Formulas · Mathematics

Statistics

3 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 10

Mean of grouped data

x ˉ = \frac{\sum f _{i} x _{i}}{\sum f _{i}}

What each letter means
$x ˉ$	the mean (average)
$\sum f_{i} x_{i}$	the sum of fᵢ × xᵢ over all the classes (each class mark times its frequency)
$\sum f_{i}$	the sum of the frequencies: how many values there are in all

For data in classes, treat every value in a class as its class mark (the middle of the class), then take the average as usual: total of (class mark × frequency) divided by the total frequency.

Why it works

The mean is the total of all the values divided by how many there are. A class holding fᵢ values, all counted as xᵢ, contributes fᵢ × xᵢ to the total. The answer is an estimate, since the values inside a class are assumed to sit at its middle.

When to use it

Finding the average from a frequency table: marks in ranges, wages in bands, ages of patients, family sizes.

How to use it

Find each class mark: (lower limit + upper limit) ÷ 2.
Multiply each class mark by its frequency, and add the products (Σfᵢxᵢ).
Add the frequencies (Σfᵢ) and divide.

To remember: Weight each middle by its count, then share out.

Other forms

Same formula, another formAssumed mean method: pick a class mark a, take dᵢ = xᵢ − a, and x̄ = a + Σfᵢdᵢ ÷ Σfᵢ. With the book's marks table and a = 47.5, Σfᵢdᵢ = 435, so x̄ = 47.5 + 14.5 = 62.
Same formula, another formStep deviation method: with class size h, uᵢ = (xᵢ − a) ÷ h and x̄ = a + (Σfᵢuᵢ ÷ Σfᵢ) × h. For the same table, Σfᵢuᵢ = 29 and x̄ = 47.5 + 29/30 × 15 = 62.

Worked examples

Story

In a survey of 50 families, 10 have 2 members, 15 have 3, 20 have 4 and 5 have 5. What is the mean family size?

Σfᵢxᵢ = 10 × 2 + 15 × 3 + 20 × 4 + 5 × 5 = 20 + 45 + 80 + 25 = 170.
$x ˉ = \frac{170}{50} = 3.4$

Answer: 3.4

Picture

Daily wages of 50 workers: ₹100 to 120 (12 workers), 120 to 140 (14), 140 to 160 (8), 160 to 180 (6), 180 to 200 (10). Find the mean daily wage.

The class marks are 110, 130, 150, 170 and 190.
Σfᵢxᵢ = 1320 + 1820 + 1200 + 1020 + 1900 = 7260.
$\frac{7260}{50} = 145.2$

Answer: ₹145.20

Direct

In a frequency table, Σfᵢxᵢ = 1860 and Σfᵢ = 30. Find the mean.

$x ˉ = \frac{1860}{30} = 62$

Answer: 62

Reverse

The mean of this distribution is 18. Find the missing frequency f: 11 to 13 (7), 13 to 15 (6), 15 to 17 (9), 17 to 19 (13), 19 to 21 (f), 21 to 23 (5), 23 to 25 (4).

Try it first, then show the working

The class marks are 12, 14, 16, 18, 20, 22 and 24.
Σfᵢ = 44 + f and Σfᵢxᵢ = 752 + 20f.
752 + 20f = 18(44 + f) = 792 + 18f, so 2f = 40 and f = 20.

Answer: 20

Exam

Marks of 30 students: 10 to 25 (2), 25 to 40 (3), 40 to 55 (7), 55 to 70 (6), 70 to 85 (6), 85 to 100 (6). Find the mean marks.

Try it first, then show the working

The class marks are 17.5, 32.5, 47.5, 62.5, 77.5 and 92.5.
Σfᵢxᵢ = 35 + 97.5 + 332.5 + 375 + 465 + 555 = 1860, and Σfᵢ = 30.
$\frac{1860}{30} = 62$

Answer: 62

Common mistake: Adding the class marks without multiplying by the frequencies, or dividing by the number of classes instead of the total frequency.

Class 10

Median of grouped data

M_{e} = l + \frac{\frac{n}{2} - c f}{f} \times h

What each letter means
$M_{e}$	the median (the middle value)
$l$	the lower limit of the median class
$n$	the total frequency (how many values)
$c f$	the cumulative frequency of the class before the median class
$f$	the frequency of the median class
$h$	the class size (width)

The median class is the one holding the middle value (the n/2-th). Inside it, the formula goes the right fraction of the way across, assuming its values are spread evenly.

Why it works

Before the median class there are cf values; we still need n/2 − cf more. The median class holds f values spread over a width h, so each value takes h ÷ f of the width, and n/2 − cf of them take (n/2 − cf) ÷ f × h. On a cumulative frequency graph (an ogive) this is reading off the straight line across the class.

When to use it

The middle value of grouped data, especially when a few very large or small values would pull the mean away (incomes, house prices, lifetimes).

How to use it

Make the cumulative frequency column and find n/2.
The median class is the first class whose cumulative frequency reaches n/2.
Read l, cf (of the class before), f and h, and put them in.

To remember: Start of the class, plus the share still needed, times the width.

Other forms

Same formula, another formOn a less-than ogive, the median is the x-value where the curve reaches n/2: the formula is the straight line across the median class.
Rearranged $c f = \frac{n}{2} - \frac{( M _{e} - l ) f}{h}$ A missing cumulative frequency, from a known median.

Worked examples

Story

Weekly pocket money of 40 students: ₹0 to 20 (5), 20 to 40 (8), 40 to 60 (12), 60 to 80 (10), 80 to 100 (5). Find the median.

n/2 = 20. The cumulative frequencies are 5, 13, 25, 35, 40, so the median class is 40 to 60.
$M_{e} = 40 + \frac{20 - 13}{12} \times 20 = 40 + \frac{35}{3} = \frac{155}{3}$
That is about ₹51.67.

Answer: 155/3

Picture

On a less-than ogive for 30 students, the curve passes through (55, 12) and (70, 18). Read off the median.

The median is where the cumulative frequency is n/2 = 15, between those two points.
$55 + \frac{15 - 12}{18 - 12} \times 15 = 62.5$
This is exactly the median formula with l = 55, cf = 12, f = 6 and h = 15.

Answer: 62.5

Direct

The median class is 55 to 70, with l = 55, n = 30, cf = 12, f = 6 and h = 15. Find the median.

$M_{e} = 55 + \frac{15 - 12}{6} \times 15$
$= 55 + 7.5 = 62.5$

Answer: 62.5

Reverse

For 60 values in classes 0 to 10 (5), 10 to 20 (x), 20 to 30 (20), 30 to 40 (15), 40 to 50 (y), 50 to 60 (5), the median is 28.5. Find x and y.

Try it first, then show the working

The median 28.5 is in the class 20 to 30, so 28.5 = 20 + (30 − (5 + x)) ÷ 20 × 10.
8.5 = (25 − x) ÷ 2, so x = 8.
All frequencies add to 60: 5 + 8 + 20 + 15 + y + 5 = 60, so y = 7.

Answer: x = 8, y = 7

Exam

Marks of 30 students: 10 to 25 (2), 25 to 40 (3), 40 to 55 (7), 55 to 70 (6), 70 to 85 (6), 85 to 100 (6). Find the median, and compare it with the mean (62) and the mode (52).

Try it first, then show the working

n/2 = 15. Cumulative frequencies: 2, 5, 12, 18, 24, 30, so the median class is 55 to 70.
$M_{e} = 55 + \frac{15 - 12}{6} \times 15 = 62.5$
Median 62.5, mean 62 and mode 52: half the students scored below 62.5, the average was 62, and the most crowded class centres near 52.

Answer: 62.5

Common mistake: Taking cf as the cumulative frequency of the median class itself (it must be the class before), or using n instead of n/2.

Class 10

Mode of grouped data

M_{o} = l + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h

What each letter means
$M_{o}$	the mode (the most frequent value)
$l$	the lower limit of the modal class (the class with the highest frequency)
$f_{1}$	the frequency of the modal class
$f_{0}$	the frequency of the class just before it
$f_{2}$	the frequency of the class just after it
$h$	the class size (width)

The mode lies in the class with the most values. Inside it, it leans towards whichever neighbouring class is bigger.

Why it works

The modal class sticks up above its neighbours by f1 − f0 on the left and f1 − f2 on the right. The mode divides the class width in the ratio of those two steps, so it goes (f1 − f0) ÷ ((f1 − f0) + (f1 − f2)) of the way across, and the bottom of that fraction is 2f1 − f0 − f2. On a histogram it is where the lines joining the top corners of the tallest bar to its neighbours cross.

When to use it

The most typical value of grouped data: the commonest shoe size range, the busiest age group, the most usual family size.

How to use it

Find the modal class: the one with the highest frequency.
Read l, h, f1 and the neighbouring frequencies f0 and f2.
Put them in the formula.

To remember: Lean towards the bigger neighbour.

Other forms

Same formula, another formThe same formula written with the two steps: M = l + (f1 − f0) ÷ ((f1 − f0) + (f1 − f2)) × h.
Special case $M_{o} = l + \frac{h}{2}$ When the two neighbouring classes are equal (f0 = f2): the mode is the middle of the class.

Worked examples

Story

A survey of 20 households found family sizes 1 to 3 (7 families), 3 to 5 (8), 5 to 7 (2), 7 to 9 (2), 9 to 11 (1). Find the modal family size.

The modal class is 3 to 5 (8 families).
$M_{o} = 3 + \frac{8 - 7}{16 - 7 - 2} \times 2 = 3 + \frac{2}{7} = \frac{23}{7}$
That is about 3.286.

Answer: 23/7

Picture

On a histogram, the tallest bar (40 to 55, height 7) has neighbours of heights 3 and 6. Join each top corner of the tallest bar to the top corner of the neighbour beside it; where do the lines cross?

The crossing point divides the bar's width in the ratio of the two steps, 7 − 3 = 4 and 7 − 6 = 1.
It is 4/5 of the way across the 15-wide class: 40 + 12 = 52, the mode.

Answer: at 52

Direct

Marks of 30 students: the modal class is 40 to 55 with frequency 7; the classes before and after have 3 and 6. Find the mode.

$M_{o} = 40 + \frac{7 - 3}{14 - 3 - 6} \times 15$
$= 40 + \frac{4}{5} \times 15 = 40 + 12 = 52$

Answer: 52

Reverse

The mode is 52, the modal class is 40 to 55 with frequency 7, and the class after it has 6. What is the frequency of the class before it?

Try it first, then show the working

52 = 40 + (7 − f0) ÷ (8 − f0) × 15, so 12(8 − f0) = 15(7 − f0).
96 − 12f0 = 105 − 15f0, so 3f0 = 9 and f0 = 3.

Answer: 3

Exam

Ages of hospital patients: 5 to 15 (6), 15 to 25 (11), 25 to 35 (21), 35 to 45 (23), 45 to 55 (14), 55 to 65 (5). Find the mode.

Try it first, then show the working

The modal class is 35 to 45 (23 patients), with 21 before and 14 after.
$35 + \frac{23 - 21}{46 - 21 - 14} \times 10 = 35 + \frac{20}{11}$
That is about 36.8 years.

Answer: 36.8 years (405/11)

Common mistake: Mixing up f0 and f2 (f0 is before the modal class, f2 after), or using the highest frequency itself as the mode.