Standard Deviation — Class 11 Statistics Notes (GSEB)

Standard Deviation — Squaring the Deviations

Words marked with a dotted underline carry the question they usually appear in — tap them to reveal it.

“Uncontrolled variation is the enemy of quality.”— Edward Deming

Chapter 3 handed us a problem: Σ(x − x̄) = 0. Every positive deviation is cancelled by a negative one, so a plain average of deviations is always zero and says nothing. The mean deviation page fixed it with absolute values. This page fixes it the other way — by squaring each deviation before averaging it:

Mean of the squared deviations = variance (s²)

Squaring kills every minus sign and keeps the numbers usable in algebra — the step absolute values refuse to take.

Positive square root of that mean = standard deviation (s)

The root brings the answer back to the data's own unit — kg stays kg, runs stay runs.

“the positive square root of the mean of the squares of the deviations measured from the mean”— Karl Pearson

Variance vs standard deviation

Variance (s²) is the mean of the squared deviations and carries the squared unit; the standard deviation (s) is its positive square root and carries the data's own unit. Variance is the working number — s is the answer you quote.

Variance — raw data

s^2 = \frac{\Sigma(x - \bar{x})^2}{n}

Standard deviation — raw data

s = \sqrt{\frac{\Sigma(x - \bar{x})^2}{n}}

Grouped — direct

s = \sqrt{\frac{\Sigma f(x - \bar{x})^2}{n}}

Alternative — Σx² form

s = \sqrt{\frac{\Sigma fx^2}{n} - \bar{x}^2}

Short-cut — assumed mean A

s = \sqrt{\frac{\Sigma d^2}{n} - \left(\frac{\Sigma d}{n}\right)^2}

Short-cut grouped — × c form

s = \sqrt{\frac{\Sigma fd^2}{n} - \left(\frac{\Sigma fd}{n}\right)^2} \times c

The measure everything else is built on

Among all the measures of dispersion, the standard deviation is the most important and widely used measure — the only one that allows further algebraic combination, which is why two groups can be merged into a combined standard deviation, and why the coefficient of variation (the next page) is built on it rather than on the range.

Three Forms, One Value — Which to Use

Direct, alternative and short-cut are three roads up the same hill. The book puts it plainly — the value of the standard deviation remains same whichever form you use. Pick by asking two questions: does the mean come out fractional, and do you even need the mean?

1.

Use when: the mean is a whole number — every deviation stays exact

2.

Illustration 12 — runs 52, 58, 40, 60, 54, 38, 48 → x̄ = 50, Σ(x − x̄)² = 432 → s² = 61.71 (runs)², s = 7.86 runs.

3.

You see every squared deviation in its own column — the honest, show-your-work form the walkthrough below builds step by step.

Worked Walkthrough — Illustration 12

One full example first, so the picture and the lab below have something to imitate: seven innings of runs, solved by the direct method.

Solved Example

Problem

Illustration 12 — Runs scored by a batsman in 7 innings: 52, 58, 40, 60, 54, 38, 48. Find the variance and the standard deviation.

Solution

s² = 61.71 (runs)², s = 7.86 runs

This exact problem is the lab's default below — walk it through the five checks yourself and every method will agree on 7.86 runs.

Signed → Squared → One Root

The same seven innings, told as a picture: signed deviations around x̄ = 50, each one flipped and grown into its own square, the seven areas pooled into one average square, and finally the side of that square — the standard deviation.

Step 1 — each innings’ runs minus x̄ = 50: 2, 8, −10, 10, 4, −12, −2. Four bars go up, three go down, and the signed sum is 0 — the same zero-sum problem mean deviation solved with absolute values.

Why the average square answers the question

Areas never cancel — a deviation of −12 becomes 144, not −144. Share the total area (432) equally over the 7 innings and you get 61.71 (runs)²: the variance. Take its side, √61.71 = 7.86 runs, and you are back to a distance in the data's own unit: the standard deviation.

Stats Problem Lab — Now You Drive

The chapter's one question generator — other topic pages link here for practice. Generate a GSEB-level problem (or paste your own data), switch between Range, Quartile Deviation, Mean Deviation, Standard Deviation and Coefficient of Variation, pick any method, and pass the locked checks: enter n → build the table → locate the mean → substitute → answer. The collapsed “Other methods” panel then proves every form lands on the same value.

Stats Problem Lab — Standard Deviation· Runs scored by a batsman (runs) — raw data

Measure

New problem of the chosen type

The steps stay the same every time: enter n → build the table → locate what the formula needs → substitute → answer. Book presets reproduce the illustration answers exactly.

Method

Change A, c or the method and the table regenerates — the final measure never changes. That is the point of the exercise.

Step 1 — Given + setup

x = 52, 58, 40, 60, 54, 38, 48 — count the observations (or total the frequency column).

s² = Σ(x − x̄)² / n

n = Σf

Some Useful Results — y = bx + a

Multiply every observation by b and every deviation is stretched by |b|; add a constant a and the whole distribution (including the mean) slides without stretching at all. So:

Scaling multiplies the spread; shifting moves it. The variance, being a square, picks up b² rather than |b|.
Measure of the series y = bx + a	What happens to it
Range	R_y = \|b\| × R_x
Quartile deviation	QD_y = \|b\| × QD_x
Mean deviation	MD_y = \|b\| × MD_x
Standard deviation	s_y = \|b\| × s_x
Variance	s_y² = b² × s_x²
The additive constant a	Changes nothing — every observation and the mean shift together, so each deviation x − x̄ stays exactly where it was.

Units follow the same rule

The same data written in different units is just a × b transformation: kilograms to grams multiplies s by 1000 and the variance by 10⁶. That is why a variance can only ever be compared with a variance of the same squared unit.

Traps and Quick Tricks

Trap — add, then multiply

The SD of the series k₁, k₂, … , kₙ is 5. What is the SD of (i) k₁ + 2, k₂ + 2, … and (ii) 3k₁, 3k₂, …?

SD of (k₁ + 2) series = 5 (unchanged); of 3k series = 15

Adding slides every value with its mean — deviations untouched. Multiplying stretches every deviation by 3.

Trap — running a CV backwards

Mean 10 and CV 60% — the coefficient is a percentage of the mean, so run it in reverse:

σ = 10 × 60 / 100 = 6

variance = σ² = 36

The question asks for the variance, not σ — squaring 6 is the half of it students forget.

Trap — transform the whole pair

Mean and SD of x are 5 and 2. If y = 3x + 4, what are the mean and SD of y?

y = 3x + 4 with mean 5, sd 2 → mean 19, sd 6

The mean runs the whole rule (3 × 5 + 4 = 19); the spread only takes the multiplier (3 × 2 = 6).

Units, Limits and a Bit of History

Two things variance can never do

Carry a plain unit.Weights in kg give a variance in (kg)², heights in cm give (cm)² — squaring the deviations squares the unit too. The standard deviation takes the root back, so only s speaks in the data's own unit.Survive open-ended classes.A class like “150 and above” has no upper limit, hence no mid-value — and every grouped variance formula runs through mid-values. For such distributions the quartile deviation remains the only working measure.

Quote the unit with the number

variance = 61.71 (runs)²

standard deviation = 7.86 runs

Writing “variance = 61.71” without its squared unit, or comparing a (kg)² variance against a (marks)² one, is a statement without meaning.

Karl Pearson, 1893

Karl Pearson coined the term “standard deviation” in 1893, and in 1911 founded the world's first university statistics department at University College, London. The measure carries his coinage to this day — as does the coefficient of variation on the next page.

Zero spread, real spread

Variance is zero only when every observation is equal — 20, 20, 20, 20, 20, 20, 20 gives s² = 0 and s = 0. Any spread at all, however small, pushes both above zero.

Where You Will Meet s in the Wild

Indian startup example — Rapido

Rapido promises a pickup in about 5 minutes. A city's mean wait can read a flattering 5.4 minutes while s = 3.6 minutes — so an 11-minute wait is barely 1.5 standard deviations above x̄, yet it is the ride the customer remembers. Operations teams size such promises on x̄ ± s and on the CV, never on the average alone.

Negative example — the battery that averaged well

A battery brand advertises “average life 30 hours” — and hides that s = 12 hours: a large share of cells die between 12 and 18 hours. They published the mean and never the spread, buyers judged on the headline number, and the returns followed. An average without its standard deviation is a half-told story.

Worked Example Gallery

Standard Deviation — solved practicals

9 problems

Raw data, direct method — runs (Illustration 12)

Full question

The runs scored by a batsman in his last 7 matches are given below :

52, 58, 40, 60, 54, 38, 48

Find the variance of the runs of the batsman. Also find the standard deviation.

x (runs)	x − x̄	(x − x̄)²
52	+2	4
58	+8	64
40	−10	100
60	+10	100
54	+4	16
38	−12	144
48	−2	4
Total	350 → x̄ = 50	432

s² = 61.71 (runs)², s = 7.86 runs

x̄ = 350 / 7 = 50 runs.
Σ(x − x̄)² = 4 + 64 + 100 + 100 + 16 + 144 + 4 = 432.
s² = 432 / 7 = 61.71 (runs)² → s = √61.71 = 7.86 runs.

Raw data, alternative (Σx²) method — puzzle times (Illustration 13)

Full question

The time (in minutes) taken to solve a puzzle by 5 students are 5, 8, 3, 6, 10. Compute the standard deviation of the time taken to solve the puzzle.

x (min)	x²
5	25
8	64
3	9
6	36
10	100
Total	234

s = 2.42 min

x̄ = 32 / 5 = 6.4 min — fractional, exactly the case the Σx² form is for.
s² = Σx²/n − x̄² = 234 / 5 − 6.4² = 46.8 − 40.96 = 5.84.
s = √5.84 = 2.4166 ≈ 2.42 min.

Raw data, short-cut method — share prices (Illustration 14)

Full question

The following are closing prices (in ₹) of 5 shares :

132, 147, 120, 152, 125

Find the standard deviation by short-cut method.

x (₹)	d = x − 135	d²
132	−3	9
147	+12	144
120	−15	225
152	+17	289
125	−10	100
Total	Σd = 1	Σd² = 767

s = ₹12.38

d = x − 135 → Σd = 1, Σd² = 767.
s² = Σd²/n − (Σd/n)² = 767/5 − (1/5)² = 153.4 − 0.04 = 153.36.
s = √153.36 = 12.383 ≈ ₹12.38.

Discrete frequency, direct method — absent days (Illustration 15)

Full question

The distribution of number of absent days of 15 students of a class in the month of January is given below. Find the standard deviation and coefficient of standard deviation of their number of absent days.

No. of absent days	0	1	2	3	4
No. of Students	1	3	7	3	1

x (days)	f	x − x̄	(x − x̄)²	f(x − x̄)²
0	1	−2	4	4
1	3	−1	1	3
2	7	0	0	0
3	3	+1	1	3
4	1	+2	4	4
Total	15	—	—	14

s = 0.97 days

Σfx = 30 and n = 15 → x̄ = 30 / 15 = 2 days.
f(x − x̄)² column: 4, 3, 0, 3, 4 → Σ = 14.
s² = 14 / 15 = 0.9333 → s = 0.966 ≈ 0.97 days.

Discrete frequency, short-cut method — mobiles sold (Illustration 16)

Full question

The information of number of mobile phones sold in last 35 days in a mobile shop is given below. Find the coefficient of standard deviation of number of mobile phones sold. (Use short-cut method).

No. of mobile phones sold	5	6	7	8	9	10
No. of Days	2	5	8	12	7	1

x	f	d = x − 8	fd	fd²
5	2	−3	−6	18
6	5	−2	−10	20
7	8	−1	−8	8
8	12	0	0	0
9	7	+1	+7	7
10	1	+2	+2	4
Total	35	—	−15	57

s = 1.20

A = 8 → Σfd = −15, Σfd² = 57, n = 35.
s² = 57/35 − (−15/35)² = 1.6286 − 0.1837 = 1.4449.
s = √1.4449 = 1.202 ≈ 1.20.

Continuous classes, step short-cut — marks of 200 students (Illustration 17)

Full question

Calculate the standard deviation from the following frequency distribution of marks obtained by 200 students of a school in an examination :

Marks	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70
No. of students	5	12	30	45	50	37	21

Class	f	x (mid)	d = (x − 35)/10	fd	fd²
0 – 10	5	5	−3	−15	135
10 – 20	12	15	−2	−24	48
20 – 30	30	25	−1	−30	30
30 – 40	45	35	0	0	0
40 – 50	50	45	+1	+50	50
50 – 60	37	55	+2	+74	148
60 – 70	21	65	+3	+63	189
Total	200	—	—	118	510

s = 14.84 marks

Mid-values 5, 15, …, 65 · A = 35 (middle row) · c = 10 (HCF of |x − A|).
d = (x − A)/c → Σfd = 118, Σfd² = 510.
s² = [510/200 − (118/200)²] × 10² = (2.55 − 0.3481) × 100 = 220.19.
s = √220.19 = 14.838 ≈ 14.84 marks.

'More than' cumulative wages (Illustration 18)

Full question

Find the standard deviation of the daily wages from following information of wages (in ₹) of workers of a factory.

Daily wages (₹)	More than 130	More than 150	More than 170	More than 190	More than 210	More than 230
No. of persons	150	142	116	57	14	0

Class	f	x (mid)	d = (x − 180)/20	fd	fd²
130 – 150	8	140	−2	−16	32
150 – 170	26	160	−1	−26	26
170 – 190	59	180	0	0	0
190 – 210	43	200	+1	+43	43
210 – 230	14	220	+2	+28	56
Total	150	—	—	29	157

s = ₹20.09

Successive subtraction gives the class frequencies: 8, 26, 59, 43, 14 (n = 150).
Mid-values 140, 160, 180, 200, 220 → Σfd = 29, Σfd² = 157.
s² = [157/150 − (29/150)²] × 20² = (1.0467 − 0.0374) × 400 = 403.71.
s = √403.71 = 20.09 → ₹20.09.

The trailing 'more than ₹230 = 0' row only marks the end of the table — it carries no class of its own.

Coverage of x̄ ± s — accidents per day (Illustration 24)

Full question

The information of number of accidents in 10 days on a particular road is given below. Find the mean and standard deviation of number of accidents per day. Find the percentage of days having the number of accidents lying between the limits x̄ ± s.

No. of accidents	1	2	3	4	5
No. of days	2	3	3	1	1

x (accidents)	f	f · x	f(x − 2.6)²
1	2	2	5.12
2	3	6	1.08
3	3	9	0.48
4	1	4	1.96
5	1	5	5.76
Total	10	26	14.4

x̄ = 2.6, s = 1.2 → band 1.4 to 3.8 → 60% of the days

x̄ = Σfx / n = 26 / 10 = 2.6 accidents.
s² = 14.4 / 10 = 1.44 → s = 1.2.
The band is 2.6 ± 1.2 = 1.4 to 3.8 accidents.
Days with 2 accidents (3) and 3 accidents (3) lie inside; days with 1, 4 or 5 do not → 6 of 10 days.

Correction of wrong entries (Illustration 25)

Full question

The mean number of units produced by 100 workers of a factory per day is 60 units and its standard deviation is 10 units. Later on, it was noticed that two workers have actually produced 30 and 20 units respectively but it was registered as 5 and 45 units respectively. By considering this, find corrected mean and corrected standard deviation of the number of units produced by the workers.

Quantity	From the given mean & SD	After correction
Σx	6000	6000
Σx²	370000	369250
Mean	60	60 (unchanged)
s	10	9.62

Corrected mean = 60 units, corrected s = 9.62 units

Recover the totals: Σx = 100 × 60 = 6000 · Σx² = 100(10² + 60²) = 370000.
Corrected Σx = 6000 − 5 − 45 + 30 + 20 = 6000 → the mean stays 60.
Corrected Σx² = 370000 − 25 − 2025 + 900 + 400 = 369250.
s² = 369250 / 100 − 60² = 3692.5 − 3600 = 92.5 → s = 9.617 ≈ 9.62 units.

The two wrong entries cancel in Σx here (−50 + 50), so the mean survives — but Σx² always loses the squares of the wrong entries and gains the squares of the right ones, so the SD rarely does.

Advantages and Limitations

Standard deviation — merits and demerits

Advantages

It is rigidly defined: every observation enters through its squared deviation, so two people working the same data always reach the same s.
It uses all the observations, like the mean deviation, and is therefore superior to the range and the quartile deviation.
It is the most important and widely used measure of dispersion — the only one that allows algebraic combination, which is what makes combined standard deviation possible.
Squaring removes the minus signs without throwing information away — unlike absolute values, squared deviations stay usable in further algebra.
It is comparatively stable from sample to sample, which is why the coefficient of variation and quality limits are both built on it.

Disadvantages

Its calculation is more involved than the range or the quartile deviation — deviations, squares, an average and finally a square root.
Extreme values get undue importance: squaring makes a couple of very large deviations dominate the entire answer.
It cannot be computed for open-ended class distributions, and its variance carries a squared unit that reads awkwardly next to the data.

Quick Reference

Idea	Statement
Formulas	s² = Σ(x − x̄)² / n · s = √(Σ(x − x̄)² / n)
Alternative form	s = √(Σfx² / n − x̄²)
Short-cut form	s = √(Σfd² / n − (Σfd / n)²), × c for d = (x − A)/c
Transformations	y = bx + a → s_y = \|b\| · s_x, s_y² = b² · s_x² (a has no effect)
Two names, one idea	Variance = mean of the squared deviations; standard deviation = its positive square root
Best measure	Standard deviation is the best measure of dispersion
Its own unit	SD has the same unit as the observations themselves
Unit of variance	Unit of variance = (unit of data)²
Zero spread	Variance is zero only when every observation is equal
Constant data	Constant data ⇒ s = 0

Key Takeaways

Variance s² is the mean of the squared deviations; standard deviation s is its positive square root. So what: squaring beats the Σ(x − x̄) = 0 deadlock without absolute values, and unlike MD it stays algebraically usable — hence combined SD.
Three forms — direct, alternative (Σx²) and short-cut (A, × c) — always return the same s; the form only changes how big the intermediate numbers are. So what: a fractional mean is your cue to switch to the Σx² form, and “only s needed” is your cue for the short-cut.
y = bx + a → s_y = |b| · s_x and s_y² = b² · s_x²; the additive a changes nothing. So what: every “SD of (k + 2) / of 3k / of 3x + 4” trap collapses into this one line.
Units matter: s speaks in the data’s unit, variance in the squared unit (kg → (kg)²). So what: quoting “variance 61.71” bare, or comparing (kg)² against (marks)², is a statement without meaning.
Constant data gives s = 0, and open-ended classes make s impossible. So what: a “150 and above” income table can only be read through quartile deviation — never through variance.
Anchors worth carrying: 7.86 runs (direct) · 2.42 min (alternative) · ₹12.38 (short-cut) · 14.84 marks (× c) · ₹20.09 (‘more than’) · 9.62 (correction). So what: these six reproduce in the lab above in every method, every time — which is exactly how you check your own working.