The table below gives an incomplete frequency distribution with two missing frequencies `f_(1) and f_(2)`
`|{:("Value of x","Frequency"),(0,f_(1)),(1,f_(2)),(2,4),(3,4),(4,3):}|`
The total frequency is 18 and the arithmetic mean of x is 2.
What is the value of `f_(2)`?

Options

A. 4
B. 3
C. 2
D. 1

Explanation

To find the value of \( f_2 \) in the given frequency distribution, we will follow these steps:

### Step 1: Set Up the Given Information
We have the following frequency distribution:
- \( f_1 \) for \( x = 0 \)
- \( f_2 \) for \( x = 1 \)
- \( 4 \) for \( x = 2 \)
- \( 4 \) for \( x = 3 \)
- \( 3 \) for \( x = 4 \)

The total frequency is given as \( 18 \) and the arithmetic mean of \( x \) is \( 2 \).

### Step 2: Write the Total Frequency Equation
The total frequency can be expressed as:
\[ f_1 + f_2 + 4 + 4 + 3 = 18 \]
This simplifies to:
\[ f_1 + f_2 + 11 = 18 \]
Thus, we can derive:
\[ f_1 + f_2 = 7 \quad \text{(Equation 1)} \]

### Step 3: Calculate the Weighted Sum \( \Sigma xf \)
Next, we calculate \( \Sigma xf \):
\[ \Sigma xf = (0 \cdot f_1) + (1 \cdot f_2) + (2 \cdot 4) + (3 \cdot 4) + (4 \cdot 3) \]
This simplifies to:
\[ 0 + f_2 + 8 + 12 + 12 = f_2 + 32 \]

### Step 4: Use the Mean Formula
The formula for the arithmetic mean is given by:
\[ \text{Mean} = \frac{\Sigma xf}{\Sigma f} \]
Substituting the known values:
\[ 2 = \frac{f_2 + 32}{18} \]

### Step 5: Solve for \( f_2 \)
To isolate \( f_2 \), we multiply both sides by \( 18 \):
\[ 36 = f_2 + 32 \]
Now, subtract \( 32 \) from both sides:
\[ f_2 = 36 - 32 = 4 \]

### Step 6: Substitute \( f_2 \) Back to Find \( f_1 \)
Using Equation 1:
\[ f_1 + 4 = 7 \]
Thus:
\[ f_1 = 7 - 4 = 3 \]

### Conclusion
The values of the missing frequencies are:
- \( f_1 = 3 \)
- \( f_2 = 4 \)

Therefore, the value of \( f_2 \) is \( 4 \).

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