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# Calculate the mean deviation about the mean, median, and mode for the following data.Class Interval(C.I.)Frequency(f)2-434-646-828-101

Last updated date: 13th Jun 2024
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Here, we have been told to calculate mean deviation which shows how far the values are from the middle value. We can calculate the mean deviation of the three measures of center in a numerical data set. Here we have grouped data and therefore we will need to calculate the class midpoints to get the term${f_i}{x_i}$. Further, we will use the formulae of mean, median, and mode to calculate the mean deviation. The formulae that we will use are,
Mean = $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$
Median = $l + \dfrac{{\left( {\dfrac{N}{2} - F} \right)}}{f} \times h$
Mode = $l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times h$

Complete step-by-step solution
We will calculate the mean deviation about the mean, median, and mode one by one.
Now, mean deviation is given by the formula: $\dfrac{{\sum {{f_i}\left| {{x_i} - X} \right|} }}{N}$
Where ${x_i}$= class midpoint
X = Mean of the grouped data
Therefore, we will first have to calculate Mean (X) =$\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$
We will calculate the following terms for the first class interval 2-4.
Class Midpoint =$\dfrac{{2 + 4}}{2}$
${f_i}{x_i} = 3 \times 3$$= 9$
Let us tabulate the same data for other class intervals as follows.
 Class Interval Frequency${f_i}$ Class midpoints${x_i}$ ${f_i}{x_i}$ Deviation$\left| {{x_i} - X} \right|$=$\left| {{x_i} - 5.2} \right|$ ${f_i}\left| {{x_i} - X} \right|$ 2-4 3 3 9 2.2 6.6 4-6 4 5 20 0.2 0.8 6-8 2 7 14 1.8 3.6 8-10 1 9 9 3.8 3.8 $\sum {{f_i} = 10}$ $\sum {{f_i}{x_i} = 52}$ $\sum {{f_i}\left| {{x_i} - X} \right|}$=14.8

Let us calculate mean,
Mean X=$\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$$= \dfrac{{52}}{{10}}$$= 5.2$
We will calculate the mean and also add the column for deviation i.e. $\left| {{x_i} - X} \right|$ in the same table.
Let us calculate Deviation $\left| {{x_i} - X} \right|$ and ${f_i}\left| {{x_i} - X} \right|$ or the first class interval 2-4.
$\left| {{x_i} - X} \right|$ = $\left| {9 - 5.2} \right|$ = $2.2$
${f_i}\left| {{x_i} - X} \right|$ $= 3 \times 2.2$
Similarly, we will calculate for the remaining class intervals and add it in the above table as shown.
Now we know that,
Mean Deviation = $\dfrac{{\sum {{f_i}\left| {{x_i} - X} \right|} }}{N}$
N = total number of values = $\sum {{f_i} = 10}$
Therefore,
Mean Deviation= $\dfrac{{14.8}}{{10}}$ = 1.48
Mean Deviation for the given data about Mean = 1.48
Here we will first have to calculate the Median.
Median is given by the formula:
Median=$l + \dfrac{{\left( {\dfrac{N}{2} - F} \right)}}{f} \times h$
Where, l= lower boundary of median class
N = Sum of frequencies
F = Cumulative frequency before median class
f = Frequency of median class
h = Size of median class
Let us tabulate all the data and then calculate median.
We will calculate the following terms for the first class interval 2-4.
Class midpoints = $\dfrac{{2 + 4}}{2}$ = 3
Similarly, we will then calculate for other class intervals and tabulate as follows:
 Class Interval Frequency${f_i}$ Cumulative Frequency Class midpoints${x_i}$ Deviation$\left| {{d_i}} \right|$ =$\left| {{x_i} - X} \right|$=$\left| {{x_i} - 5} \right|$ ${f_i}{d_i}$ 2-4 3 3 3 2 6 4-6 4 7 5 0 0 6-8 2 9 7 2 4 8-10 1 10 9 4 4 $\sum {{f_i} = 10}$ $\sum {{f_i}\left| {{x_i} - X} \right|}$=14

Next, we will calculate the Median.
Here, N=10
$\therefore \dfrac{N}{2} = 5$
The cumulative frequency greater than $\dfrac{N}{2}$ is 7 and corresponding class is 4-6.
Hence, we get median class is 4-6.
$\therefore l = 4,f = 4,F = 3,h = 2,N = 10$
$\therefore Median = l + \dfrac{{\left( {\dfrac{N}{2} - F} \right)}}{f} \times h$
$= 4 + \dfrac{{\left( {\dfrac{{10}}{2} - 3} \right)}}{4} \times 2$
$= 4 + \dfrac{{\left( {5 - 3} \right)}}{4} \times 2$
= 5
We will use this value of X=5 to calculate deviation $\left| {{x_i} - X} \right|$.
Now,
Let us calculate deviation $\left| {{x_i} - X} \right|$ and ${f_i}{d_i}$ for the first class interval 2-4.
Therefore,
$\left| {{x_i} - X} \right|$=$\left| {3 - 5} \right|$=$2$
${f_i}{d_i}$=$3 \times 2 = 6$
Similarly, we will calculate for the other class intervals and add it in the above table as shown.
Now, let us calculate the mean deviation.
Mean deviation about Median = $\dfrac{{\sum {{f_i}{d_i}} }}{N}$
$= \dfrac{{14}}{{10}}$
= 1.4
Mean deviation about Median is 1.4.
We will calculate Mode first.
Mode is given by the formula:
Mode=$l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times h$
Where, l= lower boundary of modal class
f = Frequency of modal class
${f_1}$ = Frequency of class preceding the modal class
${f_2}$ = Frequency of class succeeding the modal class
h = size of the class interval
Let us tabulate the data and then calculate mode.
We will calculate the following terms for the first class interval 2-4.
Class midpoints = $\dfrac{{2 + 4}}{2}$ = 3
Similarly, we will then calculate for other class intervals and tabulate as follows:

 Class Interval(C.I.) Frequency(f) Class midpoints${x_i}$ Deviation$\left| {{d_i}} \right|$ =$\left| {{x_i} - M} \right|$=$\left| {{x_i} - 4.7} \right|$ ${f_i}{d_i}$ 2-4 3 3 1.7 5.1 4-6 4 5 0.3 1.2 6-8 2 7 2.3 4.6 8-10 1 9 4.3 4.3 $\sum {{f_i} = 10}$ $\sum {{f_i}{d_i} = }$15.2

Now, we know that modal class is that class interval having highest frequency.
Here highest frequency is 4 corresponding to the class 4-6.
Therefore, the Modal class is 4-6.
Now, we will get the values for l, h, f,${f_1}$, ${f_2}$as
l=4, h=2, f=4, ${f_1}$= 3, ${f_2}$= 2.
We will substitute these values to calculate Mode.
Mode=$l + \left( {\dfrac{{f - {f_1}}}{{2f - {f_1} - {f_2}}}} \right) \times h$
= $4 + \left( {\dfrac{{4 - 3}}{{2 \times 4 - 3 - 2}}} \right) \times 2$
= $4 + \left( {\dfrac{1}{3}} \right) \times 2$
= 4.67
From this value of Mode (M) we will calculate $\left| {{x_i} - M} \right|$ in the table.
Now,
Let us calculate deviation $\left| {{x_i} - X} \right|$ and ${f_i}{d_i}$ for the first class interval 2-4.
$\left| {{x_i} - X} \right|$ = $\left| {3 - 4.7} \right|$ = 1.7
${f_i}{d_i}$= $3 \times 1.7 = 5.1$
We will add the above values in the above table.
Similarly, we will calculate for the other class intervals and add them to the above table as shown.
Now, let us calculate the mean deviation about the mode.
Mean deviation about Mode = $\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }}$ = $\dfrac{{15.2}}{{10}}$ = 1.52
Therefore, we get the Mean deviation about Mode is 1.52.

Note: We will make 3 different tables for each calculation method for Mean, Median and Mode for ease of calculation. We have to keep in mind to multiply the correct columns and terms. To verify accuracy of the solution, we can see that the values of mean, median and mode are approximately of the same value. Also the values of the mean deviations are also approximately same which verifies correctness of the solution. In case the values are not close we should recheck our calculations.