RD Sharma Chapter 24 Class 9 Maths Exercise 24.2 Solutions can be availed from here. These solutions for class 9th, i.e., Measures of Central Tendency, is the best tool for preparing for the examination. It can be a handy guide for the practitioners to guide them regarding the challenging questions they face upon attempting to solve. Upon providing solutions, it makes things easier for the students to solve the questions in quicker times.
RD Sharma Chapter 24 Class 9 Maths Exercise 24.2 Solutions- Measures of Central Tendency have been prepared based on the arithmetic mean of grouped data or discrete frequency distribution and Methods to solve Arithmetic mean – Direct Method, Short-cut Method, and step deviation method.
Go down to the article and check the PDF, which is accessible for free for the learners to prepare for the exam with various questions based on the RD Sharma Chapter 24 Class 9 Maths Exercise 24.2 Solutions.
Learn about Class 9 Chapter 24 (Measures Of Central Tendency)
Table of Contents
Download RD Sharma Chapterṣ 24 Class 9 Maths Exercise 24.2 Solutions PDF
Solutions for Class 9 Maths Chapter 24 Measure of Central Tendency Exercise 24.2
Important Definitions RD Sharma Chapter 24 Class 9 Maths Exercise 24.2 Solutions
The Measures of Central Tendency is based on the following two topics-
- Arithmetic mean of grouped data or discrete frequency distribution
- Methods to solve Arithmetic mean – Direct Method, Short-cut Method, step deviation method
Arithmetic mean of grouped data or discrete frequency distribution
The method employed here is called the arithmetic mean, which is the total of the entire values divided by the total number of cases. When it comes to grouped data, the summation of the entire value can be obtained upon doing multiplication of the total number of occurrences or the percentage of occurring with the variable value.
Methods to solve Arithmetic mean – Direct Method, Short-cut Method, step deviation method
The direct method, short-cut method, and step deviation methods are the different methods one can employ for solving the questions based on the arithmetic mean. However, the student might have to be asked to solve the question based on any of the mentioned above ways. They might be asked to solve the questions based on several of these ways as well.
Direct Method
In computing the arithmetic mean of a consecutive series, we use the mid-value of each class as representative of that class (as it is assumed that the frequencies of the class are concern on mid-point) multiply the multiple mid-values by their similar frequencies and total of the products is divided by the total of the frequencies.
Formula-
X=fXf
Short-Cut Method
Taking an estimated mean and calculating differences of the given variates or observations from it executes the calculation of mean easier. This average is usually taken to be a regular round number in the center of the range of the observations so that the variations can be easily taken by subtraction.
Formula-
X=A+fdN
step deviation method
The Step Deviation Method is used to Computing the Mean.
Formula-
X=A+fd’fi
Examples of RD Sharma Chapter 24 Class 9 Maths Exercise 24.2 Solutions
Ques: Determine the value of p for the following distribution, whose average is 16.6.
x |
8 |
12 |
15 |
p |
20 |
25 |
30 |
f |
12 |
16 |
20 |
24 |
16 |
8 |
4 |
Solution-
x |
f |
fx |
8 |
12 |
96 |
12 |
16 |
192 |
15 |
20 |
300 |
p |
24 |
24p |
20 |
16 |
320 |
25 |
8 |
200 |
30 |
4 |
120 |
N= 100 fx = 24p + 1228 |
Formula to determine average:
Mean (x)= fxN
= (24p + 1228)/100
= average = 16.6 (given)
= So, (24p + 1228)/100 = 16.6
= 24p + 1228 = 1660
= 24p = 1660 – 1228 = 432
= p = 432/24 = 18
= The value of p is 18.
Ques- Determine the lost value of p for the following distribution whose average is 12.58.
x |
5 |
8 |
10 |
12 |
p |
20 |
25 |
f |
2 |
5 |
8 |
22 |
7 |
4 |
2 |
Solution-
x |
f |
fx |
5 |
2 |
10 |
8 |
5 |
40 |
10 |
8 |
80 |
12 |
22 |
264 |
p |
7 |
7p |
20 |
8 |
160 |
25 |
2 |
50 |
N= 50 fx = 7p + 524 |
Formula to determine average-
= Mean (x)= fxN
= (7p + 524)/50
= average = 12.58 (given)
= So, (7p + 524)/50 = 12.58
= 7p + 524 = 12.58 x 50
= 7p + 524 = 629
= 7p = 629 – 524 = 105
= p = 105/7 = 15
= The value of p is 15.
Ques- Determine the lost frequency (p) for the following distribution whose average is 7.68.
x |
3 |
5 |
7 |
9 |
11 |
13 |
f |
6 |
8 |
15 |
p |
8 |
4 |
Solution-
x |
f |
fx |
3 |
6 |
18 |
5 |
8 |
40 |
7 |
15 |
105 |
9 |
p |
9p |
11 |
8 |
88 |
13 |
4 |
52 |
N= p+41 fx = 9p + 303 |
Formula to determine average-
= (9p + 303)/(p+41)
= Average = 7.68 (given)
= So, (9p + 303)/(p+41) = 7.68
= 9p + 303 = 7.68 (p + 41)
= 9p + 303 = 7.68p + 314.88
= 9p − 7.68p = 314.88 − 303
= 1.32p = 11.88
= or p = (11.881)/(1.32) = 9
= The value of p is 9.