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 coefficient of variance ?

Options

A. `(200)/(3)`
B. `(50sqrt5)/(9)`
C. `(600)/(sqrt5)`
D. 150

Explanation

To solve the problem, we need to find the missing frequencies \( f_1 \) and \( f_2 \) and then calculate the coefficient of variation. Here’s a step-by-step solution:

### Step 1: Set up the frequency distribution
We have the following frequency distribution:

| Value of x | Frequency |
|------------|-----------|
| 0 | \( f_1 \) |
| 1 | \( f_2 \) |
| 2 | 4 |
| 3 | 4 |
| 4 | 3 |

The total frequency is given as 18. Therefore, we can write the equation:

\[ f_1 + f_2 + 4 + 4 + 3 = 18 \]

### Step 2: Simplify the equation
Calculating the sum of the known frequencies:

\[ f_1 + f_2 + 11 = 18 \]

Now, we can isolate \( f_1 + f_2 \):

\[ f_1 + f_2 = 18 - 11 = 7 \quad \text{(Equation 1)} \]

### Step 3: Use the mean to find another equation
The arithmetic mean is given as 2. The formula for the mean is:

\[ \text{Mean} = \frac{\sum (x \cdot f)}{\sum f} \]

Calculating \( \sum (x \cdot f) \):

\[ \sum (x \cdot f) = 0 \cdot f_1 + 1 \cdot f_2 + 2 \cdot 4 + 3 \cdot 4 + 4 \cdot 3 \]

This simplifies to:

\[ 0 + f_2 + 8 + 12 + 12 = f_2 + 32 \]

Now, substituting into the mean formula:

\[ 2 = \frac{f_2 + 32}{18} \]

### Step 4: Solve for \( f_2 \)
Multiplying both sides by 18:

\[ 36 = f_2 + 32 \]

Now, isolating \( f_2 \):

\[ f_2 = 36 - 32 = 4 \quad \text{(Equation 2)} \]

### Step 5: Substitute \( f_2 \) back into Equation 1
Now we substitute \( f_2 = 4 \) into Equation 1:

\[ f_1 + 4 = 7 \]

Thus,

\[ f_1 = 7 - 4 = 3 \]

### Step 6: Finalize the frequency distribution
Now we have:

| Value of x | Frequency |
|------------|-----------|
| 0 | 3 |
| 1 | 4 |
| 2 | 4 |
| 3 | 4 |
| 4 | 3 |

### Step 7: Calculate the standard deviation
We need to calculate the standard deviation. First, we calculate \( \sum (x^2 \cdot f) \):

\[ \sum (x^2 \cdot f) = 0^2 \cdot 3 + 1^2 \cdot 4 + 2^2 \cdot 4 + 3^2 \cdot 4 + 4^2 \cdot 3 \]

Calculating this gives:

\[ 0 + 4 + 16 + 36 + 48 = 104 \]

Now, we can calculate the variance:

\[ \text{Variance} = \frac{\sum (x^2 \cdot f)}{n} - \left(\frac{\sum (x \cdot f)}{n}\right)^2 \]

Where \( n = 18 \):

\[ \text{Variance} = \frac{104}{18} - \left(\frac{36}{18}\right)^2 \]

Calculating:

\[ \text{Variance} = \frac{104}{18} - 4 = \frac{104 - 72}{18} = \frac{32}{18} = \frac{16}{9} \]

Thus, the standard deviation \( \sigma \) is:

\[ \sigma = \sqrt{\frac{16}{9}} = \frac{4}{3} \]

### Step 8: Calculate the coefficient of variation
The coefficient of variation (CV) is given by:

\[ \text{CV} = \frac{\sigma}{\text{Mean}} \times 100 \]

Substituting the values:

\[ \text{CV} = \frac{\frac{4}{3}}{2} \times 100 = \frac{4}{3} \times \frac{1}{2} \times 100 = \frac{200}{3} \]

### Final Answer
The coefficient of variation is \( \frac{200}{3} \).

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