The standard deviation of 17 numbers is zero. Then
(a) the numbers are in geometric progression with

Q: Wait, if the standard deviation is zero, does that mean the mean is also zero?

Not necessarily! The mean can be any value. Standard deviation measures the spread around the mean, not the value of the mean itself. If all numbers are $5$, the mean is $5$ and the standard deviation is $0$.

Q: Why can't we have a geometric progression with $r=1$?

The option specifically states $r \neq 1$. If $r=1$, the terms would be $a, a, a...$, which would have a standard deviation of zero. But since the option restricts $r$ to be something other than $1$, those terms must be distinct, making the standard deviation non-zero.

Q: Is it possible for the sum of squares to be zero if we have negative numbers?

Remember, we are squaring the differences $(x_i - \bar{x})^2$. Even if $(x_i - \bar{x})$ is negative, its square is positive. The only way a sum of non-negative terms equals zero is if every single term is zero.

Question

The standard deviation of 17 numbers is zero. Then

Options:

(A) the numbers are in geometric progression with common ratio not equal to one.

(B) eight numbers are positive, eight are negative and one is zero.

(C) either (a) or (b)

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the world of statistics, standard deviation, denoted by $\sigma$, is the heartbeat of your data. It quantifies how much your data points deviate from the average. If $\sigma$ is large, your data is scattered and unpredictable.

However, if $\sigma = 0$, the data exhibits zero dispersion. Let us examine the mathematical definition:

$$
\sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2}
$$

If $\sigma = 0$, then the entire expression under the square root must be zero. This leads us to the fundamental equation:

$$
\sum_{i=1}^{n} (x_i - \bar{x})^2 = 0
$$

### The Mathematical Reality

Here is the critical insight: we are summing up squares. In the realm of real numbers, the square of any value is always non-negative—it is either positive or zero.

If you add up a collection of non-negative numbers and the result is zero, there is only one logical conclusion: every single term in that sum must be zero. There is no other way to achieve a sum of zero.

This implies that for every single observation $x_i$, the difference $(x_i - \bar{x})$ must be zero. Consequently:

$$
x_1 = x_2 = \dots = x_{17} = \bar{x}
$$

**All seventeen numbers must be identical.** They are all equal to the mean.

### Debunking the Distractors

Now, let us evaluate the potential scenarios with this new clarity. Option (a) suggests the numbers are in a geometric progression with a common ratio $r 
eq 1$.

If $r 
eq 1$, the terms of the progression would be $a, ar, ar^2, \dots$, which are clearly distinct. Distinct numbers possess a spread, resulting in a non-zero standard deviation. Thus, Option (a) is impossible.

Option (b) suggests we have eight positive numbers, eight negative numbers, and one zero. These values are distinct and possess varying magnitudes. A set containing values like $5$ and $-5$ cannot have a standard deviation of zero because they are not the same.

Therefore, Option (b) is also incorrect. We are left with the only logical conclusion: none of the provided descriptions fit the reality of zero dispersion.

**The correct answer is (d).**

Whenever you encounter $\sigma = 0$, do not overcomplicate the scenario. It simply means your data has collapsed into a single, uniform value.

Animated Solution for Mathematics - Statistics: The standard deviation of 17 numbers is zero. Then

Select Answer:

Visualized Solution

Understanding Standard Deviation $σ$

Condition for $σ = 0$

Analyzing the Sum of Squares

Equality of All Observations

Visualizing Zero Dispersion

Evaluating Option (a)

Why Option (a) is Incorrect

Evaluating Option (b)

Why Option (b) is Incorrect

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Mathematical Reality

Debunking the Distractors

Similar Questions

Let $X_{1}, X_{2}, \dots, X_{18}$ be eighteen observation such that $\sum_{i = 1}^{18} (X_{i} - α) = 36$ and $\sum_{i = 1}^{18} (X_{i} - β)^{2} = 90$ , where $α$ and $β$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of $∣ α - β ∣$ is

A data consists of $n$ observations: $x_{1}, x_{2}, ....., x_{n}$ . If $\sum_{i = 1}^{n} (x_{i} + 1)^{2} = 9 n$ and $\sum_{i = 1}^{n} (x_{i} - 1)^{2} = 5 n$ , then the standard deviation of this data is :

Let $X = {x \in N : 1 \leq x \leq 17}$ and $Y = {a x + b : x \in X and a, b \in R, a > 0}$ . If mean and variance of elements of $Y$ are 17 and 216 respectively then $a + b$ is equal to :

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:

The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $σ^{2}$ respectively. If the variance of all the 30 numbers in the two sets is 13, then $σ^{2}$ is equal to

Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $x$ . If $\sum_{i = 1}^{10} (x_{i} - p) = 3$ and $\sum_{i = 1}^{10} (x_{i} - p)^{2} = 9$ where $0 \neq = p \in R$ , then the standard deviation of these observations is

Let the mean and variance of 12 observations be $\frac{9}{2}$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $\frac{m}{n}$ , where $m$ and $n$ are coprime, then $m + n$ is equal to

If $\sum_{i = 1}^{n} (x_{i} - a) = n$ and $\sum_{i = 1}^{n} (x_{i} - a)^{2} = na$ , $(n, a > 1)$ then the standard deviation of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ is:

If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :

If $\sum_{i = 1}^{9} (x_{i} - 5) = 9$ and $\sum_{i = 1}^{9} (x_{i} - 5)^{2} = 45$ , then the standard deviation of the 9 items $x_{1}, x_{2}, \dots, x_{9}$ is :