CBSE Notes For Class 10 Maths Chapter 13 Statistics

1.0Introduction to Statistics 

This chapter of CBSE Notes Class 10 covers the representation, interpretation, and analysis of data that is accomplished using measures, such as mean, median, mode, and graphical representations like histograms and cumulative frequency curves.

2.0Download CBSE Notes for Class 10 Maths Chapter 13: Statistics - Free PDF!!

Ready to ace your Class 10 Maths in Statistics chapter? These free PDF notes are your key to understanding mean, median, mode, and more, making complex concepts clear and exam-ready.

3.0CBSE Class 10 Maths Chapter 13 Statistics - Revision Notes

What is Statistics? 

Statistics is a branch of mathematics concerned with how to collect, organise, and possibly interpret numerical data. In this chapter, we shall focus on:

• Organising raw data
• Finding measures of central tendency: Mean, Median, and Mode
• Visual data representation (through graphs)

Measures of Central Tendencies

In mathematics, a measure of central tendency is a statistical value that represents the centre or typical value of a data set. It consists of three mean measures: mean, median, and mode. Each of these measures provides a summary representation of the data.

1. Mean: 

For grouped data, the mean is a weighted average of the midpoints of the class intervals where frequency for each class is used as the weight. The mean is calculated with three methods: 

• Direct Method: The direct method is quite straightforward, in which it sums all values in the data set and divides by the total number of observations.

$\text{ Mean }=\frac{\sum fi\cdot xi}{\Sigma f}$

• Assumed Mean Method: Assumed mean simplifies calculations by choosing a central value (assumed mean) and adjusting deviations of each observation from this assumed value.

$di=xi-a$

$\bar{d}=\frac{\Sigma fi\cdot di}{\Sigma f}$

$\text{ Mean }=a+\frac{\Sigma fi.di}{\Sigma f}$

• The Step Deviation Method: This is obtained by adjusting the assumed mean based on the known standard deviation and is often applied in cases where data is known to have a known variance or spread.

$ui=\frac{xi-a}{h}=\frac{di}{h}$

$\bar{u}=\frac{\Sigma fi\cdot ui}{\Sigma f}$

$\text{ Mean }=a+h\times \frac{\Sigma fi\cdot ui}{\Sigma f}$

Where 

• f is the frequency of each class. 
• h is class width. 
• a is the assumed mean frequency. 
• d is the deviation of the data point. 
• x_i is the class mark. 
• u is the step deviation frequency. 

2. Mode: 

The mode of grouped data is the value that occurs most often in the dataset. It can be found for grouped data by simply identifying which class contains the highest frequency and then using a formula to calculate the mode.

$\text{ Mode }=L+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

where:

• L is the lower boundary of the modal class,
• f₁ is the frequency of the modal class,
• f₀​ is the frequency of the class before the modal class,
• f₂​ is the frequency of the class after the modal class,
• h is the class width

3. Median: 

The median for grouped data is the value that splits the data into two halves. It is found by obtaining the cumulative frequency corresponding to the middle position of the total data.

$\text{ Median }=L+\left(\frac{\frac{N}{2}-CF}{f}\right)\times h$

where:

• L is the lower boundary of the median class,
• N is the total frequency,
• CF is the cumulative frequency of the class before the median class,
• f is the frequency of the median class,
• h is the class width.

4.0Solved Problem

Question 1: Consider the following frequency distribution of marks obtained by class 10^th in maths. 

Find the Mean of the following data with all the three methods mentioned above and also find the mode and median. 

Solution: 

$\text{ Mean from direct method }=\frac{\Sigma f_i\cdot x_i}{\Sigma f}=\frac{2305}{55}=42$

Let us assume a = 45

$\text{ Assumed mean method }=a+\frac{\Sigma fi.di}{\Sigma f}=45+\frac{-170}{55}=42$

$\text{ Step deviation method }=a+h\times \frac{\Sigma fi.ui}{\Sigma f}=45+10\times \frac{-17}{55}=42$

Let, fo = 0, f₁= 15, f₂=13, L = 20

$\text{ Mode }=L+\left(\frac{f1-f0}{2f1-f0-f2}\right)\times h=20+\left(\frac{15-0}{2\times 15-0-13}\right)\times 10=28.8$

Let N = 55, N/2 = 27.5 hence, f = 13, L = 30, CF = 15

$\text{ Median }=L+\left(\frac{\frac{N}{2}-CF}{f}\right)\times h=30+\left(\frac{27.5-15}{13}\right)\times 10=39.62$

Question 2: Find the median for the following frequency distribution. 

Solution: 

N = 80, N/2 = 40, modal class = 30-40, f = 30, L = 30, cf = 27

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

$=30+(\frac{40-27}{30})\times 10$

$=30+13/3$

$=103/3$

$=34.33$

5.0Key Features of CBSE Maths Notes for Class 10 Chapter 13

• The notes are aligned with the latest syllabus of CBSE.
• A step-by-step guide is provided along with solved problems to get a better understanding of concepts. 
• The language used is simple and easy to understand, making the notes ideal for self-learning.