Statistics

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high importance~3 Q in Tier 128 formulas⚡ 10 shortcuts5 subtopics

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All subtopics·Subtopic 4 of 5

Range, variance and standard deviation

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⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Averages say where the data sits; dispersion says how widely it spreads.

Range is the quick measure. Variance and standard deviation (SD) are the exam measures: SD is the typical distance of values from the mean, in the same unit as the data.

Range: the quick spread

Range $=$ highest value $-$ lowest value. Values 12, 5, 9, 17, 3 give range $17 - 3 = 14$ .

The range ignores everything in between, so exams use it as a warm-up, not the main dish.

Tip: A new maximum or a new minimum changes the range; a middle value never can.

Variance and SD: distance from the mean

Take each value's distance from the mean, square, average, then take the square root.

\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}, \qquad \sigma = \sqrt{\sigma^2}

$\sigma$ is the standard deviation, read "sigma". Values 2, 4, 6, 8, 10 have mean 6 and deviations $-4, -2, 0, 2, 4$ :

\sigma^2 = \frac{16+4+0+4+16}{5} = 8, \qquad \sigma = \sqrt{8} = 2\sqrt{2} \approx 2.83

Rule: The variance is in squared units; the SD is in the data's own units. That is why we take the root back.

The shift-and-scale rule

This one rule answers most SD questions:

Add the same number to every value: SD unchanged.
Multiply every value by $k$ : SD becomes $|k| \times$ SD. Variance becomes $k^2 \times$ variance.

Data with SD 7, then every value $\times 3 + 5$ : new SD $= 3 \times 7 = 21$ . The mean shifts to $3\bar{x}+5$ , but the 5 never touches the SD.

Watch: Adding 5 tempts everyone into $21 + 5 = 26$ . Adding never changes the spread.

Ready-made spreads

Some lists have a variance you should not compute from scratch:

First $n$ naturals $1, 2, \dots, n$ : $\sigma^2 = \dfrac{n^2 - 1}{12}$ .
An arithmetic progression $a, a+d, \dots$ : variance $\dfrac{(n^2-1)d^2}{12}$ , independent of $a$ .
Two values $p$ and $q$ : mean $\dfrac{p+q}{2}$ , SD $\dfrac{|p-q|}{2}$ .

First 5 naturals: $\sigma^2 = \dfrac{25-1}{12} = 2$ , so $\sigma = \sqrt{2}$ . The two values 9 and 5 have mean 7 and SD $\dfrac{4}{2} = 2$ .

Tip: Multiples of $k$ are an AP with $d = k$ : the first five multiples of 3 have SD $3\sqrt{2}$ .

Coefficient of variation: fair comparison

Two teams score in different ranges. Who is more consistent? Compare spread relative to the mean:

CV = \frac{\sigma}{\bar{x}} \times 100\%

Batsman A: mean 50, SD 5, so $CV = 10\%$ . Batsman B: mean 40, SD 3, so $CV = 7.5\%$ . Lower CV means steadier, so B is more consistent even though A's SD is larger.

Sums the shortcuts love

\sum x = n\bar{x}, \qquad \sum x^2 = n\left(\bar{x}^2 + \sigma^2\right)

With $n = 100$ , mean 50, SD 5: $\sum x = 5000$ and $\sum x^2 = 100(2500 + 25) = 252500$ .

Watch: $\sigma$ must be squared before it joins $\bar{x}^2$ ; students add $50 + 5$ and square.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

SD of a small list

How to spot it:

Five to eight plain values; the SD (or variance) is asked.

\sigma = \sqrt{\frac{\sum (x-\bar{x})^2}{n}}

Method

Compute the mean.
Find each deviation from the mean.
Square, add, divide by n, take the root.

Why it works:

Squaring stops deviations cancelling; the root returns to the data's unit.

Try this

Find the standard deviation of 2, 4, 6, 8, 10.

Show solution

Mean $= 6$ ; deviations $-4, -2, 0, 2, 4$ .
Sum of squares $= 16 + 4 + 0 + 4 + 16 = 40$ .
Variance $= 40 \div 5 = 8$ ; SD $= \sqrt{8} = 2\sqrt{2}$ .

Answer

$2\sqrt{2}$

Type 2very common3 practice Q

SD after a shift and scale

How to spot it:

Every value is multiplied and/or added to; the new SD (or mean) is asked.

\text{SD}(kx+c) = |k|\,\text{SD}(x)

Method

Note the original SD.
Multiply the SD by the multiplier's size.
Ignore any added constant.

Why it works:

Spread grows only when values are pulled apart, and only scaling does that.

Try this

The SD of a set of values is 7. Each value is multiplied by 3 and then 5 is added. The new SD is:

Show solution

Multiply: $3 \times 7 = 21$ .
Adding 5 leaves the SD unchanged.
New SD $= 21$ .

Answer

Type 3common3 practice Q

Standard lists: naturals, APs, pairs

How to spot it:

The data is the first n naturals, an AP, or just two values.

\sigma^2 = \frac{n^2-1}{12} \;\text{(first n naturals)}

Method

Match the list to a known family.
Apply the ready variance; scale by the common difference if any.
For two values, SD is half the gap.

Why it works:

These lists have fixed shapes, so their spreads are fixed too.

Try this

Find the SD of the first five natural numbers 1, 2, 3, 4, 5.

Show solution

$n = 5$ : variance $= \dfrac{25 - 1}{12} = 2$ .
SD $= \sqrt{2}$ .

Answer

$\sqrt{2}$

Type 4common2 practice Q

Range and coefficient of variation

How to spot it:

Consistency between two sets is asked, or the range after adding a value.

CV = \frac{\sigma}{\bar{x}} \times 100\%

Method

Find mean and SD for each set.
Compute CV for each.
Smaller CV wins on consistency.

Why it works:

Spread only means something next to the level it spreads from.

Try this

Batsman A scores at mean 50 with SD 5; batsman B at mean 40 with SD 3. Who is more consistent?

Show solution

$CV_A = \dfrac{5}{50} \times 100 = 10\%$ .
$CV_B = \dfrac{3}{40} \times 100 = 7.5\%$ .
$7.5\% < 10\%$ , so B is steadier.

Answer

Batsman B

Type 5occasional

Sums from mean and SD

How to spot it:

n, mean and SD are given; the sum of values or of squares is asked.

\sum x^2 = n(\bar{x}^2 + \sigma^2)

Method

Sum of values: $n\bar{x}$ .
Square the mean and the SD.
Add, multiply by n.

Why it works:

Variance is the mean of squares minus the square of the mean, rearranged.

Try this

A set of 100 values has mean 50 and SD 5. Find the sum of the squares of the values.

Show solution

$\sum x = 100 \times 50 = 5000$ .
$\sum x^2 = 100(50^2 + 5^2) = 100 \times 2525$ .
$= 252500$ .

Answer

252500

Formula sheet

Range

R = \text{max} - \text{min}

Variance

\sigma^2 = \frac{\sum (x-\bar{x})^2}{n}

Average of squared distances from the mean.

Standard deviation

\sigma = \sqrt{\sigma^2}

Shift and scale

\text{SD}(kx + c) = |k|\,\text{SD}(x)

Adding c changes nothing; multiplying scales the SD.

First n naturals

\sigma^2 = \frac{n^2-1}{12}

Two values

\text{SD} = \frac{|p-q|}{2}

Coefficient of variation

CV = \frac{\sigma}{\bar{x}} \times 100\%

Sum of squares

\sum x^2 = n(\bar{x}^2 + \sigma^2)

Shortcuts that save time

⚡ Skip the squares for symmetric lists

Values spaced evenly around their mean cancel in pairs: 2,4,6,8,10 gives squared deviations 16,4,0,4,16. Write only the distinct squares.

Example

Find the SD of 2, 4, 6, 8, 10.

Show solution

Mean $= 6$ ; deviations $-4,-2,0,2,4$ .
Squares: $16, 4, 0, 4, 16$ ; sum $= 40$ .
Variance $= 8$ , SD $= 2\sqrt{2}$ .

Answer

$2\sqrt{2}$

⚡ AP spread without listing

For k, 2k, 3k, ..., nk use variance k²(n²−1)/12. No squaring of long lists.

Example

Find the SD of 3, 6, 9, 12, 15.

Show solution

Multiples of 3: $k = 3$ , $n = 5$ .
Variance $= 9 \times \dfrac{24}{12} = 18$ .
SD $= \sqrt{18} = 3\sqrt{2}$ .

Answer

$3\sqrt{2}$

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Adding a constant to the data and changing the SD.

Adding shifts every value equally, so the spread stays the same.

Mistake 02

Forgetting to square k: variance of kx is k·variance.

Variance of kx is k² times the variance; SD is |k| times.

Mistake 03

Comparing consistency by SD alone when means differ.

Use the coefficient of variation, SD divided by the mean.

Mistake 04

Reporting the variance as the SD.

Take the square root of the variance for the SD.

Mistake 05

Dividing by n−1 in exam SD questions.

Exam data is the whole set, so divide by n; n−1 is for samples in statistics theory.

Quick revision

Read this the night before the exam.

Range $=$ max $-$ min.
$\sigma^2 = \dfrac{\sum(x-\bar{x})^2}{n}$ , $\sigma$ is its square root.
SD( $kx + c$ ) $= |k|\,$ SD( $x$ ); adding c never changes the spread.
First $n$ naturals: $\sigma^2 = \dfrac{n^2-1}{12}$ .
Two values: SD $= \dfrac{|p-q|}{2}$ .
$CV = \dfrac{\sigma}{\bar{x}} \times 100\%$ ; lower CV means more consistent.
$\sum x^2 = n(\bar{x}^2 + \sigma^2)$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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