1. Key Concepts and Formulas

• Standard Deviation ($\sigma$) and Variance ($\sigma^2$): These are measures of the spread or dispersion of data points in a distribution. Variance is the average of the squared differences from the mean, and standard deviation is the non-negative square root of the variance.
• Range of Variates: For a given frequency distribution, if all variates ($x_i$) are confined within a finite closed interval $[m, M]$, where $m$ is the absolute minimum possible value and $M$ is the absolute maximum possible value any variate can take.
• Fundamental Inequality for Standard Deviation: For any frequency distribution where the variates $x_i$ are in the interval $[m, M]$, the standard deviation $\sigma$ is bounded by: $0 \le \sigma \le \frac{1}{2}(M - m)$ This inequality implies that the variance $\sigma^2$ is bounded by: $\sigma^2 \le \frac{1}{4}(M - m)^2$ This powerful inequality allows us to determine the maximum possible spread of a dataset by only knowing its minimum and maximum possible values, without needing to know the individual data points or their frequencies.

2. Step-by-Step Solution

Step 1: Determine the absolute minimum ($m$) and maximum ($M$) values for the variates.

We are given the following conditions for the frequency distribution:

• Variates: $x_1, x_2, x_3, \ldots, x_{15}$
• Frequencies: $f_1, f_2, f_3, \ldots, f_{15}$
• Ordering and Bounds: $0 < x_1 < x_2 < x_3 < \ldots < x_{15} = 10$
• Total Frequency: $\sum_{i=1}^{15} f_i > 0$ (This ensures a non-empty dataset, making statistical calculations valid).

Our goal is to find the tightest possible interval $[m, M]$ that encompasses all possible values of the variates $x_i$.

• Identify the Maximum Value ($M$): The condition $x_{15} = 10$ explicitly states that the largest variate in our distribution is 10. Since all other variates are strictly less than $x_{15}$ (i.e., $x_i < 10$ for $i < 15$), the absolute maximum value any variate can take is 10. Therefore, we set $M = 10$.

• Identify the Minimum Value ($m$): The condition $0 < x_1$ tells us that the smallest variate ($x_1$) is strictly greater than 0. This means $x_1$ cannot be 0, but it can be arbitrarily close to 0 (e.g., 0.1, 0.001, 0.000001, etc.). For the purpose of establishing the theoretical lower bound for the interval, we consider the limit point. As $x_1$ can approach 0 infinitely closely, the absolute minimum value that the variates can conceptually take or approach is 0. Therefore, we set $m = 0$.

So, the variates $x_i$ effectively lie within the interval $(0, 10]$. For the variance inequality, we use the theoretical bounds $m=0$ and $M=10$.

Step 2: Apply the standard deviation inequality to find its upper bound.

Now that we have identified $m=0$ and $M=10$, we can substitute these values into the fundamental inequality for standard deviation: $0 \le \sigma \le \frac{1}{2}(M - m)$

Substitute $M=10$ and $m=0$: $0 \le \sigma \le \frac{1}{2}(10 - 0)$

Calculate the difference $(M - m)$, which represents the range of the variates: $0 \le \sigma \le \frac{1}{2}(10)$

Perform the multiplication to find the upper bound: $0 \le \sigma \le 5$ This result indicates that the standard deviation ($\sigma$) for this distribution must be a value between 0 (inclusive) and 5 (inclusive). In other words, $\sigma \in [0, 5]$.

Step 3: Evaluate the given options to find which value cannot be the standard deviation.

The problem asks us to identify which of the given options cannot be the standard deviation. We have established that the standard deviation $\sigma$ must satisfy $0 \le \sigma \le 5$.

Let's check each option against this permissible range:

• (A) 6: This value is greater than 5. Since the maximum possible standard deviation is 5, a standard deviation of 6 is impossible for this distribution.
• (B) 1: This value lies within the range $[0, 5]$ (i.e., $0 \le 1 \le 5$). This is a possible value for the standard deviation.
• (C) 4: This value lies within the range $[0, 5]$ (i.e., $0 \le 4 \le 5$). This is a possible value for the standard deviation.
• (D) 2: This value lies within the range $[0, 5]$ (i.e., $0 \le 2 \le 5$). This is a possible value for the standard deviation.

Therefore, the only value among the options that the standard deviation cannot be is 6.

3. Common Mistakes & Tips

• Misinterpreting $m$ and $M$: Be careful when the interval is open or semi-open. For $x_1 > 0$, the absolute minimum $m$ for the inequality is 0, even though $x_1$ itself never equals 0. The inequality uses the theoretical bounds of the interval that contains all possible data values.
• Forgetting Non-negativity: Standard deviation ($\sigma$) is always a non-negative quantity. The lower bound of its range is always 0.
• Conditions for Equality: The maximum standard deviation $\sigma = \frac{1}{2}(M-m)$ is achieved when the data points are concentrated at the two extreme values, $m$ and $M$, with equal frequencies (e.g., half the observations are $m$ and the other half are $M$). This confirms that the upper bound derived is indeed attainable.

4. Summary

The problem required us to determine which value cannot be the standard deviation for a given frequency distribution. By utilizing the fundamental inequality $0 \le \sigma \le \frac{1}{2}(M - m)$, we first identified the absolute minimum ($m=0$) and maximum ($M=10$) possible values for the variates from the given conditions ($0 < x_1$ and $x_{15}=10$). Substituting these values into the inequality, we found that the standard deviation $\sigma$ must lie in the range $[0, 5]$. Comparing this with the given options, only 6 falls outside this permissible range, making it an impossible value for the standard deviation.

5. Final Answer

The standard deviation cannot be 6. The final answer is $\boxed{A}$