A data consists of n observations : `x_(1), x_(2),……, x_(n)`. If `Sigma_(i=1)^(n)(x_(i)+1)^(2)=11n and Sigma_(i=1)^(n)(x_(i)-1)^(2)=7n`, then the variance of this data is

Options

A. 5
B. 8
C. 6
D. 7

Answer

To solve the problem, we need to find the variance of the given data based on the provided summation equations. Let's break it down step by step.

### Step 1: Understand the Given Equations
We have two equations:
1. \(\sum_{i=1}^{n} (x_i + 1)^2 = 11n\)
2. \(\sum_{i=1}^{n} (x_i - 1)^2 = 7n\)

### Step 2: Expand the Equations
We will expand both equations using the formula \((a \pm b)^2 = a^2 \pm 2ab + b^2\).

1. For the first equation:
\[ \sum_{i=1}^{n} (x_i + 1)^2 = \sum_{i=1}^{n} (x_i^2 + 2x_i + 1) = \sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i + n \]
Therefore, we can write:
\[ \sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i + n = 11n \]
Simplifying gives:
\[ \sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i = 10n \quad \text{(Equation 1)} \]

2. For the second equation:
\[ \sum_{i=1}^{n} (x_i - 1)^2 = \sum_{i=1}^{n} (x_i^2 - 2x_i + 1) = \sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i + n \]
Therefore, we can write:
\[ \sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i + n = 7n \]
Simplifying gives:
\[ \sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i = 6n \quad \text{(Equation 2)} \]

### Step 3: Subtract the Two Equations
Now we will subtract Equation 2 from Equation 1:
\[ \left(\sum_{i=1}^{n} x_i^2 + 2\sum_{i=1}^{n} x_i\right) - \left(\sum_{i=1}^{n} x_i^2 - 2\sum_{i=1}^{n} x_i\right) = 10n - 6n \]
This simplifies to:
\[ 4\sum_{i=1}^{n} x_i = 4n \]
Dividing both sides by 4 gives:
\[ \sum_{i=1}^{n} x_i = n \]

### Step 4: Calculate the Mean
The mean \(\bar{x}\) is given by:
\[ \bar{x} = \frac{\sum_{i=1}^{n} x_i}{n} = \frac{n}{n} = 1 \]

### Step 5: Substitute Mean into Equation for Variance
The variance \(Var(X)\) is given by:
\[ Var(X) = \frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n} \]
Using the relationship:
\[ \sum_{i=1}^{n} (x_i - \bar{x})^2 = \sum_{i=1}^{n} (x_i - 1)^2 \]
From Equation 2, we know:
\[ \sum_{i=1}^{n} (x_i - 1)^2 = 7n \]
Thus:
\[ Var(X) = \frac{7n}{n} = 7 \]

### Final Answer
The variance of the data is:
\[ \boxed{7} \]