Consider the following for the next items that follow: A grouped frequency distribution is given below: Weekly wages in Rupees (Rs.) Numbers of workers 2050 - 2550 5 2550 - 3050 10 3050 - 3550 k 3550 - 4050 8 4050 - 4550 2 4550 - 5050 10
If average weekly wages earned by a worker is Rs. 3,520, then what is the value of k?
15
Finding Missing Frequency in Grouped Frequency Distribution
The problem provides a grouped frequency distribution of weekly wages for workers and the average weekly wage earned by a worker. We are asked to find the value of the missing frequency, denoted by 'k', in one of the wage groups.
To solve this, we will use the formula for calculating the mean (average) of a grouped frequency distribution.
Mean Formula for Grouped Data
The mean (\(\bar{x}\)) for grouped data is calculated using the formula:
\( \bar{x} = \frac{\sum (f_i \times m_i)}{\sum f_i} \)
Where:
- \(f_i\) is the frequency of the i-th class (number of workers in a wage group).
- \(m_i\) is the midpoint of the i-th class (the average wage within a group).
- \(\sum (f_i \times m_i)\) is the sum of the products of each frequency and its corresponding midpoint.
- \(\sum f_i\) is the sum of all frequencies (total number of workers).
Calculating Midpoints and \(f_i \times m_i\)
First, we need to find the midpoint (\(m_i\)) for each class interval. The midpoint is the average of the lower and upper limits of the class interval.
- For 2050 - 2550: \(m_1 = \frac{2050 + 2550}{2} = \frac{4600}{2} = 2300\)
- For 2550 - 3050: \(m_2 = \frac{2550 + 3050}{2} = \frac{5600}{2} = 2800\)
- For 3050 - 3550: \(m_3 = \frac{3050 + 3550}{2} = \frac{6600}{2} = 3300\)
- For 3550 - 4050: \(m_4 = \frac{3550 + 4050}{2} = \frac{7600}{2} = 3800\)
- For 4050 - 4550: \(m_5 = \frac{4050 + 4550}{2} = \frac{8600}{2} = 4300\)
- For 4550 - 5050: \(m_6 = \frac{4550 + 5050}{2} = \frac{9600}{2} = 4800\)
Now, let's organize the data and calculate \(f_i \times m_i\) for each class in a table.
| Weekly Wages (Rs.) | Number of workers (\(f_i\)) | Midpoint (\(m_i\)) | \(f_i \times m_i\) |
|---|---|---|---|
| 2050 - 2550 | 5 | 2300 | \(5 \times 2300 = 11500\) |
| 2550 - 3050 | 10 | 2800 | \(10 \times 2800 = 28000\) |
| 3050 - 3550 | k | 3300 | \(k \times 3300 = 3300k\) |
| 3550 - 4050 | 8 | 3800 | \(8 \times 3800 = 30400\) |
| 4050 - 4550 | 2 | 4300 | \(2 \times 4300 = 8600\) |
| 4550 - 5050 | 10 | 4800 | \(10 \times 4800 = 48000\) |
Summing Frequencies and Products
Next, we calculate the sum of frequencies (\(\sum f_i\)) and the sum of the products (\(\sum f_i \times m_i\)).
Sum of frequencies: \( \sum f_i = 5 + 10 + k + 8 + 2 + 10 = 35 + k \)
Sum of products \(f_i \times m_i\): \( \sum (f_i \times m_i) = 11500 + 28000 + 3300k + 30400 + 8600 + 48000 \) \( \sum (f_i \times m_i) = (11500 + 28000 + 30400 + 8600 + 48000) + 3300k \) \( \sum (f_i \times m_i) = 126500 + 3300k \)
Solving for the Missing Frequency 'k'
We are given that the average weekly wage is Rs. 3,520. Using the mean formula, we can set up an equation:
\( 3520 = \frac{126500 + 3300k}{35 + k} \)
Now, we solve this equation for k:
Multiply both sides by \((35 + k)\):
\( 3520 \times (35 + k) = 126500 + 3300k \)
Distribute 3520 on the left side:
\( (3520 \times 35) + (3520 \times k) = 126500 + 3300k \) \( 123200 + 3520k = 126500 + 3300k \)
Subtract 3300k from both sides:
\( 123200 + 3520k - 3300k = 126500 \) \( 123200 + 220k = 126500 \)
Subtract 123200 from both sides:
\( 220k = 126500 - 123200 \) \( 220k = 3300 \)
Divide by 220:
\( k = \frac{3300}{220} \) \( k = \frac{330}{22} \) \( k = 15 \)
Thus, the value of k is 15.
Revision Table - Grouped Data Calculation
| Concept | Description | Relevance Here |
|---|---|---|
| Grouped Frequency Distribution | Data organized into class intervals with corresponding frequencies. | The dataset provided is in this format. |
| Class Midpoint (\(m_i\)) | The average of the lower and upper limits of a class interval. Represents the typical value for that interval. | Used as the value for each observation within a class to calculate the mean. |
| Frequency (\(f_i\)) | The number of observations falling into a specific class interval. | Given for most classes, with 'k' being the unknown frequency we need to find. |
| Mean of Grouped Data | An estimate of the average value in the dataset, calculated using class midpoints and frequencies. | The formula \(\frac{\sum f_i m_i}{\sum f_i}\) is central to solving this problem. |
Additional Information - Statistics Concepts
Understanding grouped frequency distributions and how to calculate measures like the mean is fundamental in statistics. Grouping data helps summarize large datasets, but it does involve some loss of individual data point precision, which is why we use midpoints to estimate.
There are other methods to calculate the mean for grouped data, such as the assumed mean method or step-deviation method. These methods simplify calculations when dealing with large numbers or wide class intervals, but the direct method (using \(\sum f_i m_i / \sum f_i\)) is straightforward for any dataset and is what we used here.
When working with grouped data, it's important that the class intervals are mutually exclusive (no overlap) and exhaustive (cover the entire range of the data). In this problem, the intervals are consecutive and cover a range suitable for weekly wages.
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3
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