Let x₁, x₂, x₃, x₄, x₅ be the observations with mean m and standard deviation s. The standard deviation of the

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x_1, x_2, x_3, x_4, x_5

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Concept understanding — Effect of Scaling Variance

Effect of Scaling Variance – First Encounter

Let’s start with a simple question: what happens to the spread of a dataset if you multiply every number by 2? Or by 0.5? Or by –3?

If you have a set of marks: 40, 50, 60, 70, 80, the variance is some number. Now imagine the teacher doubles every mark: 80, 100, 120, 140, 160. The marks are now twice as far apart from each other. The spread has clearly changed — but by how much?

That’s exactly what the effect of scaling tells you: a precise rule for how variance changes when you multiply (or divide) every observation by a constant.

Intuition first

Variance measures average squared distance from the mean. If you scale all values by a factor $k$ , two things happen:

The mean also gets scaled by $k$ (because mean is linear).
Each individual distance from the mean also gets scaled by $k$ .

But variance squares those distances. So if each distance is multiplied by $k$ , each squared distance is multiplied by $k^2$ . The average of those squared distances — the variance — therefore gets multiplied by $k^2$ .

Note

Scaling by $k$ multiplies the variance by $k^2$ , not by $k$ . This is because variance is in squared units of the original data.

The precise statement

Let $X$ be a random variable (or a dataset) with variance $\text{Var}(X)$ . Let $k$ be any real constant. Then:

\text{Var}(kX) = k^2 \cdot \text{Var}(X)

That’s it. No extra terms, no dependence on the mean. Just $k^2$ times the original variance.

What about adding a constant?

This is a separate but related idea: if you add a constant $c$ to every value, the spread doesn’t change — the whole distribution just shifts. So:

\text{Var}(X + c) = \text{Var}(X)

Combining both: for a linear transformation $Y = aX + b$ ,

\text{Var}(Y) = a^2 \cdot \text{Var}(X)

The constant $b$ has no effect on variance.

A quick example

Suppose the variance of heights (in cm) is 25. Convert to metres: divide by 100, i.e., multiply by $0.01$ .

\text{Var}(\text{height in m}) = (0.01)^2 \times 25 = 0.0001 \times 25 = 0.0025

That’s a tiny number — but it’s correct, because metres are a larger unit, so the spread in metres is much smaller numerically.

Common mistake to avoid

Watch out

Do not say “variance gets multiplied by $k$ ”. It’s $k^2$ . …

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When every observation is multiplied by a constant $k$ , the standard deviation also gets multiplied by $|k|$ . So the new standard deviation is $k s$ (assuming $k>0$ ), which matches option (C).

Effect of Scaling on Spread

Standard deviation measures how spread out the data is from the mean. If you stretch or shrink every data point by the same factor $k$ , the entire distribution scales — distances between points, and distances from the mean, all multiply by $|k|$ . Since standard deviation is essentially a measure of those distances, it scales by the same factor.

This is different from what happens to the mean (which also scales by $k$ ) or the variance (which scales by $k^2$ ). The key insight: scaling changes spread proportionally.

Let’s verify this step by step.

Recall the definition of standard deviation

For observations $x_1, x_2, \dots, x_n$ with mean $m$ , the standard deviation $s$ is:

s = \sqrt{ \frac{1}{n} \sum_{i=1}^n (x_i - m)^2 }

Here $n = 5$ , but the formula works for any $n$ .

What happens to the mean when we multiply by $k$ ?

The new observations are $y_i = k x_i$ . Their mean is:

\bar{y} = \frac{1}{5} \sum_{i=1}^5 k x_i = k \cdot \frac{1}{5} \sum_{i=1}^5 x_i = k m

So the mean also scales by $k$ .

Now compute the new standard deviation

Let $s'$ be the standard deviation of the $y_i$ :

s' = \sqrt{ \frac{1}{5} \sum_{i=1}^5 (y_i - \bar{y})^2 } = \sqrt{ \frac{1}{5} \sum_{i=1}^5 (k x_i - k m)^2 }

Factor $k$ out of each term inside the square:

s' = \sqrt{ \frac{1}{5} \sum_{i=1}^5 k^2 (x_i - m)^2 } = \sqrt{ k^2 \cdot \frac{1}{5} \sum_{i=1}^5 (x_i - m)^2 }

Take the square root …

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Effect of Scaling on Spread

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