For the frequency distribution:
`{:("Variate (x): " x_(1)" "x_(2)" "x_(3).....x_(15)),("Frequency (f): "f_(1)" "f_(2)" "f_(3).....f_(15)):}`
where `0lt x_(1)ltx_(2)ltx_(3) lt ….ltx_(15)=10 and sum_(i=1)^(15)f_(i) gt 0`, the standard deviation cannot be :

Options

A. 1
B. 4
C. 6
D. 2

Answer

To solve the problem regarding the standard deviation for the given frequency distribution, we will follow these steps:

### Step 1: Understand the given information
We have a frequency distribution with variates \(x_1, x_2, \ldots, x_{15}\) such that:
- \(0 < x_1 < x_2 < x_3 < \ldots < x_{15} = 10\)
- The sum of frequencies \(f_i\) (where \(i\) ranges from 1 to 15) is greater than 0.

### Step 2: Identify the bounds for the variates
From the information given:
- The maximum value (upper bound) \(M\) is \(x_{15} = 10\).
- The minimum value (lower bound) \(m\) is \(x_1 > 0\), so we can consider \(m\) to be just above 0.

### Step 3: Use the formula for variance
The variance \(\sigma^2\) can be calculated using the formula:
\[ \sigma^2 \leq \frac{1}{4}(M - m)^2 \]
Substituting the values of \(M\) and \(m\):
\[ \sigma^2 \leq \frac{1}{4}(10 - 0)^2 = \frac{1}{4} \times 100 = 25 \]

### Step 4: Determine the standard deviation
The standard deviation \(\sigma\) is the square root of the variance:
\[ \sigma \leq \sqrt{25} = 5 \]
Thus, the standard deviation can take values in the range \(0 < \sigma \leq 5\).

### Step 5: Identify which standard deviation cannot be
Since \(\sigma\) must be non-negative and cannot exceed 5, we can conclude:
- Possible values for \(\sigma\) are in the range \(0 < \sigma \leq 5\).
- Therefore, values like 1, 2, 4, and 5 are possible, but 6 is not possible.

### Conclusion
The standard deviation cannot be 6.