Measure of Central Tendency

In this post, we will discuss about properties of Arithmetic mean with Examples.

Arithmetic Mean

The arithmetic mean, often simply referred to as the mean, is a statistical measure that represents the central value of a dataset. The arithmetic mean is calculated by summing all the values in the dataset and then dividing by the total number of observations in the data.

The Sum of Deviations From the Mean is Zero

Property 1: The sum of deviations taken from the mean is always equal to zero. Mathematically $\sum\limits_{i=1}^n (x_i-\overline{x}) = 0$

Consider the ungrouped data case.

Obs. No.	$X$	$X_i-\overline{X}$
1	81	-19
2	100	0
3	96	-4
4	108	8
5	90	-10
6	102	2
7	104	4
8	103	3
9	100	0
10	109	9
11	91	-9
12	116	16
Total	$\sum X_i = 1200$	$\sum\limits_{i=1}^n (X_i-\overline{X})=0$

For grouped data $\overline{X} = \sum\limits_{i=1}^k f_i(X_i -\overline{X}) =0$, where for grouped data $\overline{X} =\frac{\sum\limits_{i=1}^n M_i f_i}{\sum\limits_{i=1}^n f_i}$. Suppose, we have the following grouped data

Classes	$f$	$M$	$fM$	$f_i(M_i – \overline{X})$
65 – 85	9	75	675	$9\times (75 – 123) = -432$
85 – 105	10	95	950	$10\times (95 – 123) = -280$
105 – 125	17	115	1955	$17\times (115 – 123) = -136$
125 – 145	10	135	1350	$10\times (135 – 123) = 120$
145 – 165	5	155	775	$5\times (155 – 123) = 160$
165 – 185	4	175	700	$4\times (175 – 123) = 208$
185 – 205	5	195	975	$5\times (195 – 123) = 360$
Total	$\Sigma f = n = 60$		$\Sigma fM = 7380$	$\sum\limits_{i=1}^k f_i(X_i -\overline{X}) =0$

Mean = $\overline{X} = \frac{\Sigma fM}{\Sigma f} = \frac{7380}{60} = 123$ .

The Combined Mean of Different Data Sets

Property 2: If there are different sets of data say $k$ then the combined mean/ average is

\begin{align*}
\overline{X}_c &= \frac{n_1 \overline{x}_1 + n_2\overline{x}_2 +\cdots + n_k \overline{x}_k }{n_1+n_2\cdots + n_k}\\
&=\frac{\Sigma x_1 + \Sigma x_2 + \cdots + \Sigma x_k}{n_1+n_2+\cdots + n_k}
\end{align*}

Suppose, we have data of $k$ groups.

Obs. No.	$X_1$	$X_2$	$X_3$	$X_4$	$X_5$
1	81	40	92	107	113
2	100	30	95	110	94
3	96	22	99	114	93
4	108	51	94	109	119
5	90		101	116	105
6	102		103	118
7	104		100	115
8	103		102
9	100		101
10	109
11	91
12	116
Sum	1200	143	887	789	524

For \begin{align*}
\overline{X}_1 &= \frac{\sum\limits_{i=1}^n X_1}{n_1} = \frac{1200}{12} = 100\\
\overline{X}_2 &= \frac{\sum\limits_{i=1}^n X_2}{n_2} = \frac{143}{4} = 35.8\\
\overline{X}_3 &= \frac{\sum\limits_{i=1}^n X_3}{n_3} = \frac{887}{9} = 98.6\\
\overline{X}_4 &= \frac{\sum\limits_{i=1}^n X_4}{n_4} = \frac{789}{7} = 112.7\\
\overline{X}_5 &= \frac{\sum\limits_{i=1}^n X_5}{n_5} = \frac{524}{5} = 104.8\\
\overline{X}_c &= \frac{n_1\overline{X}_1 + n_2 \overline{X}_2 + \cdots + n_5 \overline{X}_5}{n_1+n_2+n_3+n_4+n_5}\\
&=\frac{12\times 100 + 4\times 35.8 + 9\times 98.6 + 7\times 112.7 + 5\times 104.8}{12+4+9+7+5} =\frac{3543.5}{37} = 95.7703
\end{align*}

For combined mean, not all the data set needs to be ungrouped or grouped. It may be possible that some data sets are ungrouped and some data sets are grouped.

Sum Squared Deviations from the Mean are Always Minimum

Property 3: The sum of the squared deviations of the observations from the arithmetic mean is minimum, which is less than the sum of the squared deviations of the observations from any other values. In other words, the sum of squared deviations from the mean is less than the sum of squared deviations from any other value. Mathematically,

For Ungrouped Data: $\Sigma (X_i – \overline{X})^2 < \Sigma (X_i – A)^2$

For Grouped Data: $\Sigma f(X_i – \overline{x})^2 < \Sigma f(M_i – A)^2$

where $A$ is any arbitrary value, also known as provisional mean. For this condition, $A$ is not equal to the arithmetic mean.

Note that the difference between the sum of deviations and the sum of squared deviations is that in the sum of deviations we take the difference (subtract) of each observation from the mean and then sum all the differences. In the sum of squared deviations, we take the difference of each observation from the mean, then take the square of all the differences, and then sum all the resultant values at the end.

From the above calculations, it can observed that $\Sigma (X_i – \overline{X})^2 < \Sigma (X_i – 90)^2 < \Sigma (X_i – 99)^2$.

No Resistant to Outliers

Property 4: The arithmetic mean is not resistant to outliers. It means that the arithmetic mean can be misleading if there are extreme values in the data.

Arithmetic Mean is Sensitive to Outliers

Property 5: The arithmetic mean is sensitive to extreme values (outliers) in the data. If there are a few very large or very small values, they can significantly influence the mean.

The Affect of Change in Scale and Origin

Property 6: If a constant value is added or subtracted from each data point, the mean will be changed by the same amount.
Similarly, if a constant value is multiplied or divided by each data point, the mean will be multiplied or divided by the same constant.

Unique Value

Property 7: For any given dataset, there is only one unique arithmetic mean.

In summary, the arithmetic mean is a widely used statistical measure (a measure of central tendency) that provides a central value for a dataset.
However, it is important to be aware of the properties of arithmetic mean and its limitations, especially when dealing with data containing outliers.

FAQs about Arithmetic Mean Properties

1. Explain how the sum of deviation from the mean is zero.
2. What is meant by unique arithmetic mean for a data set?
3. What is arithmetic mean?
4. How combined mean of different data sets can be computed, explain.
5. Elaborate Sum of Squared Deviation is minimum?
6. What is the impact of outliers on arithmetic mean?
7. How does a change of scale and origin change the arithmetic mean?

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Introduction to Median of Ungrouped Data

The post is about calculating the median ungrouped data. The median is the most central point (middlemost central value) of the data/set of observations, with the condition that the data or set of observations should be arranged in ascending or descending order. The median divides the data into two equal parts. That is the main objective of the median.

It is important to note that the criteria for finding the median for grouped and ungrouped data are different.

The primary and secondary data can be defined as:

1. Primary data, also called raw or ungrouped data, does not undergo any statistical procedure/method, which is not in the form of frequency distribution.
2. Secondary data may also be called group data if it is in the form of frequency distribution.

Let us discuss how to find the median for ungrouped data.

There are two cases for ungrouped data. These cases are based on no of observations which is $n$

When the number of observations is odd (Say $n$ i.e. $n$ is odd), and when the number of observations is even (Say $n$ i.e. $n$ is even).

Median Calculations

The data below contains the odd number of observations.

Observation No. (Ascending Order)	1st	2nd	3rd	4th	5th	6th	7th	8th	9th	10th	11th
Data Values	81	89	90	96	100	102	103	104	108	109	118
(Descending Order)	11	10	9	8	7	6	5	4	3	2	1

Since the number of observations is odd ($n = 11$), the central value after arranging in ascending order will be the 6^th value. and the 6^th value is 102. That is the median is 102 for the above data.

The position of the median can be located mathematically, as follows:

\begin{align*}
\tilde{x} &= \left( \frac{n+1}{2} \right)th\,\, \text{value}\\
&=\frac{11+1}{2} = 6th\,\, \text{value}
\end{align*}

The value at the 6th position (from sorted data) is 102. The $\tilde{x}$ can be read as “x-tild” which is the notation of the median.

Median for Even Numbers of Observations

Consider the following data that contains an even number of observations.

Observation No.	1	2	3	4	5	6	7	8	9	10
Data Values	81	100	96	108	90	102	104	103	109	89

Data after sorting (either in ascending or descending order) is

Observations No.	1st	2nd	3rd	4th	5th	6th	7th	8th	9th	10th
x	81	89	90	96	100	102	103	104	108	109

Since $n=10$ which is even, the central position (that is median) lies between the 5th value and the 6th value. This central value is the average of the 5th and 6th values (from the sorted data). The average of these two central observations is called the median. The two central positions are 100 and 102, take the average of these two numbers and find the median.

$$Median = \frac{100+102}{2} = 101$$

Median Formula for Large Data Sets

The median formula for large or small data sets can be represented mathematically.

• For large data sets one can find the median of data mathematically. The formula for both odd number of observations and even numbers of observations is different.

The point to remember when computing the median is that

• For an odd number of observations, the median is the centermost value after sorting the data
• For an even number of observations, the median is the average of two central values after sorting the data

\begin{align*}
\tilde{x} &= \frac{1}{2} \left[ \left(\frac{n}{2}th \, \, value \right)+ \left(\frac{n}{2}+1 \right)the \,\, value \right]\quad \quad \text{(When observations are even)}\\
&= \frac{n+1}{2} \quad \quad \text{(when observations are odd)}
\end{align*}

The other way of the median formula is

Consider, a data set containing 157 observations. To compute the median, first of all, you need to sort the data in either ascending or descending order. The formula for this data will be

$$\tilde{x} = \frac{n+1}{2} = \frac{157+1}{2}=79th$$.

The 79th observation in the sorted data will be the median of the data.

In case, if there are even number of observations (say $n=396$, the median will be

\begin{align*}
\tilde{x} &= \frac{1}{2}\left[\left(\frac{n}{2}\right)th + \left(\frac{n+1}{2}\right)th \right]\\
&=\frac{1}{2} \left[\frac{396}{2}th + \frac{396}{2}+1 \right]\\
&= \frac{1}{2} \left[198th + 199th\right]
\end{align*}

The average of 198th value and 199th value from the sorted data will be the median of the data.

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Introduction to Geometric Mean

The geometric mean (GM) is a way of calculating an average, but instead of adding values like the regular (arithmetic) mean, it multiplies them and then takes a root. The geometric mean is defined as the $n$th root of the product of $n$ positive values.

If we have two observations let’s say 9 and 4, then the geometric mean is the square root of the product of these values, which is 6 ($\sqrt{9\times 4}=6$. If there are three values let’s say  3, 9, and 3 then the geometric average will be the $sqrt[3]{3\times 9 \times 3} = 3$. In a similar pattern, mathematically, for $n$ number of observations ($x_1, x_2, \cdots, x_n$) then the Geometric Average Formula will be

$$GM = (x_1 \times x_2 \times x_3 \times \cdots \times x_n)^{\frac{1}{n} }$$

Geometric Mean Example

Suppose we have the following set of values $x=32, 36, 36, 37, 39, 41, 45, 46, 48$. The Computation of Geometric Mean will be

\begin{align*}
GM &= (32\times 36 \times 36 \tmies 37 \times 39 \times 41 \times 45 \times 46 \times 48)^{\frac{1}{9}}\\
&=(243790484520960)^{\frac{1}{9}} = 39.7
\end{align*}

For a large number of observations one can compute the GM by taking the log of all observations using the following formula:

$$GM = antilog \left[\frac{\sum\limits_{i=1}^n log\, x}{n} \right]$$

$x$	$log\, x$
32	Log 32 = 1.5051
36	log 36 = 1.5563
36	log 36 = 1.5563
37	log 37 = 1.5682
39	log 39 = 1.5911
41	log 41 = 1.6128
45	log 45 = 1.6532
46	log 46 = 1.6628
48	log 48 = 1.6812
Total	14.3870

\begin{align*}
GM &= antilog \left[ \frac{\sum\limits_{i=1}^n log\, x}{n} \right]\\
&= antilog \left[\frac{14.3870}{9}\right] = antilog [1.5986]\\
&= 38.7
\end{align*}

One important point that should be remembered is that if any value in the data set is zero or negative then the GM cannot be computed.

Geometric Mean for Grouped Data

The GM for grouped data can also be computed using the following formula:

$$GM = antilog \left[ \frac{\Sigma f\times log\, x}{\Sigma f} \right]$$

Suppose, we have the following frequency distribution as follows:

Classes	Frequency
65 to 84	9
85 to 104	10
105 to 124	17
125 to 144	10
145 to 164	5
165 to 184	4
185 to 204	5
Tota	60

The GM of the above frequency distribution can be performed as follows

\begin{align*}
GM &= antilog \left[ \frac{124.2471}{60} \right]\\
&=antilog (2.0708) = 117.4
\end{align*}

The GM is particularly useful when dealing with rates of change or ratios, such as growth rates in investments. That is because geometric mean considers how things are multiplied over time, rather than simply added.

Use and Application of Geometric Mean

Geometric Mean is useful in situations like:

• Investment returns: When one looks at average investment growth, one wants to consider how much one’s money is multiplied over time, not just the change each year. That is why the GM is better suited for this scenario.
• Rates of change: Similar to investment returns, if something is increasing or decreasing by a percentage each time, the GM is a more accurate measure of the overall change.
• Growth Rates: When dealing with percentages or ratios that change over time (like investment returns or population growth), the geometric mean provides a more accurate picture of the overall change compared to the arithmetic mean.
• Proportional Changes: It is helpful for situations where changes are multiplied, not added. For example, if a recipe calls for doubling all ingredients, the geometric mean of the original quantities represents the final amount.

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Introduction to Weighted Averages

The multipliers or sets of numbers that express more or less relative importance of various observations (data points) in a data set are called weights.

The weighted arithmetic mean (simply called weighted average or weighted mean) is similar to an ordinary arithmetic mean except that instead of each data point contributing equally to the final average, some data points contribute more than others. Weighted means are useful in a wide variety of scenarios. Weighted averages are used when there are a bunch of values, but some of those values are more important or contribute more to the overall result.

Example of Weighted Average

For example, a student may use a weighted mean to calculate his/her percentage grade in a course. In such an example, the student would multiply the weight of all assessment items in the course (e.g., assignments, exams, sessionals, quizzes, projects, etc.) by the respective grade that was obtained in each of the categories.

As an example, suppose in a course there are a total of 60 marks, while the distribution of marks is as follows, Assignment-1 has a weightage of 10%, Assignment-2 has a weightage of 10%, the mid-term examination has a weightage of 30% and the final term examination have the weightage of 50%. The scenario is described in the table below:

Assessment Item	Weight ($w_i$)	Grades ($x_i$)	Marks	Weighted Marks ($w_ix_i$)
Midterm	10 %	70 %	6	7 %
Assignment # 2	10 %	65 %	6	6.5 %
Midterm Examination	30 %	70 %	12	21 %
Final Term Examination	50 %	85 %	30	42.5 %
	100 %	290 %	60	77 %

Weighted Average Formula

Mathematically, the weighted average forma is given as

$$\overline{x}_w = \frac{\sum\limits_{i=1}^n w_i x_i}{\sum\limits_{i=1}^n w_i}$$

Another Example

Consider another example: Suppose we have monthly expenditures of a family on different items with their quantity

Items	Weights ($w_i$)	Expenses ($x_i$)	Weighted Expenses $w_ix_i$
Food	7.5	290	2175
Rent	2.0	54	108
Clothing	1.5	96	144
Fuel and light	1.0	75	75
Misc	0.5	75	37.5
Total	12.5	590	2539.5

The average expenses will be: $AM = \frac{590}{5} = 118$.

However, the weighted average of the scenario will be $\overline{x}_w = \frac{\sum\limits_{i=1}^n w_i x_i}{\sum\limits_{i=1}^n w_i} = \frac{2539.5}{12.5}=203.16$

Keeping in mind the importance of weight, the average monthly expenses of a family was 203.16, not 118.

Note that in a frequency distribution, the computation of relative frequency (rf) is also related to the concept of weighted averages.

Classes	Frequency	Mid point ($X$)	rf	Percentage
65-84	9	74.5	$\frac{9}{60} = 0.15$	15
85-104	10	94.5	$\frac{10}{60} = 0.17$	17
105-124	17	114.5	$\frac{10}{60} = 0.28$	28
125-144	10	134.5	$\frac{10}{60} = 0.17$	17
145-164	5	154.5	$\frac{5}{60} = 0.08$	8
165-184	4	174.5	$\frac{4}{60} =0.07$	7
185-204	5	194.5	$\frac{5}{60} =0.08$	8
Total	60

Some Real-World Examples of Weighted Averages

• Calculating class grade: Different assignments might have different weights (e.g., exams worth more than quizzes). A weighted mean considers these weights to determine the overall grade.
• Stock market performance: A stock index might use a weighted average to reflect the influence of large companies compared to smaller ones.
• Customer Satisfaction: Finding the average customer satisfaction score when some customers’ feedback might hold more weight (e.g., frequent buyers).
• Average Customer Spending: if some customers buy more frequently.
• Expected Value: Determining the expected value of outcomes with different probabilities.

The following are some important questions. What is the importance of weighted mean? Describe its advantages and disadvantages. What is an average? What are the qualities of a good average? What does Arithmetic mean? Describe the advantages and disadvantages of Arithmetic mean. In which situations do we apply arithmetic mean?

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