In the previous section we completed the discussion on Frequency Polygons. In this section we will see 'Measures of Central Tendency'. So first we have to see what 'Central Tendency' is. We will start with 'Averages'. Consider the following example:
■ A farmer has 5 plants of a particular kind. They give a special fruit which is considered very costly. The yield from each plant in the previous year was as follows:
• First plant yielded 10 fruits
• Second plant yielded 7 fruits
• Third plant yielded 13 fruits
• Fourth plant yielded 20 fruits
• Fifth plant yielded 15 fruits
What is the average yield?
Solution:
We have done such problems in our earlier classes. We have seen that:
Average of a certain number of observations = Sum of all observations⁄Total number of observations.
In this problem, the sum of all the observations = 10 +7 + 13 +20 +15 = 65
So the average in this problem is 65⁄5 = 13
But all the plants will not give the same number of fruits. Some plants give more, and some plants give less. In such a situation, we take the average. For a preliminary estimate, the average can be taken as equal to the yield from a single plant.
It may be noted that, many detailed calculations involving complex theories are involved before finalising any business activity. We will study more about them in higher classes on ‘Business administration’.
In science and engineering problems, another term is used for ‘average’. It is the ‘mean’. It is denoted as x . It is read as ‘x bar’. So we can write:
■ Mean of a certain number of observations = x = Sum of all observations⁄Total number of observations.
2. So the sum of 5 observations is x1 + x2 + x3 + x4 + x5
3. We can write: x = (x1 +x2 +x3 +x4 +x5 )⁄5 = 65⁄5 = 13
4. Now we will see a method to write the formula in an even simpler way:
• The subscripts 1, 2, 3, . . . can be denoted by the letter ‘i’. So:
♦ when i = 1, xi denotes the first observation x1
♦ when i = 2, xi denotes the second observation x2
♦ - - -
♦ - - -
♦ when i = 5, xi denotes the fifth observation x5
5. The Greek symbol Σ (for the letter Sigma) is used for summation.
In computer spread sheet programs also, this symbol indicates summation. An example is shown below:
So, the sum x1 + x2 + x3 + x4 + x5 is written as:
It is read as: 'The sum of xi as i varies from 1 to 5'
6. Similarly consider a case of 30 observations. We want x1 + x2 + x3 + . . . + x30. It can be written as:
It is read as: 'The sum of xi as i varies from 1 to 30'
7. Now consider a case of n observations. We want x1 + x2 + x3 + . . . + xn. It can be written as:
It is read as: 'The sum of xi as i varies from 1 to n'
8. Now coming back to mean in our present problem, we can write:
9. In general, for n observations, we can write:
Eq.25.1:
Rs. 350, 400, 350, 450 and 450.
What is the mean daily wage of a worker in this group?
Solution:
1. We have:
2. First let us calculate the numerator:
= 350 +400 +350 +450 +450 = 2000
3. The denominator = number of observations = n = 5
4. So x = 2000⁄5 = Rs.400
What is the mean daily wage of a worker in this group?
Solution:
Note that the raw data is given to us in the form of a frequency distribution table. This is because there are 20 observations, and such a large number of observations should always be presented in the form of a frequency distribution table. Let us start our calculations to find the mean x:
1. We have:
2. First let us calculate the numerator:
This is the sum of all observations. As in the previous example, the summation can be quickly done, if we group repeating values. So we can write:
= (2 × 300) + (4 × 300) + . . . + (4 × 500)
3. The above grouping and summation can be conveniently written in a tabular form:
So we get the summation result as Rs.8200
4. The denominator = number of observations = n = 20
5. So x = 8200⁄20 = Rs.410
This can be written in short form as:
Sum of all observations =
2. Now, the total number of observations is obviously equal to the sum of all frequencies. This is in fact seen as the 'total' at the bottom of the frequency column. So we can write:
Total number of observations = f1 + f2 + f3 + . . . + fn
This can be written in short form as:
Total number of observations =
3. We have: = x = Sum of all observations⁄Total number of observations.
So we can write:
Eq.25.2:
In the next section we will see some solved examples.
■ A farmer has 5 plants of a particular kind. They give a special fruit which is considered very costly. The yield from each plant in the previous year was as follows:
• First plant yielded 10 fruits
• Second plant yielded 7 fruits
• Third plant yielded 13 fruits
• Fourth plant yielded 20 fruits
• Fifth plant yielded 15 fruits
What is the average yield?
Solution:
We have done such problems in our earlier classes. We have seen that:
Average of a certain number of observations = Sum of all observations⁄Total number of observations.
In this problem, the sum of all the observations = 10 +7 + 13 +20 +15 = 65
So the average in this problem is 65⁄5 = 13
Let us see one application of the above calculated average:
A friend of the farmer would like to grow the same fruits. He wants to know whether it would be profitable or not. He has space to grow 8 such plants. But just by having more plants, one cannot guarantee profit. Because, more number of plants will require more manure, more labour hours (especially if the plants need special care) etc., So the friend wants to know how many fruits he would get from one plant. But all the plants will not give the same number of fruits. Some plants give more, and some plants give less. In such a situation, we take the average. For a preliminary estimate, the average can be taken as equal to the yield from a single plant.
It may be noted that, many detailed calculations involving complex theories are involved before finalising any business activity. We will study more about them in higher classes on ‘Business administration’.
■ Mean of a certain number of observations = x = Sum of all observations⁄Total number of observations.
Now we will write this in the form of a simple formula.
1. In our present problem, we have 5 observations. We will first denote them as x1, x2, . . . , x52. So the sum of 5 observations is x1 + x2 + x3 + x4 + x5
3. We can write: x = (x1 +x2 +x3 +x4 +x5 )⁄5 = 65⁄5 = 13
4. Now we will see a method to write the formula in an even simpler way:
• The subscripts 1, 2, 3, . . . can be denoted by the letter ‘i’. So:
♦ when i = 1, xi denotes the first observation x1
♦ when i = 2, xi denotes the second observation x2
♦ - - -
♦ - - -
♦ when i = 5, xi denotes the fifth observation x5
5. The Greek symbol Σ (for the letter Sigma) is used for summation.
In computer spread sheet programs also, this symbol indicates summation. An example is shown below:
So, the sum x1 + x2 + x3 + x4 + x5 is written as:
6. Similarly consider a case of 30 observations. We want x1 + x2 + x3 + . . . + x30. It can be written as:
7. Now consider a case of n observations. We want x1 + x2 + x3 + . . . + xn. It can be written as:
8. Now coming back to mean in our present problem, we can write:
Eq.25.1:
Consider the following problem:
The daily wages of five friends who work in a factory are:Rs. 350, 400, 350, 450 and 450.
What is the mean daily wage of a worker in this group?
Solution:
1. We have:
2. First let us calculate the numerator:
= 350 +400 +350 +450 +450 = 2000
3. The denominator = number of observations = n = 5
4. So x = 2000⁄5 = Rs.400
Let us see if there is an easier method:
In step 2, we did the summation. Some values are occurring more than once. We can group such values as follows:
5. (2 × 350) + 400 + (2 × 450) = 700 + 400 + 900 = 2000 (The same sum as in step 2)
6. Now let us see the raw data in a tabular form:
7. We can see that, the frequency values can help us to group the values as in step (5) so that, summation can be done quickly
Another example:
The daily wages of 20 workers in a factory are:
Solution:
Note that the raw data is given to us in the form of a frequency distribution table. This is because there are 20 observations, and such a large number of observations should always be presented in the form of a frequency distribution table. Let us start our calculations to find the mean x:
1. We have:
2. First let us calculate the numerator:
This is the sum of all observations. As in the previous example, the summation can be quickly done, if we group repeating values. So we can write:
= (2 × 300) + (4 × 300) + . . . + (4 × 500)
3. The above grouping and summation can be conveniently written in a tabular form:
4. The denominator = number of observations = n = 20
5. So x = 8200⁄20 = Rs.410
From the above two examples, it is clear that the frequency helps us to group observations and thus to speed up the summation. We can write this:
1. Sum of all observations = f1x1 + f2x2 + f3x3 + . . . + fnxnThis can be written in short form as:
Sum of all observations =
2. Now, the total number of observations is obviously equal to the sum of all frequencies. This is in fact seen as the 'total' at the bottom of the frequency column. So we can write:
Total number of observations = f1 + f2 + f3 + . . . + fn
This can be written in short form as:
Total number of observations =
3. We have: = x = Sum of all observations⁄Total number of observations.
So we can write:
Eq.25.2:
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