Consider the following distribution: Marks obtained No. of students More than or equal to zero 63 More than or equal to 10 58 More than or equal to 20 55 More than or equal to 30 51 More than or equal to 40 48 More than or equal to 50 42
The frequency of class 30-40 is:
3
Calculating Frequency from More Than or Equal To Cumulative Distribution
The given table provides a "more than or equal to" cumulative frequency distribution. This means that the number listed against a certain mark represents the total number of students who obtained marks greater than or equal to that specific mark.
To find the frequency of a particular class interval, say c – d, where c > d, in a "more than or equal to" distribution, we subtract the number of students who scored "more than or equal to c" from the number of students who scored "more than or equal to d".
Let's represent the given data:
| Marks Obtained ($\ge$) | Number of students |
|---|---|
| 0 | 63 |
| 10 | 58 |
| 20 | 55 |
| 30 | 51 |
| 40 | 48 |
| 50 | 42 |
We are asked to find the frequency of the class 30-40.
The number of students who scored "more than or equal to 30" marks is 51.
The number of students who scored "more than or equal to 40" marks is 48.
The students who scored marks in the range 30-40 (i.e., $\ge$ 30 but < 40) are those who scored $\ge$ 30 minus those who scored $\ge$ 40.
Frequency of class 30-40 = (Number of students with marks $\ge$ 30) - (Number of students with marks $\ge$ 40)
Frequency of class 30-40 = $51 - 48 = 3$
Let's convert the entire distribution to a simple frequency distribution for better understanding:
- Class 0-10: Students with marks $\ge$ 0 minus students with marks $\ge$ 10 = $63 - 58 = 5$
- Class 10-20: Students with marks $\ge$ 10 minus students with marks $\ge$ 20 = $58 - 55 = 3$
- Class 20-30: Students with marks $\ge$ 20 minus students with marks $\ge$ 30 = $55 - 51 = 4$
- Class 30-40: Students with marks $\ge$ 30 minus students with marks $\ge$ 40 = $51 - 48 = 3$
- Class 40-50: Students with marks $\ge$ 40 minus students with marks $\ge$ 50 = $48 - 42 = 6$
- Class 50 and above: Students with marks $\ge$ 50 = $42$
| Class Interval | Frequency |
|---|---|
| 0-10 | 5 |
| 10-20 | 3 |
| 20-30 | 4 |
| 30-40 | 3 |
| 40-50 | 6 |
| 50 and above | 42 |
From the simple frequency distribution, the frequency of the class 30-40 is 3.
Thus, the frequency of class 30-40 is 3.
Revision Table: Cumulative Frequency vs. Simple Frequency
Here's a quick summary of how to convert "more than or equal to" cumulative frequency to simple frequency:
| Concept | Description | Calculation for Class c-d (c > d) |
|---|---|---|
| More Than or Equal To Cumulative Frequency (MTECF) | Number of observations ≥ a certain value. Decreases as the value increases. | MTECF at d = Number of observations ≥ d MTECF at c = Number of observations ≥ c |
| Simple Frequency | Number of observations within a specific class interval. | Frequency of c-d = (MTECF at d) - (MTECF at c) |
Additional Information: Types of Cumulative Frequency Distributions
There are two main types of cumulative frequency distributions:
- Less Than Cumulative Frequency: The cumulative frequency against a value represents the total number of observations less than that value. This frequency increases as the value increases. To find the simple frequency of a class a-b (a < b), you subtract the 'less than' cumulative frequency at 'a' from the 'less than' cumulative frequency at 'b'.
- More Than or Equal To Cumulative Frequency: As discussed in this solution, the cumulative frequency against a value represents the total number of observations greater than or equal to that value. This frequency decreases as the value increases. To find the simple frequency of a class a-b (a < b), you subtract the 'more than or equal to' cumulative frequency at 'b' from the 'more than or equal to' cumulative frequency at 'a'.
Understanding how to convert between simple frequency and cumulative frequency distributions is fundamental in statistics for creating various graphical representations like histograms, frequency polygons, and ogives.
Important Questions from Classification of Data
A set of annual numerical data, comparable over the years, is given for the last 12 years.
Consider the following statements:
1. The data is best represented by a broken line graph, each corner (turning point) representing the data of one year.
2. Such a graph depicts the chronological change and also enables one to make a short-term forecast.
Which of the above statements is/are correct?Consider the following statements:
Statement 1: Range is not a good measure of dispersion.
Statement 2: Range is highly affected by the existence of extreme values.
Which one of the following is correct in respect of the above statements?
Data can be represented in which of the following forms?
1. Textual form
2. Tabula form
3. Graphical form
Select the correct answer using the code given below.Diagrammatic representation of data includes which of the following?
1. Bar diagram
2. Pie-diagram
3. Pictogram
Select the correct answer using the code given below:The data collected from which one of the following methods isnot a primary data?
