Class 9 RD Sharma Solutions Chapter 24 Measures of Central Tendency Exercise 24.3 (original) (raw)

Last Updated : 24 Sep, 2024

Exercise 24.3 of Chapter 24 (Measures of Central Tendency) in the Class 9 RD Sharma textbook focuses on calculating the median for grouped data. This exercise introduces students to the concept of median in frequency distributions and teaches them how to determine the median class and apply the median formula for grouped data. Students will practice interpreting frequency tables, cumulative frequency, and using interpolation to find the precise median value.

What are the Measures of Central Tendency?

Measures of central tendency are statistical tools used to describe the typical or central value in a dataset, providing a concise summary of large amounts of information. The three primary measures are the mean (average), median, and mode. The mean is calculated by summing all values and dividing by the number of data points, offering a comprehensive view that is susceptible to outliers. The median represents the middle value in an ordered dataset, making it more robust against extreme values. The mode identifies the most frequently occurring value, particularly useful for categorical data. Each measure has its strengths and is chosen based on the data's nature and the analysis goals. These measures collectively offer insights into data distribution and help in making informed decisions across various fields, from economics to social sciences.

Question 1. Find the median of the following data:

83, 37, 70, 29, 45, 63, 41, 70, 34, 54

**Solution:

Ascending order sequence of the numbers :

29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Total number of terms, supposedly, n = 10 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space ^{th}) value}{2} \\ = \frac{\frac{10}{2}^{th} \space value + (\frac{10}{2}+1\space ^{th}) value}{2} \\ = \frac{5^{th} value+ 6^{th} value}{2} \\ = \frac{45+54}{2}

= 99/2

= 49.5

Question 2. Find the median of the following data:

133, 73, 89, 108, 94, 104, 94, 85, 100, 120

**Solution:

Ascending order sequence of the numbers :

73, 85, 89, 94, 94, 100, 104, 108, 120, 133

Total number of terms, supposedly, n = 10 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space^{th}) value}{2} \\ = \frac{\frac{10}{2}^{th} \space value + (\frac{10}{2}+1\space ^{th}) value}{2} \\ = \frac{5^{th} value+ 6^{th} value}{2} \\ = \frac{94+100}{2}

= 194/2

= 97

Question 3. Find the median of the following data:

31, 38, 27, 28, 36, 25, 35, 40

**Solution:

Ascending order sequence of the numbers :

25, 27, 28, 31, 35, 36, 38, 40

Total number of terms, supposedly, n = 8 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space ^{th}) value}{2} \\ = \frac{\frac{8}{2}^{th} \space value + (\frac{8}{2}+1\space ^{th}) value}{2} \\ = \frac{4^{th} value+ 5^{th} value}{2} \\ = \frac{31+35}{2}

= 66/2

= 33

Question 4. Find the median of the following data:

15, 6, 16, 8, 22, 21, 9, 18, 25

**Solution:

Ascending order sequence of the numbers :

6, 8, 9, 15, 16, 18, 21, 22, 25

Total number of terms, supposedly, n = 9 (odd)

Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{9+1}{2}^{th} term

= 5th term

= 16

Question 5. Find the median of the following data:

41, 43, 127, 99, 71, 92, 71, 58, 57

**Solution:

Ascending order sequence of the numbers :

41, 43, 57, 58, 71, 71, 92, 99, 127

Total number of terms, supposedly, n = 9 (odd)

Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{9+1}{2}^{th} term

= 5th term

= 71

Question 6. Find the median of the following data:

25, 34, 31, 23, 22, 26, 35, 29, 20, 32

**Solution:

Ascending order sequence of the numbers :

20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Total number of terms, supposedly, n = 10 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space ^{th}) value}{2} \\ = \frac{\frac{10}{2}^{th} \space value + (\frac{10}{2}+1\space ^{th}) value}{2} \\ = \frac{5^{th} value+ 6^{th} value}{2} \\ = \frac{26+29}{2}

= 55/2

= 27.5

Question 7. Find the median of the following data:

12, 17, 3, 14, 5, 8, 7, 15

**Solution:

Ascending order sequence of the numbers :

3, 5, 7, 8, 12, 14, 15, 17

Total number of terms, supposedly, n = 8 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space^{th}) value}{2} \\ = \frac{\frac{8}{2}^{th} \space value + (\frac{8}{2}+1\space ^{th}) value}{2} \\ = \frac{4^{th} value+ 5^{th} value}{2} \\ = \frac{8+12}{2}

= 20/2

= 10

Question 8. Find the median of the following data:

92, 35, 67, 85, 72, 81, 56, 51, 42, 69

**Solution:

Ascending order sequence of the numbers :

35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Total number of terms, supposedly, n = 10 (even)

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space ^{th}) value}{2} \\ = \frac{\frac{10}{2}^{th} \space value + (\frac{10}{2}+1\space ^{th}) value}{2} \\ = \frac{5^{th} value+ 6^{th} value}{2} \\ = \frac{67+69}{2}

= 136/2

= 68

Question 9. Numbers 50, 42, 35, 2x + 10, 2x - 8, 12, 11, 8 are written in descending order and their median is 25, find x.

**Solution:

Ascending order sequence of the numbers :

8, 11, 12, 2x - 8, 2x+10, 35, 42, 50

Total number of terms, supposedly, n = 8 (even)

Given,

Median = 25

Median =

\\ \frac{\frac{n}{2}^{th} \space value + (\frac{n}{2}+1\space ^{th}) value}{2} \\ 25 = \frac{\frac{8}{2}^{th} \space value + (\frac{8}{2}+1\space ^{th}) value}{2} \\25 = \frac{4^{th} value+ 5^{th} value}{2} \\25 = \frac{2x-8 + 2x+10}{2} \\25= \frac{4x+2}{2}

48 = 4x

x = 12

Question 10. Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33.If 92 is replaced by 99 and 41 by 43 in the new data, find the new median?

**Solution:

Ascending order sequence of the numbers :

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

n = 11 (odd)

Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{11+1}{2}^{th} term

= 6th term

= 58

If we replace 92 by 99 and 41 and 43, then the new sequence of numbers are :

33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99

New Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{11+1}{2}^{th} term

= 6th term

= 58

Question 11. Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median?

**Solution:

Arranging the data given in ascending order :

41, 43, 57, 58, 61, 71, 92, 99, 127

Number of terms, supposedly n = 9 (odd)

Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{9+1}{2}^{th} term

= 5th term

= 61

On replacing the value 58 by 85, we get,

41, 43, 57, 61, 71, 85, 92, 99, 127

New Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{9+1}{2}^{th} term

= 5th term

= 71

Question 12. The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42,30. Find the median in weight. If the 44 kg is replaced by 446kg and 27 kg by 25kg. Find the new median.

**Solution :

Arranging the data in ascending order,

27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45

Total number of terms = 15 (n = odd)

Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{15+1}{2}^{th} term

= 8th term

= 35 kg

On replacing 44 kg by 46 kg and 27 kg by 25 kg respectively, we get,

25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45,46

New Median =

\\ \frac{n+1}{2}^{th} term \\ = \frac{15+1}{2}^{th} term

= 8th term

= 35 kg

Question 13. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.

29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

**Solution :

Given,

Median = 63

Total number of terms, supposedly n = 10 (even)

Now,

Median =

\\ \frac{\frac{n}{2}^{th}\space term + (\frac{n}{2}+1^{th}) term}{2} \\ 63 = \frac{\frac{10}{2}^{th}\space term + (\frac{10}{2}+1^{th}) term}{2} \\ 63 = \frac{5^{th} + 6^{th} term}{2} \\ 63 = \frac{x+x+2}{2} \\ 63 =\frac{2x+2}{2}

63 = x+1

x = 62

Summary

Exercise 24.3 of Chapter 24 on Measures of Central Tendency delves into the calculation of median for grouped data, a crucial skill in statistical analysis. This exercise builds upon students' understanding of median in ungrouped data and extends it to more complex frequency distributions. Through a series of problems, students learn to identify the median class, calculate cumulative frequencies, and apply the median formula for grouped data. The exercise emphasizes the importance of interpolation in finding a precise median value within the median class interval. By working with various real-world scenarios such as age distributions, salary ranges, and test scores, students gain practical experience in interpreting and analyzing grouped data.