ARITHMETIC MEAN Lecture 05

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ARITHMETIC MEAN

Lecture 05

Merits and Demerits of Arithmetic Mean

Merits of Arithmetic mean

i. It is rigorously defined.

ii. It is based on all the observations in the series.

iii. It helps in the direct comparison.

iv. It helps in the further statistical analyses.

v. It is easy to understand and interpret.

Demerits of Arithmetic Mean

i. It is unduly affected by extreme values.

ii. It is sometimes unrealistic.

iii. It is a suitable average for quantitative data.

iv. It cannot be located by graphical method.

Example

The mean of 5 numbers is 12.7. What extra number must be added to bring the mean up to 13.1?

Solution: The mean of 5 numbers is 12.7

Example 3.12:

The mean score of cricket players in three matches was 55 runs.

i. How many runs did the players score together? 

After four matches, the mean score was 61 runs.

ii. How many runs did the player score in the fourth match?

Solution: The mean score in 3 matches is 55

The total score in 3 matches is 155

ii. Let X4 represent the fourth match.

The mean score in 4 matches is 62

The score in the fourth match was 79.

 PROPERTIES OF ARITHMETIC MEAN

i. The arithmetic mean of a constant is constant.

Proof: Let X1 = a, X2 = a,..., Xn = a

For example: The arithmetic mean of 10, 10, 10, 10, 10

ii. The sum of the deviations of the X’s from the mean is zero.

Proof: Let X1, X2,..., Xnbe the values of a set of data and X¯ their mean.

iii. If the mean of n observations and if each observation is increased by a constant “a” (or decreased by a), then the mean of new observations is X + a (or. X - a).

Proof: Let X1, X2,..., Xn be the values of a set of data, and their mean is given by:

Now that each observation is increased by a, then the new observations will be:

Then the arithmetic mean for new observations is given by:

Example 3.5: The arithmetic mean of 10 values is 25, if 5 are added to each value. Find the new arithmetic mean.

Solution: We using the above-derived result

iii..          If X¯ is the mean of n observations and if each observation is multiplied by a constant “a” (or divided by a), then the mean of new observations is aX¯ (or.X¯/ a)

Proof: Let X1, X2,..., Xn be the values of a set of data, and their mean X¯ is given by:

. Now that each observation is increased by a, then the new observations will be:

Then the arithmetic mean for new observations is given by:

Similarly, if each observation is divided by a constant “a”, then the new arithmetic mean is

Example 3.6: The arithmetic mean of 15 observations is 64 if each observation is multiplied by 6. Find the new arithmetic mean.

Solution: We are using the above-derived results

iv. The sum of squared deviations of the X’s from mean is a minimum.

Where “a” is an arbitrary value other than the mean.

Proof: Let X1, X2,..., Xn  be the values of a set of data, and  X¯ is their mean.

Example 3.7: For the data given blow:

10, 25, 30, 35, and 40.

Show that

Solution: let a = 30

Thus, 

Combined Arithmetic Mean (Pooled Mean)

The arithmetic mean of several sets of data may combine into a single mean for the combined sets of data and can be computed if we know the mean and the number of observations in each group of the data. The combined arithmetic mean is denoted by Xp. 

For two sets of data, the combined mean is;

For k sets of data, the combined mean is;

Example 3.8: Find the mean for the entire group of workers for the following data.

	Group 1	Group 2
Mean wages	5000	7000
Number of workers	200	120

Solution:

Weighted Mean 

In arithmetic mean, equal weights are assigned to each number. In weighted mean, we assign different weights to each value of the data. Weights are assigned according to the importance of the value. The weighted arithmetic mean is denoted by Xw and can help in the decision where some things are more important.

Let X1, X2,..., Xn be the values of data, and w1, w2,..., wn are their respective weights. Then the weighted arithematic mean can be defined as:

Example 3.9: A student obtained 70 marks in assignment, 80 in quiz, and 65 in class test. A professor assigns 30%, 20%, and 50% weights to assignments, quizzes, and class tests, respectively. Find the average score of the student.

Solution: In this example, we want to calculate an average score that is based on different percentage values for several categories.

Example 3.10: A traffic warden is trying to work out the mean number of parking tickets he has issued per day. He produced the table below but has accidentally rubbed out some of the numbers.

Tickets per day	Frequency	No. of tickets frequency
0	1	?
1	?	1
2	10	?
3	7	?
4	?	20
5	2	?
6	?	?
Total	26	72

Find the missing numbers and calculate the mean.

Solution:

Tickets per day	Frequency	No. of tickets frequency
0	1	0
1	1	1
2	10	20
3	7	21
4	5	20
5	2	10
6	0	0
Total	26	72

• Read: Geometric Mean