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

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

$\text{Median for odd number of terms}=(\frac{n+1}{2}{)}^{th}observation$

Median if the number of observations (n) is even

$\text{Median for even number of terms}=\frac{(\frac{n}{2}{)}^{th}observation+(\frac{n}{2}+1{)}^{th}observation}{2}$

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: 

$Median=l+\frac{(\frac{n}{2}-cf)}{f}\times h$

Here, 

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

$h=\text{Upper Limit -Lower limit}$

• 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: 

$3\times Median=Mode+2\times Mean$

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.

Sol.

(i)

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​−f2​f1​−f0​​}×h=125+{40−13−1420−13​}×20=125+
137×20​
=125+13140​=125+10.76=135.76 units
(iii)

N=68,a=135, h=20 and Σfi​ui​=7
By step-deviation method.
Mean =a+h× N1​×Σfi​ui​
=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.

Sol.

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.

Sol.

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.

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 :

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 :

Find the median life time of a lamp.

Sol.

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:

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.

(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​−f2​f1​−f0​​}×h=7+{80−30−1640−30​}×3=7+3430​=7.88
(iii) Here, a=8.5, h=3,n=100 and Σfi​ui​=−6.

Mean =a+h× N1​×Σfi​ui​
=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.

Sol.