Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers

NCERT Exemplar · Q35

Q. Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is (A) 6.5 (B) 2.87 (C) 3.87 (D) 8.25

Tripura TbseMCQ· 1mImportance★★★★★est

88% · 79/90 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $k^2$ . …

Start your 14-day free trial to unlock the full solution →

Adding a constant to every data point shifts the mean by the same constant but leaves the variance unchanged. The variance of the original set {1,…,10} is 8.25, so the variance after adding 1 is also 8.25 — option (D).

The key insight here is about how variance behaves under a shift. Variance measures spread — how far numbers are from their mean. If you add the same number to every value, the entire distribution slides left or right, but the distances between points stay exactly the same. So the spread doesn't change.

Let’s verify this with the actual numbers.

Original set: $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$

After adding 1, we get: $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$ .
Find the mean of the new set

Sum of new numbers = $2+3+\cdots+11$

This is an arithmetic series from 2 to 11. Number of terms $n=10$ , first term $a=2$ , last term $l=11$ .

Sum = $\frac{n}{2}(a+l) = \frac{10}{2}(2+11) = 5 \times 13 = 65$

Mean $\bar{x} = \frac{65}{10} = 6.5$
Compute variance

Variance formula for a population (since these are all the numbers, not a sample):

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

Let’s list the deviations $(x_i - \bar{x})$ :

$2 - 6.5 = -4.5$
$3 - 6.5 = -3.5$
$4 - 6.5 = -2.5$
$5 - 6.5 = -1.5$
$6 - 6.5 = -0.5$
$7 - 6.5 = 0.5$
$8 - 6.5 = 1.5$
$9 - 6.5 = 2.5$
$10 - 6.5 = 3.5$
$11 - 6.5 = 4.5$

Square each:

$(-4.5)^2 = 20.25$
$(-3.5)^2 = 12.25$
$(-2.5)^2 = 6.25$
$(-1.5)^2 = 2.25$
$(-0.5)^2 = 0.25$
$(0.5)^2 = 0.25$
$(1.5)^2 = 2.25$
$(2.5)^2 = 6.25$
$(3.5)^2 = 12.25$
$(4.5)^2 = 20.25$

Sum of squares = $20.25 + 12.25 + 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 + 12.25 + 20.25$ …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ34 — Standard deviations for first 10 natural numbers is (A) 5.5 (B) 3.87 (…→Next questionQ36 — Consider the first 10 positive integers. If we multiply each number by…

Q.Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is (A) 6.5 (B) 2.87 (C) 3.87 (D) 8.25

🎓Unlock everything free for 14 days

Q.Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. If 1 is added to each number, the variance of the numbers so obtained is
(A) 6.5
(B) 2.87
(C) 3.87
(D) 8.25

Unlock everything free for 14 days