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.

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

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.