If `x lt 6` and 17.5 is the mode of the following frequency distribution.
`{:("Class-interval:", 0-5,5-10,10-15,15-20,20-25),("Frequency:", 5,2,3,6,x):}`
Then, x =

Options

A. 3
B. 2
C. 4
D. 5

Explanation

To find the value of \( x \) in the given frequency distribution where the mode is \( 17.5 \), we can follow these steps:

### Step 1: Identify the Modal Class
The modal class is the class interval with the highest frequency. The given frequencies are:
- For \( 0-5 \): 5
- For \( 5-10 \): 2
- For \( 10-15 \): 3
- For \( 15-20 \): 6
- For \( 20-25 \): \( x \)

Since \( x < 6 \), the maximum frequency is \( 6 \) for the class interval \( 15-20 \). Thus, the modal class is \( 15-20 \).

### Step 2: Define the Parameters for the Mode Formula
The mode can be calculated using the formula:
\[ \text{Mode} = l + \frac{\delta_1}{\delta_1 + \delta_2} \cdot i \]
Where:
- \( l \) = lower limit of the modal class = \( 15 \)
- \( i \) = width of the class interval = \( 5 \) (since \( 20 - 15 = 5 \))
- \( \delta_1 \) = frequency of modal class - frequency of pre-modal class
- \( \delta_2 \) = frequency of modal class - frequency of post-modal class

### Step 3: Calculate \( \delta_1 \) and \( \delta_2 \)
- Frequency of modal class \( (15-20) \) = \( 6 \)
- Frequency of pre-modal class \( (10-15) \) = \( 3 \)
- Frequency of post-modal class \( (20-25) \) = \( x \)

Calculating \( \delta_1 \):
\[ \delta_1 = 6 - 3 = 3 \]

Calculating \( \delta_2 \):
\[ \delta_2 = 6 - x \]

### Step 4: Substitute Values into the Mode Formula
Given that the mode is \( 17.5 \):
\[ 17.5 = 15 + \frac{3}{3 + (6 - x)} \cdot 5 \]

### Step 5: Simplify the Equation
Subtract \( 15 \) from both sides:
\[ 2.5 = \frac{3}{3 + 6 - x} \cdot 5 \]
\[ 2.5 = \frac{3 \cdot 5}{9 - x} \]
\[ 2.5(9 - x) = 15 \]

### Step 6: Solve for \( x \)
Distributing \( 2.5 \):
\[ 22.5 - 2.5x = 15 \]
Rearranging gives:
\[ 22.5 - 15 = 2.5x \]
\[ 7.5 = 2.5x \]
Dividing both sides by \( 2.5 \):
\[ x = \frac{7.5}{2.5} = 3 \]

### Conclusion
Thus, the value of \( x \) is \( 3 \).