In the previous section,
we saw the basics about mean deviation.
We saw some solved examples also. In this section, we will see a few more solved examples. Later in this section, we will see mean deviation in the case of continuous frequency distribution.
Solved Example 15.4
Find the mean deviation about the mean for the following data:
 |
| Table 15.8 |
1. We are asked to find the mean deviation about the mean.
• So our first aim is to find the mean $(\bar{x})$. For that, we can use the columns I, II and III of table 15.9 below:
 |
| Table 15.9 |
•
Based on the table, we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{300}{40}~=~7.5$
2. Now we can calculate the mean deviation. For that, we can use columns II, IV and V of the table.
• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
•
Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{92}{40}~=~2.3$
Solved Example 15.5
Find the mean deviation about the median for the following data:
 |
| Table 15.10 |
Solution:
1. We are asked to find the mean deviation about the median.
• So our first aim is to find the median (M). For that, we can use the columns I, II and III of the table 15.11 below. The observations are arranged in ascending order.
 |
| Table 15.11 |
•
Based on the table, we can write:
Total number of observations = n = ∑fi = 30
•
"30" is an even number. So the median will be the average of the values at:
$\frac{n}{2}~\text{and}~\left(\frac{n}{2} + 1 \right)$
♦ That is., the average of the values at:
$\frac{30}{2}~\text{and}~\left(\frac{30}{2} + 1 \right)$
♦ That is., the average of the values at:
15 and 16,
•
From the column for cumulative frequency, we see that:
♦ the value at the 15th position is 13.
♦ the value at the 16th position is also 13.
•
So we get: M = average of 13 and 13 = 13
2. Now we can calculate the mean deviation. For that, we can use columns II, IV and V of the table.
• We have:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - M|)}}{\sum{f_i}}$
•
Substituting the values, we get:
$\text{M.D.(M)}~=~\frac{149}{30}~=~4.97$
Calculating mean deviation about mean for continuous frequency distribution
•
We know that, when the data is large, we arrange the observations into various groups. It is called continuous frequency distribution.
•
We have seen the method to calculate mean and median in such cases. So now we will see the method to calculate the following items:
♦ mean deviation about mean for a continuous frequency distribution.
♦ mean deviation about median for a continuous frequency distribution.
•
The method is similar to what we saw in the previous section. There we dealt with discrete values. But here we will be dealing with groups of values. These groups are called class intervals.
•
So for each class interval, we choose a value to represent that class interval. Usually we choose the midpoint as the representative value.
• For example, if the class interval is 30-35, then the representative value will be the midpoint, which is:
$\frac{30 + 35}{2}~=~\frac{65}{2}~=~32.5$
•
Once the representative value is fixed, the procedure is the same as before.
•
The following solved example will demonstrate the method:
Solved Example 15.6
Find the mean deviation about the mean for the following data:
 |
| Table 15.12 |
Solution:
1. We are asked to find the mean deviation about the mean.
•
So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.13 below:
 |
| Table 15.13 |
•
$f_i x_i$ is calculated in the column IV. So we can write:
$\bar{x}~=~\frac{\sum{f_i x_i}}{\sum{f_i}}~=~\frac{1800}{40}~=~45$
2. Now we can calculate the mean deviation. For that, we can use columns II, V and VI.
• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
•
Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{400}{40}~=~10$
Shortcut method for calculating mean deviation about $\bar{x}$
• In our earlier classes, we have seen a shortcut method to find $\bar{x}$. It is called assumed mean method (details here).
• We have also seen the modification of that method. It is called step-deviation method (details here). Using that method, we calculated $\bar{x}$ very easily.
♦ If $\bar{x}$ can be calculated very easily,
♦ it means that,
♦ mean deviation about $\bar{x}$ can also be calculated very easily.
• So let us apply the step-deviation method to the solved example 6 that we saw above. It will be our next solved example:
Solved Example 15.7
Find the mean deviation about the mean for the data in table 15.12.
Solution:
1. We are asked to find the mean deviation about the mean.
•
So our first aim is to find the mean $(\bar{x})$. For that, we want $f_i x_i$.
• For calculating $f_i x_i$, we want $x_i$.
• $x_i$ values are the midpoint values. They are calculated in the third column of the table 15.14 below:
 |
| Table 15.14 |
•
Now we reduce the sizes of all $x_i$ values. For that, we consider a middle value as the assumed mean. For our present case, we consider 45 as the assumed mean (a). Then we subtract 'a' from all $x_i$ values. The values obtained after subtraction are denoted as $d_i$. This is calculated in column IV.
• Next, we reduce the size of $d_i$. This is achieved by dividing with a common factor (h). For our present case, 10 is the common factor. The values obtained after division are denoted as $u_i$. This is calculated in column V.
• Finally, $f_i u_i$ is calculated in the column VI. So we can write:
$\bar{u}~=~\frac{\sum{f_i u_i}}{\sum{f_i}}~=~\frac{0}{40}~=~0$
• Using $\bar{u}$, we can calculate $\bar{x}$.
$\bar{x} = a + h \bar{u} = (45 + 10 \times 0) = 45$
2. Now we can calculate the mean deviation. For that, we can use columns II, VII and VIII.
• We have:
$\text{M.D.}(\bar{x})~=~\frac{\sum{(f_i |x_i - \bar{x}|)}}{\sum{f_i}}$
•
Substituting the values, we get:
$\text{M.D.}(\bar{x})~=~\frac{400}{40}~=~10$
While using the step-deviation method for calculating the “mean deviation about the mean”, we must keep an important fact in our minds. It can be written in 2 steps:
1. Calculation of the “mean deviation about the mean” involves two parts.
• In the first part, we calculate the mean ($\bar{x}$).
• In the second part, we calculate the mean deviation about that mean.
2. The step-deviation is a shortcut method for the first part only.
• For the second part, we need to follow the usual procedure.
Calculating mean deviation about median for continuous frequency distribution
•
We know how to find the median (M) of a continuous frequency distribution (details here).
• Once we find the M, we can find the required mean deviation as usual.
• The following solved example demonstrates the procedure.
Solved Example 15.8
Find the mean deviation about the median for the following data:
 |
| Table 15.15 |
Solution:
1. We are asked to find the mean deviation about the median.
•
So our first aim is to find the median (M). It can be done in steps:
(i) In the column II of table 15.16 below, we see that, $\sum{f_i}$ is 50. That means, there are a total of 50 observations. So we can write: n = 50.
(ii) '50' is an even number. So the median will be the average of the following two values:
♦ The value at the $\frac{n}{2}$ position.
♦ The value at the $\frac{n}{2} + 1$ position.
(iii) Let us calculate those positions:
♦ $\frac{n}{2}~=~\frac{50}{2}~=~25$
♦ $\frac{n}{2} + 1~=~(25 + 1)~=~26$
(iv) In the table 15.16 below, consider the classes with cumulative frequencies 13 and 28.
 |
| Table 15.16 |
• It is clear that, positions 25 and 26 occur inside the class 20-30.
• So 20-30 is the median class.
(v) But we do not know the actual values inside the classes. So we do not know the exact values at 25 and 26 positions. To find the median in such a situation , we use the formula:
$M~=~l + \left[\frac{\frac{n}{2}~-~cf}{f} \right]h$
• Where:
♦ l = lower limit of the median class
♦ n = number of observations
♦ cf = cumulative frequency of the class preceding the median class
♦ f = frequency of the median class
♦ h = width of the class interval (assuming all classes are of the same width)
(vi) In our present case:
♦ The median class is 20-30. So l = 20
♦ n = total number of observations = 50
♦ cf = cumulative frequency of class 10-20 = 13
♦ f = frequency of class 20-30 = 15
♦ h = width of class intervals = 10
• Substituting these values, we get:
$M~=~20 + \left[\frac{\frac{50}{2}~-~13}{15} \right]10 ~=~28$
2. Now we can calculate the mean deviation. For that, we can use columns II, IV, V and VI of the table.
• We have:
$\text{M.D.(M)}~=~\frac{\sum{(f_i |x_i - M|)}}{\sum{f_i}}$
•
Substituting the values, we get:
$\text{M.D.(M)}~=~\frac{508}{50}~=~10.16$
Link to a few more solved examples is given below:
In the next section, we will see Variance and Standard deviation.
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