If the values of the variable (i.e., observations or variates) be x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) and their corresponding frequencies are f\(_{1}\), f\(_{2}\), f\(_{3}\), f\(_{4}\), ....., f\(_{n}\) then the mean of the data is given by
Mean = A (or \(\overline{x}\)) = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3} + x_{4}f_{4} + .... + x_{n}f_{n}}{ f_{1} + f_{2} + f_{3} + f_{4} + ..... + f_{n}}\)
Symbolically, A = \(\frac{\sum{x_{i}. f_{i}}}{\sum f_{i}}\); i = 1, 2, 3, 4, ...., n.
In words,
Mean = \(\frac{\textbf{Sum of products of the Variables and their corresponding Frequencies}}{\textbf{Total Frequency}}\)
This is the formula for finding the mean of the grouped data by direct method.
For Example:
The number of Mobile sold is given in the table below. Find the mean of the number of Mobile sold.
Number of Mobile Sold |
2 |
5 |
6 |
10 |
12 |
Number of Shops |
6 |
10 |
8 |
1 |
5 |
Solution:
Here, x\(_{1}\) = 2, x\(_{2}\) = 5, x\(_{3}\) = 6, x\(_{4}\) = 10, x\(_{5}\) = 12.
f\(_{1}\) = 6, f\(_{2}\) = 10, f\(_{3}\) = 8, f\(_{4}\) = 1, f\(_{5}\) = 5.
Therefore, mean = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3} + x_{4}f_{4} + x_{5}f_{5}}{f_{1} + f_{2} + f_{3} + f_{4} + f_{5}}\)
= \(\frac{2 × 6 + 5 × 10 + 6 × 8 + 10 × 1 + 12 × 5}{6 + 10 + 8 + 1 + 5}\)
= \(\frac{12 + 50 + 48 10 + 60}{30}\)
= \(\frac{180}{30}\)
= 6.
Therefore, mean number of Mobile sold is 6.
Short-cut method for finding the mean of grouped data:
We know that the direct method of finding mean for grouped data gives
mean A = \(\frac{\sum{x_{i}. f_{i}}}{\sum f_{i}}\)
where x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) are variates and f\(_{1}\), f\(_{2}\), f\(_{3}\), f\(_{4}\), ....., f\(_{n}\) are their corresponding frequencies.
Let a = a number taken as assumed mean from which the diviation of the variate is d_{i} = x_{i} - a.
Then, A =\(\frac{\sum{(a + d_{i})f_{i}}}{\sum f_{i}}\)
= \(\frac{\sum{af_{i}} + \sum{d_{i}f_{i}}}{\sum f_{i}}\)
= \(\frac{a\sum{f_{i}} + \sum{d_{i}f_{i}}}{\sum f_{i}}\)
= a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)
Therefore, A = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\), where d_{i} = x_{i} - a.
For Example:
Find the mean of the following distribution using the short-cut method.
Variate |
20 |
40 |
60 |
80 |
100 |
Frequency |
15 |
22 |
18 |
30 |
16 |
Solution:
Putting the calculated values in a tabular form, we have the following.
Variate |
Frequency |
Deviation d_{i} from assumed mean a = 60, i.e., (x_{i} - a) |
d_{i}x_{i} |
20 |
15 |
-40 |
-600 |
40 |
22 |
-20 |
-440 |
60 |
18 |
0 |
0 |
80 |
30 |
20 |
600 |
100 |
16 |
40 |
640 |
\(\sum f_{i}\) = 101 |
\(\sum d_{i}f_{i}\) = 200 |
Therefore, mean A = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)
= 60 + \(\frac{200}{101}\)
= 61\(\frac{99}{101}\)
= 61.98.
Solved Examples on Mean of Grouped Data or Mean of the Arrayed Data:
1. A class has 20 students whose ages (in years) are as follows.
14, 13, 14, 15, 12, 13, 13, 14, 15, 12, 15, 14, 12, 16, 13, 14, 14, 15, 16, 12
Find the mean ago of the students of the class.
Solution:
In the data, only five different numbers appear respectively. So, we write the frequencies of the variates as below.
Age (in years) (x\(_{i}\)) |
12 |
13 |
14 |
15 |
16 |
Total |
Number of Students (f\(_{i}\)) |
4 |
4 |
6 |
4 |
2 |
20 |
Therefore, mean A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3} + x_{4}f_{4} + x_{5}f_{5}}{ f_{1} + f_{2} + f_{3} + f_{4} + f_{5}}\)
= \(\frac{12 × 4 + 13 × 4 + 14 × 6 + 15 × 4 + 16 × 2}{4 + 4 + 6 + 4 + 2}\)
= \(\frac{48 + 52 + 84 + 60 + 32}{20}\)
= \(\frac{276}{20}\)
= 13.8
Therefore, the mean age of the students of the class = 13.8 years.
2. The weights (in kg) of 30 boxes are as given below.
40, 41, 41, 42, 44, 47, 49, 50, 48, 41, 43, 45, 46, 47, 49, 41, 40, 43, 46, 47, 48, 48, 50, 50, 40, 44, 44, 47, 48, 50.
Find the mean weight of the boxes by preparing a frequency table of the arrayed data.
Solution:
The frequency table for the given data is
Weight (in Kg) (x_{i}) |
Tally Mark |
Frequency (f_{i}) |
x_{i}f_{i} |
40 |
/// |
3 |
120 |
41 |
//// |
4 |
164 |
42 |
/ |
1 |
42 |
43 |
// |
2 |
86 |
44 |
/// |
3 |
132 |
45 |
/ |
1 |
45 |
46 |
// |
2 |
92 |
47 |
//// |
4 |
188 |
48 |
//// |
4 |
192 |
49 |
// |
2 |
98 |
50 |
//// |
4 |
200 |
\(\sum f_{i}\) = 30 |
\(\sum x_{i}f_{i}\) = 1359 |
By formula, mean = \(\frac{\sum{x_{i}f_{i}}}{\sum f_{i}}\)
= \(\frac{1359}{30}\)
= 45.3.
Therefore, the mean weight of the boxes = 45.3 kg.
3. Four variates are 2, 4, 6 and 8. The frequencies of the first three variates are 3, 2 and 1 respectively. If the mean of the variates is 4 then find the frequency of the fourth variate.
Solution:
Let the frequency of the fourth variate (8) be f. Then,
mean A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3} + x_{4}f_{4}}{ f_{1} + f_{2} + f_{3} + f_{4}}\)
⟹ 4 = \(\frac{2 × 3 + 4 × 2 + 6 × 1 + 8 × f}{3 + 2 + 1 + f}\)
⟹ 4 = \(\frac{6 + 8 + 6 + 8f}{6 + f}\)
⟹ 24 + 4f = 20 + 8f
⟹ 4f = 4
⟹ f = 1
Therefore, the frequency of 8 is 1.
4. Find the mean of the following data.
Variate (x)
1
2
3
4
5
Cumulative Frequency
3
5
9
12
15
Solution:
The frequency table and calculations involved in finding the mean are given below.
Variate (x_{i}) |
Cumulative Frequency |
Frequency (f_{i}) |
x_{i}f_{i} |
1 |
3 |
3 |
3 |
2 |
5 |
2 |
4 |
3 |
9 |
4 |
12 |
4 |
12 |
3 |
12 |
5 |
15 |
3 |
15 |
\(\sum f_{i}\) = 15 |
\(\sum x_{i}f_{i}\) = 46 |
Therefore, mean = \(\frac{\sum{x_{i}f_{i}}}{\sum f_{i}}\)
= \(\frac{46}{15}\)
= 3.07.
5. Find the mean mark from the following frequency table by using the short-cut method.
Marks Obtained |
30 |
35 |
40 |
45 |
50 |
Number of Students |
45 |
26 |
12 |
10 |
7 |
Solution:
Taking the assumed mean a = 40, the calculations will be as follows.
Marks Obtained (x_{i}) |
Number of Students (f_{i}) |
Deviation d_{i} = x_{i} - a = x_{i} - 40 |
d_{i}f_{i} |
30 |
45 |
-10 |
-450 |
35 |
26 |
-5 |
-130 |
40 |
12 |
0 |
0 |
45 |
10 |
5 |
50 |
50 |
7 |
10 |
70 |
\(\sum f_{i}\) = 100 |
\(\sum d_{i}f_{i}\) = -460 |
Therefore, mean = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)
= 40 + \(\frac{-460}{100}\)
= 40 - 4.6
= 35.4.
Therefore, the mean mark is 35.4.
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