Statistics: Class 11 Maths · Chapter 13

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 22 Oct 2026 · Updated 22 Oct 2026

78 min
1 simulation
33 practice questions
25 MCQs

What you will learn in Statistics

Explain why a measure of central tendency alone cannot describe a data set, and compare data sets by their spread.
Find and interpret the range of a data set and recognise its weaknesses.
Calculate the mean deviation about the mean and about the median for raw data and for discrete and continuous frequency distributions.
Find the median of a grouped distribution from cumulative frequencies and from the median-class formula.
Use step deviations (an assumed mean and a change of scale) to simplify the arithmetic for means and standard deviations.
Define variance and standard deviation, derive their computing formulas and calculate them for ungrouped and grouped data.
Predict how the mean, variance and standard deviation change when every observation is shifted or scaled, and correct them after a recording error.

Original companion text aligned to NCERT Mathematics Class 11, Chapter 13, pp. 257–288. Not affiliated with or endorsed by NCERT.

13.1 Introduction

Two buses, Route P and Route Q, run from the same stop in Pune to Meera's college. For eight mornings she notes how many minutes she waits for each. Route P: 15, 3, 21, 11, 1, 19, 7, 11. Route Q: 12, 10, 11, 13, 9, 11, 12, 10. On the morning of an exam, which bus should she trust?

Start with the averages from earlier classes, the measures of central tendency. For $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ the mean is their total shared equally:

mean of n observations

\overset{ˉ}{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}

(13.1)

where

$\sum_{i = 1}^{n} x_{i}$: the sum $x_{1} + x_{2} + \dots + x_{n}$

For the median, first arrange the observations in increasing (or decreasing) order. Then:

if $n$ is odd, the median is the $(\frac{n + 1}{2})$ th observation;
if $n$ is even, it is the mean of the $\frac{n}{2}$ th and the $(\frac{n}{2} + 1)$ th observations.

Each route's eight waits add up to 88 minutes, so both means are 11 minutes. In order, Route P reads 1, 3, 7, 11, 11, 15, 19, 21 and Route Q reads 9, 10, 10, 11, 11, 12, 12, 13. The 4th and 5th values are 11 and 11 in both lists, so both medians are 11 as well. By their averages, the two routes are identical.

3/3Both routes have mean 11 and median 11, yet Q is far more predictable.

Fig. 13.1Same centre, very different spread.NCERT Fig 13.1 · p. 258

Two number lines from 0 to 24 minutes. In the first step, Route P's eight waits appear as round dots at 1, 3, 7, 11, 11, 15, 19 and 21, the two elevens stacked; they are spread across most of the line. In the second step, Route Q's waits appear on the lower line as square dots at 9, 10, 10, 11, 11, 12, 12 and 13, packed tightly together with the repeated values stacked. In the third step a dashed vertical line at 11 crosses both plots and is labelled mean = median = 11, showing that the same centre hides very different spreads.

Yet anyone who rode both buses would call them very different. Route Q's waits were never more than 2 minutes from 11. Route P once made her wait 21 minutes and once only 1. The values of Q huddle around the centre while those of P are strewn across the line (Fig. 13.1). An average tells us where the data sit. We also need a number that tells us how widely they are spread.

Definition

Measure of dispersion

A single number that describes how spread out (scattered) a set of observations is, usually about a central value.

ExampleThe range, the mean deviation and the standard deviation.

This chapter builds such measures of dispersion (प्रकीर्णन की माप) and shows how to compute them for raw lists and for frequency tables.

13.2 Measures of Dispersion

There is no single way to put a number on spread. Some measures look only at the two extremes; others ask how far every observation lies from a centre such as the mean or the median. The ones in common use are:

the range;
the quartile deviation;
the mean deviation;
the standard deviation, together with its square, the variance.

Of these, the quartile deviation is left for later courses; this chapter develops the other three in turn.

13.3 Range

The quickest way to sense spread is to look at the two extremes. Route P's waits ran from 1 to 21 minutes; Route Q's from 9 to 13.

Definition

Range

The difference between the largest and the smallest observation of a data set.

ExampleFor Route P, $21 - 1 = 20$ minutes.

range

Range = x_{\max} - x_{\min}

(13.2)

So Route P has a range (परिसर) of 20 minutes and Route Q a range of 4 minutes: on this measure P is five times as scattered. The range is in the same units as the data.

3/3If one wait of 13 becomes 40, Q's range jumps to 31.

Fig. 13.2The range uses only the two extreme values.

Two number lines from 0 to 40 minutes. In the first step a bracket over Route P's dots runs from 1 to 21 and is labelled range 20. In the second step a bracket over Route Q's tightly packed dots runs from 9 to 13 and is labelled range 4. In the third step one of Route Q's dots, the 13, slides far to the right to 40, as if the bus broke down that day; the bracket stretches from 9 to 40 and is labelled range 31, although the other seven dots have not moved.

The range is easy to find, but it listens to only two observations. Suppose Route Q broke down on day 4 and Meera waited 40 minutes instead of 13. Seven of the eight mornings are unchanged, yet the range leaps from 4 to 31 (Fig. 13.2). The range also says nothing about how the values sit around the mean or the median. For that we need measures built from the distance of every observation from a centre.

Quick check

1 of 3

The highest temperatures in Jaisalmer over a week were 41, 44, 39, 46, 43, 40 and 45 °C. Find the range.

°C

Show answer

Answer 7 °C

46 - 39 = 7

°C. The order of the days does not matter.

13.4 Mean Deviation

Choose a central value $a$ , usually the mean or the median. The deviation (विचलन) of an observation $x$ from $a$ is $x - a$ : positive for values above $a$ and negative for values below it. The obvious next move is to average the deviations. About the mean, this fails completely. In the sums below, $\sum$ runs over all the observations.

deviations about the mean

\begin{aligned} \sum (x_{i} - \overset{ˉ}{x}) \\ = \sum x_{i} - n \overset{ˉ}{x} \\ = n \overset{ˉ}{x} - n \overset{ˉ}{x} = 0 \end{aligned}

(13.3)

The second line splits the sum; the third uses $\sum x_{i} = n \overset{ˉ}{x}$ , which is just the definition of the mean. The positive deviations exactly cancel the negative ones, because the mean is the balance point of the data. An average deviation of zero would claim that every data set has no spread, which is useless.

The cure is to ignore the signs. The absolute value $∣ x - a ∣$ is the distance between $x$ and $a$ on the number line, and distances never cancel.

Definition

Mean deviation about a

The mean of the absolute deviations $∣ x_{i} - a ∣$ of the observations from a fixed value $a$ ; it is written M.D. $(a)$ .

mean deviation about a

\begin{aligned} M.D. (a) \\ = \frac{1}{n} \sum ∣ x_{i} - a ∣ \end{aligned}

(13.4)

Any central value can play the role of $a$ , but in practice we use the mean $\overset{ˉ}{x}$ or the median. From now on $M$ stands for the median. The mean deviation (माध्य विचलन) is measured in the units of the data.

13.4.1 Mean deviation for ungrouped data

For raw observations $x_{1}, x_{2}, \dots, x_{n}$ the work has four steps:

find the centre $a$ , the mean $\overset{ˉ}{x}$ or the median $M$ ;
subtract $a$ from every observation;
take absolute values, turning each deviation into a distance $∣ x_{i} - a ∣$ ;
add the distances and divide by $n$ .

3/4The distances 5, 2, 1, 2, 6 add up to 16.

Fig. 13.3Signed deviations cancel; distances do not.

A number line from 0 to 14 carries five dots at 2, 5, 6, 9 and 13 and a marker at the mean 7. In the second step five horizontal arrows start at the mean and end at each dot: three point left and are labelled minus 5, minus 2 and minus 1, two point right and are labelled plus 2 and plus 6; the left arrows total 8 and so do the right ones. In the third step the three left arrows flip over to point right, so all five are now distances 5, 2, 1, 2 and 6 with total 16. In the fourth step a thick bar from the mean of length 3.2 is labelled M.D. = 16 ÷ 5 = 3.2.

Written with the two usual centres, the definition becomes (Fig. 13.3):

about the mean and about the median

\begin{aligned} M.D. (\overset{ˉ}{x}) \\ = \frac{1}{n} \sum ∣ x_{i} - \overset{ˉ}{x} ∣ \\ M.D. (M) \\ = \frac{1}{n} \sum ∣ x_{i} - M ∣ \end{aligned}

(13.5)

Example 13.1NCERT Example 1 · p. 261

Mean deviation of quiz scores

Eight friends score 14, 9, 17, 6, 11, 20, 8 and 11 in a 20-mark quiz. Find the mean deviation about the mean.

Try it yourself first. Then reveal the solution one step at a time.

With a little practice you can skip writing the signed deviations and list the distances straight away.

Example 13.2NCERT Example 2 · p. 261

A longer list

Sixteen children logged these minutes of outdoor play on one day: 25, 40, 10, 35, 45, 15, 30, 45, 20, 50, 30, 5, 40, 55, 25, 10. Find the mean deviation about the mean.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.3NCERT Example 3 · p. 262

About the median

Nine students read 42, 37, 55, 48, 29, 61, 50, 38 and 45 pages of a novel in one week. Find the mean deviation about the median.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Find the mean deviation about the mean of 3, 7, 8, 10, 12.

Show answer

Answer 2.4

The distances from 8 are 5, 1, 0, 2, 4, with total 12, so M.D.

= \frac{12}{5} = 2.4.

Averaging the signed deviations would give 0.

13.4.2 Mean deviation for grouped data

Large data sets usually arrive as frequency tables, in one of two forms:

a discrete frequency distribution lists each distinct value $x_{i}$ with its frequency $f_{i}$ , the number of times it occurs;
a continuous frequency distribution sorts the values into class intervals without gaps, such as 0–20, 20–40, 40–60, and gives the frequency of each class.

Discrete frequency distribution

Let the distinct values $x_{1}, x_{2}, \dots, x_{n}$ occur $f_{1}, f_{2}, \dots, f_{n}$ times, and let $N = \sum f_{i}$ be the total frequency. A value that occurs $f_{i}$ times adds $f_{i} x_{i}$ to the total and $f_{i} ∣ x_{i} - a ∣$ to the sum of distances.

(i) About the mean. The mean and the mean deviation are

mean of a frequency distribution

\begin{aligned} \overset{ˉ}{x} & = \frac{1}{N} \sum f_{i} x_{i} \\ N & = \sum f_{i} \end{aligned}

(13.6)

about the mean

\begin{aligned} M.D. (\overset{ˉ}{x}) \\ = \frac{1}{N} \sum f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣ \end{aligned}

(13.7)

(ii) About the median. Write the values in increasing order and add a cumulative frequency (c.f.) column, the running total of the frequencies. The c.f. of a value is the last position it fills when all $N$ observations are listed in order. The median is in position $\frac{N + 1}{2}$ ; for even $N$ it lies halfway between positions $\frac{N}{2}$ and $\frac{N}{2} + 1$ . So look for the first value whose c.f. reaches that position (Fig. 13.4). In most tables both middle positions fall on the same value.

about the median

\begin{aligned} M.D. (M) \\ = \frac{1}{N} \sum f_{i} ∣ x_{i} - M ∣ \end{aligned}

(13.8)

4/4Size 7 fills positions 17 to 30, so $M = 7$ .

Fig. 13.4Cumulative frequency finds the middle position.NCERT Table 13.2 · p. 264

Bars stand over shoe sizes 4 to 10. In the first step their heights are the frequencies 2, 5, 9, 14, 10, 6 and 4. In the second step each bar grows to its cumulative frequency, giving a rising staircase 2, 7, 16, 30, 40, 46 and 50. In the third step a dashed horizontal line at 25, half of the 50 observations, crosses the staircase. In the fourth step the bar for size 7 is highlighted: its cumulative frequency jumps from 16 to 30, so positions 17 to 30, including the 25th and 26th, are all size 7, and the median is 7.

Example 13.4NCERT Example 4 · p. 263

Books borrowed in a term

A school library recorded how many books each of 40 students borrowed in one term. Find the mean deviation about the mean. Try filling in the working table first.

Table 13.1Books borrowed (working table)

Books $x_{i}$	Students $f_{i}$	$f_{i} x_{i}$	$∣ x_{i} - \overset{ˉ}{x} ∣$	$f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣$
4	3
6	8
8	8
10	12
12	5
14	4

Your entries are saved automatically.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.5NCERT Example 5 · p. 264

Shoe sizes sold in a day

A shoe shop in Agra sold 50 pairs in one day. Find the mean deviation about the median size.

Table 13.2

Size	Pairs sold
4	2
5	5
6	9
7	14
8	10
9	6
10	4

Try it yourself first. Then reveal the solution one step at a time.

Continuous frequency distribution

In a continuous frequency distribution the individual values are lost: we know only how many fall in each class. A value equal to an upper limit, such as 40 for the class 20–40, is counted in the next class, 40–60.

To compute anything we agree to treat every observation in a class as if it sat at the mid-point of the class, its class mark. For 20–40 the class mark is $\frac{20 + 40}{2} = 30.$ The errors made this way partly cancel, because some values in a class lie above its mid-point and some below (Fig. 13.5).

2/3Each class becomes a pile of values at its mid-point.

Fig. 13.5Every value in a class is treated as its mid-point.NCERT p. 265

In the first step a histogram has six touching bars over the classes 0–20, 20–40, 40–60, 60–80, 80–100 and 100–120 kilometres with frequencies 2, 10, 15, 11, 8 and 4. In the second step the bars fade to dashed outlines and each class is replaced by a single spike at its mid-point, 10, 30, 50, 70, 90 and 110, of the same height, showing every value of a class treated as its class mark. In the third step a triangle under the axis marks the mean 60, the balance point of the spikes.

(i) About the mean. Replace each class by its class mark $x_{i}$ and work exactly as for a discrete distribution, with (13.6) and (13.7).

Example 13.6NCERT Example 6 · p. 266

Distances ridden by delivery riders

A food-delivery company in Bengaluru logged the distance (in km) each of its 50 riders covered on one day. Find the mean deviation about the mean.

Table 13.3Riders' distances (working table)

Distance (km)	Riders $f_{i}$	Mid-point $x_{i}$	$f_{i} x_{i}$	$∣ x_{i} - \overset{ˉ}{x} ∣$	$f_{i} ∣ x_{i} - \overset{ˉ}{x} ∣$
0–20	2
20–40	10
40–60	15
60–80	11
80–100	8
100–120	4

Your entries are saved automatically.

Try it yourself first. Then reveal the solution one step at a time.

A shortcut for the mean: step deviations. The products $f_{i} x_{i}$ grow large quickly. Two moves make the numbers small. First, shift the origin: measure each mid-point from a convenient value $a$ near the middle, called the assumed mean. Second, change the scale: if all the shifted values share a factor $h$ (often the class width), divide by it. The results $d_{i}$ are the step deviations (Fig. 13.6).

3/3Dividing by $h = 20$ leaves the small numbers $- 2$ to $3$ .

Fig. 13.6Shift the origin, then shrink the scale.NCERT Figs 13.3, 13.4 · p. 267

Three parallel rulers share the same six tick positions. The top ruler is labelled mid-points x and reads 10, 30, 50, 70, 90 and 110. In the second step a lower ruler appears, labelled subtract 50, reading −40, −20, 0, 20, 40 and 60, with the zero under the old 50 highlighted as the new origin. In the third step a third ruler appears, labelled divide by 20, reading −2, −1, 0, 1, 2 and 3: the same points on a smaller scale.

mean by step deviations

\overset{ˉ}{x} = a + h \cdot \frac{\sum f_{i} d_{i}}{N}

(13.9)

where

$d_{i}$: the step deviation $\frac{x_{i} - a}{h}$
$a$: the assumed mean
$h$: the common factor

Why does this work? From $d_{i} = \frac{x_{i} - a}{h}$ we get $x_{i} = a + h d_{i}$ . Multiplying by $f_{i}$ , adding and dividing by $N$ gives $\overset{ˉ}{x} = a + h \overset{ˉ}{d} :$ shifting the origin shifts the mean, and changing the scale scales it.

Example

Example 13.6 by step deviations

Find the mean distance of the riders in Example 13.6 by the step-deviation method.

Try it yourself first. Then reveal the solution one step at a time.

The shortcut changes only how you find $\overset{ˉ}{x}$ . The distances $∣ x_{i} - \overset{ˉ}{x} ∣$ are still measured on the original scale, so the mean deviation is again 22 km.

(ii) About the median. For a continuous distribution the median is estimated inside one class. The median class is the class in which the c.f. first reaches $\frac{N}{2}$ . Suppose $C$ observations come before it, it starts at $l$ , it has width $h$ and holds $f$ observations, spread evenly across it. We need $\frac{N}{2} - C$ of those $f$ observations, so we go that fraction of the way across the class (Fig. 13.7):

median of a continuous distribution

M = l + \frac{\frac{N}{2} - C}{f} \times h

(13.10)

where

$l$: lower limit of the median class
$C$: c.f. of the class before it
$f$: frequency of the median class
$h$: width of the median class

4/4 $M = 50 + \frac{6}{15} \times 10 = 54$

Fig. 13.7The median sits the right fraction of the way through its class.NCERT p. 269

A strip at the top shows the 60 observations in order as three blocks: 24 below 50, 15 in the class 50–60 and 21 above 60. In the second step a marker at position 30, half of 60, falls inside the middle block. In the third step the middle block is enlarged below as a bar from 50 to 60 holding 15 evenly spaced dots, and the first 6 dots are highlighted, because 30 minus 24 is 6. In the fourth step an arrow runs from 50 to 54 along the bar, six fifteenths of the width 10, and is labelled M = 54.

Then measure the distances of the class marks from $M$ and use (13.8).

Example 13.7NCERT Example 7 · p. 269

Ages at a yoga camp

The ages of 60 people at a yoga camp in Rishikesh are grouped below. Find the mean deviation about the median age.

Table 13.4Ages (working table)

Age (years)	People $f_{i}$	c.f.	Mid-point $x_{i}$	$∣ x_{i} - M ∣$	$f_{i} ∣ x_{i} - M ∣$
20–30	3
30–40	5
40–50	16
50–60	15
60–70	11
70–80	10

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Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The values 1, 2 and 3 occur 5, 3 and 2 times. Find the mean deviation about the mean.

Show answer

Answer 0.7

\overset{ˉ}{x} = \frac{5 + 6 + 6}{10} = 1.7.

The distances 0.7, 0.3, 1.3 weighted by 5, 3, 2 give

3.5 + 0.9 + 2.6 = 7

, so M.D.

= \frac{7}{10} = 0.7.

Practice · 12 questions

Practice 13.1: questions with answers

NumericalEasyNCERT Exercise 13.1, Q1
A student's scores in eight weekly spelling tests were 5, 9, 12, 18, 7, 10, 14, 5. How far, on average, are they from their mean? Give the mean deviation about the mean.
Show solution
The total is 80, so $\overset{ˉ}{x} = 10$ .
The distances from 10 are 5, 1, 2, 8, 3, 0, 4, 5, with total 28.
M.D. $(\overset{ˉ}{x}) = \frac{28}{8} = 3.5.$
NumericalEasyNCERT Exercise 13.1, Q2
Ten students score 62, 48, 75, 55, 81, 59, 66, 70, 52 and 62 out of 100 in a test. Find the mean deviation about the mean.
Show solution
$\overset{ˉ}{x} = \frac{630}{10} = 63.$
The distances from 63 are 1, 15, 12, 8, 18, 4, 3, 7, 11, 1, with total 80.
M.D. $(\overset{ˉ}{x}) = \frac{80}{10} = 8$ marks.
NumericalMediumNCERT Exercise 13.1, Q3
A shop recorded these daily sales of a magazine over twelve days: 21, 13, 26, 16, 31, 22, 19, 25, 15, 28, 20, 24. Work out the mean deviation about the median.
Show solution
In order: 13, 15, 16, 19, 20, 21, 22, 24, 25, 26, 28, 31. The median is $\frac{21 + 22}{2} = 21.5.$
The distances from 21.5 are 8.5, 6.5, 5.5, 2.5, 1.5, 0.5, 0.5, 2.5, 3.5, 4.5, 6.5, 9.5, with total 52.
M.D. $(M) = \frac{52}{12} \approx 4.33.$
NumericalMediumNCERT Exercise 13.1, Q4
The weights (kg) of ten members of a kabaddi club are 64, 38, 55, 47, 73, 52, 49, 58, 44 and 52. Find the mean deviation about the median.
kg
Show solution
In order: 38, 44, 47, 49, 52, 52, 55, 58, 64, 73. The 5th and 6th values are both 52, so $M = 52$ .
The distances are 14, 8, 5, 3, 0, 0, 3, 6, 12, 21, with total 72.
M.D. $(M) = \frac{72}{10} = 7.2$ kg.
NumericalMediumNCERT Exercise 13.1, Q5
For the frequency table below, work out the mean deviation about the mean.
Table 13.5
$x_{i}$ $f_{i}$
3 2
6 8
9 12
12 4
15 4
Show solution
$N = 30$ and the products $f_{i} x_{i}$ are 6, 48, 108, 48, 60, with total 270, so $\overset{ˉ}{x} = 9$ .
The distances $∣ x_{i} - 9 ∣$ are 6, 3, 0, 3, 6; times $f_{i}$ : 12, 24, 0, 12, 24, total 72.
M.D. $(\overset{ˉ}{x}) = \frac{72}{30} = 2.4.$
NumericalMediumNCERT Exercise 13.1, Q6
How far, on average, do the values in this table lie from their mean? Find the mean deviation about the mean.
Table 13.6
$x_{i}$ $f_{i}$
20 4
40 12
60 14
80 15
100 5
Show solution
$N = 50$ and $\sum f_{i} x_{i} = 80 + 480 + 840 + 1200 + 500 = 3100,$ so $\overset{ˉ}{x} = 62$ .
The distances from 62 are 42, 22, 2, 18, 38; times $f_{i}$ : 168, 264, 28, 270, 190, total 920.
M.D. $(\overset{ˉ}{x}) = \frac{920}{50} = 18.4.$
NumericalMediumNCERT Exercise 13.1, Q7
For the frequency table below, work out the mean deviation about the median.
Table 13.7
$x_{i}$ $f_{i}$
4 9
6 3
8 12
11 4
13 8
16 4
Show solution
The c.f. column is 9, 12, 24, 28, 36, 40. The 20th and 21st values lie in positions 13 to 24, which are all 8, so $M = 8$ .
The distances from 8 are 4, 2, 0, 3, 5, 8; times $f_{i}$ : 36, 6, 0, 12, 40, 32, total 126.
M.D. $(M) = \frac{126}{40} = 3.15.$
NumericalMediumNCERT Exercise 13.1, Q8
Using the median as the centre, find the mean deviation of the values in this table.
Table 13.8
$x_{i}$ $f_{i}$
12 2
18 12
24 10
28 3
33 3
Show solution
The c.f. column is 2, 14, 24, 27, 30. Positions 15 to 24 hold 24, so $M = 24$ .
The distances are 12, 6, 0, 4, 9; times $f_{i}$ : 24, 72, 0, 12, 27, total 135.
M.D. $(M) = \frac{135}{30} = 4.5.$
NumericalHardNCERT Exercise 13.1, Q9
The daily earnings of 50 street vendors in a market are grouped below. Find the mean deviation about the mean.
Table 13.9
Earnings (₹) Vendors
100–200 5
200–300 8
300–400 12
400–500 9
500–600 6
600–700 6
700–800 4
₹
Show solution
Mid-points: 150, 250, …, 750. With $a = 450$ and $h = 100$ , $d_{i}$ runs from $- 3$ to $3$ and $\sum f_{i} d_{i} = - 15 - 16 - 12 + 0 + 6 + 12 + 12 = - 13.$
$\overset{ˉ}{x} = 450 + 100 \times \frac{- 13}{50} = 424.$
The distances from 424 are 274, 174, 74, 26, 126, 226, 326; times $f_{i}$ : 1370, 1392, 888, 234, 756, 1356, 1304, total 7300.
M.D. $(\overset{ˉ}{x}) = \frac{7300}{50} = 146,$ that is, ₹146.
NumericalMediumNCERT Exercise 13.1, Q10
The weights of 60 girls in a sports academy are grouped below. Find the mean deviation about the mean.
Table 13.10
Weight (kg) Girls
40–44 3
44–48 9
48–52 13
52–56 16
56–60 13
60–64 6
kg
Show solution
$\sum f_{i} x_{i} = 126 + 414 + 650 + 864 + 754 + 372 = 3180,$ so $\overset{ˉ}{x} = \frac{3180}{60} = 53$ kg.
The distances from 53 are 11, 7, 3, 1, 5, 9; times $f_{i}$ : 33, 63, 39, 16, 65, 54, total 270.
M.D. $(\overset{ˉ}{x}) = \frac{270}{60} = 4.5$ kg.
NumericalMediumNCERT Exercise 13.1, Q11
Using the median as the centre, work out the mean deviation for the grouped data below.
Table 13.11
Class Frequency
10–20 4
20–30 8
30–40 13
40–50 10
50–60 9
60–70 6
Show solution
The c.f. column is 4, 12, 25, 35, 44, 50. $\frac{N}{2} = 25$ is reached in 30–40, so $l = 30$ , $C = 12$ , $f = 13$ , $h = 10$ .
$M = 30 + \frac{25 - 12}{13} \times 10 = 40.$
The mid-points 15, 25, …, 65 are 25, 15, 5, 5, 15, 25 from 40; times $f_{i}$ : 100, 120, 65, 50, 135, 150, total 620.
M.D. $(M) = \frac{620}{50} = 12.4.$
NumericalHardNCERT Exercise 13.1, Q12
The ages of 80 visitors to a science museum are grouped below. The classes have gaps (15 and 16 are in different classes). Find the mean deviation about the median age.
Table 13.12
Age (years) Visitors
11–15 5
16–20 12
21–25 17
26–30 20
31–35 18
36–40 6
41–45 2
years
Show solution
Ages are whole numbers, so 15 and 16 are neighbours and the true boundary between the first two classes is 15.5. In the same way the classes become 10.5–15.5, 15.5–20.5, …, 40.5–45.5. The mid-points 13, 18, …, 43 do not change.
The c.f. column is 5, 17, 34, 54, 72, 78, 80. $\frac{N}{2} = 40$ falls in 25.5–30.5, with $C = 34$ , $f = 20$ , $h = 5$ .
$M = 25.5 + \frac{40 - 34}{20} \times 5 = 27.$
The distances from 27 are 14, 9, 4, 1, 6, 11, 16; times $f_{i}$ : 70, 108, 68, 20, 108, 66, 32, total 472.
M.D. $(M) = \frac{472}{80} = 5.9$ years.

$x_{i}$	$f_{i}$
3	2
6	8
9	12
12	4
15	4

$x_{i}$	$f_{i}$
20	4
40	12
60	14
80	15
100	5

$x_{i}$	$f_{i}$
4	9
6	3
8	12
11	4
13	8
16	4

$x_{i}$	$f_{i}$
12	2
18	12
24	10
28	3
33	3

Earnings (₹)	Vendors
100–200	5
200–300	8
300–400	12
400–500	9
500–600	6
600–700	6
700–800	4

Weight (kg)	Girls
40–44	3
44–48	9
48–52	13
52–56	16
56–60	13
60–64	6

Class	Frequency
10–20	4
20–30	8
30–40	13
40–50	10
50–60	9
60–70	6

Age (years)	Visitors
11–15	5
16–20	12
21–25	17
26–30	20
31–35	18
36–40	6
41–45	2

13.4.3 Limitations of mean deviation

The mean deviation is easy to understand, but it has three weak points.

The answer depends on which centre we choose. The total distance is smallest about the median, $\sum ∣ x_{i} - M ∣ \leq \sum ∣ x_{i} - \overset{ˉ}{x} ∣,$ so the mean deviation about the mean is never the smaller of the two.
For wildly scattered data no centre describes the values well. In 1, 2, 3, 40, 41 the median is 3, far from two of the five values, so a spread measured from it tells us little.
Absolute values are awkward in algebra. Unlike $(x - a)^{2}$ , the expression $∣ x - a ∣$ cannot be expanded, so there is no simple rule for the mean deviation of combined or transformed data.

SimulationWhere is the sum of distances smallest?

A coordinate grid from 0 to 14 across, with five fixed points A, B, C, D and E on the x-axis at 1, 2, 3, 7 and 12, the observations. A sixth point M, the centre, starts at the mean 5 and can be dragged in half-unit steps; the panel lists the distances MA, MB, MC, MD and ME. At 5 they total 18, a mean deviation of 3.6. Sliding M to the median 3 lowers the total to 16, a mean deviation of 3.2, and moving M either way from 3 raises it again.

Fig. 13.8The observations 1, 2, 3, 7, 12 sit on the

x

-axis. Slide the centre M along the axis and add up the five distances.

Try this

M starts at the mean, 5. Add the five distances, then drag M to the median, 3, and add again. Which total is smaller?
Put M at 2, then at 4. Does the total rise on both sides of 3?
Keep M on the $x$ -axis: off the axis the distances are no longer $∣ x_{i} - a ∣$ .

So we want a measure that still uses the distance of every observation from the centre but behaves well in algebra. Squaring the deviations does exactly that.

Quick check

1 of 2

For the data 1, 2, 3, 7, 12, the sum

\sum ∣ x_{i} - a ∣

is smallest when

a

3, the median5, the mean6.5, the middle of the range1, the smallest value

Show answer

Answer (A) 3, the median

The sums are 18 at 5, 16 at 3, 19.5 at 6.5 and 20 at 1. The sum of distances is always least about the median.

13.5 Variance and Standard Deviation

Squaring is the other way to stop deviations cancelling, since no square is negative. The first candidate is the total $\sum (x_{i} - \overset{ˉ}{x})^{2} .$ It behaves sensibly in one respect: it is 0 exactly when every value equals the mean.

But a total has a flaw that is easy to miss: it grows just because more observations are added. Two sets with the same mean, 30, expose it. Set A has five observations: 10, 20, 30, 40, 50. Set B has twenty-five: every whole number from 18 to 42. For A the squared deviations are 400, 100, 0, 100, 400, with total 1000. For B the deviations run from −12 to 12, so

\begin{aligned} \sum (y_{i} - 30)^{2} \\ = 2 (1^{2} + \dots + 12^{2}) \\ = 2 \times \frac{12 \cdot 13 \cdot 25}{6} \\ = 1300 \end{aligned}

using $1^{2} + 2^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6} .$ By the sums, B looks more scattered than A. Yet every value of B lies within 12 of the mean, while A reaches 20 away (Fig. 13.9). B's total is larger because it adds up five times as many squares.

3/3Dividing by $n$ : A gives 200, B gives 52.

Fig. 13.9A sum grows with the number of observations; a mean does not.NCERT Figs 13.5, 13.6 · p. 273

Two dot plots on a scale from 0 to 60. Set A has five dots at 10, 20, 30, 40 and 50; set B has twenty-five dots, one at every whole number from 18 to 42. In the second step two bars compare the sums of squared deviations: A's bar is 1000 and B's longer bar is 1300, which wrongly suggests B is more spread out. In the third step the bars change to the mean of the squares: A's becomes 200 and B's shrinks to 52, so A is now correctly shown as the more scattered set.

The remedy is the one we used for distances: take the mean. The mean of the squared deviations is $\frac{1000}{5} = 200$ for A and $\frac{1300}{25} = 52$ for B, which now ranks the sets correctly.

Definition

Variance

The average of the squared deviations of the observations from their mean; it is written $σ^{2}$ (read sigma squared).

variance

σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{ˉ}{x})^{2}

(13.11)

The variance (प्रसरण) of set A is 200 and that of set B is 52.

13.5.1 Standard Deviation

Variance has one drawback: its units are the squares of the data's units. Waiting times in minutes give a variance in square minutes, which nobody can picture. Taking the positive square root brings us back to the original units.

Definition

Standard deviation

The positive square root of the variance, $σ = \sqrt{σ^{2}} .$ It measures spread about the mean in the units of the data.

standard deviation

σ = \sqrt{\frac{1}{n} \sum (x_{i} - \overset{ˉ}{x})^{2}}

(13.12)

The standard deviation (मानक विचलन) has a neat picture behind it (Fig. 13.10). Draw a square on each deviation. The variance is the average area of these squares, and $σ$ is the side of a square with that average area.

3/3Mean area $σ^{2} = \frac{80}{5} = 16,$ so $σ = 4$ .

Fig. 13.10Variance is the average square; σ is its side.

A number line from 2 to 18 carries dots at 4, 8, 10, 12 and 16 and a marker at the mean 10. In the second step a square is drawn on each deviation, between each dot and the mean: two large squares of side 6 and area 36 for 4 and 16, two small squares of side 2 and area 4 for 8 and 12, and nothing for 10. In the third step these squares fade and one bold square of area 16, the average of 36, 4, 0, 4 and 36, stands over the mean; its side, 4, is the standard deviation.

Example 13.8NCERT Example 8 · p. 274

Variance of puzzle times

Eight students take 12, 18, 21, 15, 27, 24, 9 and 18 minutes to finish a sudoku. Find the variance and the standard deviation.

Try it yourself first. Then reveal the solution one step at a time.

Back to the two buses. Route P's deviations from 11 are $4, - 8, 10, 0, - 10, 8, - 4, 0,$ so $σ_{P}^{2} = \frac{360}{8} = 45$ and $σ_{P} \approx 6.71$ minutes. Route Q gives $σ_{Q}^{2} = \frac{12}{8} = 1.5$ and $σ_{Q} \approx 1.22$ minutes. One number now says what the dot plots showed: Route Q is far more dependable.

InteractiveSpread lab

Two modes. In Dot plot, eight dots on a number line from 0 to 24 can be dragged or moved with the arrow keys; buttons load the waiting times of Route P, of Route Q, or of Route Q with one late bus at 24 minutes. A panel gives the mean, median, range, mean deviation about the mean and about the median, variance and standard deviation, and an overlay can show each distance from the mean, or a bar chart of the squared deviations with a dashed line at their average, the variance. In Shift and stretch, sliders set k from −3 to 3 and c from −6 to 6 for the data 2, 4, 5, 6, 8; two dot plots compare x with kx + c, each with a band of one standard deviation either side of its mean, and the panel checks that the new mean is k times the old mean plus c and the new standard deviation is |k| times the old one.

Fig. 13.11Drag the dots to change the data, or shift and stretch a data set, and watch the measures of spread respond.

13.5.2 Standard deviation of a discrete frequency distribution

When the value $x_{i}$ occurs $f_{i}$ times, its squared deviation is counted $f_{i}$ times. With $N = \sum f_{i}$ :

discrete distribution

\begin{aligned} σ = \sqrt{V}, where \\ V = \frac{1}{N} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2} \end{aligned}

(13.13)

Example 13.9NCERT Example 9 · p. 275

Passengers boarding a bus

On one trip a city bus made 40 stops. The table shows how many stops had $x$ passengers boarding. Find the variance and the standard deviation.

Table 13.13

Passengers $x_{i}$	Stops $f_{i}$
2	1
5	5
8	10
11	10
14	7
17	5
20	2

Try it yourself first. Then reveal the solution one step at a time.

13.5.3 Standard deviation of a continuous frequency distribution

For classes, use the class marks as the $x_{i}$ and proceed as for a discrete distribution. When $\overset{ˉ}{x}$ is not a whole number, though, every deviation is a messy decimal. An equivalent formula avoids deviations altogether.

DerivationA formula without deviations

Expand the square and split the sum into three parts:

\begin{aligned} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2} \\ = \sum f_{i} x_{i}^{2} \\ - 2 \overset{ˉ}{x} \sum f_{i} x_{i} \\ + {\overset{ˉ}{x}}^{2} \sum f_{i} \end{aligned}

Now $\sum f_{i} x_{i} = N \overset{ˉ}{x}$ and $\sum f_{i} = N$ , so the last two parts are $- 2 N {\overset{ˉ}{x}}^{2} + N {\overset{ˉ}{x}}^{2} = - N {\overset{ˉ}{x}}^{2} .$ Dividing by $N$ :

mean of squares minus square of mean

\begin{aligned} σ^{2} & = \frac{1}{N} \sum f_{i} x_{i}^{2} \\ - {\overset{ˉ}{x}}^{2} \end{aligned}

(13.14)

Next put $\overset{ˉ}{x} = \frac{1}{N} \sum f_{i} x_{i}$ and multiply through by $N^{2}$ :

computing formula

\begin{aligned} N^{2} σ^{2} \\ = N \sum f_{i} x_{i}^{2} \\ - (\sum f_{i} x_{i})^{2} \end{aligned}

(13.15)

So $σ$ is $\frac{1}{N}$ times the square root of the right-hand side. For raw data every $f_{i} = 1$ and $N = n$ , and (13.14) reads $σ^{2} = \frac{1}{n} \sum x_{i}^{2} - {\overset{ˉ}{x}}^{2} :$ the mean of the squares minus the square of the mean.

Example 13.10NCERT Example 10 · p. 277

Speeds on a highway

A speed camera on a highway recorded the speeds of 40 vehicles. Find the mean, the variance and the standard deviation.

Table 13.14

Speed (km/h)	Vehicles
20–30	2
30–40	5
40–50	7
50–60	7
60–70	9
70–80	8
80–90	2

Try it yourself first. Then reveal the solution one step at a time.

Example 13.11NCERT Example 11 · p. 278

Push-ups in a fitness test

In a fitness test, 50 students did the numbers of push-ups shown. Find the standard deviation using (13.15).

Table 13.15

Push-ups $x_{i}$	Students $f_{i}$
5	3
10	7
15	12
20	14
25	9
30	5

Try it yourself first. Then reveal the solution one step at a time.

13.5.4 Shortcut method to find variance and standard deviation

When the values are large, even $f_{i} x_{i}^{2}$ is heavy. Code the data as in the step-deviation method, with an assumed mean $A$ and a scale $h$ .

DerivationCoding the data

\begin{aligned} y_{i} = \frac{x_{i} - A}{h}, so \\ x_{i} = A + h y_{i} \end{aligned}

Multiply by $f_{i}$ , add over all $i$ and divide by $N$ :

coded mean

\overset{ˉ}{x} = A + h \overset{ˉ}{y}

(13.16)

Subtracting, $x_{i} - \overset{ˉ}{x} = h (y_{i} - \overset{ˉ}{y}) .$ So each squared deviation of $x$ is $h^{2}$ times the matching one of $y$ ; weighting by $f_{i}$ and averaging gives $σ_{x}^{2} = h^{2} σ_{y}^{2},$ that is,

σ scales with h

σ_{x} = h σ_{y}

(13.17)

Finally apply (13.15) to the coded values $y_{i}$ :

shortcut formula

\begin{aligned} N^{2} σ_{x}^{2} \\ = h^{2} [N \sum f_{i} y_{i}^{2} \\ - (\sum f_{i} y_{i})^{2}] \end{aligned}

(13.18)

In words: shifting the origin by $A$ leaves the spread alone, and changing the scale by $h$ multiplies $σ$ by $h$ . Example 13.13 and Example 13.15 show the same effects for any data.

Example 13.12NCERT Example 12 · p. 280

Example 13.10 by the shortcut

Find the mean and the standard deviation of the speeds in Example 13.10 by the shortcut method.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

For a distribution,

N = 10

\sum f_{i} x_{i} = 60

and

\sum f_{i} x_{i}^{2} = 400

. Find the variance.

Show answer

Answer 4

\overset{ˉ}{x} = 6

, so

σ^{2} = 40 - 36 = 4.

Forgetting to square the mean gives 34.

Practice · 10 questions

Practice 13.2: questions with answers

1
NumericalEasyNCERT Exercise 13.2, Q1
Find the mean and the variance of 14, 9, 17, 6, 11, 15, 10, 6.
1. the mean
  Show solution
  $\overset{ˉ}{x} = \frac{88}{8} = 11.$
2. the variance
  Show solution
  The deviations from 11 are $3, - 2, 6, - 5, 0, 4, - 1, - 5;$ their squares 9, 4, 36, 25, 0, 16, 1, 25 add up to 116.
  $σ^{2} = \frac{116}{8} = 14.5.$
ProofHardNCERT Exercise 13.2, Q2
Find the mean and the variance of the first $n$ odd natural numbers $1, 3, 5, \dots, 2 n - 1.$
Show solution
Answermean $n$ ; variance $\frac{n^{2} - 1}{3}$
The sum of the first $n$ odd numbers is $n^{2}$ , so $\overset{ˉ}{x} = \frac{n^{2}}{n} = n .$
$\begin{aligned} \sum (2 k - 1)^{2} \\ = 4 \sum k^{2} - 4 \sum k + n \\ = \frac{n (2 n - 1) (2 n + 1)}{3} \end{aligned}$
after simplifying (check it with $n = 2$ : $1 + 9 = 10$ ).
$σ^{2} = \frac{4 n^{2} - 1}{3} - n^{2} = \frac{n^{2} - 1}{3}$
For example, 1, 3, 5, 7, 9 has mean 5 and variance $\frac{24}{3} = 8$ . (The same method gives variance $\frac{n^{2} - 1}{12}$ for $1, 2, \dots, n$ .)
3
NumericalEasyNCERT Exercise 13.2, Q3
Find the mean and the variance of the first 9 multiples of 6.
1. the mean
  Show solution
  The nine values are evenly spaced, so the mean is the middle one, 30.
2. the variance
  Show solution
  The deviations are $\pm 24, \pm 18, \pm 12, \pm 6$ and 0. The sum of squares is $2 (576 + 324 + 144 + 36) = 2160.$
  $σ^{2} = \frac{2160}{9} = 240.$
NumericalMediumNCERT Exercise 13.2, Q4
Find the mean and the variance of this distribution.
Table 13.16
$x_{i}$ $f_{i}$
4 2
8 3
11 8
15 10
20 11
23 3
27 3
Show solution
Answer $\overset{ˉ}{x} = 16$ , $σ^{2} = 34.4$
The products $f_{i} x_{i}$ are 8, 24, 88, 150, 220, 69, 81, with total 640, so $\overset{ˉ}{x} = 16$ .
The squared deviations are 144, 64, 25, 1, 16, 49, 121; times $f_{i}$ : 288, 192, 200, 10, 176, 147, 363, total 1376.
$σ^{2} = \frac{1376}{40} = 34.4.$
NumericalMediumNCERT Exercise 13.2, Q5
The heights (cm) of 30 students are given below. Find the mean and the variance.
Table 13.17
Height (cm) Students
143 1
146 4
148 4
150 8
153 7
155 4
160 2
Show solution
Answer $\overset{ˉ}{x} = 151$ cm, $σ^{2} = 15.4$
With $y_{i} = x_{i} - 150$ : $- 7, - 4, - 2, 0, 3, 5, 10.$ Then $\sum f_{i} y_{i} = 30$ and $\sum f_{i} y_{i}^{2} = 492$ .
$\overset{ˉ}{x} = 150 + \frac{30}{30} = 151$ cm.
Shifting does not change the variance: $σ^{2} = \frac{492}{30} - 1^{2} = 15.4.$
NumericalMediumNCERT Exercise 13.2, Q6
The weights (g) of 60 eggs are given below. Find the mean and the standard deviation by the shortcut method.
Table 13.18
Weight (g) Eggs
42 5
44 6
46 8
48 11
50 15
52 13
54 2
Show solution
Answer $\overset{ˉ}{x} = 48.4$ g, $σ \approx 3.26$ g
$y_{i} = \frac{x_{i} - 48}{2}$ runs $- 3$ to $3$ . Then $\sum f_{i} y_{i} = 12$ and $\sum f_{i} y_{i}^{2} = 162$ .
$\overset{ˉ}{x} = 48 + 2 \times \frac{12}{60} = 48.4$ g.
$\begin{aligned} 60^{2} σ^{2} & = 4 [60 (162) - 12^{2}] \\ = 4 \times 9576 \end{aligned}$
$σ^{2} = 10.64$ and $σ \approx 3.26$ g.
NumericalMediumNCERT Exercise 13.2, Q7
Find the mean and the variance of this distribution.
Table 13.19
Class Frequency
0–40 1
40–80 10
80–120 13
120–160 11
160–200 3
200–240 2
Show solution
Answer $\overset{ˉ}{x} = 111$ , $σ^{2} = 2079$
Mid-points 20, 60, …, 220; $y_{i} = \frac{x_{i} - 100}{40}$ is $- 2, - 1, 0, 1, 2, 3$ . Then $\sum f_{i} y_{i} = 11$ and $\sum f_{i} y_{i}^{2} = 55$ .
$\overset{ˉ}{x} = 100 + 40 \times \frac{11}{40} = 111.$
$40^{2} σ^{2} = 40^{2} [40 (55) - 11^{2}],$ so $σ^{2} = 2200 - 121 = 2079.$
8
NumericalMediumNCERT Exercise 13.2, Q8
Find the mean and the variance of this distribution.
Table 13.20
Class Frequency
10–20 6
20–30 7
30–40 17
40–50 11
50–60 9
1. the mean
  Show solution
  $\sum f_{i} x_{i} = 90 + 175 + 595 + 495 + 495 = 1850,$ so $\overset{ˉ}{x} = 37$ .
2. the variance
  Show solution
  The squared deviations from 37 are 484, 144, 4, 64, 324; times $f_{i}$ : 2904, 1008, 68, 704, 2916, total 7600.
  $σ^{2} = \frac{7600}{50} = 152.$
NumericalHardNCERT Exercise 13.2, Q9
The heights (cm) of 60 saplings in a nursery are grouped below. Find the mean, the variance and the standard deviation by the shortcut method.
Table 13.21
Height (cm) Saplings
120–124 2
124–128 5
128–132 8
132–136 18
136–140 13
140–144 7
144–148 7
Show solution
Answer $135.6$ cm; $37.44$ ; $σ \approx 6.12$ cm
$y_{i} = \frac{x_{i} - 134}{4}$ runs $- 3$ to $3$ . $\sum f_{i} y_{i} = - 6 - 10 - 8 + 0 + 13 + 14 + 21 = 24$ and $\sum f_{i} y_{i}^{2} = 150$ .
$\overset{ˉ}{x} = 134 + 4 \times \frac{24}{60} = 135.6$ cm.
$60^{2} σ^{2} = 16 [60 (150) - 24^{2}] = 16 \times 8424,$ so $σ^{2} = 37.44$ and $σ \approx 6.12$ cm.
NumericalHardNCERT Exercise 13.2, Q10
The lengths (mm) of 80 nails from a box were sorted into the classes below, which have gaps. Find the mean length and the standard deviation.
Table 13.22
Length (mm) Nails
20–24 8
25–29 15
30–34 29
35–39 21
40–44 7
Show solution
Answer $32.25$ mm; $σ \approx 5.47$ mm
Lengths are whole millimetres, so the true boundaries lie halfway between classes, at 24.5, 29.5 and so on. The classes become 19.5–24.5, …, 39.5–44.5, with the same mid-points 22, 27, 32, 37, 42. Take $A = 32$ , $h = 5$ : $y_{i}$ is $- 2, - 1, 0, 1, 2$ .
$\sum f_{i} y_{i} = - 16 - 15 + 0 + 21 + 14 = 4$ and $\sum f_{i} y_{i}^{2} = 96$ .
$\overset{ˉ}{x} = 32 + 5 \times \frac{4}{80} = 32.25$ mm.
$80^{2} σ^{2} = 25 [80 (96) - 16] = 25 \times 7664,$ so $σ^{2} \approx 29.94$ and $σ \approx 5.47$ mm.

Height (cm)	Students
143	1
146	4
148	4
150	8
153	7
155	4
160	2

Weight (g)	Eggs
42	5
44	6
46	8
48	11
50	15
52	13
54	2

Class	Frequency
0–40	1
40–80	10
80–120	13
120–160	11
160–200	3
200–240	2

Class	Frequency
10–20	6
20–30	7
30–40	17
40–50	11
50–60	9

Height (cm)	Saplings
120–124	2
124–128	5
128–132	8
132–136	18
136–140	13
140–144	7
144–148	7

Length (mm)	Nails
20–24	8
25–29	15
30–34	29
35–39	21
40–44	7

Miscellaneous Examples

These examples run the formulas backwards and show how spread reacts when every observation is changed in the same way (Fig. 13.12).

3/3 $2 x$ : the mean becomes 10, $σ$ becomes 4.

Fig. 13.12Adding moves the data; multiplying stretches it.

Three number lines from 0 to 16 share vertical grid lines. The top one shows the data 2, 4, 5, 6 and 8 with mean 5 and a shaded band from 3 to 7, one standard deviation of 2 on each side. In the second step the middle line shows each value plus 6: 8, 10, 11, 12 and 14; the band slides to 9 to 13 but keeps its width, so sigma is still 2. In the third step the bottom line shows each value doubled: 4, 8, 10, 12 and 16; the mean doubles to 10 and the band widens to 6 to 14, so sigma becomes 4 and the variance grows from 4 to 16.

Example 13.13NCERT Example 13 · p. 282

Celsius to Fahrenheit

A weather station's noon temperatures over 15 days have variance 6, measured in °C. A visitor converts every reading to Fahrenheit with $F = 1.8 C + 32$ . What is the variance of the Fahrenheit readings?

Try it yourself first. Then reveal the solution one step at a time.

Example 13.14NCERT Example 14 · p. 283

Two lost observations

Five friends play a quiz. Their scores have mean 6 and variance 8.8. Meera remembers three of the scores, 2, 7 and 9, but not the other two. Find them.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.15NCERT Example 15 · p. 284

Shifting every observation

A teacher adds the same number $c$ of grace marks to every score $x_{1}, x_{2}, \dots, x_{n}$ in a class test ( $c$ could even be negative, a penalty). Prove that the variance of the scores is unaffected.

Try it yourself first. Then reveal the solution one step at a time.

Example 13.16NCERT Example 16 · p. 284

Fixing a copying error

A clerk typing 40 daily rainfall readings (mm) into a spreadsheet entered 21 for a reading that was really 37. The spreadsheet reported a mean of 22 and a standard deviation of 6.6. Find the correct mean and standard deviation.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

The variance of

x_{1}, \dots, x_{n}

is 5. What is the variance of

- 2 x_{1} + 7, \dots, - 2 x_{n} + 7

- 3

10

20

27

Show answer

Answer (C) $20$

(- 2)^{2} \times 5 = 20

. A variance is never negative, and 10 forgets to square the factor.

Recap

Statistics: key points and summary

A measure of central tendency gives the centre; a measure of dispersion gives the spread. Data with equal means and medians can be spread very differently (Section 13.1).
Range = largest − smallest. It is quick but uses only two values (Section 13.3).
Deviations about the mean add up to zero, so we use distances $∣ x_{i} - a ∣$ or squares $(x_{i} - \overset{ˉ}{x})^{2}$ ((13.3)).
Raw data: M.D. $(a) = \frac{1}{n} \sum ∣ x_{i} - a ∣,$ about $\overset{ˉ}{x}$ or $M$ (Section 13.4.1).
Grouped data: M.D. $(a) = \frac{1}{N} \sum f_{i} ∣ x_{i} - a ∣$ with $N = \sum f_{i}$ ; classes are replaced by their class marks (Section 13.4.2).
The median of grouped data comes from cumulative frequencies; for classes, $M = l + \frac{N / 2 - C}{f} \times h$ ((13.10)).
Step deviations $d_{i} = \frac{x_{i} - a}{h}$ give $\overset{ˉ}{x} = a + h \overset{ˉ}{d}$ ((13.9)).
The sum of distances is least about the median, and absolute values are hard to handle algebraically (Section 13.4.3).
Variance $σ^{2} = \frac{1}{N} \sum f_{i} (x_{i} - \overset{ˉ}{x})^{2};$ standard deviation $σ = \sqrt{σ^{2}},$ in the units of the data (Section 13.5).
Computing form: $σ^{2} = \frac{1}{N} \sum f_{i} x_{i}^{2} - {\overset{ˉ}{x}}^{2},$ or $N^{2} σ^{2} = N \sum f_{i} x_{i}^{2} - (\sum f_{i} x_{i})^{2}$ ((13.15)).
Shortcut: with $y_{i} = \frac{x_{i} - A}{h},$ $\overset{ˉ}{x} = A + h \overset{ˉ}{y}$ and $σ_{x} = h σ_{y}$ (Section 13.5.4).
For $y = k x + c$ : $\overset{ˉ}{y} = k \overset{ˉ}{x} + c,$ $σ_{y}^{2} = k^{2} σ_{x}^{2};$ adding $c$ leaves the spread unchanged (Miscellaneous Examples).

Timeline

How statistics grew

c. 300 BCE
Arthashastra
Kautilya's treatise on running the Mauryan state describes village officials who keep records of households, births and deaths.
c. 1595
Ain-i-Akbari
Abul Fazl's record of Akbar's empire sets out land surveys, crop yields and revenue, province by province.
1662
John Graunt
Studies London's weekly records of births and deaths and finds steady patterns in them, founding the study of population data.
1713
Jacob Bernoulli
His Ars Conjectandi, published after his death, proves an early law of large numbers.
1880s
Francis Galton
Uses measurements of thousands of people to study inheritance, introducing regression and correlation.
1894
Karl Pearson
First uses the term standard deviation in print; he later founded the first university statistics department, in London.
1918
R. A. Fisher
Introduces the word variance; his methods reshape statistics in genetics and agriculture.
1931
P. C. Mahalanobis
Founds the Indian Statistical Institute in Kolkata; India marks his birthday, 29 June, as Statistics Day.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise 13.1, Q4	Practice 13.1 · Q4
NCERT Exercise 13.1, Q5	Practice 13.1 · Q5
NCERT Exercise 13.1, Q6	Practice 13.1 · Q6
NCERT Exercise 13.1, Q7	Practice 13.1 · Q7
NCERT Exercise 13.1, Q8	Practice 13.1 · Q8
NCERT Exercise 13.1, Q9	Practice 13.1 · Q9
NCERT Exercise 13.1, Q10	Practice 13.1 · Q10
NCERT Exercise 13.1, Q11	Practice 13.1 · Q11
NCERT Exercise 13.1, Q12	Practice 13.1 · Q12
NCERT Exercise 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise 13.2, Q8	Practice 13.2 · Q8
NCERT Exercise 13.2, Q9	Practice 13.2 · Q9
NCERT Exercise 13.2, Q10	Practice 13.2 · Q10
NCERT Miscellaneous Exercise, Q1	Miscellaneous exercises · Q1
NCERT Miscellaneous Exercise, Q2	Miscellaneous exercises · Q2
NCERT Miscellaneous Exercise, Q3	Miscellaneous exercises · Q3
NCERT Miscellaneous Exercise, Q4	Miscellaneous exercises · Q4
NCERT Miscellaneous Exercise, Q5	Miscellaneous exercises · Q5
NCERT Miscellaneous Exercise, Q6	Miscellaneous exercises · Q6

Quick check: MCQs on Statistics

13.1 Introduction3 questions
13.3 Range3 questions
13.4.1 Mean deviation for ungrouped data3 questions
Continuous frequency distribution4 questions
13.4.3 Limitations of mean deviation2 questions
13.5.1 Standard Deviation4 questions
13.5.4 Shortcut method to find variance and standard deviation3 questions
Miscellaneous Examples3 questions

Glossary · 14

Key terms (English–Hindi)

assumed meanकल्पित माध्यkalpit madhya: A convenient value near the middle of the data from which deviations are measured to simplify arithmetic.First seen in the chapter →

class markवर्ग चिह्नvarg chihn: The mid-point of a class interval, used to stand for every observation in that class.First seen in the chapter →
continuous frequency distributionसतत बारंबारता बंटनsatat barambarta bantan: A table giving the number of observations in each of a set of class intervals that have no gaps between them.First seen in the chapter →
cumulative frequencyसंचयी बारंबारताsanchayi barambarta: The running total of the frequencies up to and including a value or class.First seen in the chapter →

deviationविचलनvichalan: The difference $x - a$ between an observation and a fixed value $a$ ; it can be positive or negative.First seen in the chapter →
discrete frequency distributionअसतत बारंबारता बंटनasatat barambarta bantan: A table listing each distinct value with the number of times it occurs.First seen in the chapter →

mean deviationमाध्य विचलनmadhya vichalan: The mean of the distances $∣ x_{i} - a ∣$ of the observations from a central value $a$ , usually the mean or the median.First seen in the chapter →
measure of central tendencyकेंद्रीय प्रवृत्ति की मापkendriya pravritti ki maap: A single value, such as the mean or the median, around which the data are centred.First seen in the chapter →
measure of dispersionप्रकीर्णन की मापprakeernan ki maap: A number that describes how widely the observations are scattered, such as the range or the standard deviation.First seen in the chapter →
median classमाध्यिका वर्गmadhyika varg: The class in which the cumulative frequency first reaches half the total frequency.First seen in the chapter →

rangeपरिसरparisar: Largest observation minus smallest observation.First seen in the chapter →

standard deviationमानक विचलनmanak vichalan: The positive square root of the variance, written $σ$ ; it has the units of the data.First seen in the chapter →
step deviationपद विचलनpad vichalan: A deviation from the assumed mean divided by a common factor $h$ : $d = \frac{x - a}{h}$ .First seen in the chapter →

varianceप्रसरणprasaran: The mean of the squared deviations from the mean, written $σ^{2}$ .First seen in the chapter →

Statistics: Class 11 Maths · Chapter 13

What you will learn in Statistics

13.1 Introduction

13.2 Measures of Dispersion

13.3 Range

13.4 Mean Deviation

13.4.1 Mean deviation for ungrouped data

13.4.2 Mean deviation for grouped data

Discrete frequency distribution

Continuous frequency distribution

Practice 13.1: questions with answers

13.4.3 Limitations of mean deviation

13.5 Variance and Standard Deviation

13.5.1 Standard Deviation

13.5.2 Standard deviation of a discrete frequency distribution

13.5.3 Standard deviation of a continuous frequency distribution

13.5.4 Shortcut method to find variance and standard deviation

Practice 13.2: questions with answers

Miscellaneous Examples

Statistics: key points and summary

Practice questions with answers (NCERT “Miscellaneous exercises”)

Find your NCERT exercise question

Quick check: MCQs on Statistics

Key terms (English–Hindi)

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