**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, a**nd 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 |

**Study notes of Maths & Science**

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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