Statistics

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One number hides how spread out the data is

◆FRAME

Two batsmen can share the exact same average — 40 runs an innings. Even so, they can play very differently. One scores close to 40 almost every time out. The other swings wildly, some innings a duck, some a century. Those wild swings just happen to balance at 40. The average cannot tell the two batsmen apart. It was never built to.

That is the gap this chapter closes. A measure of CENTRAL TENDENCY — mean, median, mode — always drops information on the way to one summary number. A measure of DISPERSION reports how much was dropped. It shows how far, on average, the real values scatter around that summary. Two data sets can share one mean and still disagree completely on spread. A mean quoted alone is only half the picture.

One number says where the centre is. This chapter asks how far the data strays from it.

Both groups have the mean 6, marked by the dotted line in the same place on each chart. Read against the shared scale, group A piles almost everything onto the mean itself while group B puts most of its weight at the two ends. One number cannot tell these two apart, which is what a measure of dispersion is for.

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The simplest measure of spread: the range

▲CONCEPT

The simplest measure of dispersion needs no formula at all. A weather report gives a RANGE each day — the day's highest reading minus its lowest. That is the whole idea: the biggest value in a data set, minus the smallest, with nothing else to compute.

Range uses only two numbers from the whole data set, however big that set is. It says nothing about how the values in between behaved. Did the temperature climb smoothly all afternoon? Did it spike and drop twice before settling? Range cannot tell you. Two very different days can still post the same range, if their highs and lows happen to match, even when everything between those readings looked nothing alike.

Fast and honest about what it ignores — every value except the two extremes.

Set A puts four of its six values between 6 and 9; set B has nothing at all between 4 and 10. Read the dots and these are two different data sets; read the brackets and they are the same one.

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Why deviations from the mean can’t just be averaged

▲CONCEPT

Go back to that batsman. Try the obvious next step: measure spread by seeing how far each score sits from the average, then average THAT. Take each score. Subtract the mean. See what the deviations add up to.

They add up to zero. Always. For any data set with mean $\overline{x}$:

$\sum_{i=1}^n (x_i - \overline{x}) = 0.$

This is not a property of well-behaved data. It holds for every data set, however wild the scores are. The reason lies in how the mean itself is defined: the point where the scores above it and the scores below it balance out exactly. A score above average contributes a positive deviation. A score below contributes a negative one. The two kinds cancel exactly, every time, by the mean's own construction.

The raw deviations always cancel — which is exactly why they cannot be averaged directly to measure spread.

Three pieces on the left, two on the right, and both bars still stop at the same mark, 9. Move any one value and the mean slides until the two sides balance again — that balancing is what the mean is, not something this particular data happens to do.

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Mean deviation about the mean

▲CONCEPT

A factory checks packaged product weights. It does not care whether a packet ran over the labelled target or under it — only how far off it ran, on average. That is precisely the problem the last section's cancelling deviations create. MEAN DEVIATION ABOUT THE MEAN fixes it directly: make every deviation positive FIRST, using an absolute value, before averaging:

$1/n \sum_{i=1}^n |x_i - \overline{x}| \cdot$

The absolute value bars are not decoration. They stop a packet that ran 5 grams over, and a packet that ran 5 grams under, from cancelling each other out in the average. Their signed deviations would have done exactly that.

Same deviations as before — only now forced positive before they are averaged, so they cannot cancel.

Read the second line under each bar to see exactly what was thrown away: the minus signs, and nothing else. The two green bars are the values that sat above the mean, stood up beside the red ones — that standing-up is the whole of the operation.

Worked example

Find the mean deviation about the mean of $2, 4, 6, 8, 10$

$\overline{x} = (2+4+6+8+10)/5 = 30/5 = 6$
find the mean first — every deviation is measured against this one number
deviations: $-4, -2, 0, 2, 4$
subtract the mean from each value, in order
absolute deviations: $4, 2, 0, 2, 4$
drop every negative sign — this is the step that stops cancelling
$(4+2+0+2+4)/5 = 12/5 = 2.4$
average the absolute deviations — this single number is the mean deviation about the mean

Four of the five values sit away from the mean; the value 6 sits on it, so its distance is 0 and it adds nothing to the total. Laid end to end the five distances reach 12. The lower bar is that same 12, re-cut into five equal shares — one share is 2.4, and that is the mean deviation. Dividing by 5 is not a step in a formula here; it is the length being shared out.

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Mean deviation about the median

▲CONCEPT

A delivery company wants to promise one "estimated delivery time" to every customer. That promise should minimize how much customers complain about lateness or earliness, across every past delivery. The number that does this best is not the mean delivery time. It is the MEDIAN.

The same mean-deviation idea works with the median $M$ in place of the mean:

$1/n \sum_{i=1}^n |x_i - M| \cdot$

Of every possible fixed number a company could promise, the median makes this average absolute distance SMALLEST. Mean deviation about the median is never larger than mean deviation about any other fixed point, including the mean itself. That is not a coincidence of this one example. It is a property of the median, true for every data set.

Same formula, one reference point changed — and that one change is provably the best choice available.

The curve changes direction only above a data value — 3, 4, 5, 9 — and runs perfectly straight between them. A straight stretch has no low point of its own, so the best reference point is always one of the values in the data, never a number in the gap between two of them.

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Variance: squaring instead of taking the absolute value

▲CONCEPT

Two data sets can share one mean and still disagree completely on how far their values wander from it — the same gap this chapter opened with. VARIANCE measures that wandering directly. It squares each deviation instead of taking its absolute value, then averages:

$\sigma^2 = 1/n \sum_{i=1}^n (x_i - \overline{x})^2.$

Squaring, like the absolute value before it, makes every term non-negative. The deviations still cannot cancel. But squaring is a smooth operation in a way absolute value is not — it has no sharp corner at zero. That single difference is why nearly every further idea in statistics builds on variance, not on mean deviation.

Squaring also does something absolute value does not: it punishes big deviations harder than small ones. Take two deviations, $2$ and $6$ — three times as large. Their absolute values stay in that same 3-to-1 ratio. Their squares are $4$ and $36$ — now in a 1-to-9 ratio. One badly-off value pulls variance up far harder than it pulls mean deviation up, which is exactly why variance is the sharper alarm for a few extreme values sitting inside an otherwise calm data set.

Same non-negative trick as mean deviation — a different tool for making it, with much further reach.

Nine is not a fixed penalty. Stretch the far deviation to four times the near one and the big square holds sixteen copies of the small one, not nine — the further out a value sits, the harder squaring leans on it.

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Standard deviation: back to the data’s own units

▲CONCEPT

Suppose a data set is measured in one unit — say, metres. Squaring every deviation to compute variance also squares the unit: the result comes out in SQUARE METRES, a unit that no longer matches the original quantity. Nobody reports a spread in square metres.

Standard deviation is the positive square root of variance:

$\sigma = \sqrt{1/n \sum_{i=1}^n (x_i - \overline{x})^2} \cdot$

STANDARD DEVIATION fixes exactly this. The square root undoes the squaring's unit distortion, landing back in the data's own original unit — the same one the values themselves were measured in, whatever it happens to be.

Variance does the averaging in squared units; the square root undoes that, back to the unit the data was measured in.

Variance and standard deviation are the same fact in two shapes. The square's area is 8, and an area cannot be laid against marks, grams or runs — there is nothing on the axis to compare it with. Its side can: 2.83 reaches from the mean out past one data point and stops short of the next. Three of the five values fall inside one standard deviation of the mean, and you can check that against the band rather than take it on trust.

Worked example

Find the variance and standard deviation of $2, 4, 6, 8, 10$

$\overline{x} = 6$
the same mean already found for this data set's mean deviation
squared deviations: $16, 4, 0, 4, 16$
square each deviation from the mean instead of taking its absolute value
$\sigma^2 = (16+4+0+4+16)/5 = 40/5 = 8$
average the squared deviations — this is the variance
$\sigma = \sqrt{8} = 2\sqrt{2} \approx 2.83$
take the positive square root to return to the data's own units

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A shortcut formula for variance

▲CONCEPT

Computing every individual deviation before squaring it works. But it is not the only way to reach the same number. Variance has an equivalent SHORTCUT form:

$\sigma^2 = 1/n \sum_{i=1}^n x_i^2 - (\overline{x})^2.$

In words: the mean of the squares, minus the square of the mean. A calculator or a spreadsheet can keep a running sum of $x_i^2$ alongside a running sum of $x_i$. Those two totals alone give a variance directly, with no second pass through the data. Both formulas always return the same number for the same data. The shortcut is a faster route to that number, not a different one.

Same variance, reached from two different directions — one of them faster once the sum of squares is already on hand.

Worked example

Recompute the variance of $2, 4, 6, 8, 10$ with the shortcut formula

$\overline{x} = 6$
the mean is needed either way — only the route to variance changes
$\sum x_i^2 = 4+16+36+64+100 = 220$
square every value first this time, and sum the squares directly
$\sigma^2 = 220/5 - 6^2 = 44 - 36 = 8$
the mean of the squares, minus the square of the mean
$\sigma^2 = 8$
matches the direct-deviation computation exactly, as the algebra guarantees it must

The shortcut works with two numbers the direct route never produces: 44 and 36. Laid from the same zero, the part of 44 that 36 does not cover is 8 units long, and the direct route's own answer is a bar of exactly that length. The dashed lines carry the leftover piece down onto it. Both routes are doing the same subtraction; only one of them makes you compute five deviations first.

Every square here is built from one value on its own — 2 gives 4, 10 gives 100 — so the total can grow as the data arrives, one number at a time. The direct route cannot draw a single square until the mean 6 is known, because its squares stand on deviations instead. That is where the shortcut's speed actually comes from: not fewer steps, but no second pass through the data.

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Variance from a frequency table

▲CONCEPT

Real exam or survey results almost never arrive as one long list of scores. They arrive as a FREQUENCY TABLE: so many students scored this mark, so many scored that one. Re-listing every response separately, before applying the formulas above, would be pure wasted work.

Variance reweights the same idea by frequency instead. For a value $x_i$ occurring $f_i$ times:

$\sigma^2 = 1/N \sum_i f_i (x_i - \overline{x})^2.$

Here $N = \sum_i f_i$ is just the total count of observations, however they are grouped. Each deviation is counted $f_i$ times through the weight $f_i$, instead of being listed and squared $f_i$ separate times. The table format changes. The underlying question does not: how far, on average, do the values disagree with the mean?

One reweighted formula reads straight off the table nobody has time to unpack into a list.

The data arrives as a table rather than a list: the value 6 occurred five times, the value 4 three times, and so on. The mean is not the middle of the five values — it is each value weighted by how often it occurred, which here works out to exactly 6. Variance for grouped data weights each squared deviation the same way.

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Comparing consistency: the coefficient of variation

▲CONCEPT

A student compares consistency across a 100-mark theory paper and a 30-mark practical exam. The two raw standard deviations cannot be compared directly. The scales are too different for that comparison to mean anything. The subject with bigger numbers will almost always show the bigger standard deviation, whatever its actual consistency.

The COEFFICIENT OF VARIATION fixes this. It reports standard deviation as a PERCENTAGE of the mean:

$\sigma/\overline{x} \times 100.$

Unlike standard deviation or variance, this carries no units at all. Grams, rupees, or marks all cancel out of the ratio. That is exactly what makes it possible to compare two data sets that use different units, or that simply have very different means. It is the theory-versus-practical comparison the student actually needs.

Standard deviation answers "how spread out." Coefficient of variation answers "how spread out, relative to the size of the numbers involved" — a fairer question across two different scales.

The lower axis carries no unit at all: it reads in percent, and both means stack at 100. That is what lets two data sets measured in different things — marks against grams, runs against rupees — be laid on the same axis at all.

Worked example

Compare the consistency of two groups by coefficient of variation

Group A: mean $=20$, standard deviation $=4$
the two numbers a coefficient-of-variation comparison always starts from
$(4/20) \times 100 = 20$ percent
standard deviation as a percentage of the mean, for Group A
Group B: mean $=50$, standard deviation $=8$
Group B's raw standard deviation is larger than Group A's
$(8/50) \times 100 = 16$ percent
the same ratio, computed for Group B
Group B is the more consistent group
despite the larger raw standard deviation, Group B's smaller CV means less spread relative to its own mean

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Every measure, one question

■RECAP

Every measure in this chapter answers one question: how much do the real values disagree with a single summary number?

Range uses only the two extremes. It ignores everything between them. Mean deviation forces every deviation positive with an absolute value, before averaging — so the raw deviations' guaranteed cancelling cannot hide the spread. Variance and standard deviation force the same positivity by squaring instead. That is a smoother operation, which is why nearly all further statistical theory builds on them, not on mean deviation. The coefficient of variation strips away units entirely, so two differently scaled data sets can be compared on consistency alone.

In practice, let the question decide the tool. Want the fastest possible sense of spread, with no computation to speak of? Range. Want every value counted the same way, whether it ran over or under the centre? Mean deviation. Want a measure smooth enough for further theory to build on? Variance, or its square root, standard deviation. Want to compare two data sets that live on entirely separate scales? Coefficient of variation, since it carries no units at all.

Every measure here exists because the raw sum of deviations is always zero and, by itself, says nothing.

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Two traps this chapter sets

✕MISCONCEPTION

THE TRAP. Measure a data set's spread by averaging the raw deviations, $x_i - \overline{x}$, directly. No absolute value, no squaring needed.

THE REALITY. That sum is always exactly zero, for every data set, no matter how spread out it is. It carries no information about spread at all. A wildly scattered data set and a perfectly uniform one both give the same answer, zero, if the raw deviations are averaged unmodified. This is exactly why mean deviation takes an absolute value, and variance squares each deviation, before either one is averaged.

Skip the absolute value or the square, and the "spread" you compute is zero every time — not because the data has no spread, but because the method cannot see it.

Averaging the deviations two ways

Weaker. Take the data 2, 4, 6, 8, 10. The mean is 6, so the deviations are $-4$, $-2$, $0$, $2$ and $4$. Average them the obvious way: $(-4 - 2 + 0 + 2 + 4) / 5 = 0$. The answer says this data has no spread. Now run the same method on 1, 1, 1, 1, 1. The answer is 0 again. Run it on 0, 0, 0, 0, 30. Still 0. Three data sets that could hardly be less alike, and one number for all three. This is not a measure of spread that happens to be inaccurate. It is a measure that cannot tell any two data sets apart, ever.

Stronger. Same data, one change: drop the sign before averaging. The absolute deviations are 4, 2, 0, 2 and 4, and their average is $12 / 5 = 2.4$. Run the same three data sets through this version. For 1, 1, 1, 1, 1 the answer is 0 — correct, because that data really has no spread. For 0, 0, 0, 0, 30 the answer is 9.6. Three data sets, three different answers, and they come out in the right order. The absolute value is not tidying up the formula. It is the step that makes the formula able to see anything at all.

Three data sets that could hardly be less alike, and one shared mean. Average their raw deviations and all three give 0 — the red stroke is a single mark because it is a single answer, and a measure that returns the same number for every data set has told you nothing about any of them. Drop the signs first and the three separate: 1.2, 2.4 and 7.2, in the order the dots above already suggested. The absolute value is not tidying up the formula; it is what makes the formula able to see anything at all.

✕MISCONCEPTION

THE TRAP. To decide which of two data sets is "more consistent," compare their raw standard deviations directly. The smaller one wins.

THE REALITY. Standard deviation carries the same scale as the data itself. Comparing it directly is only fair when the two data sets share the same, or a similar, mean. When the means differ — the theory-versus-practical comparison from earlier in this chapter — the coefficient of variation is the fair comparison, not the raw standard deviation. A data set can carry a LARGER standard deviation and still be the MORE consistent one, once its larger mean is taken into account.

A bigger standard deviation is not automatically a less consistent data set. It can simply be a data set with bigger numbers to begin with.

Which paper did the student sit more steadily?

Weaker. One student, one term. The theory paper is out of 100: mean $62$, standard deviation $12$. The practical is out of 30: mean $20$, standard deviation $5$. Compare the two standard deviations directly and the verdict comes fast — $5$ is smaller than $12$, so the practical marks must be the steadier ones. But the two numbers were never on one scale. A 12-mark swing on a 100-mark paper and a 5-mark swing on a 30-mark paper are not the same size of swing, and nothing in that comparison noticed.

Stronger. Divide each standard deviation by its own mean before comparing. Theory: $(12 slash 62) \times 100 = 19.4$ percent. Practical: $(5 slash 20) \times 100 = 25$ percent. The verdict reverses. The theory marks are the more consistent set, even though their standard deviation is more than twice as large — because the marks they vary around are more than three times as big. The coefficient of variation carries no units, which is the only reason two papers marked out of 100 and out of 30 can be compared at all.

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Practice set

Exercise 13.1 — Range and mean deviation

practice Find the range of the data set $7, 10, 18, 25, 32$.
practice Find the mean deviation about the mean of $3, 5, 7, 7, 8$.
practice Find the mean deviation about the median of $8, 3, 10, 5, 9$.
practice For the data set $4, 4, 4, 4, 4$, what is the mean deviation about the mean?
1. $0$
2. $4$
3. $1$
4. $20$

Answers

$32 - 7 = 25$
Mean $\overline{x} = 6$; mean deviation $= 8/5 = 1.6$
Median $M = 8$; mean deviation $= 11/5 = 2.2$
$0$

The figure right after this one puts the same four problems through a second test — not which formula the wording is asking for, but which answer should already look bigger before any arithmetic starts. Between the two, each problem gets checked twice: once for what it is asking, once for whether the computed number even lands in the right range.

Look before computing. Problem 3's dots reach further from their own dashed line than problem 2's do, so its mean deviation has to be the bigger of the two — that much is settled before any arithmetic starts. The bars confirm it: 1.6 against 2.4, drawn as lengths on the same axis the dots are standing on. A rough expectation formed by eye costs nothing and it catches the slipped decimal or the divisor of 4 that a worked answer alone never would.

Exercise 13.2 — Variance and standard deviation

practice Find the variance and standard deviation of $4, 6, 6, 8$.
practice Use the shortcut formula to find the variance of $1, 2, 3, 4, 5$.
practice A frequency table shows the value $1$ occurring $2$ times, $3$ occurring $4$ times, and $5$ occurring $2$ times. Find the variance.

Answers

Mean $\overline{x} = 6$; variance $\sigma^2 = 8/4 = 2$; standard deviation $\sigma = \sqrt{2} \approx 1.41$
$\sigma^2 = 55/5 - 3^2 = 11 - 9 = 2$
Mean $\overline{x} = 3$; variance $\sigma^2 = 16/8 = 2$

The table lists three rows, so three is the number in front of you when the divisor is needed — and three is wrong. Counted as dots there are eight values, and 16 ÷ 8 = 2. The bars show the second thing this problem catches: the tallest column, four dots deep, contributes nothing at all, because it is standing on the mean. Frequency decides how many times a deviation is counted, not how large it is.

Miscellaneous — Comparing consistency

practice Brand A bulbs have mean life $1000$ hours with standard deviation $100$ hours. Brand B bulbs have mean life $1200$ hours with standard deviation $150$ hours. Which brand is more consistent?
practice Data set P has mean $10$ and standard deviation $5$. Data set Q has mean $40$ and standard deviation $6$. Which data set is relatively more variable?
1. P
2. Q
3. Both equally variable
4. Cannot be determined

Answers

$10$ percent (Brand A) versus $12.5$ percent (Brand B) — Brand A is steadier.
$50$ percent (P) versus $15$ percent (Q) — P is more variable.

Both miscellaneous problems on one scale, which is possible only because the scale has no units. Read the four sigmas along the bottom: 100, 150, 6, 5. The smallest of them, set P's 5, sits furthest right — the most variable data on the page. The bulb brands are 2.5 points apart, which is what a close call looks like; P and Q are 35 apart, which is what a landslide looks like. Neither gap could be read off the raw standard deviations at all.

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Chapter-end problems

Chapter-end problems — graded set

board-easy Find the range of the data set $12, 18, 25, 9, 30, 15$.
1. $21$
2. $-21$
3. $25$
4. $39$
board-easy Find the mean deviation about the mean of $4, 8, 12, 16$.
board-easy Find the mean deviation about the median of $3, 9, 5, 3, 12$.
board-easy Two students both average 70 marks across their exams. Which of these could reveal that they actually perform very differently, exam to exam?
1. Their mean again
2. Their mode
3. Their range
4. How many exams they sat
board-easy Find the variance of $3, 5, 5, 7$.
board-easy Use the shortcut formula to find the variance of $2, 3, 4, 5, 6$.
board-easy A data set has mean $25$ and standard deviation $5$. Find its coefficient of variation.
board-easy A frequency table has the value $2$ occurring once, $4$ occurring twice, and $6$ occurring once. Find the variance.
board-easy Find the variance of $6, 6, 6, 6$.
1. $0$
2. $6$
3. $36$
4. $1$
board-easy Which measure of spread lets you fairly compare two data sets measured on completely different scales?
1. Range
2. Mean deviation about the mean
3. Variance
4. Coefficient of variation
board-standard A city recorded these daily high temperatures (in degrees): $18, 22, 17, 25, 20, 19, 23$. Find the range.
board-standard Find the mean deviation about the mean of $10, 12, 14, 16, 18, 20$.
board-standard Find the mean deviation about the median of $5, 7, 9, 11, 13$.
board-standard A machine fills bags weighing $48, 50, 52, 49, 51$ grams. Find the standard deviation of the weights.
board-standard Use the shortcut formula to find the variance of $1, 2, 3, 4, 5, 6, 7$.
board-standard A frequency table has value $1$ (frequency $2$), $2$ (frequency $3$), $3$ (frequency $5$), $4$ (frequency $3$), and $5$ (frequency $2$). Find the variance.
board-standard Class A has mean $50$ marks with standard deviation $10$. Class B has mean $80$ marks with standard deviation $12$. Which class is relatively more consistent?
board-standard Machine X has mean output $200$ units with standard deviation $20$. Machine Y has mean output $150$ units with standard deviation $18$. Which machine is more consistent?
1. Machine X
2. Machine Y
3. Equally consistent
4. Cannot be determined
board-standard For the data set $3, 5, 9, 11$, the raw (signed) deviations from the mean sum to zero. Explain why this can never be used, by itself, to measure spread.
board-standard An investment’s monthly returns, in percent, are $2, 4, 3, 5, 6$. Find the variance and the standard deviation.
board-standard For the data set $2, 3, 3, 8, 10$, find the mean deviation about the mean and about the median, and confirm which is smaller.
board-standard Find the standard deviation of $10, 20, 30, 40, 50$ using the shortcut formula.
JEE Five numbers $8, 10, 10, 12, x$ have mean $10$. Find $x$, then find the variance of the five numbers.
JEE A frequency table has value $2$ (frequency $3$), $4$ (frequency $5$), $6$ (frequency $f$), and $8$ (frequency $2$), with mean $5$. Find $f$, then find the variance.
JEE A data set has mean $20$ and coefficient of variation $15$ percent. Find its standard deviation.
JEE A data set has six values. Five of the deviations from the mean are $-3, -1, 0, 2, 4$. Find the sixth deviation.
JEE The mean deviation about the mean of six values is $3$. Five of the absolute deviations are $5, 2, 1, 4, 3$. Find the sixth.
JEE A data set of five values has mean $6$ and variance $4$. Find the sum of the squares of the five values.
JEE A survey of $20$ households records family size: $2$ members ($4$ households), $3$ members ($6$), $4$ members ($6$), $5$ members ($4$). Find the variance of family size.
JEE Team A scores $40, 50, 60, 70, 80$ across five matches. Team B scores $20, 25, 30, 35, 40$. Compare their coefficients of variation.

Answers

$21$
$4$
$3$
Their range
$2$
$2$
$20$ percent
$2$
$0$
Coefficient of variation
$8$
$3$
$2.4$
$\sigma \approx 1.41$
$4$
$\sigma^2 \approx 1.47$
Class B
Machine X
sum $= 0$
$\sigma^2 = 2, \sigma \approx 1.41$
MD about median $= 2.6$ is smaller
$\sigma = 10\sqrt{2} \approx 14.14$
$x = 10, \sigma^2 = 1.6$
$f = 8, \sigma^2 = 29/9 \approx 3.22$
$\sigma = 3$
$-2$
$3$
$200$
$\sigma^2 = 1.05$
$23.57$ percent, equally consistent

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JEE-application problems

JEE-application problems — from the item bank

A data set has values 8, 12, 23, 34, 45, 67. Find its range.
1. -59
2. 59
3. 55
4. 75
Check your answer
1. -59 — -59 flips the correct subtraction — range is always the larger extreme minus the smaller.
2. ✓ 59 — (B) Range = 67 - 8 = 59, the largest value minus the smallest.
3. 55 — 55 uses 12, the second-smallest value, instead of 8, the true smallest.
4. 75 — 75 adds 67 and 8 instead of subtracting — range is a difference, not a sum.
A data set has values -7, -3, 2, 5, 8. Find its range.
1. -15
2. 1
3. 15
4. 12
Check your answer
1. -15 — -15 flips the correct subtraction — range cannot be negative.
2. 1 — 1 treats the smallest value, -7, as if it were 7, losing its sign entirely.
3. ✓ 15 — (C) Range = 8 - (-7) = 15, the largest value minus the smallest.
4. 12 — 12 uses 5, the second-largest value, instead of 8, the true largest.
Find the mean deviation about the mean of 3, 5, 7, 9, 11.
1. 2.4
2. 0
3. 12
4. 0.48
Check your answer
1. ✓ 2.4 — (A) Mean = 7; absolute deviations 4, 2, 0, 2, 4 sum to 12; 12 / 5 = 2.4.
2. 0 — 0 comes from averaging the SIGNED deviations, which always cancel to zero — the absolute value must come first.
3. 12 — 12 is the sum of the absolute deviations alone; the mean deviation still needs dividing by n = 5.
4. 0.48 — 0.48 divides 12 by 25 (n squared) instead of by 5 (n) — divide by the count of values, not its square.
A factory’s packets weigh 48, 50, 52, 49, 51 grams. Find the mean deviation about the mean.
1. 0
2. 6
3. 1.5
4. 1.2
Check your answer
1. 0 — 0 comes from averaging the SIGNED deviations, -2, 0, 2, -1, 1, which always cancel to zero.
2. 6 — 6 is the sum of the absolute deviations alone; it still needs dividing by n = 5.
3. 1.5 — 1.5 divides the sum of 6 by 4 (n minus one) instead of by 5 (n).
4. ✓ 1.2 — (D) Mean = 50; absolute deviations 2, 0, 2, 1, 1 sum to 6; 6 / 5 = 1.2 grams.
Find the mean deviation about the median of 4, 7, 9, 15, 20.
1. 5.2
2. 4.8
3. 0
4. 24
Check your answer
1. 5.2 — 5.2 measures distance from the mean, 11, instead of the median, 9 — the median is the reference point this formula asks for.
2. ✓ 4.8 — (B) Median = 9; absolute deviations 5, 2, 0, 6, 11 sum to 24; 24 / 5 = 4.8.
3. 0 — 0 only appears if the deviations are left signed rather than made positive first.
4. 24 — 24 is the sum of the absolute deviations alone; it still needs dividing by n = 5.
Find the variance of 5, 7, 9, 11, 13.
1. 2.83
2. 40
3. 8
4. 10
Check your answer
1. 2.83 — 2.83 is the square root of the correct variance — that is the standard deviation, not the variance itself.
2. 40 — 40 is the sum of the squared deviations alone; variance still needs dividing by n = 5.
3. ✓ 8 — (C) Mean = 9; squared deviations 16, 4, 0, 4, 16 sum to 40; 40 / 5 = 8.
4. 10 — 10 divides the sum of 40 by 4 (n minus one) instead of by 5 (n).
Find the standard deviation of 1, 2, 3, 4, 5, 6, 7.
1. 2
2. 4
3. 28
4. 2.16
Check your answer
1. ✓ 2 — (A) Mean = 4; squared deviations sum to 28; variance = 28/7 = 4; standard deviation = square root of 4 = 2.
2. 4 — 4 is the variance — the standard deviation still needs its square root, giving 2.
3. 28 — 28 is the sum of squared deviations alone, with no division and no square root applied at all.
4. 2.16 — 2.16 takes the square root after dividing by 6 (n minus one) instead of by 7 (n).
Use the shortcut formula to find the variance of 4, 8, 12, 16, 20.
1. 320
2. 164
3. 880
4. 32
Check your answer
1. 320 — 320 adds 176 and 144 (the mean squared) instead of subtracting — the shortcut formula subtracts the square of the mean.
2. 164 — 164 subtracts 12 (the mean itself) instead of 144 (the mean squared).
3. 880 — 880 is the sum of the squares alone, with no averaging or subtraction applied at all.
4. ✓ 32 — (D) Mean = 12; sum of squares = 880; 880/5 - 12 squared = 176 - 144 = 32.
Use the shortcut formula to find the variance of 3, 6, 9, 12, 15.
1. 180
2. 18
3. 90
4. 495
Check your answer
1. 180 — 180 adds 99 and 81 (the mean squared) instead of subtracting — the shortcut formula subtracts the square of the mean.
2. ✓ 18 — (B) Mean = 9; sum of squares = 495; 495/5 - 9 squared = 99 - 81 = 18.
3. 90 — 90 subtracts 9 (the mean itself) instead of 81 (the mean squared).
4. 495 — 495 is the sum of the squares alone, with no averaging or subtraction applied at all.
A frequency table has value 2 (frequency 1), 4 (frequency 2), and 6 (frequency 1). Find the variance.
1. 8
2. 2.67
3. 2
4. 1.41
Check your answer
1. 8 — 8 is the weighted sum of squared deviations alone; it still needs dividing by N = 4.
2. 2.67 — 2.67 divides 8 by 3 (N minus one) instead of by 4 (N), the total frequency.
3. ✓ 2 — (C) N = 4; mean = 4; weighted squared deviations sum to 8; 8 / 4 = 2.
4. 1.41 — 1.41 is the square root of the correct variance — that is the standard deviation, not the variance.
A frequency table has value 5 (frequency 2), 10 (frequency 4), and 15 (frequency 2). Find the variance.
1. 12.5
2. 100
3. 14.29
4. 3.54
Check your answer
1. ✓ 12.5 — (A) N = 8; mean = 10; weighted squared deviations sum to 100; 100 / 8 = 12.5.
2. 100 — 100 is the weighted sum of squared deviations alone; it still needs dividing by N = 8.
3. 14.29 — 14.29 divides 100 by 7 (N minus one) instead of by 8 (N), the total frequency.
4. 3.54 — 3.54 is the square root of the correct variance — that is the standard deviation, not the variance.
A data set has mean 25 and standard deviation 5. Find its coefficient of variation.
1. 5
2. 500
3. 0.2
4. 20
Check your answer
1. 5 — 5 is the standard deviation alone; the coefficient of variation still needs it divided by the mean.
2. 500 — 500 multiplies the standard deviation by 100 directly, skipping the divide-by-mean step entirely.
3. 0.2 — 0.2 is the correct ratio, standard deviation over mean, but never multiplied by 100.
4. ✓ 20 — (D) Coefficient of variation = (5/25) times 100 = 20 percent.
A data set has mean 40 and standard deviation 6. Find its coefficient of variation.
1. 6
2. 15
3. 0.15
4. 666.67
Check your answer
1. 6 — 6 is the standard deviation alone; the coefficient of variation still needs it divided by the mean.
2. ✓ 15 — (B) Coefficient of variation = (6/40) times 100 = 15 percent.
3. 0.15 — 0.15 is the correct ratio, standard deviation over mean, but never multiplied by 100.
4. 666.67 — 666.67 divides the mean by the standard deviation instead of the standard deviation by the mean — the ratio is inverted.
A data set has mean 20 and coefficient of variation 15 percent. Find its standard deviation.
1. 0.75
2. 300
3. 3
4. 1.33
Check your answer
1. 0.75 — 0.75 divides 15 by 20 instead of converting 15 percent to a fraction and multiplying by the mean.
2. 300 — 300 multiplies 15 (not 0.15) directly by the mean, forgetting the percent-to-fraction step.
3. ✓ 3 — (C) Standard deviation = coefficient of variation (as a fraction) times mean = 0.15 times 20 = 3.
4. 1.33 — 1.33 divides the mean by the raw percentage number instead of using the correct formula.
A data set has mean 50 and coefficient of variation 8 percent. Find its standard deviation.
1. 4
2. 0.16
3. 400
4. 6.25
Check your answer
1. ✓ 4 — (A) Standard deviation = coefficient of variation (as a fraction) times mean = 0.08 times 50 = 4.
2. 0.16 — 0.16 divides 8 by 50 instead of converting 8 percent to a fraction and multiplying by the mean.
3. 400 — 400 multiplies 8 (not 0.08) directly by the mean, forgetting the percent-to-fraction step.
4. 6.25 — 6.25 divides the mean by the raw percentage number instead of using the correct formula.
A data set has five values. Four of the deviations from the mean are -5, -2, 1, 3. Find the fifth deviation.
1. -3
2. 0
3. 5
4. 3
Check your answer
1. -3 — -3 is the sum of the four known deviations itself — the fifth deviation must cancel that sum, not repeat it.
2. 0 — 0 wrongly assumes four deviations already sum to zero on their own — the fifth one is exactly what makes the total zero.
3. 5 — 5 comes from mis-adding -5, -2, 1 and 3; recompute that sum carefully before negating it.
4. ✓ 3 — (D) The five deviations must sum to zero. -5-2+1+3 = -3, so the fifth deviation is 3.
The mean deviation about the mean of five values is 4. Four of the absolute deviations are 6, 2, 5, 3. Find the fifth.
1. 20
2. 4
3. 16
4. 0.8
Check your answer
1. 20 — 20 is the full total of all five absolute deviations — the fifth one alone must still be isolated by subtraction.
2. ✓ 4 — (B) Total of all five = 4 times 5 = 20. Known four sum to 16. Fifth = 20 - 16 = 4.
3. 16 — 16 is the sum of the four known absolute deviations alone, before subtracting it from the full total.
4. 0.8 — 0.8 divides the mean deviation by n instead of first multiplying by n to recover the full total.
A data set has values 12, 15, 18, 21, 24. Find the sum of the raw (signed) deviations from the mean.
1. 18
2. -6
3. 0
4. 90
Check your answer
1. 18 — 18 is the mean itself — the question asks for the sum of the deviations from it, not the mean.
2. -6 — -6 is only the smallest deviation, from the value 12 — all five deviations must be summed together.
3. ✓ 0 — (C) Mean = 18; deviations -6, -3, 0, 3, 6 sum to 0, as they always do for any data set.
4. 90 — 90 is the sum of the raw values — the deviations, each value minus the mean, sum to something else entirely.
Class X has mean 80 and standard deviation 12. Class Y has mean 20 and standard deviation 4. Which class is relatively more consistent?
1. Class X
2. Class Y
3. Equally consistent
4. Cannot be determined
Check your answer
1. ✓ Class X — (A) Coefficient of variation for X = (12/80) times 100 = 15 percent; for Y = (4/20) times 100 = 20 percent. Class X is the more consistent class, despite its larger raw standard deviation.
2. Class Y — Class Y’s raw standard deviation, 4, is smaller than Class X’s, 12 — but the means differ too much for that raw comparison to be fair.
3. Equally consistent — The two coefficients of variation are not equal — computing them, Class X’s is 15 percent and Class Y’s is 20 percent.
4. Cannot be determined — The coefficient of variation is exactly the tool built to settle this — it can and should be computed here.
Ten shops report daily sales, in thousands of rupees, 20 (4 shops), 25 (3 shops), 35 (3 shops). Find the standard deviation of daily sales.
1. 39
2. 6.58
3. 26
4. 6.24
Check your answer
1. 39 — 39 is the variance — the standard deviation still needs its square root, about 6.24.
2. 6.58 — 6.58 takes the square root after dividing by 9 (N minus one) instead of by 10 (N).
3. 26 — 26 is the mean daily sales, not a measure of spread at all.
4. ✓ 6.24 — (D) N = 10; mean = 26; variance = 390/10 = 39; standard deviation = square root of 39, about 6.24.

Nothing here is a new formula. The same reading settles three more items in this bank: a coefficient of variation fixes a standard deviation, and signed deviations must total zero.

Three shops nine from the mean contribute 81 times what three shops one away do: distance decides, headcount only scales it. Then 390 over 10 shops is 39, and its root is 6.24.