Number of telephone cells received in 245 succesive one minute intervals at an exchange is given belwo in the following frequency distribution.
`{:("Number of calls",0,1,2,3,4,5,6,7),("Frequency",14,21,25,43,51,40,39,12):}`
What is the median of the distribution?

Options

A. 3.5
B. 4
C. 4.5
D. 5

Explanation

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

### Step 1: Organize the Data
We have the following frequency distribution:

| Number of Calls (x) | Frequency (f) |
|----------------------|---------------|
| 0 | 14 |
| 1 | 21 |
| 2 | 25 |
| 3 | 43 |
| 4 | 51 |
| 5 | 40 |
| 6 | 39 |
| 7 | 12 |

### Step 2: Calculate the Cumulative Frequency (cf)
We will calculate the cumulative frequency by adding the frequencies cumulatively.

- For x = 0: cf = 14
- For x = 1: cf = 14 + 21 = 35
- For x = 2: cf = 35 + 25 = 60
- For x = 3: cf = 60 + 43 = 103
- For x = 4: cf = 103 + 51 = 154
- For x = 5: cf = 154 + 40 = 194
- For x = 6: cf = 194 + 39 = 233
- For x = 7: cf = 233 + 12 = 245

Now we can summarize the cumulative frequency:

| Number of Calls (x) | Frequency (f) | Cumulative Frequency (cf) |
|----------------------|---------------|----------------------------|
| 0 | 14 | 14 |
| 1 | 21 | 35 |
| 2 | 25 | 60 |
| 3 | 43 | 103 |
| 4 | 51 | 154 |
| 5 | 40 | 194 |
| 6 | 39 | 233 |
| 7 | 12 | 245 |

### Step 3: Find n and n/2
The total number of observations (n) is 245.

Now, we calculate n/2:
\[ n/2 = 245 / 2 = 122.5 \]

### Step 4: Identify the Median Class
We need to find the class where the cumulative frequency is greater than or equal to 122.5.

From the cumulative frequency table:
- The cumulative frequency just greater than 122.5 is 154 (for x = 4).

Thus, the median class is the class corresponding to x = 4.

### Step 5: Calculate the Median
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 = 4
- \( n \) = total frequency = 245
- \( cf \) = cumulative frequency of the class preceding the median class = 103
- \( f \) = frequency of the median class = 51
- \( h \) = class width = 1 (since the classes are 0-1, 1-2, etc.)

Substituting the values:
\[ \text{Median} = 4 + \left(\frac{122.5 - 103}{51}\right) \times 1 \]
\[ = 4 + \left(\frac{19.5}{51}\right) \]
\[ = 4 + 0.3824 \approx 4.38 \]

### Final Answer
The median of the distribution is approximately **4.38**.