Given the following frequency distribution with some missing frequencies .
`{:("Class",10-20,20-30,30-40,40-50,50-60,60-70,70-80),(" "f," "180," "-," "34," "180," "136," "-," "50):}`
If the total frequency is 685 and median is 42.6 then the missing frequencies are

Options

A. 81,24
B. 80,25
C. 82,23
D. 83,22

Answer

To find the missing frequencies in the given frequency distribution, we will follow these steps:

### Step 1: Set up the frequency distribution table
We have the following classes and frequencies:

| Class | Frequency (f) |
|---------|---------------|
| 10-20 | 180 |
| 20-30 | f1 |
| 30-40 | 34 |
| 40-50 | 180 |
| 50-60 | 136 |
| 60-70 | f2 |
| 70-80 | 50 |

### Step 2: Write the equation for total frequency
The total frequency is given as 685. Therefore, we can write the equation:

\[ 180 + f1 + 34 + 180 + 136 + f2 + 50 = 685 \]

### Step 3: Simplify the equation
Adding the known frequencies:

\[ 180 + 34 + 180 + 136 + 50 = 580 \]

Thus, we can rewrite the equation as:

\[ 580 + f1 + f2 = 685 \]

### Step 4: Solve for f1 + f2
Rearranging gives us:

\[ f1 + f2 = 685 - 580 \]
\[ f1 + f2 = 105 \quad \text{(Equation 1)} \]

### Step 5: Use the median to find another equation
The median is given as 42.6. Since the median falls in the class 40-50, we can use the median formula:

\[ \text{Median} = L + \left( \frac{N/2 - F}{f} \right) \times h \]

Where:
- \( L = 40 \) (lower boundary of the median class)
- \( N = 685 \) (total frequency)
- \( F = 180 + f1 + 34 = 214 + f1 \) (cumulative frequency before the median class)
- \( f = 180 \) (frequency of the median class)
- \( h = 10 \) (class width)

### Step 6: Substitute values into the median formula
Substituting into the median formula gives:

\[ 42.6 = 40 + \left( \frac{685/2 - (214 + f1)}{180} \right) \times 10 \]

Calculating \( N/2 \):

\[ N/2 = 342.5 \]

Now substituting this into the equation:

\[ 42.6 = 40 + \left( \frac{342.5 - (214 + f1)}{180} \right) \times 10 \]

### Step 7: Solve for f1
Rearranging the equation:

\[ 42.6 - 40 = \left( \frac{342.5 - 214 - f1}{180} \right) \times 10 \]
\[ 2.6 = \left( \frac{128.5 - f1}{180} \right) \times 10 \]
\[ 2.6 = \frac{1285 - 10f1}{180} \]

Multiplying both sides by 180:

\[ 468 = 1285 - 10f1 \]
\[ 10f1 = 1285 - 468 \]
\[ 10f1 = 817 \]
\[ f1 = 81.7 \approx 82 \]

### Step 8: Find f2 using Equation 1
Now, substituting \( f1 \) back into Equation 1:

\[ f1 + f2 = 105 \]
\[ 82 + f2 = 105 \]
\[ f2 = 105 - 82 \]
\[ f2 = 23 \]

### Final Answer
The missing frequencies are:
- \( f1 = 82 \)
- \( f2 = 23 \)