Overview
Test Series
Assumed mean method finds the actual mean of the data by first assuming a mean value. The term “mean” refers to the average value of a set of data, which is derived by dividing the total number of counts by all the data. The mean, or total average, is easily determined by adding all the numbers together, then dividing by the entire number of numbers. The mean value in a set of data is a determined average that lies halfway between the highest and lowest values.
Assumed mean method gives us smaller numbers to work with making calculations easier and is thus suitable if your data set has large values. When calculating the mean using the direct mean method, you obtain significantly bigger numbers. The likelihood of making calculating errors is decreased when utilizing the assumed mean approach, also known as a shift of origin because it gives you smaller numbers to work with (as well as negative numbers that lower the sum). You can obtain even lower values by utilizing an assumed mean, a shifted origin and a scale change.
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We can calculate mean using the assumed mean method by following the below steps:
Quantitative Data can be categorized into 2 categories:
Grouped data are data formed by aggregating individual observations of a variable into groups so that a frequency distribution of these groups serves as a convenient means of summarizing or analyzing the data. Grouped data is of the form 1−10,11−20,21,−30 and so on. Hence, we find its class mark. The midpoint of each class interval is the definition of a class mark. Calculating a class
mark is as follows:
Where fi is class frequency. The frequency of a class interval is the number of observations that occur in a particular predefined interval. So, for example, if 30 people of weight 55 to 60 appear in our study’s data, the frequency for the 55 – 60 interval is 30.
The Step Deviation Method is a simplified way to calculate the mean (average) of grouped data in statistics. It is especially helpful when the class intervals and frequencies are large and difficult to handle manually.
When direct or assumed mean methods become lengthy due to large numbers, the Step Deviation Method reduces the calculations by scaling down the data, making it quicker and easier to find the mean.
Step deviation method is the extended version of the shortcut or assumed method for calculating the mean of large values. These deviation values can be divided by a common factor that has been scaled down to a smaller amount. Change of origin or scale method is another name for the step deviation method.
Learn about Rolle’s Theorem
Now let’s see some solved examples on the assumed mean method.
Now let’s see some solved examples on the assumed mean method.
Example 1: Find the mean of the following data using direct method, assumed mean method and step deviation method. \(40, 50, 55, 78, 58\).
Solution:
\(x\) |
\(d = x – A (40)\) |
\(d^\prime = \frac{d}{c(10)}\) |
40 |
0 |
0 |
50 |
10 |
1.0 |
55 |
15 |
1.5 |
78 |
38 |
3.8 |
58 |
18 |
1.8 |
281 |
81 |
8.1 |
Direct Method:
\(\bar{x}=\frac{\sum{x}}{N}=\frac{281}{5}\)
\(\bar{x}=56.2\)
Assumed Mean Method:
\(\bar{x}=A+\frac{\sum{d}}{N}=40+\frac{81}{5}=40+16.2\)
\(\bar{x}=56.2\)
Step Deviation Method:
\(\bar{x}=A+\frac{\sum{d^\prime}}{N}\times{c}=40+\frac{8.1}{5}\times{10}\)
\(\bar{x}=40+1.62\times{10}\)
\(\bar{x}=40+16.2\)
\(\bar{x}=56.2\)
Example 2: Calculate the arithmetic mean for the following data using Assumed Mean Method.
Marks |
Number of students |
65 |
6 |
70 |
11 |
75 |
3 |
80 |
5 |
85 |
4 |
90 |
7 |
95 |
10 |
100 |
4 |
Solution: Now we have to use the formula given above to find the arithmetic mean using Assumed Mean Method.
Take the assumed mean \(A = 80\)
\(x\) |
\(f\) |
\(d = x – A\) |
\(f_d\) |
65 |
6 |
-15 |
-90 |
70 |
11 |
-10 |
-110 |
75 |
3 |
-5 |
-15 |
80 |
5 |
0 |
0 |
85 |
4 |
5 |
20 |
90 |
7 |
10 |
70 |
95 |
10 |
15 |
150 |
100 |
4 |
20 |
80 |
Total |
\(N = 50\) |
\({\sum}fd = 115\) |
– |
Arithmetic Mean \(= A + [\frac{{\sum}fd}{N]\)
\(= 80 + (\frac{115}{50})\)
\(= 80 + (\frac{23}{10})\)
\(= 80 + 2.3\)
\(= 82.3\)
Example 3: Consider the following data set and calculate the mean using Assumed Mean Method.
\(f_i\) |
\(x_i\) |
3 |
149 |
2 |
156 |
4 |
153 |
6 |
159 |
5 |
147 |
Solution: Let the assumed mean for this data be \(a = 150
Mean for this data, [latex]\bar{x} = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}
[latex]f_i\) |
\(x_i\) |
\(d_i=x_i–a=x_i–150\) |
\(f_id_i\) |
\(3\) |
\(149\) |
\(149–150=−1\) |
\(3{\times}−1=−3\) |
\(2\) |
\(156\) |
\(156–150=6\) |
\(2{\times}6=12\) |
\(4\) |
\(153\) |
\(153–150=3\) |
\(4{\times}3=12\) |
\(6\) |
\(159\) |
\(159–150=9\) |
\(6{\times}9=54\) |
\(5\) |
\(147\) |
\(147–150=−3\) |
\(5{\times}−3=−15\) |
\(\sum {{{f}}_{{i}}} = 3 + 2 + 4 + 6 + 5 = 20\) |
\(\sum {{{f}}_{{i}}}{{{x}}_{{i}}}= -3 + 12 + 12 + 54 – 15 = 60\) |
\(\Rightarrow\bar{x} = 150 + \frac{{60}}{{20}}\)
\(\Rightarrow\bar{x} = 150 + 3\)
\(\therefore\bar{x} = 153\)
Example 4: The following table shows the weight of 15 students. Calculate using Assumed Mean Method:
Weight ( in kg ) |
Number of students |
47 |
6 |
48 |
4 |
49 |
2 |
50 |
2 |
51 |
1 |
Solution: Let the assumed mean be \(A = 49\)
Weight ( in kg ) \(x_i\) |
Number of students \(f_i\) |
\(d_i=x_i−A=x_i−49\) |
\(f_id_i\) |
47 |
6 |
-2 |
-12 |
48 |
4 |
-1 |
-4 |
49 |
2 |
0 |
0 |
50 |
2 |
1 |
2 |
51 |
1 |
2 |
2 |
– |
\({\sum}f_i=15\) |
– |
\( {\sum}f_id_i=−12\) |
We have, \({\sum}f_i = 15, {\sum}f_id_i = −12\) and A = 49\)
We know,
\(Mean = A + \frac{\sum_{i=1}^nf_id_i}{\sum_{i=1}^nf_i}\)
\(= 49 + \frac{-12}{15}\)
\(= 49 – 0.8\)
\(= 48.02\)
Hence, mean weight = \(48.2\) kg.
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