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Mean deviation for continuous frequency distribution is one of the most used methods of data representation in statistical math. It specifies how required data is away from the centre of the data set.
- Mean Deviation is a measurement of the average deviation from the mean value of the data set.
- It is calculated with respect to the central tendency of sample space.
- The deviation is defined as the difference between the observed value and the expected value of data.
- Data representation makes it easy to know the frequency of data.
There are many data representation methods, such as tabular, graphical, and so on.
- The central value can be mean, median or mode, depending upon the type of data set.
- Biologists use it to differentiate between the weight of animals.
- Mathematically, it can be represented as:
[Σ |X – µ|]/N
Where
- Σ : addition of data values
- X : each value in the sample space
- µ : mean of the sample space
- N : number of data values
Key Terms: Mean deviation for Continuous Frequency Distribution, Data, Mean Deviation, Continuous Frequency Distribution, Step Deviation Method, Standard Deviation, Mean
Mean Deviation for Grouped Data
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Mean deviation is a data representation method used to calculate the deviation for the given data set. Students often confuse mean deviation with standard deviation.
- Frequency is the occurrence number of an observation within a specific interval.
- In continuous frequency distribution, no gaps are left between the classes.
- Moreover, each class has a respective frequency in the table.
In other words, the class intervals should be arranged so that the frequency is continuously distributed with no gaps or skipping of numbers in between.
- Therefore, the class intervals must be mutually exclusive and exhaustive.
- The marking of the first class is marked by the beginning of the next class.
Mean Deviation for Grouped DataExample: Below-provided example will help you to understand the concept of frequency distribution of continuous type.
This table represents the age group of children living in a locality. As it can be observed, the representation is continuously going on.
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Mean deviation for Grouped data
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Steps to Calculate Mean Deviation of Continuous Frequency Distribution
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Certain steps need to be followed in order to calculate the mean deviation of continuous frequency distribution. These steps are as follows:
Step 1
As we have to calculate mean deviation and mean literally means the mid-point. Therefore, we simply assume that the frequency is centered at the midpoint in each class,as our first step.
- Therefore, we are calculating the mean of these mid-points at which the frequency is focused.
- Continuing with the above-mentioned table the mid points are as follow:
| Age Group | Mid-points | Number of people |
|---|---|---|
| 2-4 | 3 | 54 |
| 4-6 | 5 | 34 |
| 6-8 | 7 | 30 |
| 8-10 | 9 | 25 |
Table 2
- Now, the mean would be calculated with the help of a formula. Formula for mean deviation continuous frequency distribution is-
Mean Deviation for Continuous Frequency Distribution
Step 2
Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (I)
Step 3
The mean deviation formula for continuous frequency distribution is as follows:
Mean Deviation for Continuous Frequency Distribution
These formulas can be used to find the mean deviation continuous frequency distribution.
Read More:
| Class 12 Mathematics Related Concept | ||
|---|---|---|
| Statistical Inference | Sampling Error Formula | Mode of Grouped Data |
| Equally Likely Events | Frequency Distribution Table | Interquartile range |
Mean Deviation for Continuous Frequency Distribution Formula
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The step deviation method is essentially useful to calculate the mean deviation for class intervals without a hussle and in a lesser time. The mean value or mid-value is assumed in this method.
Mean Deviation for Continuous Frequency Distribution
where
- A is the assumed mean in this formula
- h is the common factor
- d which means derivation which is equal to (xi-a)/h.
Example of Mean Deviation for Continuous Frequency Distribution FormulaExample: Determine the mean deviation for the data values 7, 2, 8, 9, 4, 6. Solution: Given data values are 7, 2, 8, 9, 4, 6. We know that the procedure to calculate the mean deviation. First, find the mean for the given data: Mean, µ = ( 7+2+8+9+4+6)/6 µ = 36/6 µ = 6 Therefore, the mean value is 6. Now, subtract each mean from the data value, and ignore the minus symbol if any
Now, the obtained data set is 1, 4, 2, 3, 2, 0. Finally, find the mean value for the obtained data set Therefore, the mean deviation is = (1+4 + 2+ 3+ 2+0) /6 = 12/6 = 2 Hence, the mean deviation for 7, 2, 8, 9, 4, 6 is 2. |
Things to Remember
- Mean deviation for Continuous Frequency Distribution is helpful for accurate and continuous quantitative data.
- It is not the same as standard deviation as they both are different from each other.
- A continuous frequency distribution series is when there is zero gap between the class intervals.
- The mean is used to measure central tendency.
- The Step Deviation method can save you from the complicated process of finding mean deviation.
- It is used in various business and economic activities.
- The value fluctuates the least number of times.
Solved Questions
Ques. Find the mean deviation from the mean for the following continuous frequency distribution series. (5 marks)
| Sales in Rs. Thousand | Number of companies |
|---|---|
| 40-50 | 5 |
| 50-60 | 15 |
| 60-70 | 25 |
| 70-80 | 30 |
| 80-90 | 20 |
| 90-100 | 5 |
Ans. Class intervals i.e., Sales are already provided. Frequency is also provided in the question. Now,we would write the mid-points or mean values as our first step.
| Sales in Rs. Thousand | Mean Sales | Number of companies |
|---|---|---|
| 40-50 | 45 | 5 |
| 50-60 | 55 | 15 |
| 60-70 | 65 | 25 |
| 70-80 | 75 | 30 |
| 80-90 | 85 | 20 |
| 90-100 | 95 | 5 |
After finding these mid-points which are nothing else than the mid value of each class interval, we will apply the formula to find the mean deviation now.
Mean = ∑fixi/∑fi = 7100/700 = 71
Formula for Deviation from mean =
∑fi\(\mid\)xi − x- \(\mid\) / ∑fi
Putting the values into the formula...
∑fi\(\mid\)xi − x-\(\mid\) / ∑fi =
1040/100 = 10.4
Therefore, mean deviation is 10.4
Ques. Find the mean deviation about the mean for the following data: 39, 72, 48, 41, 43, 55, 60, 45, 43. (3 marks)
Ans. To find the mean deviation in this question, firstly we need to find out the mean by applying the mean formula i.e., adding all the values and then dividing this sum with the number of values.
Sum= 39+72+48+41+43+55+60+45+43 = 446
No. of values = 9
446/9 = 49.5
Applying Mean deviation formula
∑|x+xi|/N
= 10.5+22.5+1.5+8.5+6.5+5.5+10.5+4.5+6.5 divided by 9
= 76.5 divided by 9
= 8.5
Thus, mean deviation is 8.5
Ques. The mean of the data, in a continuous frequency distribution is given as 21. If each and every observation of the data is increased by a value of 5, then find the new mean. (2 marks)
Ans. Each data is increased by 5 which is an equal amount. Therefore, the mean will also be increased by 5.
Hence, new mean= 21 + 5 = 26
Ques. Calculate the mean deviation about the mean for the following data. (5 marks)
| Class interval (C.I.) | Frequency (f) |
|---|---|
| 2-4 | 3 |
| 4-6 | 4 |
| 6-8 | 2 |
| 8-10 | 1 |
Ans. In this question, mean deviation needs to be calculated. As already known, Mean deviation indicates how far the values are from the middle value. One can calculate the mean deviation in any given numerical data set. But, first we need to calculate the class midpoints i.e., mean to get the term fixi.
Then, we will be applying the formula of mean first in order to calculate the mean deviation. After this, we will calculate the mean deviation about the mean.
The formulae of mean is given by,
Mean(X) = ∑fixi/∑fi
Mean deviation is given by the formula: ∑fi|xi−X|N
Where xi = class midpoint
X = Mean of the grouped data
| Class Interval | Mid-points | Frequency | fixi |
|---|---|---|---|
| 2-4 | 3 | 3 | 9 |
| 4-6 | 5 | 4 | 20 |
| 6-8 | 7 | 2 | 14 |
| 8-10 | 9 | 1 | 9 |
| ∑fi= 10 | ∑fixi= 52 |
Now, we will calculate mean,
Mean X= ∑fixi/∑fi
=52/10
=5.2
One more column for mean deviation i.e. |xi−X| will also be added in the same table.
Let us calculate Deviation |xi−X|
and fi|xi−X| for the first class interval 2-4.
|xi−X|
= |9−5.2|
= 2.2
fi|xi−X| = 3 × 2.2
The remaining class intervals can also be calculated and added to the table likewise.
Now we know the formula of Mean Deviation = ∑fi|xi−X|N
Where, N = total number of values = ∑fi=10
Therefore,
Mean Deviation= 14.8/10
= 1.48
Hence, Mean Deviation for the given data about Mean = 1.48
Ques. The mean deviation of the numbers 3, 4, 5, 6, 7 from mean is (3 marks)
a) 25
b) 5
c) 1.2
d) 9
Ans. C) 1.2
Explanation: We can calculate mean by formula, mean = ∑xin , where, xi is the data given and n
is the number of terms. The mean deviation about the mean is calculated by ∑|xi−M|n. Now, put the value of mean in the formula and mean deviation about mean will be there.
After following these steps, mean deviation about mean is 1.2
Hence, the correct option is C.
Note: Don't forget to double check the values while solving these types of questions.
Ques. Mean deviation of 6, 8, 12, 15, 10, 9 through mean is. (2 marks)
a) 10
b) 2.33
c) 2
d) None of these
Ans. b) 2.33
Explanation: Find out the mean first and then go for mean deviation about the mean. The formulas for calculating mean and mean deviation are provided above. You would have to use the same method that was used in question 5.
After following the steps, the mean deviation of given observations is 2.33.
Ques. Find the mean deviation about the mean for the data 4,7,8,9,10,12,13,17. (3 marks)
Ans. Step 1: Firstly, we will find the value of ∑xi which can be calculated by adding all the data given. Therefore,
∑xi = 4+7+8+9+10+12+13+17
∑xi = 80
Step 2: We will find the value of the mean with the help of the formula.
N = 8
X = 80/8
X = 10
Therefore, the mean is 10.
Step 3: Now, we will use the mean deviation formula to find the deviation about mean.
Step 4: The mean deviation about the mean is 3.
Ques. Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44. (3 marks)
Ans. Step 1: Firstly, we will find the value of ∑xi which can be calculated by adding all the data given.
Step 2: Then, we will find the value of the mean with the help of the formula to calculate the mean.
Step 3: Now, we will use the mean deviation formula to find the deviation about mean.
Step 4: Therefore, mean deviation would be 8.4
Ques. Find the mean deviation about the mean for the following data: 30, 72, 48, 40, 43, 50, 60, 45, 42. (3 marks)
Ans. To find the mean deviation in this question, firstly we need to find out the mean by applying the mean formula i.e., adding all the values and then dividing this sum with the number of values.
Sum= 30+72+48+40+43+50+60+45+42 = 430
No. of values = 9
430/9 = 47.77
Applying Mean deviation formula
∑|x+xi|/N
= 17.77+24.23+0.23+7.77+2.23+4.7+12.23+2.7+5.77divided by 9
= 77.77 divided by 9
= 8.62
Thus, mean deviation is 8.62
Ques. The mean of the data, in a continuous frequency distribution is given as 30. If each and every observation of the data is increased by a value of 10, then find the new mean. (2 marks)
Ans. Each data is increased by 10 which is an equal amount. Therefore, the mean will also be increased by 10.
Hence, new mean= 30 + 10 = 40
Ques. Find the mean deviation about the mean for the data 4,5,6,9,10,12,13. (3 marks)
Ans. Step 1: Firstly, we will find the value of ∑xi which can be calculated by adding all the data given. Therefore,
∑xi = 4+5+6+9+10+12+13
∑xi = 59
Step 2: We will find the value of the mean with the help of the formula.
N = 7
X = 59/7
X = 8.42
Therefore, the mean is 8.42
Step 3: Now, we will use the mean deviation formula to find the deviation about mean.
Step 4: The mean deviation about the mean is 5.
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