Mean of Ungrouped Data

The mean of data indicate how the data are distributed around the central part of the distribution. That is why the arithmetic numbers are also known as measures of central tendencies.


Mean Of Raw Data:       

The mean (or arithmetic mean) of n observations (variates) x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) is given by

Mean = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + .... + x_{n}}{n}\)

In words, mean = \(\frac{\textbf{Sum of the Variables}}{\textbf{Total Number of Variates}}\)

Symbolically, A = \(\frac{\sum x_{i}}{n}\); i = 1, 2, 3, 4, ...., n.

Solved Example:

A student scored 80%, 72%, 50%, 64% and 74% marks in five subjects in an examination. Find the mean percentage of marks obtained by him.

Solution:

Here, observations in percentage are

x\(_{1}\) = 80, x\(_{2}\) = 72, x\(_{3}\) = 50, x\(_{4}\) = 64, x\(_{5}\) = 74.

Therefore, their mean A = \(\frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5}}{5}\)

                                   = \(\frac{80 + 72 + 50 + 64 + 74}{5}\)

                                   = \(\frac{340}{5}\)

                                   = 68.

Therefore, mean percentage of marks obtained by the student was 68%.

 

Mean of Arrayed Data:

If the values of the variable (i.e., observations or variates) be x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) and their corresponding frequencies are f\(_{1}\), f\(_{2}\), f\(_{3}\), f\(_{4}\), ....., f\(_{n}\) then the mean of the data is given by

Mean = A (or \(\overline{x}\)) = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4} + .... + x_{n}f_{n}}{ f_{1} +  f_{2} + f_{3} + f_{4} + ..... + f_{n}}\)

In words, mean = \(\frac{\textbf{Sum of products of the Variables and their corresponding Frequencies}}{\textbf{Total Frequency}}\)

Symbolically, A = \(\frac{\sum{x_{i}. f_{i}}}{\sum f_{i}}\); i = 1, 2, 3, 4, ...., n.


Solved Example:

A class has 20 students whose ages (in years) are as follows.

14, 13, 14, 15, 12, 13, 13, 14, 15, 12, 15, 14, 12, 16, 13, 14, 14, 15, 16, 12

Find the mean ago of the students of the class.

Solution:

In the data, only five different numbers appear respectively. So, we write the frequencies of the variates as below.

Age (in years)

(x\(_{i}\))

12

13

14

15

16

Total

Number of Students

(f\(_{i}\))

4

4

6

4

2

20

Therefore, mean A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4} + x_{5}f_{5}}{ f_{1} +  f_{2} + f_{3} + f_{4} + f_{5}}\)

= \(\frac{12 × 4 + 13 × 4 + 14 × 6 + 15 × 4 + 16 × 2}{4 + 4 + 6 + 4 + 2}\)

= \(\frac{48 + 52 + 84 + 60 + 32}{20}\)

= \(\frac{276}{20}\)

= 13.8

Therefore, the mean age of the students of the class = 13.8 years.

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9th Grade Math

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