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MEAN,MODE AND MEDIAN NCERT
Mean, mode and median are the terms of statistics which represent the central tendency of a data, The single value of the mean, mode or median tells us about the state of data. All of these three are important in solving statistical problems in economics, science, commerce, and other fields.
Mean- The mean of data represents the most common value of the data, it is calculated as follows.
When data is given as an individual series- Indidual series means when each value of the variable is given without showing its frequency.
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Example-
Find out the average weight of 5 people whose weights are given as follows.
45 kg, 50 kg, 40 kg,60 kg and 55 kg
So, the most common weight or average weight of 5 people in the given data is 50
Disadvantages of mean– The mean is unable to represent the data in the condition when data is skewed (much differences between larger values and the rest of the individual values of the data), then the median would be truer to represent the data, see the following example.
| maths marks of 5 friends in a test | 5 | 60 | 95 | 10 | 65 |
Median- Median is called the middle term of the data provided data should be ascending order, the mean of the above data is 47 which is not representing the data properly so here median of the data will the best to show the nature of data of maths marks of 5 friends in a test.
| maths marks of 5 friends in a test | 5 | 10 | 60 | 65 | 95 |
n = 5 (odd number)
m = 60
The median of this data is 60 is representing the given data in a better way because of the difference between the largest value and median is lesser than the difference between the largest value and mean.
When the individual data is given and the number of observations are even
Direct Method for evaluating the Mean
Discreet series- When frequencies are also given with the values of variable then such a series is known as discreet series, see the example
| x | Frequency |
| 50 | 6 |
| 40 | 8 |
| 70 | 5 |
| 60 | 4 |
| 80 | 5 |
Mean of such a data is given as follows
Solution-
| x | f | fx |
| 50 | 6 | 300 |
| 40 | 8 | 320 |
| 70 | 5 | 350 |
| 60 | 4 | 240 |
| 80 | 5 ∑f= 28 | 400 ∑fx=1610 |
Determining the Median in case of discreet series-
| x | f | cf |
| 40 | 8 | 8 |
| 50 | 6 | 14 |
| 60 | 4 | 18 |
| 70 | 5 | 23 |
| 80 | 5 N=28 | 28 |
Therefore the median of the above data is 55.
The other ways of calculating mean are following
Assumed mean method for evaluating the Mean
Assuming an observation as a mean(A) and then forming the third column and then writing the deviation(d) each observation about the mean that is equal to the difference between observation and the assumed mean keeping in view the sign convention of the numbers, thereafter forming 4 th column and evaluating fd the product of deviation and their corresponding frequencies. After the table is formed then calculate the value of mean from the following formula.
∑fd is the addition of the product of deviation and corresponding frequencies and ∑f is the sum of total frequencies of individual observations.
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Step deviation method for evaluating Mean
In this method, one more column is added d’ as compared to the assumed mean method, in this method deviation(d) is reduced by dividing the common factor(h )of all the deviations.
After evaluating d’ one more column of fd’ is also incorporated in the table and then calculate the mean from the following formula.
When the continuous series is given-
| Class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 5 | 10 | 20 | 4 | 1 |
Solution-
| Sr.n0 | Class interval | f(Frequency) | x(Classmark) | fx |
| 1 | 10-20 | 5 | 15 | 75 |
| 2 | 20-30 | 10 | 25 | 250 |
| 3 | 30-40 | 20 | 35 | 700 |
| 4 | 40-50 | 4 | 45 | 180 |
| 5 | 50-60 | 1 | 55 | 55 |
| ∑f=40 | ∑fx=1260 |
Other methods- (i) Assumed mean method- Already discussed above
Step deviation method– Already discussed above
Median of the data when continuous series is given-
| Class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| Frequency | 5 | 10 | 20 | 4 | 1 |
Solution- Median is calculated by using the following formula when continuous series is given
| Class interval | Frequency(f) | Cumulative frequency(cf) |
| 10-20 | 5 | 5 |
| 20-30 | 10 | 15(cf) |
| 30-40 | 20(f) | 35 |
| 40-50 | 4 | 39 |
| 50-60 | 1 | 40(N) |
The median group is N/2 th term of the series i.e N/2=40/2 = 20 th term which lies in 35 th term so the median group is 30-40
L = 30 (Lower limit of median class)
f =20 (Frequency of median class)
Hence the median of the given series is 32.5
The mode – In a series, the term with the highest frequency is known as mode of the series.
Mode of an individual series- When the individual measurements of the variable are given without showing their frequency then the terms with high frequency is known as a mode of the series.
Example- The performances of a student in 5 tests are given as follows, find the mode of data.
| Tests | i | ii | iii | iv | v |
| Percentage | 4o | 30 | 50 | 40 | 60 |
40 has the highest frequency so the mode of the above data is 40
Mode in a discreet series- When the frequency is also given of each term then such a series is known as discreet series.
| Age of students in a class(in years) | 13 | 15 | 16 | 14 | 12 |
| Number of students | 7 | 6 | 4 | 15 | 8 |
In the given data most numbers of students are of the age 14 so the mode of the data is 14.
Mode of the continuous series- Find the mode of the following series
| The age of employs(in years) | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| The number of employs | 30 | 50 | 25 | 15 | 2 |
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Solution –
| The ages of employs(in years) | frequency(f) |
| 20-30 | 30 |
| 30-40 | 50 |
| 40-50 | 25 |
| 50-60 | 15 |
| 60-70 | 2 |
The mode(M ) of continuous series is calculated as follows
The class 30-40 has the highest frequency so it is mode group of above data
L = 40(lower limit of mode group)
f0= 30(preceding frequency of mode group)
f1= 50( frequency of mode group)
f2= 25 (successor frequency of mode group)
i= Class interval of mode group
M = 30+ 4.44
M = 34.44
Hence the mode of the series is 34.44
The relationship between Mean,Mode and Median
Mode = 3×Median – 2×Mean
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