient and the coefficient of variation.

Questions 4 and 5

A number like $\sigma$ sums up spread in one value. A box and whisker plot (or box plot) shows it at a glance, and it is the easiest way to compare two or more sets of data side by side. It is drawn from five numbers, the five-number summary:

• the smallest value,
• the lower quartile Q₁,
• the median $M_d$,
• the upper quartile Q₃,
• the largest value.

On a scale, a box is drawn from Q₁ to Q₃ with a line across it at the median, and a whisker (a line) from each end of the box out to the smallest and to the largest value. The five numbers cut the data into four parts with about 25% of the data in each: the left whisker, the two halves of the box, and the right whisker.

To draw a box plot:

1. Find the five-number summary: arrange the data in ascending order, find the smallest and largest values, Q₁, the median and Q₃.
2. Draw a scale that covers the smallest and largest values, and mark the five numbers on it with short upright lines.
3. Join the lines at Q₁ and Q₃ into a box, with the median’s line inside it, and draw the whiskers from the box out to the smallest and the largest values.

Questions 1 and 2

Question 3

The box plot shows the weights (kg) of the students of a secondary school.

Questions 4 to 7

Here are Lalitpur’s twelve monthly averages of fine dust in 2023 (µg/m³), in ascending order:
13.7, 14.2, 14.8, 27.1, 30.9, 44.2, 44.4, 53.6, 74, 79, 79.7, 80.1

• The middle: $\overline{X} = \frac{555.7}{12} \approx 46.3$; the median, the 6.5th item, is $\frac{44.2 + 44.4}{2} = 44.3$.
• The quartiles: $\frac{N + 1}{4} = 3.25$. $Q_1 = 14.8 + 0.25(27.1 - 14.8) \approx 17.9$ and Q₃, the 9.75th item, = 74 + 0.75(79 − 74) = 77.75. So $\text{Q.D.} \approx \frac{77.75 - 17.88}{2} \approx 29.9$, coefficient $\approx 0.63$.
• The deviations: the mean deviation from the mean is about 22.5 µg/m³ (coefficient 0.49), and the standard deviation $\sigma \approx 25.6$ µg/m³, so the C.V. is about 55%.

What it means: a coefficient of variation of 55% says that a typical month was more than half the mean away from it; the box covering almost the whole range says the year had no “usual” air, just a clean monsoon and a dirty dry season. One average, 46, hides that, and both halves of the story matter: the dry months are when people with weak lungs most need to take care, and they are also the months to work on, by cutting the smoke from kilns, vehicles and burning.

Grade 9’s final exam follows CDC’s Specification Grid and Model Question for Optional Mathematics, Grade 9 (2082). There Statistics has 7 marks of the 75 (10 teaching hours):

• Group A: one multiple-choice question, 1 mark (understanding).
• Statistics’ own question: one question in three parts, 3 marks (knowledge 1, understanding 1, higher ability 1).
• A question joining areas: parts of one question, 3 marks (application).

In the grid’s model paper the multiple-choice question shows a box plot of weights and asks for the lower quartile (read off the box’s left edge). Statistics’ own question gives a coffee shop’s customers over seven days, 15, 22, 18, 25, 20, 16, 23, and asks for the formula for the place of Q₃, then Q₃ itself (23, the 6th item in order), and then whether a standard deviation can be negative. And a question joining areas takes the function f(x) = 3x − 2 for x = 1, 2, 3, 4, 5, whose values are 1, 4, 7, 10 and 13, and asks for their standard deviation.