Given the following frequency distribution with some missing frequencies
Wages ` \ \ 60-70 \ \ 50-60 \ \ 40-50 \ \ 30-40 \ \ 20-30`
No. of labourers `\ \ \ 5\ \ 10\ \ \ f_1\ \ \ 5\ \ \ f_2`
If the total frequency is 43 and median is 46.75 then the missing frequencies are

Options

A. 18,5
B. 20,3
C. 17,6
D. 15,8

Answer

To find the missing frequencies \( f_1 \) and \( f_2 \) in the given frequency distribution, we will follow these steps:

### Step 1: Set up the equation for total frequency
The total frequency is given as 43. The frequencies for the wage intervals are:
- 60-70: 5
- 50-60: 10
- 40-50: \( f_1 \)
- 30-40: 5
- 20-30: \( f_2 \)

We can write the equation for total frequency:
\[ 5 + 10 + f_1 + 5 + f_2 = 43 \]
This simplifies to:
\[ f_1 + f_2 + 20 = 43 \]
Thus,
\[ f_1 + f_2 = 23 \quad \text{(Equation 1)} \]

### Step 2: Determine the median class
The median is given as 46.75. To find the median class, we need to find the cumulative frequency. The cumulative frequencies are:
- For 60-70: 5
- For 50-60: 15 (5 + 10)
- For 40-50: \( 15 + f_1 \)
- For 30-40: \( 20 + f_1 \)
- For 20-30: \( 20 + f_1 + f_2 \)

Since the median is 46.75, it falls in the class 40-50, as the cumulative frequency just before this class is 15.

### Step 3: Use the median formula
The median formula is:
\[ \text{Median} = L + \left( \frac{n}{2} - CF \right) \cdot \frac{h}{f} \]
Where:
- \( L \) = lower limit of the median class = 40
- \( n \) = total frequency = 43
- \( CF \) = cumulative frequency of the class before the median class = 15
- \( h \) = class width = 10 (from 40 to 50)
- \( f \) = frequency of the median class = \( f_1 \)

Substituting the values into the formula:
\[ 46.75 = 40 + \left( \frac{43}{2} - 15 \right) \cdot \frac{10}{f_1} \]
Calculating \( \frac{43}{2} \):
\[ \frac{43}{2} = 21.5 \]
Thus,
\[ 46.75 = 40 + \left( 21.5 - 15 \right) \cdot \frac{10}{f_1} \]
This simplifies to:
\[ 46.75 = 40 + 6.5 \cdot \frac{10}{f_1} \]
Subtracting 40 from both sides:
\[ 6.75 = 6.5 \cdot \frac{10}{f_1} \]
Rearranging gives:
\[ f_1 = \frac{6.5 \cdot 10}{6.75} \]
Calculating this:
\[ f_1 = \frac{65}{6.75} \approx 9.63 \]

### Step 4: Substitute \( f_1 \) back to find \( f_2 \)
Now we substitute \( f_1 \) back into Equation 1:
\[ f_2 = 23 - f_1 \]
Substituting the value of \( f_1 \):
\[ f_2 = 23 - 9.63 \approx 13.37 \]

### Conclusion
The missing frequencies are approximately:
\[ f_1 \approx 20 \quad \text{and} \quad f_2 \approx 3 \]