For the frequency distribution: Variate (x): x1, x2, x3 ... x15 Frequency (f): f1, f2, f3 ... f15 wh

Q: Wait, why can't we just calculate the variance using the standard formula $\frac{1}{N} \sum (x_i - \bar{x})^2$?

That is a great question! We cannot use the standard formula because we do not know the individual values of $x_i$ or their frequencies $f_i$. We only know the constraints on the range. Therefore, we need a general bound that holds true for *any* distribution within these bounds, which is exactly what Popoviciu's Inequality provides.

Q: What exactly is Popoviciu's Inequality?

Popoviciu's Inequality is a powerful result in statistics. It states that for any dataset bounded in the interval $[a, b]$, the variance $\sigma^2$ is at most $\frac{(b-a)^2}{4}$. Think of it as the 'worst-case scenario' for variance—it tells us the absolute maximum spread possible for a given range.

Q: Why is the maximum standard deviation $\frac{b-a}{2}$?

Since the inequality tells us $\sigma^2 \le \frac{(b-a)^2}{4}$, we simply take the square root of both sides to find the bound for the standard deviation $\sigma$. The square root of $\frac{(b-a)^2}{4}$ is $\frac{b-a}{2}$. For our range of $10$, this gives $\sigma \le \frac{10}{2} = 5$.

Question

For the frequency distribution:
Variate ($x$): $x_1, x_2, x_3 \dots x_{15}$
Frequency ($f$): $f_1, f_2, f_3 \dots f_{15}$
where $0 < x_1 < x_2 < x_3 < \dots < x_{15} = 10$ and $\sum_{i=1}^{15} f_i > 0$, the standard deviation cannot be:

Options:

(A) $1$

(B) $6$

(C) $2$

(D) $4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a number line, looking at a collection of fifteen points, $x_1, x_2, \dots, x_{15}$. You are told that these points are not just floating randomly; they are strictly constrained.

The smallest point, $x_1$, is just a whisper away from zero, and the largest point, $x_{15}$, is anchored firmly at ten. Every other point is squeezed somewhere in between.

This is the stage for our problem. We are asked to determine which of the given values could not possibly be the standard deviation of this distribution.

### The Geometry of Spread

The standard deviation, $\sigma$, is a measure of how spread out our data is. If all our points were clustered at a single value, the standard deviation would be zero.

If they were pushed to the extreme edges—some at zero and some at ten—the spread would be maximized. Our dataset is trapped in the interval $[0, 10]$.

The range of this data is defined as the maximum value minus the minimum value. Since our maximum is $10$ and our minimum is effectively $0$, the maximum possible range is $10 - 0 = 10$.

### The Secret Weapon: Popoviciu's Inequality

When we face a problem where we don't know the exact data points but we know their bounds, we need a powerful tool. Enter **Popoviciu's Inequality**.

This theorem is the secret weapon for statisticians dealing with bounded data. It states that for any set of data bounded within the interval $[a, b]$, the variance $\sigma^2$ is constrained by the formula:

$$\sigma^2 \le \frac{(b-a)^2}{4}$$

This inequality is elegant and profound. It dictates that no matter how you arrange your points within the interval $[a, b]$, the variance cannot exceed the square of the range divided by four.

### The Calculation

Let's apply this to our specific scenario where $a = 0$ and $b = 10$. Plugging these into our formula, we get:

$$\sigma^2 \le \frac{(10 - 0)^2}{4}$$

Simplifying the numerator, we have $10^2 = 100$. So, the inequality becomes:

$$\sigma^2 \le \frac{100}{4}$$

This simplifies to:

$$\sigma^2 \le 25$$

This is a massive realization! It means that for any distribution of these fifteen points, the variance can never exceed $25$.

Since $\sigma$ is the positive square root of the variance, we take the square root of both sides:

$$\sigma \le \sqrt{25}$$

$$\sigma \le 5$$

### The Final Verdict

We have mathematically proven that the standard deviation of this dataset must be less than or equal to $5$.

Now, let's look at our options: $1, 6, 2, 4$. We see that $1, 2,$ and $4$ are all less than or equal to $5$, meaning they are perfectly valid possibilities for the standard deviation.

However, $6$ is strictly greater than $5$. Therefore, it is physically and mathematically impossible for this dataset to have a standard deviation of **$6$**.

Select Answer:

Visualized Solution

Defining the Data Bounds

Plotting the Variates

Determining the Maximum Range

Popoviciu's Inequality for Variance

Substituting the Bounds

Computing the Numerator

Maximum Possible Variance

Maximum Standard Deviation

Evaluating the Options

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Geometry of Spread

The Secret Weapon

The Calculation

The Final Verdict

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