Statistics

Grade 9 found averages of single numbers. Start there. Here are the marks (out of 30) of 27 students of Grade 10 in a second terminal examination in Mathematics:

25, 15, 30, 22, 27, 12, 25, 30, 22, 24, 15, 23, 19, 27, 28, 17, 19, 22, 25, 15, 14, 13, 28, 26, 18, 20, 22

Look at (b): more students are above the mean than below it. That is not a mistake. A few low marks (12, 13, 14) are far below the rest and pull the mean down; the median, 22, only cares how many marks lie on each side. Keep that in mind: the mean and the median answer slightly different questions, and this unit uses both.

Activity 1: data that repeats

When the same value comes many times, don’t add Rs 30 six times: multiply. Write each value x beside its frequency f, the number of times it occurs, and add up fx:

Amount, x (Rs)	Students, f	fx
25	3	75
30	6	180
35	7	245
40	4	160
45	4	180
50	1	50
	N = 25	$\sum fx = 890$

$$\overline{X} = \frac{f_1 x_1 + f_2 x_2 + \cdots + f_n x_n}{f_1 + f_2 + \cdots + f_n} = \frac{\sum fx}{\sum f} = \frac{\sum fx}{N} = \frac{890}{25} = 35.6$$

On average a student brings Rs 35.60. $\sum$ (the Greek capital sigma) means “add up all the”, and $N = \sum f$ is the total number of values. A table of values with their frequencies is a discrete series; the plain list in 13.0 is an individual series.

Activity 2: data in groups

When there are many different values, a table of every value would be long and tell you little. So we put them in class intervals (classes, for short; each one is a class interval). Here are the marks (out of 50) of 40 Grade 9 students in Mathematics:

25, 10, 31, 22, 37, 42, 45, 37, 32, 34, 45, 40, 29, 27, 28, 17, 19, 22, 25, 33, 15, 14, 13, 28, 36, 38, 41, 42, 39, 25, 24, 31, 21, 22, 25, 26, 35, 36, 39, 49

The lowest is 10 and the highest 49, so classes 10 wide fit: 10–20, 20–30, 30–40, 40–50. Go through the list once, making a tally mark in each mark’s class. A class 10–20 includes 10 but not 20: a mark of 20 goes in 20–30. (This is how CDC’s tables and the SEE write classes; a series written this way is called a continuous series.)

Marks	Tally	Students, f
10–20	⫽⫽⫽⫽ I	6
20–30	⫽⫽⫽⫽ ⫽⫽⫽⫽ IIII	14
30–40	⫽⫽⫽⫽ ⫽⫽⫽⫽ III	13
40–50	⫽⫽⫽⫽ II	7
	Total	40

This is a frequency table (or frequency distribution), and the number in each class is that class’s frequency.

Now we have lost the exact marks: we know 6 students scored somewhere from 10 to just under 20, not which marks. The best single guess for each of them is the middle of the class, its mid-value:

$$m = \frac{\text{lower limit} + \text{upper limit}}{2}$$

For 10–20, $m = \frac{10 + 20}{2} = 15$. Then treat the 6 students as if each scored 15, and find the mean as in a discrete series with m in place of x.

Marks	f	Mid-value m	fm
10–20	6	15	90
20–30	14	25	350
30–40	13	35	455
40–50	7	45	315
	N = 40		$\sum fm = 1210$

$$\overline{X} = \frac{\sum fm}{N} = \frac{1210}{40} = 30.25$$

This is the direct method. The 40 marks themselves add up to 1,199, a true mean of 29.975: the grouped mean, 30.25, is an estimate, a little off because the marks inside each class aren’t exactly centered on its mid-value. With many values the two are usually very close.

The short-cut (assumed mean) method

The products fm can get large. A trick makes the numbers smaller. Pick any mid-value as an assumed mean A (one in the middle of the table is best), find each mid-value’s deviation from it, d = m − A, and then

$$\overline{X} = A + \frac{\sum fd}{N}$$

With A = 25 for the same marks:

Marks	f	m	d = m − 25	fd
10–20	6	15	−10	−60
20–30	14	25	0	0
30–40	13	35	10	130
40–50	7	45	20	140
	N = 40			$\sum fd = 210$

$$\overline{X} = 25 + \frac{210}{40} = 25 + 5.25 = 30.25$$

The same answer, and it always is, whatever A you pick. Why? Every m is A + d, so $\sum fm = \sum f(A + d) = A \sum f + \sum fd = AN + \sum fd$. Divide by N:

$$\frac{\sum fm}{N} = A + \frac{\sum fd}{N}$$

The assumed mean is a guess, and $\frac{\sum fd}{N}$ is the correction to it: here the guess 25 was 5.25 too low.

Worked examples: the mean

Question 1

Find the mean of each of these.

Question 2

Find the mean of each table by the direct method and by the short-cut method.

Question 3

Find the unknown value.

Question 4

Question 5

Make a frequency table from the data and find the mean.

Question 6

Find the mean.

Activity 3: the middle mark

The median is the middle value of data arranged in order: as many values below it as above. For N values in an individual or discrete series it is the

$$\left(\frac{N + 1}{2}\right)\text{th item}$$

With an even N that falls between two items (the 8.5th of 16), and the median is the average of the two.

Activity 4: the median of grouped data

In a continuous series we know how many values are in each class, not what they are. So we find the class the middle value is in, and then estimate where in the class it is.

First, a less-than cumulative frequency table. Beside each class write how many values are less than its upper limit.

Marks	f		cf
0–8	6	less than 8	6
8–16	10	less than 16	16
16–24	16	less than 24	32
24–32	18	less than 32	50
32–40	12	less than 40	62
40–48	10	less than 48	72
48–56	8	less than 56	80

Second, where the median is. In a continuous series the median is the $\left(\frac{N}{2}\right)$th item: here the $\frac{80}{2} =$ 40th.

Third, the median class, the first whose cf reaches 40: 24–32 (32 marks are below 24, 50 below 32).

Last, the formula.

$$M_d = L + \frac{\dfrac{N}{2} - cf}{f} \times h$$

where L is the lower limit of the median class, cf the cumulative frequency of the class before it, f the frequency of the median class and h its width (the class size; some books and exam papers write i for it).

$$\begin{aligned} M_d &= 24 + \frac{40 - 32}{18} \times 8 \\ &= 24 + 3.56 \\ &= 27.56 \end{aligned}$$

The median mark is about 27.56.

Why does it work? 32 students are below 24, so the 40th is the 8th of the 18 students in 24–32. Spread those 18 evenly across the 8 marks of the class, and the 8th of them is $\frac{8}{18}$ of the way across: $24 + \frac{8}{18} \times 8$. That is all the formula says. Draw the cumulative frequencies as a graph and the same reading appears as a picture:

Activity 5: classes with gaps

Here the classes have gaps: one ends at 6 and the next starts at 7. (They were made for heights measured to the nearest foot.) Close the gaps by a correction factor:

$$\text{correction} = \frac{\text{lower limit of the 2nd class} - \text{upper limit of the 1st}}{2} = \frac{7 - 6}{2} = 0.5$$

Subtract it from every lower limit and add it to every upper limit: 4–6 becomes 3.5–6.5, 7–9 becomes 6.5–9.5, and so on. A tree of 6.4 ft, recorded as 6, belongs in the first class; one of 6.6 ft, recorded as 7, in the second.

Height (ft)	f	cf
3.5–6.5	2	2
6.5–9.5	3	5
9.5–12.5	10	15
12.5–15.5	7	22
15.5–18.5	4	26
18.5–21.5	3	29
21.5–24.5	2	31

N = 31; the median is the $\frac{31}{2} =$ 15.5th item, in 12.5–15.5.

$$\begin{aligned} M_d &= 12.5 + \frac{15.5 - 15}{7} \times 3 \\ &= 12.5 + 0.21 \\ &= 12.71 \end{aligned}$$

The median height is about 12.71 ft. Without the correction, L would be 13 and h 2, and the answer wrong.

Question 1

Find the median.

Question 2

Find the median.

Question 3

Find the missing frequency.

Question 4

Find the median.

Question 5

Activity 6: the most common value

The value that occurs most often is the mode, $M_o$. It is the only one of the three averages that works for things that aren’t numbers (the most common blood group, the best-selling shoe size, the favorite momo filling), and data can have one mode, two, or none (when every value occurs once).

Activity 7: the mode of grouped data

In a continuous series the class with the highest frequency, the modal class, contains the mode. Where in it? Nearer the neighbor with more values. The formula:

$$M_o = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

where L is the lower limit of the modal class, f₁ its frequency, f₀ the frequency of the class before it, f₂ of the class after it, and h the width.

Look at the denominator: 2f₁ − f₀ − f₂ = (f₁ − f₀) + (f₁ − f₂). The modal class is taller than its left neighbor by f₁ − f₀ and its right neighbor by f₁ − f₂, and the mode divides the class in that ratio: if the left neighbor is almost as tall as the modal bar, f₁ − f₀ is small and the mode sits near the left end.

Question 1

Find the mode.

Question 2

Find the mode.

Question 3

Find the mode.

Activity 8: quarters of the class

The median cuts ordered data into two halves. The quartiles cut it into four quarters: a quarter of the values lie below the first quartile Q₁ (also called the lower quartile), half below the median (the second quartile Q₂), and three quarters below the third quartile Q₃ (the upper quartile). For an individual or discrete series, as in Grade 9:

$$Q_1 = \left(\frac{N + 1}{4}\right)\text{th item}, \qquad Q_3 = \left(\frac{3(N + 1)}{4}\right)\text{th item}$$

Quartiles of grouped data

For a continuous series the method is the median’s, with $\frac{N}{4}$ or $\frac{3N}{4}$ in place of $\frac{N}{2}$:

1. Make the less-than cumulative frequency table.
2. Q₁ is the $\left(\frac{N}{4}\right)$th item and Q₃ the $\left(\frac{3N}{4}\right)$th.
3. The quartile’s class is the first whose cf reaches that position.
4. Then

$$Q_1 = L + \frac{\dfrac{N}{4} - cf}{f} \times h \qquad Q_3 = L + \frac{\dfrac{3N}{4} - cf}{f} \times h$$

with L, cf (of the class before), f and h all belonging to the quartile’s own class, which is usually not the median’s.

Question 1

Find Q₁ and Q₃.

Question 2

Find Q₁ and Q₃.

Question 3

Question 4

Find Q₁ and Q₃.

Question 5

Half of the 2,84,160 students graded in SEE 2082 got a GPA below about 3.14, and the top quarter got about 3.49 or more. So a GPA of 3.25 is in the top half, but not the top quarter.

That comes straight from this unit’s formulas, with NEB’s bands as the classes:

GPA	1.6–2.0	2.0–2.4	2.4–2.8	2.8–3.2	3.2–3.6	3.6–4.0
Students	7	5,190	55,977	94,222	80,372	48,392
cf	7	5,197	61,174	1,55,396	2,35,768	2,84,160

The median is the 1,42,080th student, in 2.8–3.2 (61,174 below 2.8):

$$\begin{aligned} M_d &= 2.8 + \frac{142080 - 61174}{94222} \times 0.4 \\ &= 2.8 + 0.34 \\ &= 3.14 \end{aligned}$$

Q₃ is the 2,13,120th, in 3.2–3.6: $3.2 + \frac{213120 - 155396}{80372} \times 0.4 = 3.49$. Read them off the ogive yourself:

The mean, from mid-values 1.8, 2.2, …, 3.8, is $\frac{896791.2}{284160} \approx 3.16$, and the mode $2.8 + \frac{94222 - 55977}{2 \times 94222 - 55977 - 80372} \times 0.4 \approx 3.09$. All three agree that the typical graded student got a little over 3.

CDC’s Specification Grid for Compulsory Mathematics gives Statistics and Probability 11 marks in two questions. The statistics question, as in the grid’s model paper, is 6 marks in four parts: knowledge (1), understanding (2), application (2) and higher ability (1). Here is one of our own in that shape.