Consider the following distribution Class Frequency 0 - 20 17 20 - 40 28 40 - 60 32 60 - 80 F 80 - 100 19 If the mean of the above distribution is 50 then what is the value of f?
24
Finding Missing Frequency in Grouped Data
This problem asks us to find the value of a missing frequency (denoted by 'f') in a given grouped frequency distribution. We are provided with the class intervals, their corresponding frequencies (except for one), and the mean of the entire distribution. To solve this, we will use the formula for the mean of grouped data.
The formula for the mean (\(\bar{x}\)) of a grouped frequency distribution is:
\[\bar{x} = \frac{\sum (f_i \times x_i)}{\sum f_i}\]
Where:
- \(\sum (f_i \times x_i)\) is the sum of the products of each class frequency (\(f_i\)) and its corresponding class midpoint (\(x_i\)).
- \(\sum f_i\) is the sum of all frequencies.
First, let's determine the midpoint (\(x_i\)) for each class interval. The midpoint of a class interval is calculated as (Lower Limit + Upper Limit) / 2.
| Class Interval | Frequency (\(f_i\)) | Class Midpoint (\(x_i\)) | \(f_i \times x_i\) |
|---|---|---|---|
| 0 - 20 | 17 | \[(0 + 20) / 2 = 10\] | \[17 \times 10 = 170\] |
| 20 - 40 | 28 | \[(20 + 40) / 2 = 30\] | \[28 \times 30 = 840\] |
| 40 - 60 | 32 | \[(40 + 60) / 2 = 50\] | \[32 \times 50 = 1600\] |
| 60 - 80 | f | \[(60 + 80) / 2 = 70\] | \[f \times 70 = 70f\] |
| 80 - 100 | 19 | \[(80 + 100) / 2 = 90\] | \[19 \times 90 = 1710\] |
Next, we need to calculate the sum of frequencies (\(\sum f_i\)) and the sum of the products (\(\sum (f_i \times x_i)\)).
Sum of frequencies (\(\sum f_i\)):
\[\sum f_i = 17 + 28 + 32 + f + 19\] \[\sum f_i = 96 + f\]
Sum of products (\(\sum (f_i \times x_i)\)):
\[\sum (f_i \times x_i) = 170 + 840 + 1600 + 70f + 1710\] \[\sum (f_i \times x_i) = 4320 + 70f\]
We are given that the mean (\(\bar{x}\)) of the distribution is 50. Now we can substitute the values into the mean formula:
\[50 = \frac{4320 + 70f}{96 + f}\]
To solve for 'f', we can cross-multiply:
\[50 \times (96 + f) = 4320 + 70f\]
Distribute the 50 on the left side:
\[50 \times 96 + 50 \times f = 4320 + 70f\] \[4800 + 50f = 4320 + 70f\]
Now, rearrange the equation to isolate the terms with 'f' on one side and the constant terms on the other:
\[4800 - 4320 = 70f - 50f\] \[480 = 20f\]
Finally, divide by 20 to find the value of 'f':
\[f = \frac{480}{20}\] \[f = 24\]
Thus, the value of the missing frequency 'f' is 24.
Revision Table: Key Concepts
Let's quickly summarise the key components used in solving this problem:
| Concept | Description | Relevance to Problem |
|---|---|---|
| Grouped Frequency Distribution | Data organized into class intervals with corresponding frequencies. | The data provided is in this format. |
| Class Midpoint (\(x_i\)) | The central value of a class interval, calculated as the average of its lower and upper limits. | Required for calculating the mean of grouped data. |
| Mean of Grouped Data (\(\bar{x}\)) | A measure of central tendency for grouped data, calculated using the formula \(\frac{\sum (f_i \times x_i)}{\sum f_i}\). | The primary formula used to find the missing frequency. |
| Missing Frequency (f) | An unknown frequency value within the distribution that needs to be determined. | The objective of the problem. |
Additional Information: Calculating Mean Methods
While we used the direct method to calculate the mean in this problem, there are other methods available for calculating the mean of grouped data, especially useful when the numbers are large:
- Assumed Mean Method: This method simplifies calculations by choosing an assumed mean (A) from the class midpoints and calculating deviations (\(d_i = x_i - A\)). The formula is \(\bar{x} = A + \frac{\sum (f_i \times d_i)}{\sum f_i}\).
- Step-Deviation Method: This is a further simplification of the assumed mean method, used when class sizes are equal. It involves calculating step-deviations (\(u_i = \frac{x_i - A}{h}\)), where 'h' is the class size. The formula is \(\bar{x} = A + \left(\frac{\sum (f_i \times u_i)}{\sum f_i}\right) \times h\).
These alternative methods can help reduce the complexity of calculations, particularly in exams without calculators. However, the direct method is conceptually straightforward and sufficient when the values are manageable, as seen in this problem. Understanding these methods provides a broader perspective on handling grouped data.
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