Find the median from the following frequency distribution :
`{:(x,3,4,5,6,7,8,9,10,11),(f,13,90,81,117,66,27,6,2,2):}`

Answer

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

### Step 1: Create a frequency table
We will organize the data into a frequency table that includes the values (x) and their corresponding frequencies (f).

| x | f |
|----|----|
| 3 | 13 |
| 4 | 90 |
| 5 | 81 |
| 6 | 117|
| 7 | 66 |
| 8 | 27 |
| 9 | 6 |
| 10 | 2 |
| 11 | 2 |

### Step 2: Calculate the cumulative frequency
Next, we will calculate the cumulative frequency (CF) for each value. This is done by adding the frequency of each value to the cumulative frequency of the previous value.

| x | f | Cumulative Frequency (CF) |
|----|----|----------------------------|
| 3 | 13 | 13 |
| 4 | 90 | 103 (13 + 90) |
| 5 | 81 | 184 (103 + 81) |
| 6 | 117| 301 (184 + 117) |
| 7 | 66 | 367 (301 + 66) |
| 8 | 27 | 394 (367 + 27) |
| 9 | 6 | 400 (394 + 6) |
| 10 | 2 | 402 (400 + 2) |
| 11 | 2 | 404 (402 + 2) |

### Step 3: Find the total frequency (N)
Now, we will find the total frequency (N) by summing all the frequencies.

N = 404

### Step 4: Calculate N/2
To find the median, we need to calculate N/2.

N/2 = 404 / 2 = 202

### Step 5: Locate the median class
We will now locate the median class, which is the class where the cumulative frequency is greater than or equal to 202.

From our cumulative frequency table:
- CF for x = 5 is 184 (less than 202)
- CF for x = 6 is 301 (greater than 202)

Thus, the median class is the class corresponding to x = 6.

### Step 6: Determine the median
Since the median lies in the class where the cumulative frequency first exceeds 202, we can conclude that the median of the given frequency distribution is:

**Median = 6**