Let S be the standard deviation of n observations. Each of the n observations is multiplied by a con

Q: Wait, why do we reject the negative root here? If $c$ is negative, shouldn't the standard deviation be negative?

That is a fantastic question! Remember, standard deviation is a measure of 'spread' or 'dispersion.' By definition, it is a distance, and distances in the physical and mathematical world are always non-negative. When we take the square root of $c^2 S^2$, we get $|c|S$. If $c$ is negative, the scaling factor for the spread is the magnitude of $c$, not $c$ itself. We are measuring the width of the distribution, not the direction of the shift!

Q: Why can't we just use energy conservation or some other physics principle?

I love the enthusiasm for physics! However, this is a problem of pure statistical mechanics—the mechanics of data. We must rely on the algebraic definition of variance. Think of the variance formula as the 'energy' of the distribution; we are simply transforming the coordinate system of our data points, and the algebra dictates the result.

Q: Does the mean shift affect the variance? I thought variance was independent of the mean.

Variance is independent of the *absolute* value of the mean (shifting the whole distribution by a constant $k$ doesn't change the spread). However, here we are *scaling* the data, not shifting it. Because the mean is also scaled by $c$, the entire distribution is stretched. The variance is absolutely affected by this scaling factor $c$.

Question

Let $S$ be the standard deviation of $n$ observations. Each of the $n$ observations is multiplied by a constant $c$. Then the standard deviation of the resulting number is

Options:

(A) $s$

(B) $cs$

(C) $s\sqrt{c}$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to peel back the curtain on one of the most elegant properties of statistics: how scaling data affects its spread.

Imagine you have a set of data points, $x_1, x_2, \dots, x_n$. They have a mean, $\bar{x}$, and a standard deviation, $S$.

Now, imagine you multiply every single one of these points by a constant, $c$. We want to determine how this transformation affects the standard deviation.

### The Heartbeat of Data: Variance

Before we dive into the algebra, let's ground ourselves in the definition. The variance, $S^2$, is the average squared deviation from the mean:

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

Think of this as the 'energy' of the distribution. It tells us how far, on average, our data points wander from the center.

If $S^2$ is large, the data is scattered; if $S^2$ is small, the data is tightly packed.

### The Transformation

Now, we perform a transformation. We create a new set of observations, $y_i$, where $y_i = c \cdot x_i$.

Visually, if you imagine these points on a number line, you are stretching the entire line by a factor of $c$. If $c > 1$, the points fly apart; if $0 < c < 1$, they are squished together.

The new mean, $\bar{y}$, is simply the average of these new points:

$$\bar{y} = \frac{1}{n} \sum_{i=1}^{n} (c x_i)$$

Because $c$ is a constant, we can pull it out of the summation:

$$\bar{y} = c \left( \frac{1}{n} \sum_{i=1}^{n} x_i ight) = c \bar{x}$$

The center of our distribution has shifted by exactly the same factor $c$.

### The Algebraic Dance

Now, let's calculate the new variance, $\sigma_y^2$. We plug our new variables into the definition:

$$\sigma_y^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \bar{y})^2$$

Substituting $y_i = cx_i$ and $\bar{y} = c\bar{x}$, we get:

$$\sigma_y^2 = \frac{1}{n} \sum_{i=1}^{n} (cx_i - c\bar{x})^2$$

Here is where the magic happens. We can factor out the $c$ from inside the parenthesis:

$$\sigma_y^2 = \frac{1}{n} \sum_{i=1}^{n} [c(x_i - \bar{x})]^2$$

Remember your exponent rules, where $(c \cdot a)^2 = c^2 \cdot a^2$. The expression becomes:

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

Since $c^2$ does not depend on the index $i$, we can pull it completely out of the summation:

$$\sigma_y^2 = c^2 \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 ight)$$

Look closely at the term inside the parenthesis. That is our original variance, $S^2$. We have arrived at the conclusion:

$$\sigma_y^2 = c^2 S^2$$

### Final Calculation

We are almost there. We have the new variance, but the question asks for the new standard deviation. We take the square root of both sides:

$$	ext{New S.D.} = \sqrt{\sigma_y^2} = \sqrt{c^2 S^2}$$

This simplifies to **$|c|S$**.

Assuming $c$ is positive, the new standard deviation is simply **$cS$**. The spread has scaled exactly by the factor $c$.

Animated Solution for Mathematics - Statistics: Let $S$ be the standard deviation of $n$ observations. Each of the $n$ observations is multiplied by a constant $c$ . Then the standard deviation of the resulting number is

Select Answer:

Visualized Solution

The Original Distribution

Standard Deviation Formula

Scaling the Observations

Calculating the New Mean

Setting up the New Variance

Substituting Transformed Values

Factoring out the Constant

Isolating the Original Variance

The New Standard Deviation

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Heartbeat of Data

The Transformation

The Algebraic Dance

Final Calculation

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