Variance of 3n observation is 4 , mean of first 2n observations is 6 and mean of next n observations is 3 . If 1 is added in first 2n observation and 1 is subtracted from last n observations than find new variance

Answer

To solve the problem step by step, we need to calculate the new variance after modifying the observations as described in the question.

### Step 1: Understand the given data
- Variance of \(3n\) observations is \(4\).
- Mean of the first \(2n\) observations is \(6\).
- Mean of the next \(n\) observations is \(3\).

### Step 2: Calculate the summation of the observations
1. **For the first \(2n\) observations:**
\[ \text{Mean} = \frac{\text{Sum of first } 2n \text{ observations}}{2n} = 6 \implies \text{Sum} = 6 \times 2n = 12n \]

2. **For the next \(n\) observations:**
\[ \text{Mean} = \frac{\text{Sum of next } n \text{ observations}}{n} = 3 \implies \text{Sum} = 3 \times n = 3n \]

3. **Total sum of all \(3n\) observations:**
\[ \text{Total Sum} = 12n + 3n = 15n \]

### Step 3: Calculate the new observations after modifications
1. **Adding \(1\) to the first \(2n\) observations:**
\[ \text{New Sum for first } 2n \text{ observations} = 12n + 2n = 14n \]

2. **Subtracting \(1\) from the last \(n\) observations:**
\[ \text{New Sum for last } n \text{ observations} = 3n - n = 2n \]

3. **Total new sum of all observations:**
\[ \text{New Total Sum} = 14n + 2n = 16n \]

### Step 4: Calculate the new mean
\[ \text{New Mean} = \frac{\text{New Total Sum}}{3n} = \frac{16n}{3n} = \frac{16}{3} \]

### Step 5: Calculate the new variance
1. **Using the formula for variance:**
\[ \text{Variance} = \frac{\sum (x_i^2)}{N} - \left(\frac{\sum x_i}{N}\right)^2 \]
Given that the variance of the original \(3n\) observations is \(4\):
\[ 4 = \frac{\sum (x_i^2)}{3n} - \left(\frac{15n}{3n}\right)^2 \]
\[ 4 = \frac{\sum (x_i^2)}{3n} - 5^2 \]
\[ 4 = \frac{\sum (x_i^2)}{3n} - 25 \]
\[ \frac{\sum (x_i^2)}{3n} = 29 \implies \sum (x_i^2) = 29 \times 3n = 87n \]

2. **Calculating the new variance:**
\[ \text{New Variance} = \frac{\sum (y_i^2)}{3n} - \left(\frac{16n}{3n}\right)^2 \]
- For the first \(2n\) observations:
\[ \sum (y_i^2) = \sum (x_i^2) + 2n + 2 \times \sum x_i = 87n + 2n + 2 \times 12n = 87n + 2n + 24n = 113n \]
- For the last \(n\) observations:
\[ \sum (y_i^2) = \sum (x_i^2) - 2n + n = 87n - n = 86n \]

- Total:
\[ \sum (y_i^2) = 113n + 86n = 199n \]

- Now, substituting back:
\[ \text{New Variance} = \frac{199n}{3n} - \left(\frac{16}{3}\right)^2 \]
\[ = \frac{199}{3} - \frac{256}{9} \]
\[ = \frac{597 - 256}{9} = \frac{341}{9} \]

### Final Answer:
The new variance is \(\frac{341}{9}\).