Statement-1 :If `a le x_i le b` where `x_i` denotes the value of x in the `i^"th"` case for i=1,2,….n, then `(b-a)^2 le` Variance (x) .
Statement-2 : S.D. `le` range (b-a)

Answer

To solve the problem, we need to analyze the two statements given in the question regarding variance and standard deviation.

### Step 1: Understand the Given Statements
We have two statements:
- **Statement 1**: If \( a \leq x_i \leq b \) for \( i = 1, 2, \ldots, n \), then \( (b - a)^2 \leq \text{Variance}(x) \).
- **Statement 2**: Standard Deviation \( \leq \) Range \( (b - a) \).

### Step 2: Analyze Statement 1
1. **Definition of Variance**: The variance of a set of values \( x_1, x_2, \ldots, x_n \) is defined as:
\[ \text{Variance}(x) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
where \( \bar{x} \) is the mean of the values.

2. **Range**: The range of the data is defined as \( b - a \).

3. **Mean Location**: Since \( a \leq x_i \leq b \), the mean \( \bar{x} \) must also lie between \( a \) and \( b \). Thus, the deviations \( (x_i - \bar{x}) \) will also be bounded.

4. **Bounding the Variance**: The maximum deviation from the mean \( \bar{x} \) can be at most \( b - a \). Therefore:
\[ (x_i - \bar{x})^2 \leq (b - a)^2 \]
for all \( i \).

5. **Summing Deviations**: When we sum these squared deviations:
\[ \sum_{i=1}^{n} (x_i - \bar{x})^2 \leq n(b - a)^2 \]
Dividing by \( n \) gives:
\[ \text{Variance}(x) \leq \frac{(b - a)^2}{n} \cdot n = (b - a)^2 \]
This contradicts Statement 1, which claims \( (b - a)^2 \leq \text{Variance}(x) \). Thus, **Statement 1 is incorrect**.

### Step 3: Analyze Statement 2
1. **Standard Deviation**: The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{\text{Variance}(x)} \]

2. **Using the Result from Statement 1**: Since we found that \( \text{Variance}(x) \leq (b - a)^2 \), we can take the square root:
\[ \sigma \leq \sqrt{(b - a)^2} = |b - a| = b - a \]
Therefore, **Statement 2 is correct**.

### Conclusion
- **Statement 1**: Incorrect
- **Statement 2**: Correct