In the previous section we saw an example of frequency distribution table. In this section, we will see a case where grouped frequency table becomes essential. We have already seen the basics in a previous chapter here. In this chapter we will see more details.
1. The marks obtained (out of 100) by 98 students in a national level exam are given below:
2. Here the amount of data is large. So we make them into groups called class-intervals. The class-intervals will be: 30-39, 40-49, 50-59 . . . upto 90-99
• The first class-interval is 30-39 because the smallest value is 32
• The last class-interval is 90-99 because the largest value is 97
3. The size of class-intervals is called class-width
4. In any class-interval,
• the lower number is called the lower class limit
• the upper number is called the upper class limit
5. Once we decide the class-interval, the next step is this:
• Find the number of values that will fall in each class-interval.
6. For that, we use the method of tally marks. This is shown in the table below:
■ One information that we get at the first glance is:
Greater number of students fall in the class-interval 50-59. That means, the scores of the largest number of students fall between 50 and 59, both scores inclusive.
■ One information that we get when we do a little analysis is:
Number of students who scored 70 or more marks is: 15+14+10 = 39
■ In this manner, many more information can be obtained from the final presentation.
Let us now see another case:
Consider the following grouped frequency distribution table in fig.25.1(a) below:
It gives the weights of 38 students in a class. Now, if two new students of weights 35.5 kg and 40.5 kg are admitted into this class, in which interval will we include them?
Solution:
1. Take 35.5 kg. We cannot include it in the class 31-35. Because the upper limit of this interval is only 35. Our value of 35.5 falls above this class-interval 31-35.
2. We cannot include it in the class 36-40. Because the lower limit of this interval is 36. Our value of 35.5 falls below this class-interval 36-40.
3. The same happens to 40.5 kg: It falls above 36-40 and it falls below 41-45
The values 35.5 and 40.5 have no place in the table. This happens because, there are gaps in between class-intervals. Such a gap between 31-35 and 36-40 is shown in fig.25.2(a) below:
4. So we have to avoid those gaps. That means, any class should begin at the same point at which the previous class ends.
5. For that, the following procedure is adopted:
(i) Divide a gap into two equal parts
• In fig.b, the gap between 31-35 and 36-40 is divided into two equal parts.
• This is done by marking 35.5 at the exact midway between 35 and 36
(ii) Take the interval on the left side of the gap. Bring it’s upper limit to the midpoint
• In the fig.c, the upper limit 35 of the left side interval is brought up to 35.5
(iii) Take the interval on the right side of the gap. Bring it’s lower limit to the midpoint
• In the fig.c, the lower limit 36 of the right side interval is brought down to 35.5
(iv) So the gap is covered
6. This procedure should be done at all the gaps. So both the upper and lower limits of all the class-intervals will change.
7. Let us find the new limits of all the given intervals in the problem:
■ 1st class-interval is 31-35
• 2nd class-interval is 36-40
• Gap between them = lower limit of 2nd – upper limit of 1st = 36-35 = 1
• Half of the gap = 1/2 = 0.5
• New upper limit of 1st interval = 35 + 0.5 = 35.5
• New lower limit of 2nd interval = 36 - 0.5 = 35.5
• Thus the gap between 31-35 and 36-40 is avoided
■ 2nd class-interval is 36-40
• 3rd class-interval is 41-45
• Gap between them = lower limit of 3rd – upper limit of 2nd = 41-40 = 1
• Half of the gap = 1/2 = 0.5
• New upper limit of 2nd interval = 40 + 0.5 = 40.5
• New lower limit of 3rd interval = 41 - 0.5 = 40.5
• Thus the gap between 36-40 and 41-45 is avoided
8. In this way, we can find the new limits of all the given intervals. But now another problem arises: (i) The weight of the new students are 35.5 and 40.5
(ii) Consider 35.5. It is eligible to fall in two intervals. 30.5-35.5 and 35.5-40.5
(iii) In such a situation, we follow the procedure that is accepted all over the world. That is., such a value is taken into the upper of the two eligible classes.
(iv) So 35.5 is taken into 35.5-40.5
Similarly 40.5 is taken into 40.5-45.5. And not 35.5-40.5
9. The modified table is shown in fig.25.1(b)
So we can write a general method to close the gap between any two given class intervals, and thus make them continuous:
The blood groups of 30 students of a class are recorded as follows:
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Solution:
The required frequency distribution table is given below:
From the table, we can see that,
• The most common blood group is 'O'
• The rarest blood group is AB
Solved example 25.2
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution:
• The minimum distance in the raw data is 2 km. So the first interval of 0-5 is appropriate.
• It is given that class size is to be 5. So the various class-intervals are: 0-5, 5-10, 10-15, ...
• The maximum distance in the raw data is 32 km. So the last class-interval will be 30-35
• It is given that, in the class-interval 0-5, the distance 5 km should not be included. So, if there is a distance of 5 km in the raw data, it should be included in the higher class 5-10
• Similarly, if there is a distance of 10 km in the raw data, it should not be included in the interval 5-10. It should be included in the higher class 10-15 only
• Based on the above facts, the grouped frequency distribution table is constructed as shown below:
Sample calculation:
• Consider the class-interval 15-20. The distances from 15 to 20 in the raw data are:
20, 19, 17, 18, 17, 16, 15, 18, 15 and 15.
• All together, there are 10 distances. But 20 km should be included in the upper class 20-25
• So only 9 distances are eligible to be included in the class-interval 15-20
• Thus, the frequency of the class-interval 15-20 is 9
Following are some useful information that can be obtained from the tabular representation:
1. No. of engineers who travel 15 km or more to their place of work is 9 +1 +1 +2 = 13
2. No. of engineers who travel 25 km or more to their place of work is 1 +2 = 3
3. No. of engineers who travel less than 10 km to their place of work is 5 +11 = 16
In the next section we will see more solved examples.
1. The marks obtained (out of 100) by 98 students in a national level exam are given below:
2. Here the amount of data is large. So we make them into groups called class-intervals. The class-intervals will be: 30-39, 40-49, 50-59 . . . upto 90-99
• The first class-interval is 30-39 because the smallest value is 32
• The last class-interval is 90-99 because the largest value is 97
3. The size of class-intervals is called class-width
4. In any class-interval,
• the lower number is called the lower class limit
• the upper number is called the upper class limit
5. Once we decide the class-interval, the next step is this:
• Find the number of values that will fall in each class-interval.
6. For that, we use the method of tally marks. This is shown in the table below:
Greater number of students fall in the class-interval 50-59. That means, the scores of the largest number of students fall between 50 and 59, both scores inclusive.
■ One information that we get when we do a little analysis is:
Number of students who scored 70 or more marks is: 15+14+10 = 39
■ In this manner, many more information can be obtained from the final presentation.
Consider the following grouped frequency distribution table in fig.25.1(a) below:
 |
| Fig.25.1 |
Solution:
1. Take 35.5 kg. We cannot include it in the class 31-35. Because the upper limit of this interval is only 35. Our value of 35.5 falls above this class-interval 31-35.
2. We cannot include it in the class 36-40. Because the lower limit of this interval is 36. Our value of 35.5 falls below this class-interval 36-40.
3. The same happens to 40.5 kg: It falls above 36-40 and it falls below 41-45
The values 35.5 and 40.5 have no place in the table. This happens because, there are gaps in between class-intervals. Such a gap between 31-35 and 36-40 is shown in fig.25.2(a) below:
 |
| Fig.25.2 |
5. For that, the following procedure is adopted:
(i) Divide a gap into two equal parts
• In fig.b, the gap between 31-35 and 36-40 is divided into two equal parts.
• This is done by marking 35.5 at the exact midway between 35 and 36
(ii) Take the interval on the left side of the gap. Bring it’s upper limit to the midpoint
• In the fig.c, the upper limit 35 of the left side interval is brought up to 35.5
(iii) Take the interval on the right side of the gap. Bring it’s lower limit to the midpoint
• In the fig.c, the lower limit 36 of the right side interval is brought down to 35.5
(iv) So the gap is covered
6. This procedure should be done at all the gaps. So both the upper and lower limits of all the class-intervals will change.
7. Let us find the new limits of all the given intervals in the problem:
■ 1st class-interval is 31-35
• 2nd class-interval is 36-40
• Gap between them = lower limit of 2nd – upper limit of 1st = 36-35 = 1
• Half of the gap = 1/2 = 0.5
• New upper limit of 1st interval = 35 + 0.5 = 35.5
• New lower limit of 2nd interval = 36 - 0.5 = 35.5
• Thus the gap between 31-35 and 36-40 is avoided
■ 2nd class-interval is 36-40
• 3rd class-interval is 41-45
• Gap between them = lower limit of 3rd – upper limit of 2nd = 41-40 = 1
• Half of the gap = 1/2 = 0.5
• New upper limit of 2nd interval = 40 + 0.5 = 40.5
• New lower limit of 3rd interval = 41 - 0.5 = 40.5
• Thus the gap between 36-40 and 41-45 is avoided
8. In this way, we can find the new limits of all the given intervals. But now another problem arises: (i) The weight of the new students are 35.5 and 40.5
(ii) Consider 35.5. It is eligible to fall in two intervals. 30.5-35.5 and 35.5-40.5
(iii) In such a situation, we follow the procedure that is accepted all over the world. That is., such a value is taken into the upper of the two eligible classes.
(iv) So 35.5 is taken into 35.5-40.5
Similarly 40.5 is taken into 40.5-45.5. And not 35.5-40.5
9. The modified table is shown in fig.25.1(b)
Let any two consecutive class intervals with gap in between them, be denoted as the 1st and 2nd class intervals
Step 1: Gap = lower limit of 2nd – upper limit of 1st.
Step 2: New upper limit of 1st = Original upper limit + Gap⁄2
Step 3: New lower limit of 2nd = Original lower limit - Gap⁄2
Now we will see some solved examples
Solved example 25.1The blood groups of 30 students of a class are recorded as follows:
Solution:
The required frequency distribution table is given below:
From the table, we can see that,
• The most common blood group is 'O'
• The rarest blood group is AB
Solved example 25.2
The distance (in km) of 40 engineers from their residence to their place of work were found as follows:
Solution:
• The minimum distance in the raw data is 2 km. So the first interval of 0-5 is appropriate.
• It is given that class size is to be 5. So the various class-intervals are: 0-5, 5-10, 10-15, ...
• The maximum distance in the raw data is 32 km. So the last class-interval will be 30-35
• It is given that, in the class-interval 0-5, the distance 5 km should not be included. So, if there is a distance of 5 km in the raw data, it should be included in the higher class 5-10
• Similarly, if there is a distance of 10 km in the raw data, it should not be included in the interval 5-10. It should be included in the higher class 10-15 only
• Based on the above facts, the grouped frequency distribution table is constructed as shown below:
• Consider the class-interval 15-20. The distances from 15 to 20 in the raw data are:
20, 19, 17, 18, 17, 16, 15, 18, 15 and 15.
• All together, there are 10 distances. But 20 km should be included in the upper class 20-25
• So only 9 distances are eligible to be included in the class-interval 15-20
• Thus, the frequency of the class-interval 15-20 is 9
Following are some useful information that can be obtained from the tabular representation:
1. No. of engineers who travel 15 km or more to their place of work is 9 +1 +1 +2 = 13
2. No. of engineers who travel 25 km or more to their place of work is 1 +2 = 3
3. No. of engineers who travel less than 10 km to their place of work is 5 +11 = 16
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