Statistics
A cricket player played 7 matches and scored 88, 72, 90, 94, 56, 62, and 76 runs. What is the average score or mean score of the player in these 7 matches?
We know that,
“The mean is the sum of all observations divided by the number of observations”.
We can use the following formula to find the mean of the given data.
Using this formula, the mean of scores obtained by the player
= 76.86
Thus, the average score of the player is 76.86. There are many situations where the number of observations is very large and then it is not possible to use this formula to find the mean of the given data.
Let us consider such a situation.
We have the marks distribution of 100 students of a class in a Mathematics test. The distribution is given in the form of groups like 0 − 10, 10 − 20 …. To find the mean in such situations, we follow another method. This method is explained in the given video.
Let us solve some more examples to understand the concept better.
Example 1:
The following table shows the various income brackets of the employees of a company.
Income of employees (in Rupees) | 10000 − 15000 | 15000 − 20000 | 20000 − 25000 |
25000 − 30000 |
30000 − 35000 |
35000 − 40000 |
40000 − 45000 |
No. of employees |
32 |
18 |
13 |
7 |
14 |
5 |
11 |
Find the average monthly income of each employee.
Solution:
Firstly, we will calculate the class marks of each class interval and then the product of the class marks with the corresponding frequencies. By doing so, we obtain the following table.
Income of employees (in Rupees) |
No. of employees: f_{i} |
Class Mark: x_{i} |
x_{i} × f_{i} |
10000 − 15000 15000 − 20000 20000 − 25000 25000 − 30000 30000 − 35000 35000 − 40000 40000 − 45000 |
32 18 13 7 14 5 11 |
12500 17500 22500 27500 32500 37500 42500 |
400000 315000 292500 192500 455000 187500 467500 |
Total (∑) |
100 |
2310000 |
The mean can now be easily calculated using the formula.
Mean
Hence, the average income of the employees is Rs 23100.
Example 2:
The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Find the value of f_{1} and f_{2}.
Class Interval |
Frequency |
0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 100 − 120 |
5 f_{1} 10 f_{2} 7 8 |
Total |
50 |
Solution:
Firstly, let us find the class marks of each class interval and then the product of class marks with the corresponding frequencies for each class interval as shown in the following table.
Class Interval |
Frequency: f_{i} |
Class Mark: x_{i} |
x_{i} × f_{i} |
0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 100 − 120 |
5 f_{1} 10 f_{2} 7 8 |
10 30 50 70 90 110 |
50 30 f_{1} 500 70 f_{2} 630 880 |
30 + f_{1} + f_{2} = 50 |
2060 + 30 f_{1} + 70 f_{2} |
It is given that the sum of the frequencies is 50.
∴ 30 + f_{1} + f_{2} = 50
f_{1} + f_{2} = 20
f_{1 }= 20 − f_{2} … (1)
The mean is given as 62.8.
∴ = 62.8
Using equation (1),
30(20 − f_{2}) + 70 f_{2} = 1080
600 + 40 f_{2} = 1080
40 f_{2} = 1080 − 600 = 480
f_{2} =
On putting the value of f_{2} in equation (1), we obtain
f_{1 }= 20 − f_{2} = 20 − 12 = 8
Thus, the values of f_{1} and f_{2} are 8 and 12 respectively.
CBSE Board question(s) related to this lesson:
CBSE Board question(s) related to this lesson:
We know that direct method can be used to find the mean of any data given in grouped form, but the calculation in direct method becomes very tough when the data is given in the form of large numbers, because finding the product of x_{i }and f_{i} becomes difficult and time consuming.
Therefore, we introduce assumed mean method to find the mean of grouped data. This method is also known as shift of origin method. This is an easier method to find the mean as it involves less calculation.
Let us discuss an example in the video to find the mean using the assumed mean method.
Let us solve more examples using the assumed mean method....
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