A number equal to 2 times the mean and with a frequency equal to k is inserted in a data having n observation. If the new mean is `(4)/(3)` times the old mean, then the value of `(k)/(n)` is

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the Mean
Let the old mean be denoted as \( \alpha \). The sum of the observations in the dataset with \( n \) observations can be expressed as:
\[ \text{Sum of old observations} = n \alpha \]

### Step 2: Define the New Number
According to the problem, a new number equal to \( 2 \times \text{mean} \) is added, which means:
\[ \text{New number} = 2\alpha \]
This new number is added with a frequency of \( k \), so the total contribution of this new number to the sum is:
\[ \text{Contribution of new number} = k \times (2\alpha) = 2k\alpha \]

### Step 3: Calculate the New Sum
The new sum of all observations after adding the new number is:
\[ \text{New sum} = n\alpha + 2k\alpha = (n + 2k)\alpha \]

### Step 4: Calculate the New Number of Observations
The new number of observations after adding \( k \) observations is:
\[ \text{New number of observations} = n + k \]

### Step 5: Define the New Mean
The new mean can be expressed as:
\[ \text{New mean} = \frac{\text{New sum}}{\text{New number of observations}} = \frac{(n + 2k)\alpha}{n + k} \]

### Step 6: Set Up the Equation
According to the problem, the new mean is \( \frac{4}{3} \) times the old mean:
\[ \frac{(n + 2k)\alpha}{n + k} = \frac{4}{3}\alpha \]

### Step 7: Simplify the Equation
We can cancel \( \alpha \) from both sides (assuming \( \alpha \neq 0 \)):
\[ \frac{n + 2k}{n + k} = \frac{4}{3} \]

### Step 8: Cross-Multiply
Cross-multiplying gives:
\[ 3(n + 2k) = 4(n + k) \]

### Step 9: Expand and Rearrange
Expanding both sides:
\[ 3n + 6k = 4n + 4k \]
Rearranging gives:
\[ 6k - 4k = 4n - 3n \]
\[ 2k = n \]

### Step 10: Solve for \( \frac{k}{n} \)
Dividing both sides by \( n \):
\[ \frac{k}{n} = \frac{1}{2} \]

### Final Answer
Thus, the value of \( \frac{k}{n} \) is:
\[ \frac{k}{n} = 0.5 \]

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