The class marks in a frequency table are given to be 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. The class limits of the first five classes are
2.5-7.5, 7.5-12.5, 12.5-17.5, 17.5-22.5, 22.5-27.5
Understanding Class Marks and Class Limits in Frequency Tables
In a frequency table, data is grouped into classes or class intervals. Each class has a lower limit and an upper limit. The class mark is the midpoint of a class interval and is calculated as the average of the lower and upper limits of that class.
We are given a series of class marks: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50. We need to find the class limits for the first five classes.
Calculating Class Width
When class marks are given consecutively for class intervals of uniform width, the class width is the difference between any two consecutive class marks.
Let's calculate the difference between the first two class marks: Class width = 10 - 5 = 5.
We can verify this with other consecutive class marks: 15 - 10 = 5 20 - 15 = 5 ...and so on. So, the uniform class width is 5.
Relating Class Mark and Class Limits
For any class interval with lower limit \(L\) and upper limit \(U\), the class mark \(M\) and class width \(w\) are related by the following formulas:
- Class Mark (\(M\)) = \(\frac{L + U}{2}\)
- Class Width (\(w\)) = \(U - L\)
From these, we can derive formulas to find the lower and upper limits if we know the class mark and class width:
- Lower Limit (\(L\)) = \(M - \frac{w}{2}\)
- Upper Limit (\(U\)) = \(M + \frac{w}{2}\)
Finding the Class Limits for the First Five Classes
We have the class width \(w = 5\). Now we will use the derived formulas to find the class limits for the first five class marks (5, 10, 15, 20, 25).
Class 1: Class Mark = 5
- Lower Limit (\(L_1\)) = \(5 - \frac{5}{2} = 5 - 2.5 = 2.5\)
- Upper Limit (\(U_1\)) = \(5 + \frac{5}{2} = 5 + 2.5 = 7.5\)
The first class interval is 2.5 - 7.5.
Class 2: Class Mark = 10
- Lower Limit (\(L_2\)) = \(10 - \frac{5}{2} = 10 - 2.5 = 7.5\)
- Upper Limit (\(U_2\)) = \(10 + \frac{5}{2} = 10 + 2.5 = 12.5\)
The second class interval is 7.5 - 12.5.
Class 3: Class Mark = 15
- Lower Limit (\(L_3\)) = \(15 - \frac{5}{2} = 15 - 2.5 = 12.5\)
- Upper Limit (\(U_3\)) = \(15 + \frac{5}{2} = 15 + 2.5 = 17.5\)
The third class interval is 12.5 - 17.5.
Class 4: Class Mark = 20
- Lower Limit (\(L_4\)) = \(20 - \frac{5}{2} = 20 - 2.5 = 17.5\)
- Upper Limit (\(U_4\)) = \(20 + \frac{5}{2} = 20 + 2.5 = 22.5\)
The fourth class interval is 17.5 - 22.5.
Class 5: Class Mark = 25
- Lower Limit (\(L_5\)) = \(25 - \frac{5}{2} = 25 - 2.5 = 22.5\)
- Upper Limit (\(U_5\)) = \(25 + \frac{5}{2} = 25 + 2.5 = 27.5\)
The fifth class interval is 22.5 - 27.5.
Summary of First Five Class Limits
The class limits for the first five classes are:
- 2.5 - 7.5
- 7.5 - 12.5
- 12.5 - 17.5
- 17.5 - 22.5
- 22.5 - 27.5
Comparison with Options
Let's compare our calculated class limits with the given options:
| Calculated Limits | Option 1 | Option 2 | Option 3 | Option 4 |
|---|---|---|---|---|
| 2.5 - 7.5 | 3 - 7 | 2.5 - 7.5 | 1.5 - 8.5 | 2 - 8 |
| 7.5 - 12.5 | 7 - 13 | 7.5 - 12.5 | 8.5 - 11.5 | 8 - 12 |
| 12.5 - 17.5 | 13 - 17 | 12.5 - 17.5 | 11.5 - 18.5 | 12 - 18 |
| 17.5 - 22.5 | 17 - 23 | 17.5 - 22.5 | 18.5 - 21.5 | 18 - 22 |
| 22.5 - 27.5 | 23 - 27 | 22.5 - 27.5 | 21.5 - 28.5 | 22 - 28 |
The calculated class limits match the intervals provided in Option 2.
Revision Table: Key Concepts in Frequency Distribution
| Concept | Definition | How to Calculate (if applicable) |
|---|---|---|
| Frequency Distribution | A table that shows the frequency of occurrence of data values in different classes. | Organize data into classes and count occurrences. |
| Class Interval | A range of values within which the data is grouped. Represented as Lower Limit - Upper Limit. | Decided based on range of data and number of classes. |
| Lower Class Limit | The smallest value that can be included in a class interval. | \(L = M - w/2\) |
| Upper Class Limit | The largest value that can be included in a class interval. | \(U = M + w/2\) |
| Class Mark (Midpoint) | The middle value of a class interval. | \(M = (L + U)/2\) |
| Class Width (Size) | The difference between the upper and lower class limits (for exclusive classes) or the difference between two consecutive lower limits (or upper limits). | \(w = U - L\) or Difference between consecutive class marks. |
| Frequency | The number of data values that fall within a specific class interval. | Count of observations in the class. |
Additional Information: Types of Class Intervals
Class intervals can be represented in different ways, affecting how limits and boundaries are defined:
- Exclusive Method: The upper limit of one class is the lower limit of the next class. The upper limit is generally not included in the class. For example, 10-20, 20-30. A value of 20 would fall in the 20-30 class, not 10-20. This is common for continuous data. The class limits calculated in this problem (2.5-7.5, etc.) follow this pattern if interpreted strictly, where 7.5 is the upper limit of the first class and the lower limit of the second, etc.
- Inclusive Method: Both the lower and upper limits are included in the class. The upper limit of one class is not the lower limit of the next class; there is a gap between them. For example, 10-19, 20-29. This is common for discrete data. To get continuous boundaries for calculations (like finding median, quartiles), class boundaries are calculated. Class Boundary = Lower Limit - 0.5 (or half the gap) and Upper Limit + 0.5 (or half the gap). In our problem, the calculated limits 2.5-7.5 etc. are effectively class boundaries used with the exclusive method, which makes sense as consecutive class marks imply continuous data or continuous representation.
Understanding the difference between class limits and class boundaries is crucial, especially when dealing with grouped data for calculating measures like the mean, median, and mode. In the exclusive method, class limits often serve as class boundaries.
Similar Questions
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Important Questions from Classification of Data
Consider the following LPP.:
Max Z = 15x 1 + 10x 2
Subject to the constraints
4x 1 + 6x 2 ≤ 360
3x 1 + 0x 2 ≤ 180
0x 1 + 5x 2 ≤ 200
x 1, x 2 ≥ 0
The solution of the LPP using Graphical solution-technique is :
A graph of a cumulative frequency distribution is called :
Which of the following is not an example of compressed data?
A cumulative frequency distribution is given below
Class
60-62
63-65
66-68
69-71
72-74
Cumulative frequency
3
20
36
48
50
Which one of the following class has maximum frequency?
The measure of the central tendency is given by the X-coordinate of the point of intersection of the more than ogive and less than ogive is:
