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Home » Aakash Coaching » What is the relation between mean, median and mode: CBSE Class 11 Maths

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    What is the relation between mean, median and mode: CBSE Class 11 Maths

    Are you a student of Class 11 and preparing for your Maths exam? Here is all you need to know about the concept of Mean, Median and Mode

    by Team @Aakash
    Apr 26, 2022, 7:49 AM IST
    in Aakash Coaching
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    Mean Median and Mode relationship

    What is the relation between mean, median and mode: CBSE Class 11 Maths

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    Mean, median, and mode are the three fundamental measures of central tendency in statistics. Mean is the average value of a given data set, the median is the middle value of the given set of data, and mode is the value repeated the highest number of times. Let us look at these measures in detail and establish their relationship.

    Table of Contents
    What are the mean, mode, and median?
    Relation between mean, mode, and median
    Frequently asked questions

    What are mean, median, and mode?

    Mean

    Mean is the average value of a given set of data. It is calculated by dividing the sum of all the values in a given set of data by the total number of values

    Median

    It is denoted by M. It consists of three types that have their formulas:

     Basic rules for solving questions:

    • Always arrange the series: either in ascending or descending order.
    • Always calculate CF (cumulative frequency).
    • For individual series, n is equal to the number of values, and for discrete and continuous series, n is equal to the last term of CF or F
    • The answer will always be from x 
    1. Individual series: Here, all the values will be  x
    • Here first, we have to find out if the number of given values is even or odd
    • If the number of values is odd, M = size of n+1th2 term.
    • If the number of values is even, M = size of n/2th 2 term.

           2. Discrete series: Here, along with x, frequency (F)values will also be present.

    • Formula, M= size of n+12th term

           3. Continuous series: We have class intervals and frequency (F) here. Continuous series is also called frequency distribution.

    • Formula, M = size of n/2th term

                                M =       L1+n/2-CFF I

         Here I stands for class interval 

         L1 is the lower limit

    First, we must identify which of the above categories the question falls in. Then we apply the relevant formula accordingly.

    Question on Individual series:

    X = 17, 32, 35, 33, 15, 21, 41, 32, 11, 10, 20

    Calculate median?

    Ans: 

    • As we can see, it is an odd series. So, we will arrange it first in ascending order 

                10, 11, 15, 17, 20, 21, 32, 32, 33, 35, 41

    • Now let us apply the formula

                M = size of n+12th term

                    = size of 11+12th term

                    = size of 122th term

                    = size of 6th term

             Here the 6th term is 21. So, M is equal to 21.

    Question on discrete series:

    Q1: Find the median for the given set of values.

          x       F
          2       2
          3       3
          4       8
          5     10
          6     12
          7     16
          8       10
          9       8
          10       6

    First, we calculate the cumulative frequency (CF)

     

    x F CF
    2 2 2
    3 3 5
    4 8 13
    5 10 23
    6 12 35
    7 16 51
    8 10 61
    9 8 69
    10 6 75

    CF for each value is calculated by adding the value with the adjacent value.

    The formula for calculating M for discrete values = size of n+12th term

    Let us put the known values in the formula

    M = size of 75+12th term

    M = size of 762th term

        = size of 382th term

       38 is not in CF. So, we look for the next largest number, that is, 51

       Next to 51, x is equal to 7. So, M is equal to 7 here.

     

    Question on continuous series: Find the median for the given values.

    Class interval/ x Frequency
    0-10 22
    10-20 36
    20-30 46
    30-40 35
    40-50 20

    First, we calculate the cumulative frequency (CF). CF for each value is calculated by adding the value with the adjacent value.

    CI/x F CF
    0-10 22 22
    10-20 36 60
    20-30 46 106
    30-40 35 141
    40-50 20 161

    The formula for calculating M for continuous series is = size of n2 th term

                                                                                     = size of 1612th term

                                                                                    = 80.5th term

    80.5 is not any value in CF, so we take the next highest value, 106. x value next to 106 is 20-30. This means the answer is in the range of 20-30. Now how will we know the exact answer? We apply another formula

           M = L1 + n2-CFFi

    Here, we know that n2 = 8.5, CF = the value of CF above the value of CF for the value we are calculating, that is, 60, and I is the difference between the upper and lower limit of CI (30-20 = 10). Now let us apply the values in the formula.

         M = 20 + 80.5-604610

             = 20 + 20.54010

             = 20 + 2054010

             = 20 + 4.45 

             = 24.45

    So, M here is 24.45

    Mode

    Mode is denoted by Z. It consists of three types:

    1. Individual series: Has only x. Here the value that appears most of the time in the given values will be the answer.
    2. Discrete series: Here, the mode can be found by the simple and grouping methods. In the simple method, the highest frequency value is taken as the mode. For the grouping method, we use the formula
    3. Continuous series: Here also mode can be found out by two methods, simple and continuous. 

    Formula for calculating mode by simple method = L1 + F1-F02F1-F0-F2i

    Here, F1 = highest frequency

             F0 = frequency above F1

             F2 = frequency below F1

             L1 = lower limit near F1

    Question on Individual series: Calculate mode.

        X = 17, 17, 18, 18, 19, 20, 20, 21, 21, 22, 22, 22, 22, 23, 24

    Ans: 

    • The set of values is already arranged in ascending order
    • The value that is repeated the maximum number of times is 22. So, here mode is 22.

     

    Question on discrete series:

    Q1: Calculate mode by the simple method.

     

    x F
    110 2
    120 4
    130 8
    140 10
    150 5
    160 4

    The highest frequency in the given set of frequencies is 10. The value of x next to 10 is 140. So, here mode (Z) is 140.

     

    Q2: Calculate mode by grouping method.

    x F
    2 3
    3 8
    4 10
    5 12
    6 16
    7 14
    8 10
    9 8
    10 17
    11 5
    12 4
    13 1

    Grouping table:

    X F 1+2 2+3 1+2+3 2+3+4 3+4+5
    2 3 11
    3 8 18
    4 10 22 21
    5 12 28 30
    6 16 30 32
    7 14 24 42
    8 10 18
    9 8 25 40
    10 17 22 35 32
    11 5 9
    12 4 5
    13 1 10 30 26

    Analysis table:

    2 3 4 5 6 7 8 9 10 11 12 13 14
    1 ✔️
    2 ✔️ ✔️
    3 ✔️ ✔️
    4 ✔️ ✔️ ✔️
    5 ✔️ ✔️ ✔️
    6

    Total

    ✔️

    1

    ✔️

    3

    ✔️

    5

    3 1 1

    The maximum number of values is under 6. So, the mode here is 6.

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    Relation between mean, median and mode

    Formula 1: Mode = 3 Median – 2 Mean

    For example, If median = 12, mean = 8, mode = ?

    We know that mode = 3 median – 2 mean

    So, let us put the values we know in the formula

    Mode = 3 (12) – 2 (8)

    Mode = 36 – 16 = 20

    Thus, the mode is equal to 20. Very simple!

     

    Formula 2: Mode = Mean – 3 (mean – median)

    For example, If mode = 6, mean – median = 12, mean = ?

    We know that, mode = mean – 3 (mean – median)

    So, let us put the values we know in the formula

    6 = mean – 3 (12)

    6 = mean – 36

    6 + 36 = mean

    Thus, the mean is equal to 42.

     

    Formula 3: Mode = Median – ⅔ (mean – mode)

    For example, mode = 13, mean – mode = 18, median = ?

    We know that, mode = median – ⅔ (mean – mode)

    So, let us put the values we know in the formula

    13 = median – ⅔ (18)

    13 = median – 12

    13 + 12 = median

    Median = 25

    Thus, the median is equal to 25.

     

    Simple solved question: Calculate the mean, mode and median for the following set of values:

    3, 5, 7, 9, 8, 4, 3, 8, 7, 9, 7

    Ans: 

    1. We know that mode is the value repeated the maximum number of times in the given values. So, here mode is 7.
    2. Mean = Sum of all the values/ total number of values

                          = 70/11

                          = 6.36

        So, here mean is equal to 6.36

    1. To calculate the median for an odd set of values, we first need to arrange all the given numbers in ascending order.

               3, 3, 4, 5, 7, 7, 7, 8, 8, 9, 9, 9

    • Now let’s add the total number of values (n) to 1 and divide the result by 2. 

                11+½ = 12/6 = 6

    • The 6th number in the given set of numbers is the median. So, here the median is 7.

    If you go through the whole topic on mean, median and mode carefully from NCERT Solutions from Class 11 Maths and practise as many questions as possible, no one can stop you from securing full marks in the questions that come from this topic in the exam. Also, don’t forget to solve the previous years’ question papers for better understanding.

    Also See: CBSE Class 12 Term 2 Syllabus Maths: 10 High Scoring Topics for Class 12 Board Exam Maths

    FAQs

    1. Is NCERT enough for Class 11 Maths?

    NCERT Solutions for Class 11 Maths covers the syllabus’s wider and more complex parts. However, Maths requires more explanation; therefore, you  should study supplement books like RD Sharma Solutions and RS Aggarwal Solutions to better understand the concepts. Maths is better understood in a face to face class. Find Aakash Coaching Institute Near You to learn Maths from the best teachers.

    2. Can the mean and median be the same?

    Yes. In a perfectly symmetrical distribution, the mean and median are the same. However, in a symmetrical distribution with two modes (bimodal), two values have been repeated the highest number of times, and the modes are different from the mean.

    3. What are the limitations of median?

    The disadvantage of the median is that it does not consider the precise value of each observation and therefore does not effectively acknowledge all the information available in the given set of data. Unlike the mean, the median is affected by fluctuations in sampling and is not compatible with further mathematical calculation, so it is not used in many statistical tests.

    4. What is the application of mean, median, and mode?

    The mean, median and mode are widely used by insurance analysts, real estate agents, marketers, and individuals who work in Human Resources departments at companies to assess data and analyse progress. For example, marketers often use mean, median and mode to determine how their advertisements perform.

    5. What are mean, median, and mode?

    Mean, median, and mode are the three measures of central tendency. The mean is the average of the given set of numbers. Median is the middle value of the given set of numbers, and mode is the value repeated the maximum number of times.

    6. How can I score more than 95% in Maths in board exams?

    • Start the syllabus on time
    • Be attentive in the class
    • Practice as many questions as possible
    • Study a little every day
    • Form an academic timetable and follow it
    • Revise the important chapters multiple times. Check  Important Concepts and Math Concepts to know about the topics you need to focus on more
    • Complete your day’s homework within the same day. Don’t procrastinate
    • Start with the chapters you find easy and then proceed to the more difficult ones
    • Solve previous ten years’ question papers

    7. Is Class 11 Maths tough?

    Maths is a subject that needs rigorous practise and understanding of the fundamental concepts. If you are attentive in your classes and practice as many questions as possible, including previous year papers, you will find the course quite easy. It will also help you prepare ahead for your competitive exams like JEE Main Exam. Math is an important part of the JEE Main 2022 Syllabus, so understanding its concepts well will give you a headstart if you want to compete in this exam.

     

    Tags: CBSE Class 11CBSE Class 11 Maths
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