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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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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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$1, 15, 35, 53, 72$
›Reveal solutionSolution
The second dataset is just the first dataset shifted down by 2 for every value; shifting data by a constant never changes its variance.

Concept and Intuition

Variance measures spread around the mean, and shifting every value by the same constant $k$ shifts the mean by $k$ too — so every deviation from the mean is completely unchanged. This is why $\text{Var}(X+k)=\text{Var}(X)$ always, regardless of $k$ .

Step-by-Step Solution

First dataset (sorted): $1,15,35,53,64,72$ .

Second dataset (sorted): $-1,13,33,51,62,70$ .

Check the difference at each position: $1-(-1)=2,\ 15-13=2,\ 35-33=2,\ 53-51=2,\ 64-62=2,\ 72-70=2$ .

So the second dataset is exactly the first dataset with every value reduced by $2$ .

Since variance is translation-invariant, $\text{Var}(\text{second set}) = \text{Var}(\text{first set})$ .

The standard deviation of the first set is given as $x$ , so its variance is $x^2$ . …
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AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
$x_1, x_2, x_3 \ldots x_n$
›Reveal solutionSolution
This tests extracting the mean and $\sum x_i^2$ from two given sum-of-squares expressions, then using $\text{Var}=\dfrac{\sum x_i^2}{n}-\bar x^2$ . Answer: variance $=12$ .

Concept and Intuition

Expanding both given identities as perfect squares produces two linear equations in $\sum x_i$ and $\sum x_i^2$ (with $n$ as a common factor); adding and subtracting isolates each unknown cleanly, after which the standard variance formula finishes the job.

Step-by-Step Solution

$\sum(x_i+2)^2=\sum x_i^2+4\sum x_i+4n=28n$ ... (i)

$\sum(x_i-2)^2=\sum x_i^2-4\sum x_i+4n=12n$ ... (ii)

(i) $-$ (ii): $8\sum x_i=16n \Rightarrow \sum x_i=2n \Rightarrow \bar x=2$ . …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$Q. The variance of 20 observations is 5. If each one of the observations is multiplied by 2, then the variance of the resulting observations is (A) 40 (B) 80 (C) 20 (D) 10$
›Reveal solutionSolution
Multiplying every observation by a constant $k$ scales the variance by $k^2$ (variance has "squared units"). Here new variance $=4\times5=20$ .

Concept and Intuition

Variance measures squared spread from the mean. If $Y_i=kX_i$ , then $\bar Y=k\bar X$ and $Y_i-\bar Y=k(X_i-\bar X)$ , so $(Y_i-\bar Y)^2=k^2(X_i-\bar X)^2$ . Averaging gives $\mathrm{Var}(Y)=k^2\mathrm{Var}(X)$ — the mean's shift doesn't matter (variance is a "spread" measure, translation-invariant), only the scaling factor matters, and it enters squared.

Step-by-Step Solution

Given $\mathrm{Var}(X)=5$ , and $Y_i=2X_i$ for each of the 20 observations.

$\mathrm{Var}(Y)=2^2\cdot\mathrm{Var}(X)=4\times5=20$ . …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$x_1, x_2, \ldots, x_n$
›Reveal solutionSolution
Variance is invariant under a shift (adding/subtracting a constant) because it measures spread about the mean, not the values themselves.

Concept and Intuition

Variance $=\dfrac{1}{n}\sum(x_i-\bar x)^2$ depends only on how far each point is from the mean of the data, not on the absolute values. Shifting every point by $k$ shifts the mean by $k$ too, so every deviation $(x_i-\bar x)$ is exactly unchanged.

Step-by-Step Solution

Let $y_i = x_i+k$ . Then $\bar y = \bar x+k$ .

$y_i-\bar y = (x_i+k)-(\bar x+k) = x_i-\bar x$ — the deviation is unchanged. …
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AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
$5/9$
›Reveal solutionSolution
Variance scales as the square of the multiplying constant, so scaling four numbers by 5 multiplies their variance by 25, giving 225.

Concept and Intuition

Variance measures squared deviation from the mean. If every data point is multiplied by a constant $k$ , both the mean and every deviation scale by $k$ , so the squared deviations (and hence the variance) scale by $k^2$ — a well-known linear-transformation property of variance.

Step-by-Step Solution

Property: $\text{Var}(kw,kx,ky,kz) = k^2\,\text{Var}(w,x,y,z)$ .

Here $k=5$ and $\text{Var}(w,x,y,z)=9$ .

New variance $=5^2\times9=25\times9=225$ .

Common Mistakes …
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AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
$Q. The standard deviation of first 10 multiples of 4 is (A) 7 (B) 8 (C) 11.5 (D) 14$
›Reveal solutionSolution
The 10 numbers $4,8,\dots,40$ form an arithmetic progression, so the closed-form AP standard-deviation formula applies directly, giving $11.5$ .

Concept and Intuition

For an AP with common difference $d$ and $n$ terms, there is a ready-made formula for the standard deviation: $\sigma=d\sqrt{\dfrac{n^2-1}{12}}$ . This avoids computing the mean and every squared deviation by hand.

Step-by-Step Solution

The numbers are $4,8,12,\dots,40$ : an AP with $d=4$ , $n=10$ .

Apply $\sigma=d\sqrt{\dfrac{n^2-1}{12}}=4\sqrt{\dfrac{100-1}{12}}=4\sqrt{\dfrac{99}{12}}=4\sqrt{8.25}$ . …
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AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$10, 20, 30, 40$
›Reveal solutionSolution
Since each dataset is simply a scaled version of $1,2,3,4$ , the standard deviation scales with that multiplier; the smallest multiplier (1) gives the least SD.

Concept and Intuition

If a dataset $Y = kX$ (each value scaled by constant $k$ ), then $\text{SD}(Y) = |k|\cdot\text{SD}(X)$ . All four given sets are exact multiples of $\{1,2,3,4\}$ : by $10$ , by $2$ , by $3$ , and by $1$ respectively. So whichever has the smallest multiplier has the smallest SD.

Step-by-Step Solution

Recognize: $\{10,20,30,40\}=10\times\{1,2,3,4\}$ ; $\{2,4,6,8\}=2\times\{1,2,3,4\}$ ; $\{3,6,9,12\}=3\times\{1,2,3,4\}$ ; $\{1,2,3,4\}=1\times\{1,2,3,4\}$ .

Compute the base SD of $\{1,2,3,4\}$ : mean $=2.5$ ; deviations $-1.5,-0.5,0.5,1.5$ ; squared sum $=2.25+0.25+0.25+2.25=5$ ; variance $=5/4=1.25$ ; SD $=\sqrt{1.25}\approx1.118$ .

Scaled SDs: $10\times1.118\approx11.18$ (A), $2\times1.118\approx2.236$ (B), $3\times1.118\approx3.354$ (C), $1\times1.118\approx1.118$ (D).

The smallest is clearly option (D), the unscaled base set.

Common Mistakes …
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AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
$\bar{x}$
›Reveal solutionSolution
Scaling every data value by 2 scales the mean by 2 as well, giving a new mean of 20.

Concept and Intuition

The arithmetic mean is a linear operator: $\text{mean}(kx_i) = k\cdot\text{mean}(x_i)$ , because $\dfrac{1}{n}\sum(kx_i) = k\cdot\dfrac1n\sum x_i$ .

Step-by-Step Solution

Original mean: $\bar x = \dfrac1n\sum x_i = 10$ .

New data: $y_i = 2x_i$ for each observation. …
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AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
$\dfrac{X}{5}$
›Reveal solutionSolution
When every number in a set is divided by a constant, the mean is also divided by that constant. Given the new mean is $X/5$ , the original mean must be $X$ . The correct option is (B).

Effect of Scaling Variance

The mean is a linear statistic: if you scale every data point by a factor $k$ , the mean scales by exactly the same factor. Here, dividing each number by 5 is scaling by $1/5$ . So if the original mean is $\bar{X}$ , the new mean is $\bar{X}/5$ . The problem gives the new mean as $X/5$ , so we can equate and solve.

Define the original mean Let the original set of $n$ numbers be $a_1, a_2, \dots, a_n$ . Their mean is

$\bar{X} = \frac{a_1 + a_2 + \cdots + a_n}{n}.$

Apply the division by 5 Each number is divided by 5, so the new set is $\frac{a_1}{5}, \frac{a_2}{5}, \dots, \frac{a_n}{5}$ . The new mean is

$\text{New mean} = \frac{\frac{a_1}{5} + \frac{a_2}{5} + \cdots + \frac{a_n}{5}}{n} = \frac{1}{5} \cdot \frac{a_1 + a_2 + \cdots + a_n}{n} = \frac{\bar{X}}{5}.$

Equate to the given value The problem states this new mean equals $\frac{X}{5}$ . Therefore

$\frac{\bar{X}}{5} = \frac{X}{5}.$
Multiply both sides by 5:

$\bar{X} = X.$ …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$180, 198, 90, 126, 72, 144, 18, 81, 27, 54$
›Reveal solutionSolution
Factoring out the common multiple 9 shrinks the arithmetic dramatically; the underlying set has variance 42, so the original data's variance is $81\times42=3402$ .

Concept and Intuition

If every data value is scaled by a constant $k$ ( $x_i=ky_i$ ), then $\text{Var}(x) = k^2\,\text{Var}(y)$ (variance scales with the square of the multiplier, since it's built from squared deviations). Spotting a common factor before crunching the numbers keeps the numbers small and the arithmetic error-free.

Step-by-Step Solution

Data: $180,198,90,126,72,144,18,81,27,54$ . Each is $9\times$ one of $20,22,10,14,8,16,2,9,3,6$ .

Let $y_i$ be this smaller set. Sum of $y_i = 20+22+10+14+8+16+2+9+3+6=110$ , so mean $\bar y = 11$ .

Sum of $y_i^2 = 400+484+100+196+64+256+4+81+9+36 = 1630$ . …
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Effect of Scaling on Spread

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