Find the median of the following distribution:
`{:(x,f," "x,f),(10,2," "30,4),(20,3," "40,5),(50,6," "80,3),(60,5," "90,3),(70,4," "100,1):}`

Answer

To find the median of the given distribution, we will follow these steps:

### Step 1: Organize the data
We have the following values of \( x \) and their corresponding frequencies \( f \):

| \( x \) | \( f \) |
|---------|---------|
| 10 | 2 |
| 20 | 3 |
| 50 | 6 |
| 60 | 5 |
| 70 | 4 |
| 80 | 3 |
| 90 | 3 |
| 100 | 1 |

### Step 2: Calculate the cumulative frequency
We will calculate the cumulative frequency \( CF \) by adding the frequencies cumulatively:

| \( x \) | \( f \) | \( CF \) |
|---------|---------|----------|
| 10 | 2 | 2 |
| 20 | 3 | 5 |
| 50 | 6 | 11 |
| 60 | 5 | 16 |
| 70 | 4 | 20 |
| 80 | 3 | 23 |
| 90 | 3 | 26 |
| 100 | 1 | 27 |

### Step 3: Find the total frequency \( n \)
Now, we sum the frequencies to find \( n \):
\[ n = 2 + 3 + 6 + 5 + 4 + 3 + 3 + 1 = 27 \]

### Step 4: Calculate \( n/2 \)
Next, we calculate \( n/2 \):
\[ n/2 = 27/2 = 13.5 \]

### Step 5: Locate the median class
We need to find the cumulative frequency that is just greater than \( 13.5 \). From the cumulative frequency table, we see that:
- The cumulative frequency of 11 (for \( x = 50 \)) is less than 13.5.
- The cumulative frequency of 16 (for \( x = 60 \)) is greater than 13.5.

Thus, the median class is \( 60 \).

### Step 6: Apply the median formula
The median can be calculated using the formula:
\[ \text{Median} = L + \left( \frac{n/2 - CF}{f} \right) \times h \]
Where:
- \( L \) = lower boundary of the median class = 60
- \( n/2 \) = 13.5
- \( CF \) = cumulative frequency of the class before the median class = 11
- \( f \) = frequency of the median class = 5
- \( h \) = class width (assuming classes are continuous, we can take \( h = 10 \))

Substituting the values:
\[ \text{Median} = 60 + \left( \frac{13.5 - 11}{5} \right) \times 10 \]
\[ = 60 + \left( \frac{2.5}{5} \right) \times 10 \]
\[ = 60 + 0.5 \times 10 \]
\[ = 60 + 5 = 65 \]

### Final Answer
The median of the distribution is **65**.

---