Deviations Taken From the Assumed Mean: Types of Data -Embibe
  • Written By Sohan G
  • Last Modified 20-05-2022
  • Written By Sohan G
  • Last Modified 20-05-2022

Deviations Taken From the Assumed Mean: Grouped Data, Ungrouped Data, Standard Method

Deviations taken from the assumed mean: While working with data sets, we try to find a value that can represent the data. The most commonly used representation of a data set is the mean, the others being median and mode. Mean is simply the arithmetic mean of all the values in a data set.

We add all the values in the data set and divide them by the number of values. This method slightly differs when dealing with discrete data versus grouped data, although it follows the same basic principle for finding the mean. The part where we add the values becomes tedious when the values are significantly large. We can use the assumed mean method or step deviation method here to find the mean as a convenient alternative.

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Assumed Mean Method vs Standard Method

The standard method to find the mean is the simplest one that we can use. But when a data set has large values, the calculations to find its mean are tedious and time-consuming. In such cases, using the assumed mean method allows us to work with the deviations of the individual values that can be very small if the assumed mean is chosen well.

How to Choose the Assumed Mean?

You can choose any random value as the assumed mean in the assumed mean method. However, selecting an assumed value that is approximately equal to the actual mean of the data can help us reduce the large individual values of \(x_i\), to significantly smaller values of \(d_i\). A well-chosen assumed mean, one that is close to the actual mean,  will reduce the calculations and help us reach the mean quickly.
In the first example, we have this data set – \(7042,\,7058,\,6957,\,7030,\,6996,\,7033,\,7024.\)
We can observe that the mean of this data set would be close to \(7000\) and thus choose it as the assumed mean.
We have these value of \(x_i\) in the second example – \(149,\,156,\,153,\,159,\,147.\)
We can observe that the mean of this data set would be close to \(150\) and thus choose it as the assumed mean.

Finding Mean for Ungrouped Data

Following are the methods to find mean for Ungrouped Data

Standard Method

Mean of ungrouped data can be found using the formula

\({{\bar x}} = \frac{{\sum {{{x}}_{{i}}}}}{{{n}}}\)

Here,
\(\bar x =\) mean of this data
\(x_i =\) Each value of this data set
\(\sum x_i =\) Sum of all the values
\(n =\) number of values

Example:

Consider the following data set – \(7042,\,7058,\,6957,\,7030,\,6996,\,7033,\,7024\). Calculate the mean.

Solution:

Mean for this data,
\(\bar x = \frac{{\sum {x_i}}}{n}\)
\( \Rightarrow \bar x = \frac{{7042 + 7058 + 6957 + 7030 + 6996 + 7033 + 7024}}{7}\)
\( \Rightarrow \bar x = \frac{{49140}}{7}\)
\(\therefore \,\bar x = 7020\)
As the values are large in this case, it becomes time-consuming to find the mean by this method.

Assumed Mean Method

This method involves the following steps:

Step 1: Assume any value, such as the arithmetic mean of the values given. It is called the assumed mean, say \(a\), and hence, the name assumed mean method.
Step 2: Subtract \(a\) from each value \(x_i\) to get the deviations from the assumed mean called \(d_i\).
\({d_i} = {x_i} – a\)
Step 3: Find the mean of this data set using the formula,
\(\bar x = a + \frac{{\sum {d_i}}}{n}\)
Here,
\(\bar x = \) mean of this data
\(x_i =\) each value of this data set
\(\sum d_i =\) sum of all the values of \(d_i\)
\(n =\) number of values

Subtracting the assumed mean from all the values reduces the size of these values and makes the calculation easier and quicker. Let’s try applying the assumed mean method to the example that we solved previously.

Example: Consider the following data set- \(7042,\,7058,\,6957,\,7030,\,6996,\,7033,\,7024\). Calculate the mean.

Solution:

Let the assumed mean for this data be \(a = 7000\)

The deviations from the assumed mean \(d_i\) are

\(7042 – 7000\)\(7058 – 7000\)\(6957 – 7000\)\(7030 – 7000\)\(6996 – 7000\)\(7033 – 7000\)\(7024 – 7000\)
\(42\)\(58\)\(-43\)\(30\)\(-4\)\(33\)\(24\)

Mean for this data is calculated as,

\(\bar x = a + \frac{{\sum {d_i}}}{n}\)

\(\Rightarrow \bar x = 7000 + \frac{{42 + 58 – 43 + 30 – 4 + 33 + 24}}{7}\)

\(\Rightarrow \bar x = 7000 + \frac{{140}}{7}\)

\(\Rightarrow \bar x = 7000 + 20\)

\(\therefore \bar x = 7020\)

We can see that using the assumed mean method helps us make calculations quicker.

Finding Mean for Grouped Data

Following are the methods to find mean for Grouped Data

Standard Method

Mean of grouped data can be found using the formula

\(\bar x = \frac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}\)

\(\bar x = \) mean of this data
\(x_i =\) Represents each value of this data set
\(f_i =\) Represents frequency of each value of this data set
\(\sum f_i =\) sum of all the values

Example:

Consider the following data set and calculate the mean.

\(f_i\)\(3\)\(2\)\(4\)\(6\)\(5\)
\(x_i\)\(149\)\(156\)\(153\)\(159\)\(147\)

Solution: Mean for this data, \(\bar x = \frac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}\)

\(f_i\)\(x_i\)\(f_i x_i\)
\(3\)\(149\)\(3 \times 149 = 447\)
\(2\)\(156\)\(2 \times 156 = 312\)
\(4\)\(153\)\(4 \times 153 = 612\)
\(6\)\(159\)\(6 \times 159 = 954\)
\(5\)\(147\)\(5 \times 147 = 735\)
\(\sum {{{f}}_{{i}}} = 3 + 2 + 4 + 6 + 5 = 20\)\(\sum {{{f}}_{{i}}}{{{x}}_{{i}}}\)
\(= 447 + 312 + 612 + 954 + 735 = 3060\)

\( \Rightarrow \bar x = \frac{{3060}}{{20}}\)

\( \therefore \bar x = 153\)

As the values are large in this case, it becomes time-consuming to find the mean by this method.

Assumed Mean Method

This method involves the following steps:

Step 1: Assume any value as the mean of the values of \(x_i\) given. It is called the assumed mean, say \(a\). Hence, the name assumed mean method.
Step 2: Subtract \(a\) from each value \(x_i\) to get the deviations from the assumed mean, say \(d_i\).
\(d_i = x_i – a\)
Step 3: Now, we can find mean of this data set using the formula,
\(\bar x = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}\)

Here,
\(\bar x = \) mean of this data
\(x_i =\) each value of this data set
\(f_i =\) frequency of each value of this data set
\(\sum f_i =\) sum of all the values

Subtracting the assumed mean from all the values reduces the size of the values of \(x_i\) and hence, makes the calculation more straightforward and quicker.
Let’s try applying the assumed mean method to the example that we solved previously.

Example:

Consider the following data set. Calculate the mean.

\(f_i\)\(3\)\(2\)\(4\)\(6\)\(5\)
\(x_i\)\(149\)\(156\)\(153\)\(159\)\(147\)

Solution:

Let the assumed mean for this data be \(a = 150\)
Mean for this data, \(\bar x = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}\)

\(f_i\)\(x_i\)\(d_i = x_i – a = x_i – 150\)\(f_i d_i\)
\(3\)\(149\)\(149 – 150 = -1\)\(3 \times -1 = -3\)
\(2\)\(156\)\(156 – 150 = 6\)\(2 \times 6 = 12\)
\(4\)\(153\)\(153 – 150 = 3\)\(4 \times 3 = 12\)
\(6\)\(159\)\(159 – 150 = 9\)\(6 \times 9 = 54\)
\(5\)\(147\)\(147 – 150 = -3\)\(5 \times -3 = -15\)
\(\sum {{{f}}_{{i}}} = 3 + 2 + 4 + 6 + 5 = 20\)\(\sum {{{f}}_{{i}}}{{{x}}_{{i}}}\)
\(= -3 + 12 + 12 + 54 – 15 = 60\)

\( \Rightarrow \bar x = 150 + \frac{{60}}{{20}}\)

\( \Rightarrow \bar x = 150 + 3\)

\( \therefore \bar x = 153\)

As the values are large in this case, it becomes time-consuming to find the mean by this method.

Step Deviation Method

We can take the assumed mean method further by dividing all the deviations by the same number. It further diminishes the deviations and reduces the calculations even more. This method is advantageous when the deviations obtained in the assumed mean method have a common factor.

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Solved Examples – Deviations Taken From the Assumed Mean

Below are a few solved examples that can help in getting a better idea.

Q.1. Consider the following data set: \(87,\;48,\;13,\;32,\;11,\;37,\;23,\;45,\;52,\;84,\;70,\;29,\;28,\;15,\;26\). Calculate its mean by the assumed mean method.
Solution:
Let the assumed mean be \(a = 50\).
Then the deviations of \(x_i\) from \(a\) become,

\(x_i\)\(d_i = x_i – a = x_i – 50\)
\(87\)\(87 – 50 = 37\)
\(48\)\(48 – 50 = -2\)
\(13\)\(13 – 50 = -37\)
\(32\)\(32 – 50 = -18\)
\(11\)\(11 – 50 = -39\)
\(37\)\(37 – 50 = -13\)
\(23\)\(23 – 50 = -27\)
\(45\)\(45 – 50 = -5\)
\(52\)\(52 – 50 = 2\)
\(84\)\(84 – 50 = 34\)
\(70\)\(70 – 50 = 20\)
\(29\)\(29 – 50 = -21\)
\(28\)\(28 – 50 = -22\)
\(15\)\(15 – 50 = -35\)
\(26\)\(26 – 50 = -24\)

Sum of the deviations,
\(\sum {d_i} = 37 – 2 – 37 – 18 – 39 – 13 – 27 – 5 + 2 + 34 + 20 – 21 – 35 – 24\)
\(\therefore \,\sum {d_{{i}}} = – 150\)
Mean of this data,
\({{\bar x}} = {{a}} + \frac{{\sum {{{d}}_{{i}}}}}{{{n}}}\)
\( \Rightarrow \bar x = 50 + \frac{{ – 150}}{{15}}\)
\(\Rightarrow \bar x = 50 + \left( { – 10} \right)\)
\(\therefore \,\bar x = 40\)

Q.2. Consider the following data set: \(10,\,8,\,p,\,7,\,6\). It is known that its mean is \(10\). Find the values of \(p\) and \(q\).
Solution:
Let the assumed mean be \(a = 8\).
Then the deviations of \(x_i\) from \(a\) become,

\(x_i\)\(d_i = x_i – a = x_i – 8\)
\(10\)\(10 – 8 = 2\)
\(8\)\(8 – 8 = 0\)
\(p\)\(p- 8\)
\(7\)\(7 – 8 = -1\)
\(6\)\(6 – 8 = -2\)

\(\sum {{{d}}_{{i}}} = 2 + 0 + \left( {{{p}} – 8} \right) – 1 – 2\)
\(\therefore \sum {{{d}}_{{i}}} = {{p}} – 9\)
Mean of this data,
\(\bar x = a + \frac{{\sum {d_i}}}{n}\)
\( \Rightarrow 10 = 8 + \frac{{{{p}} – 9}}{5}\)
\( \Rightarrow 10 = \frac{{\left( {8 \times 5} \right) + {{p}} – 9}}{5}\)
\( \Rightarrow 10 \times 5 = 40 + p – 9\)
\( \Rightarrow 50 = 31 + p\)
\( \Rightarrow p = 50 – 31\)
\( \therefore p = 19\).

Alternate Solution:
Remember: The sum of the deviations from the actual mean is zero.
Deviations from actual mean, \(\bar x = 10\)

\(x_i\)\(x_i – \bar x = x_i – 10\)
\(10\)\(10 – 10 = 0\)
\(8\)\(8 – 10 = -2\)
\(p\)\(p – 10\)
\(7\)\(7 – 10 = -3\)
\(6\)\(6 – 10 = -4\)

Sum of the derivations from the actual mean \(= 0\)
\( \Rightarrow 0 – 2 + \left( {{\text{p}} – 10} \right) – 3 – 4 = 0\)
\( \Rightarrow p – 19 = 0\)
\( \therefore = 19\)

Q.3. Calculate the mean of the following frequency distribution by using the assumed mean method.

Frequency\(2\)\(4\)\(5\)\(3\)\(2\)\(7\)\(1\)\(4\)\(5\)\(7\)
Size\(300\)\(290\)\(280\)\(270\)\(260\)\(250\)\(240\)\(230\)\(220\)\(210\)

Solution:
Let the assumed mean for this data be \(a = 250\)
Mean for this data \(\bar x = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}\)

\(f_i\)\(x_i\)\(d_i = x_i – a = x_i – 250\)\(f_i d_i\)
\(2\)\(300\)\( 300 – 250 = 50\)\(2 \times 50 = 100\)
\(4\)\(290\)\( 290 – 250 = 40\)\(4 \times 40 = 160\)
\(5\)\(280\)\( 280 – 250 = 30\)\(5 \times 30 = 150\)
\(3\)\(270\)\( 270 – 250 = 20\)\(3 \times 20 = 60\)
\(2\)\(260\)\( 260 – 250 = 10\)\(2 \times 10 = 20\)
\(6\)\(250\)\( 250 – 250 = 0\)\(6 \times 0 = 0\)
\(2\)\(240\)\( 240 – 250 = -10\)\(2 \times -10 = -20\)
\(4\)\(230\)\( 230 – 250 = -20\)\(4 \times -20 = -80\)
\(5\)\(220\)\( 220 – 250 = -30\)\(5 \times -30 = -150\)
\(7\)\(210\)\( 210 – 250 = -40\)\(7 \times -40 = -280\)

\(\sum {{{f}}_{{i}}} = 2 + 4 + 5 + 3 + 2 + 6 + 2 + 4 + 5 + 7 = 40\)
\(\sum {{{f}}{{i}}}{{{d}}{{i}}} = 100 + 160 + 150 + 60 + 20 + 0 – 20 – 80 – 150 – 280 = – 40\)
\( \Rightarrow \bar x = 250 + \frac{{ – 40}}{{40}}\)
\( \Rightarrow \bar x = 250 + \left( { – 1} \right)\)
\(\therefore \,\bar x = 249\).

Q.4. Marks obtained by students of standard 8th in the annual exam were as follows. Find the average marks scored by a student in this class.

Marks\(0 – 10\)\(10 – 20\)\(20 – 30\)\(30 – 40\)\(40 – 50\)
Number of Students\(8\)\(15\)\(21\)\(33\)\(23\)

Solution:
In this problem we need to find the class mark.
For each class,
\({\text{Class}\;{\text{mark}}} = \frac{{{\text{Lower}\;{\text{limit}}} + {\text{Upper}\;{\text{limit}}}}}{2}\)
Let the assumed mean for this data be \(a = 30\)

ClassClass Mark, \(x_i\)\(f_i\)\(d_i = x_i – a = x_i – 30\)\(f_i d_i\)
\(0 – 10\)\(5\)\(8\)\(5 – 30 = -25\)\(8 \times -25 = -400\)
\(10 – 20\)\(15\)\(15\)\(15 – 30 = -15\)\(15 \times -15 = -225\)
\(20 – 30\)\(25\)\(21\)\(25 – 30 = -9\)\(21 \times -5 = -105\)
\(30 – 40\)\(35\)\(33\)\(35 – 30 = 5\)\(33 \times 5 = -165\)
\(40 – 50\)\(45\)\(23\)\(45 – 30 = 15\)\(23 \times 15 = -345\)

\(\sum {f_i} = 8 + 15 + 21 + 33 + 23 = 100\)
\(\sum {f_i}{d_i} = – 400 – 225 – 105 + 165 + 345 = – 220\)
\( \Rightarrow \bar x = 30 + \frac{{ – 220}}{{100}}\)
\( \Rightarrow \bar x = 30 + \left( { – 2.2} \right)\)
\(\therefore \,\bar x = 27.8\)

Q.5. Sort the following data into groups of width \(10\) and then find the mean.
\(63,\;78,\;76,\;54,\;61,\;77,\;95,\;75,\;78,\;57,\;82,\;85,\;79,\;50,\;66,\;91,\;100,\;90,\;74,\;71,\;87,\;85,\;91,\;73,\;70\)
Solution:
The data given in the problem varies from \(50\) to \(100\). We can group them in classes of width \(10\) as shown below,
For each class,
\({\text{Class}\;{\text{mark}}},{{{x}}_{{i}}} = \frac{{{\text{Lower}\;{\text{limit}}} + {\text{Upper}\;{\text{limit}}}}}{2}\)
Let the assumed mean for this data be \(a = 75\)
Deviation from the assumed mean, \(d_i = x_i – 75\)

ClassClass Mark, \(x_i\)Values to be includedFrequency, \(f_i\)\(d_i\)\(\sum f_i d_i\)
\(50 – 60\)\(55\)\(54,\,57,\,50\)\(3\)\(-20\)\(3 \times -20 = -60\)
\(60 – 70\)\(65\)\(63,\,61,\,66\)\(3\)\(-10\)\(3 \times -10 = -30\)
\(70 – 80\)\(75\)\(78,\,76,\,77,\,75,\,78,\)
\(79,\,74,\,71,\,73,\,70\)
\(10\)\(0\)\(9 \times 0 = 0\)
\(80 – 90\)\(85\)\(82,\,85,\,87,\,85\)\(4\)\(10\)\(4 \times 10 = 40\)
\(90 – 100\)\(95\)\(95,\,91,\,100,\,90,\,91\)\(5\)\(20\)\(5 \times 20 = 100\)

\(\sum {{{f}}_{{i}}} = 3 + 3 + 10 + 4 + 5 = 25\)
\(\sum {{{f}}{{i}}}{{{d}}{{i}}} = – 60 – 30 + 0 + 40 + 100 = 50\)
\( \Rightarrow \bar x = 75 + \frac{{50}}{{25}}\)
\( \Rightarrow \bar x = 75 + 2\)
\(\therefore \,\bar x = 79\)

Summary

The calculation of the mean involves dividing the sum of the values in the given data by the number of values. The standard method to find the mean is the simplest one that we can use. But when a data set has large values, the calculations to find its mean are tedious and time-consuming. Hence, the assumed mean method comes into play. The assumed mean method is a modification of the standard formula to find the mean. In this method, we assume a mean value approximately equal to the actual mean and subtract it from the given values, \(x_i\), to get the deviations from the assumed mean, \(d_i\). The formula \(\bar x = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}\) is used to find the mean. The closer the assumed mean is to the actual mean, the easier the calculation is.

Frequently Asked Questions (FAQs)

Students might be having many questions regarding the Deviations Taken From the Assumed Mean. Here are a few commonly asked questions and answers.

Q.1. How do you calculate deviation from the assumed mean?
Ans:
Consider any random value as the assumed mean. However, it is suggested that selecting a value close to the actual mean makes calculations easier. Now, subtract each value, \(x_i\), from the assumed mean \(a\) to get the deviation from the assumed mean,\(d_i\).
\(d_i = x_i – a\).

Q.2. What is the effect of change in origin on the mean?
Ans:
When we add or subtract any constant from all the values of the original data, the mean of the data also changes. Hence, in the formula for mean by the assumed mean method we add \(a\) to compensate for the change in origin caused by subtraction of \(a\) from all values.

Q.3. What is the difference between actual mean and assumed mean?
Ans:
We select a random value from which we find deviations of the given value. This arbitrary value is called the assumed mean. A value that is the average or arithmetic mean of the given values is called the actual mean.

Q.4. Where are the assumed mean method and step deviation method used?
Ans:
We use the assumed mean and step deviation methods, where the values in the given data are so large that calculations become time-consuming.

Q.5. What is the sum of deviations taken from mean in a series?
Ans:
The sum of the deviations taken from the mean in a series is zero.

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