What is covered in Exercise 13.3 of NCERT Class 10 Maths?

Exercise 13.3 focuses on finding the mode of grouped data using a specific formula and frequency distribution tables.

What is a modal class?

The modal class is the class interval with the highest frequency in a grouped frequency distribution table.

How is mode different from mean and median?

Mode represents the most frequent value, mean gives the average, and median represents the middle value of the dataset.

What is the median in statistics?

The median is the middle value of a dataset when arranged in ascending order, dividing it into two equal halves.

What are common mistakes students make while solving median questions?

Common mistakes include incorrect cumulative frequency calculation, choosing the wrong median class, and calculation errors in applying the formula.

NCERT Solutions

Class 10

Maths

Chapter 13 Statistics

Exercise 13.3

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.3 focuses on solving the questions related to the median of a dataset, a crucial concept of statistics. To calculate the median, the exercise includes forming cumulative frequency tables to interpret the grouped data. This exercise is a great help for solving the Median for simple problems as well as forming a solid base for complex real-world data analysis. In this exercise, we will also learn about the relationship between different measures of central tendencies. So, let’s start exploring this interesting topic of statistics.

1.0Download NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3: Free PDF

NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

2.0Introduction to Median

Before we get started on the formulas and procedures for obtaining the median of a data set, let's discuss the general concept of the median. The median is the middle number in a data set when it is ordered from least to greatest or from greatest to least. For ungrouped data, determining the median is simply done by identifying the middle figure. However, for grouped data, it's a little complicated since data is tabulated in class intervals.

3.0Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Key Concepts

Now that we have understood the basic idea of the median let’s explore some of the important formulas and key concepts used to solve various questions in the exercise:

Median of Ungrouped Data

The steps to determine the median of ungrouped data are quite straightforward. However, it can differ based on the number of terms, either odd or even, in a dataset. To find this, simply arrange the data in ascending or descending order. After arranging the data, simply use the following formulas to determine the median in each case:

Median if the number of observations (n) is odd

$Median for odd number of terms = (\frac{n + 1}{2})^{t h} o b ser v a t i o n$

Median if the number of observations (n) is even

$Median for even number of terms = \frac{( \frac{n}{2} ) ^{t h} o b ser v a t i o n + ( \frac{n}{2} + 1 ) ^{t h} o b ser v a t i o n}{2}$

Median of Grouped Data

To determine the median for grouped data, we use the principle of cumulative frequency and apply it to determine the median class, wherein lies the middle-most value. The formula to determine the median for grouped data can be established after determining the median class. To find the median of grouped data, we simply need to construct a cumulative frequency table for each class interval. After the formation of the table, simply use the following formula to find the median:

$M e d ian = l + \frac{( \frac{n}{2} - c f )}{f} \times h$

Here,

l = lower limit of the median class
n = total number of observations
h = class size (assuming equal class sizes), calculated as:

$h = Upper Limit -Lower limit$

cf = cumulative frequency of the class before the median class
f = frequency of the median class

Median, Mean, and Mode: Relation

In statistics, mean and mode are the primary measures of central tendency other than the median. Each of these measures gives a different insight into a dataset. For example, the mean is the average of all values of a dataset, the mode is the most frequent value of data, and the median is the middle value of data when arranged in ascending or descending order. These three values are related to one another by the following equation:

$3 \times M e d ian = M o d e + 2 \times M e an$

To further comprehend these principles and enhance your expertise, ensure practising constantly with the NCERT Solutions Class 10 Maths Chapter 13 Exercise.

4.0Benefits of Solving NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

Helps in learning how to calculate the mode of grouped data.
Practicing mode-related questions improves logical thinking and mathematical accuracy.
Strengthens the basics of statistics for higher studies and competitive exams.
This exercise helps in analyzing frequency distribution tables efficiently.

5.0NCERT Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Detailed Solution

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of consumers
65−8565−85	4
85−10585−105	5
105−125105−125	13
125−145125−145	20
145−165145−165	14
165−185165−185	8
185−205185−205	4

Sol.

(i)

Monthly Consumpti on (in units)	Number of consumers	Cumulative frequency
65−8565−85	4	4
85−10585−105	5	9
105−125105−125	13	22
125−145125−145	20	42
145−165145−165	14	56
165−185165−185	8	64
185−205185−205	4	68
Total	N=68N=68

n=68 gives 2N=34
So, we have the median class (125-145)
ℓ=125, N=68,f=20,cf=22, h=20
Median =ℓ+{f2N−cf}×h
=125+{2034−22}×20=137 units.
(ii) Modal class is (125 - 145) having maximum frequency f1=20,f0=13,f2=14, ℓ=125 and h=20
Mode =ℓ+{2f1−f0−f2f1−f0}×h=125+{40−13−1420−13}×20=125+
137×20
=125+13140=125+10.76=135.76 units
(iii)

Monthly consumption (in units)	Number of consumers fifi	Class marks xixi	ui=xiui=xi	fixifixi
65−8565−85	4	75	-3	-12
85−10585−105	5	95	-2	-10
105−125105−125	13	115	-1	-13
125−145125−145	20	135=a135=a	0	0
145−165145−165	14	155	1	14
165−185165−185	8	175	2	16
185−205185−205	4	195	3	12
Total	N=68N=68			77

N=68,a=135, h=20 and Σfiui=7
By step-deviation method.
Mean =a+h× N1×Σfiui
=135+20×681×7
=135+1735=135+2.05
=137.05 units

2. If the median of the distribution given below is 28.5 , find the values of x and y.

Class interval	Frequency
0−100−10	5
10−2010−20	x
20−3020−30	20
30−4030−40	15
40−5040−50	y
50−6050−60	5
Total	60

Sol.

Class interval	Frequency	Cumulative frequency
0−100−10	5	5
10−2010−20	xx	5+x5+x
20−3020−30	20	25+x25+x
30−4030−40	15	40+x40+x
40−5040−50	yy	40+x+y40+x+y
50−6050−60	5	45+x+y45+x+y
Total	60

Median =28.5 lies in the class-interval (20-30).
Then the median class is (20-30).
So, we have ℓ=20,f=20,cf=5+x,h=10,
N=60
Median =ℓ+{f2N−cf}×h=28.5
28.5=20+{2030−(5+x)}×10
⇒8.5=225−x⇒17=25−x⇒x=8
Find the given table, we have
i.e., x+y+45=60 or x+y=15
⇒y=15−x=15−8=7, i.e., y=7

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)	No. of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

Sol.

Age (in years)	Number of policy holders fifi	Cumulative frequency
Below 20	2=22=2	2
20−2520−25	(6−2)(6−2) =4=4	6
Median
class	25−3025−30	(24−6)(24−6) =18=18
30−3530−35	(45−24)(45−24) =21=21	24
40−4040−40	(78−45)(78−45) =21=21	78
40−4540−45	(89−78)(89−78) =11=11	89
45−5045−50	(92−89)(92−89) =3=3	92
50−5550−55	(98−92)(98−92) =6=6	98
55−6055−60	(100−98)(100−98) =2=2	100
Total	N=100N=100

Here, ℓ=35, N=100,f=33,cf=45, h=5
Median =ℓ+{f2N−cf}×h
=35+{3350−45}×5
=35+3325=35+0.76
=35.76 years.

4. The length of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table.

Length (in mm)	No. of leaves
118−126118−126	3
127−135127−135	5
136−144136−144	9
145−153145−153	12
154−162154−162	5
163−171163−171	4
172−180172−180	2

Find the median length of the leaves.

Sol.

The given series is in inclusive form. We may prepare the table in exclusive form and prepare the cumulative frequency table as given below :

Length (in mm)	No. of leaves (fi)(fi)	Cumulative frequency
117.5−126.5117.5−126.5	3	3
126.5−135.5126.5−135.5	5	8
135.5−144.5135.5−144.5	9	17
144.5−153.5144.5−153.5	12	29
153.5−162.5153.5−162.5	5	34
162.5−171.5162.5−171.5	4	38
171.5−180.5171.5−180.5	2	40
	N=40N=40

Here, N=40
⇒2N=20
The cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5.
So, the median class is 144.5-153.5.
∴ℓ=144.5, N=40,cf=17,f=12 and h=9
Therefore, median =ℓ+{f2N−cf}×h
=144.5+12(20−17)×9
=144.5+123×9
=144.5+2.25=146.75
Hence, median length of leaves is 146.75 mm .

5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours)	No. of lamps
1500−20001500−2000	14
2000−25002000−2500	56
2500−30002500−3000	60
3000−35003000−3500	86
3500−40003500−4000	74
4000−45004000−4500	62
4500−50004500−5000	48

Find the median life time of a lamp.

Sol.

Life time (in hours)	No. of lamps	cf
1500−20001500−2000	14	14
2000−25002000−2500	56	70
2500−30002500−3000	60	130
3000−35003000−3500	86	216
3500−40003500−4000	74	290
4000−45004000−4500	62	352
4500−50004500−5000	48	400

2N=2400=200
Median class =3000−3500
Median =ℓ+{f2N−cf}×h
=3000+{86200−130}×500=3406.98
hours
Hence, median life time of a lamp 3406.98 hours.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

No. of letters	No. of surnames
1−41−4	6
4−74−7	30
7−107−10	40
10−1310−13	16
13−1613−16	4
16−1916−19	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Sol.

No. of letters	No. of Surnames fifi	Cumulative frequency
1−41−4	6	6=66=6
4−74−7	30	6+306+30 =36=36
7−107−10	40	36+4036+40 =76=76	50=2N50=2N
10−1310−13	16	76+1676+16 =92=92
13−1613−16	4	92+492+4 =96=96
16−1916−19	4	96+496+4 =100=100
Total	N=100N=100

(i) Here,
ℓ=7, N=100,f=40,cf=36, h=3
Median =ℓ+{f2N−cf}×h
=7+{4050−36}×3
=7+2021=8.05
(ii) Modal class is (7-10).
ℓ=7,f1=40,f0=30,f2=16, h=3 Mode =ℓ+{2f1−f0−f2f1−f0}×h=7+{80−30−1640−30}×3=7+3430=7.88
(iii) Here, a=8.5, h=3,n=100 and Σfiui=−6.

Number of letters	fifi	mark xixi	ui=3xi−83ui=3xi−83	fiuifiui
1−41−4	6	2.5	-2	-12
4−74−7	30	5.5	-1	-30
7−107−10	40	8.5=a8.5=a	0	0
10−1310−13	16	11.5	1	16
13−1613−16	4	14.5	2	8
16−1916−19	4	17.5	3	12
Total	N=100N=100			−6−6

Mean =a+h× N1×Σfiui
=8.5+3×1001×(−6).
=8.5−10018=8.5−0.18=8.32

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	No. of students
40−4540−45	2
45−5045−50	3
50−5550−55	8
55−6055−60	6
60−6560−65	6
65−7065−70	3
70−7570−75

Sol.

Weight (in kg)	No. of students	Cumulative frequency
40−4540−45	2	2
45−5045−50	3	5
50−5550−55	8	13
55−6055−60	6	19
60−6560−65	6	25
65−7065−70	3	28
70−7570−75	2	30

2N=230=15
Median class =55−60
Median =ℓ+{f2N−cf}×h=55+{615−13}×5=56.67 kg

NCERT Class 10 Maths Ch 13 Statistics Other Exercises:

Exercise 13.1

Exercise 13.2

Exercise 13.3

Exercise 13.4

NCERT Solutions Class 10 Maths All Chapters:-

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Linear Equations in Two Variables

Chapter 4 - Quadratic Equations

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometery

Chapter 8 - Introdction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Areas Related to Circles

Chapter 12 - Surface Areas and Volumes

Chapter 13 - Statistics

Chapter 14 - Probability

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NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

1.0Download NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3: Free PDF

2.0Introduction to Median

3.0Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Key Concepts

Median of Ungrouped Data

Median if the number of observations (n) is odd

Median for odd number of terms=(2n+1​)thobservation

Median if the number of observations (n) is even

Median of Grouped Data

Median, Mean, and Mode: Relation

4.0Benefits of Solving NCERT Solutions Class 10 Maths Chapter 13 Statistics Exercise 13.3

5.0NCERT Class 10 Maths Chapter 13 Statistics Exercise 13.3 : Detailed Solution

Frequently Asked Questions

What is covered in Exercise 13.3 of NCERT Class 10 Maths?

What is a modal class?

How is mode different from mean and median?

What is the median in statistics?

What are common mistakes students make while solving median questions?

Join ALLEN!

$Median for odd number of terms = (\frac{n + 1}{2})^{t h} o b ser v a t i o n$