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# Arithmetic Mean - Measures of Central Tendency, Business Mathematics & Statistics B Com Notes | EduRev

Created by: Arshit Thakur

## B Com : Arithmetic Mean - Measures of Central Tendency, Business Mathematics & Statistics B Com Notes | EduRev

The document Arithmetic Mean - Measures of Central Tendency, Business Mathematics & Statistics B Com Notes | EduRev is a part of the B Com Course Business Mathematics and Statistics.
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MEAN : There are three types of mean :
(i) Arithmetic Mean (A.M.)
(ii) Geometric Mean (G. M.)
(iii) Harmonic Mean (H.M.) Of these the Arithmetic mean is the most commonly used. In fact, if not specifically mentioned by mean we shall always refer to arithmetic Mean (AM) and calculate accordingly.

1. Arithmetic Mean :
(i) Simple Arithmetic mean :

(Calculating mean from ungrouped data) The simple arithmetic mean of a given series of values, say, x1 , x2 ,…….. x n is defined as the sum of these values divided by their total number : thus

Note. Often we do not write xj , x means summation over all the observations.

Example 1 : Find the arithmetic mean of 3,6,24 and 48.

(ii) Weighted Arithmetic Mean : (Calculating the mean from grouped data) If the number x1 , x2 , ……. xn occur f1 , f2…….fn times respectively (i.e. occur with frequencies f1 , f2 ……..fn ) the arithmetic mean is

Where N = f is the total frequency, i.e., total number of cases. This mean x is called the weighted Arithmetic mean, with weights f1 , f2 …….f n respectively.

In particular, when the weights (or frequencies) f1 , f2……f n are all equal. We get the simple Arithmetic Mean.

Example 2 : If 5, 8, 6 and 2 occur with frequency 3, 2, 4 and 1 respectively, find the Arithmetic mean.

Arithmetic mean

Calculation of Arithmetic Mean (or simply Mean) from a grouped frequency distribution –– Continuous Series.

(i) Ordinary method (or Direct Method) In this method the mid-values of the class-intervals are multiplied by the corresponding class-frequencies. The sum of products thus obtained is divided by the total frequency to get the Mean. The mean x is given by

where x = mid-value of a class and N = total frequency

Example 3 : Calculate the mean of daily-wages of the following table :

Solution:

∴ Mean Daily Wages

(ii) Shortcut Method (Method of assumed Mean)

In this method, the mid-value of one class interval (preferably corresponding to the maximum frequency lying near the middle of the distribution) is taken as the assumed mean (or the arbitrary origin) A and the deviation from A are calculated. The mean is given by the formula :

where, d = x – A = (mid value) – (Assumed Mean).

Step deviation method :

Example 4: Compute the Arithmetic Mean of the following frequency distribution :

Solution:

∴ Arithmetic Mean

(iii) Method of Assumed mean (by using step deviations)

Calculation of A. M. from grouped frequency distribution with open ends

If in a grouped frequency distribution, the lower limit of the first class or the upper limit of the last class are not known, it is difficult to find the A.M. When the closed classes (other than the first and last class) are of equal widths, we may assume the widths of the open classes equal to the common width of closed class and hence determine the AM. But we can find Median or Mode without assumption.

Properties of Arithmetic Mean :

1. The sum total of the values fx is equal to the product of the number of values of their A.M.
e.g. Nx = fx.

2. The algebraic sum of the deviations of the values from their AM is zero. If x1 , x2……xn are the n values of the variable x and  their AM then are called the deviation of x1 , x2 ………..xn respectively from from

Algebraic sum of the deviations

Similarly, the result for a weighted AM can be deduced.

3. If group of n1 values has AM.  and another group of n2 values has AM , then A.M. ( x ) of the composite group (i.e. the two groups combined) of n1 + n2 values is given by

In general, for a group the AM is given by

Example 5: The means of two samples of sizes 50 and 100 respectively are 54.1 and 50.3. Obtain the mean of the sample size 150 obtained by combining the two sample.

Finding of missing frequency : In a frequency distribution if one (or more) frequency be missing (i.e. not known) then we can find the missing frequency provided the average of the distribution is known. The idea will be clear from the following example :

For one missing frequency :

Example 6 : The AM of the following frequency distribution is 67.45. Find the value of f3 , Let A = 67. Now using the formula.

Solution:
Calculation of missing frequency

Let A = 67. Now using the formula

For two missing frequencies :

Example 7: The A.M. of the following frequency distribution is 1.46

Find f1 and f2 ? Let, x = No. of accidents, f = No. of days

Solution:

f 2 = 200 – (46 + 76 + 25 + 10 + 5) = 38

Example 8 : Arithmetic mean of the following frequency distribution is 8.8. Find the missing frequencies :

Solution:

​

Wrong Observation :

After calculating A.M. of n observations if it is detected that one or more observations have been taken wrongly (or omitted), then corrected calculation of A.M. will be as follows : Let wrong observations x1 , y1 being taken instead of correct values x, y then corrected x = given x – (x1 + y1 ) + (x + y), in this case total no. of observations will be same.

Example 9. The mean of 20 observations is found to be 40. Later on, it was discovered that a marks 53 was misread as 83. Find the correct marks.

Wrong x = 20 × 40 = 800, Correct x = 800 – 83 + 53 = 770

Example 10. A.M. of 5 observations is 6. After calculation it has been noted that observations 4 and 8 have been taken in place of observations 5 and 9 respectively. Find the correct A.M.

Calculation of A.M. from Cumulative Frequency Distribution
At first we are to change the given cumulative frequency distribution into a general form of frequency distribution, then to apply the usual formula to compute A.M. the idea will be clear from the following examples.

Example 11 : Find A.M. of the following distributions :

(i) The difference between any two variables is 4; so the width of class-intervals will be 4. Accordingly, we get the general group frequency distribution as follows :

Solution:

Let A = 10

Let A = 7.5

(i) It is easy to calculate and simple to understand.
(ii) For counting mean, all the data are utilised. It can be determined even when only the number of items and their aggregate are known.
(iii) It is capable of further mathematical treatment.
(iv) It provides a good basis to compare two or more frequency distributions. (v) Mean does not necessitate the arrangement of data.

(i) It may give considerable weight to extreme items. Mean of 2, 6, 301 is 103 and more of the values is adequately represented by the mean 103.
(ii) In some cases, arithmetic mean may give misleading impressions. For example, average number of patients admitted in a hospital is 10.7 per day, Here mean is a useful information but does not represent the actual item.
(iii) It can hardly be located by inspection.

Example 12 : Fifty students appeared in an examination. The results of passed students are given below :

The average marks for all the students is 52. Find out the average marks of students who failed in the examination.

Let average marks of failed students

∴required average marks = 21.

Example 13. From the following frequency table, find the value of x if mean is 23.5

Class : 50–59         40–49         30–39          20 –29       10–19         0–9
frequency : x – 4       x – 2          x + 3            x + 5          x + 10        x –2

Solution:​

∴ Here, i = width of the class = 10.

 or, x = 5 (on reduction).

Example 14 : The mean salary of all employees of a company is ` 28,500.The mean salaries of male and female employees are ` 30,000 and ` 25,000 respectively. Find the percentage of males and females employed by the company.

Let number of male employees be n1 and that of female be n2 . We know

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