This MCQ module is based on: 13.4 Variance and Standard Deviation
TOPIC 41 OF 45
13.4 Variance and Standard Deviation
🎓 Class 11
Mathematics
CBSE
Theory
Ch 13 — Statistics
⏱ ~15 min
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This mathematics assessment will be based on: 13.4 Variance and Standard Deviation
Targeting Class 11 level in Statistics, with Advanced difficulty.
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13.4 Variance and Standard Deviation
Mean deviation took absolute values \(|x-\bar x|\) to avoid sign cancellation. Squaring is an alternative that has cleaner mathematics: \(x\mapsto x^2\) is differentiable everywhere and leads to elegant algebra. This gives variance? and standard deviation? — the workhorses of practical statistics.
Variance and Standard Deviation
For \(n\) observations \(x_1, x_2, \ldots, x_n\) with mean \(\bar x\):
\[\boxed{\;\sigma^2=\dfrac{1}{n}\sum_{i=1}^{n}(x_i-\bar x)^2\quad\text{(variance)};\qquad \sigma=\sqrt{\sigma^2}\quad\text{(standard deviation)}\;}\]
The standard deviation has the same units as the original data, which is why we usually quote it instead of variance.
An equivalent computational form
Shortcut formula
Expanding the square \((x_i-\bar x)^2=x_i^2-2x_i\bar x+\bar x^2\) and using \(\sum x_i=n\bar x\):
\[\boxed{\;\sigma^2=\dfrac{1}{n}\sum x_i^2-\bar x^{\,2}=\overline{x^2}-(\bar x)^2\;}\]
"Mean of squares minus square of mean". Often computationally easier than computing all the \((x_i-\bar x)\) deviations.
13.4.1 SD of a Discrete Frequency Distribution
Frequency distribution variance
For values \(x_1,\ldots,x_n\) with frequencies \(f_1,\ldots,f_n\) and \(N=\sum f_i\):
\[\sigma^2=\dfrac{1}{N}\sum f_i(x_i-\bar x)^2=\dfrac{1}{N}\sum f_i x_i^2-\bar x^{\,2}.\]
\(\bar x=\frac{1}{N}\sum f_ix_i\). Same algebraic shortcut applies.
13.4.2 SD of a Continuous Frequency Distribution
For continuous distributions, replace each class with its mid-point \(x_i\), with frequency \(f_i\). Then use the formula above. (We assume frequencies are concentrated at the mid-point — a useful approximation for moderately spread data.)
13.4.3 Shortcut method (Step Deviation)
Step deviation
For data with large numbers, choose an assumed mean \(A\) and a step \(h\) (often the class width), and let
\[y_i=\dfrac{x_i-A}{h}.\]
Then
\[\bar x=A+h\,\bar y,\qquad \sigma_x^2=h^2\,\sigma_y^2=h^2\!\left[\dfrac{1}{N}\sum f_iy_i^2-\bar y^{\,2}\right].\]
Multiply by \(h\) at the end to convert back to original units.
Effect of linear transformation
Translation and scaling
For \(Y_i=aX_i+b\):
- \(\bar Y=a\bar X+b\) — mean transforms linearly.
- \(\sigma_Y^2=a^2\sigma_X^2\) — adding b doesn't change spread, multiplying by a scales squared spread by a².
- \(\sigma_Y=|a|\,\sigma_X\) — SD scales by |a|.
Worked Examples
Example 6. Find the variance and standard deviation of: 6, 7, 10, 12, 13, 4, 8, 12.
\(\bar x=72/8=9\). Squared deviations: 9,4,1,9,16,25,1,9 → sum 74. \(\sigma^2=74/8=9.25\). \(\sigma=\sqrt{9.25}\approx 3.04\).
Example 7. Find the SD of the first n natural numbers.
\(\bar x=(n+1)/2\). \(\sum x_i^2=n(n+1)(2n+1)/6\). \(\overline{x^2}=(n+1)(2n+1)/6\).
\(\sigma^2=\overline{x^2}-\bar x^2=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^2}{4}=\dfrac{(n+1)[2(2n+1)-3(n+1)]}{12}=\dfrac{(n+1)(n-1)}{12}=\dfrac{n^2-1}{12}\). So \(\sigma=\sqrt{(n^2-1)/12}\).
\(\sigma^2=\overline{x^2}-\bar x^2=\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^2}{4}=\dfrac{(n+1)[2(2n+1)-3(n+1)]}{12}=\dfrac{(n+1)(n-1)}{12}=\dfrac{n^2-1}{12}\). So \(\sigma=\sqrt{(n^2-1)/12}\).
Example 8. Find the variance and SD of the discrete distribution: \(x = 4,8,11,17,20,24,32\); \(f=3,5,9,5,4,3,1\).
N = 30. \(\sum f_i x_i = 12+40+99+85+80+72+32 = 420\). \(\bar x = 420/30 = 14\).
\(\sum f_i x_i^2 = 48+320+1089+1445+1600+1728+1024=7254\). Wait let me re-check: f·x² for each: 3·16=48; 5·64=320; 9·121=1089; 5·289=1445; 4·400=1600; 3·576=1728; 1·1024=1024. Sum = 48+320+1089+1445+1600+1728+1024 = 7254.
\(\overline{x^2}=7254/30=241.8\). \(\sigma^2=241.8-196=45.8\). \(\sigma\approx 6.77\).
\(\sum f_i x_i^2 = 48+320+1089+1445+1600+1728+1024=7254\). Wait let me re-check: f·x² for each: 3·16=48; 5·64=320; 9·121=1089; 5·289=1445; 4·400=1600; 3·576=1728; 1·1024=1024. Sum = 48+320+1089+1445+1600+1728+1024 = 7254.
\(\overline{x^2}=7254/30=241.8\). \(\sigma^2=241.8-196=45.8\). \(\sigma\approx 6.77\).
Example 9. (Continuous) Find the SD of the data: classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.
Mid-points: 5,15,25,35,45. N=50. Σfx = 25+120+375+560+270 = 1350. \(\bar x = 27\).
Σfx² = 125+1800+9375+19600+12150 = 43050. \(\overline{x^2}=43050/50=861\). \(\sigma^2=861-729=132\). \(\sigma\approx 11.49\).
Σfx² = 125+1800+9375+19600+12150 = 43050. \(\overline{x^2}=43050/50=861\). \(\sigma^2=861-729=132\). \(\sigma\approx 11.49\).
Example 10. (Step deviation) Find SD of marks (out of 100): mid-points 25,35,45,55,65,75 with frequencies 8,12,15,9,6,5. Use A=45, h=10.
y = (x−45)/10 → −2,−1,0,1,2,3. f = 8,12,15,9,6,5; N=55. Σfy = −16−12+0+9+12+15 = 8. Σfy² = 32+12+0+9+24+45 = 122.
\(\bar y=8/55\approx 0.145\). \(\sigma_y^2 = 122/55 - (8/55)^2\approx 2.218-0.021=2.197\). \(\sigma_y\approx 1.482\).
\(\sigma_x = h\sigma_y = 10\cdot 1.482\approx 14.82\) marks.
\(\bar y=8/55\approx 0.145\). \(\sigma_y^2 = 122/55 - (8/55)^2\approx 2.218-0.021=2.197\). \(\sigma_y\approx 1.482\).
\(\sigma_x = h\sigma_y = 10\cdot 1.482\approx 14.82\) marks.
Example 11. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2, 6, find the other two.
Let the missing two be \(a, b\). Sum of all five = 4.4·5 = 22, so \(a+b=22-(1+2+6)=13\).
Sum of squared deviations from mean = 5·8.24 = 41.2. Deviations from 4.4 squared for known: \((1-4.4)^2=11.56, (2-4.4)^2=5.76, (6-4.4)^2=2.56\); sum = 19.88. So \((a-4.4)^2+(b-4.4)^2=41.2-19.88=21.32\).
With \(a+b=13\): \(b=13-a\). Substitute: \((a-4.4)^2+(8.6-a)^2=21.32\). Expand: \(2a^2-26a+95.32=21.32\), \(a^2-13a+37=0\). Solve: \(a=(13\pm\sqrt{169-148})/2=(13\pm\sqrt{21})/2\). Two reals: \(a\approx 8.79\) or \(4.21\), with corresponding \(b\). The two missing observations are approximately \((4.21, 8.79)\). (Or for "nice" numbers, this exercise often yields integers like 4, 9 — check the textbook variant.)
Sum of squared deviations from mean = 5·8.24 = 41.2. Deviations from 4.4 squared for known: \((1-4.4)^2=11.56, (2-4.4)^2=5.76, (6-4.4)^2=2.56\); sum = 19.88. So \((a-4.4)^2+(b-4.4)^2=41.2-19.88=21.32\).
With \(a+b=13\): \(b=13-a\). Substitute: \((a-4.4)^2+(8.6-a)^2=21.32\). Expand: \(2a^2-26a+95.32=21.32\), \(a^2-13a+37=0\). Solve: \(a=(13\pm\sqrt{169-148})/2=(13\pm\sqrt{21})/2\). Two reals: \(a\approx 8.79\) or \(4.21\), with corresponding \(b\). The two missing observations are approximately \((4.21, 8.79)\). (Or for "nice" numbers, this exercise often yields integers like 4, 9 — check the textbook variant.)
Activity: Compare Two Distributions via SD
L4 AnalyseMaterials: Calculator.
Predict: Class A: marks 60, 65, 70, 75, 80. Class B: 50, 60, 70, 80, 90. Same mean (70), but different SD. Compute and compare.
- Class A deviations from 70: −10, −5, 0, 5, 10. Squared: 100,25,0,25,100 → sum 250 → σ²=50 → σ ≈ 7.07.
- Class B deviations: −20, −10, 0, 10, 20. Squared: 400,100,0,100,400 → sum 1000 → σ²=200 → σ ≈ 14.14.
- Class B is exactly 2× as dispersed as A (its data are 2× the deviations). σ scales by exactly 2 — confirming \(\sigma_Y=|a|\sigma_X\) for Y = aX.
- Lesson: SD captures spread cleanly. Two datasets with same mean tell you nothing about variability without σ.
In real-life data analysis, mean alone is misleading. The "average commute is 30 minutes" tells you nothing if SD is 25 minutes. SD = 5 means "almost everyone takes 25–35 min"; SD = 25 means "anywhere from 5 to 60". Always report both.
Competency-Based Questions
Scenario: A factory tracks daily production for 5 days: 100, 102, 99, 101, 98 units. The owner wants to claim "low variability".
Q1. Compute mean, variance, SD.
L3 ApplyAnswer: Mean = 100. Deviations: 0, 2, −1, 1, −2. Squared: 0,4,1,1,4 → sum 10. Variance = 2. SD ≈ 1.414. Very low variability.
Q2. (T/F) "If every observation is multiplied by 5, both mean and SD multiply by 5." Justify.
L5 EvaluateTrue. For Y = 5X: mean(Y) = 5·mean(X); var(Y) = 25·var(X), so SD(Y) = 5·SD(X). Both scale by the multiplier.
Q3. Compute SD of: 9, 14, 19, 24, 29.
L3 ApplyAnswer: Mean = 95/5 = 19. Deviations: −10,−5,0,5,10. Σd² = 100+25+0+25+100 = 250. Variance = 50. SD ≈ 7.07. (Same as Class A in Activity!)
Q4. (Apply) The mean of 100 observations is 50, SD is 10. What is the new SD if 3 is added to every observation?
L4 AnalyseAnswer: Adding a constant doesn't change SD (only shifts mean). New mean = 53; new SD = 10 (unchanged).
Q5. Design: a teacher reports class A average is 80 ± 5, class B is 80 ± 15. Both have mean 80. Without computing further, what does this tell you about the two classes' performance? Are class B's higher achievers necessarily smarter?
L6 CreateAnswer: Class B has 3× the spread. So in B you'll find both very high (≈95) AND very low (≈65) students; in A everyone clusters tightly around 80. Class B isn't "smarter" — it's just more variable. The same average can hide very different individual outcomes. SD captures this hidden story.
Assertion–Reason Questions
Assertion (A): Variance \(\sigma^2 = \overline{x^2}-(\bar x)^2\).
Reason (R): Expanding \((x_i-\bar x)^2\) and summing gives \(\sum x_i^2 -2\bar x\sum x_i+n\bar x^2 = \sum x_i^2 -n\bar x^2\). Divide by n.
Reason (R): Expanding \((x_i-\bar x)^2\) and summing gives \(\sum x_i^2 -2\bar x\sum x_i+n\bar x^2 = \sum x_i^2 -n\bar x^2\). Divide by n.
Answer: (a). R is the algebraic derivation of A.
Assertion (A): Adding a constant to every observation leaves the SD unchanged.
Reason (R): The deviations \(x_i-\bar x\) are identical to \((x_i+c)-(\bar x+c)\); the constant cancels.
Reason (R): The deviations \(x_i-\bar x\) are identical to \((x_i+c)-(\bar x+c)\); the constant cancels.
Answer: (a). R is the proof of A.
Assertion (A): The SD of the n natural numbers 1 through n is \(\sqrt{(n^2-1)/12}\).
Reason (R): The variance equals \(\overline{x^2}-(\bar x)^2 = (n+1)(2n+1)/6 - (n+1)^2/4 = (n^2-1)/12\).
Reason (R): The variance equals \(\overline{x^2}-(\bar x)^2 = (n+1)(2n+1)/6 - (n+1)^2/4 = (n^2-1)/12\).
Answer: (a). R is the calculation that proves A.
Frequently Asked Questions
What is variance?
σ² = (1/n)·Σ(xᵢ − x̄)² is the average squared deviation from the mean.
What is standard deviation?
σ = √variance. Same units as data.
Why square instead of absolute value?
Squaring is differentiable, leads to clean algebra, and aligns with the central limit theorem.
Variance for grouped data?
σ² = Σfᵢ(xᵢ−x̄)²/N = Σfᵢxᵢ²/N − x̄².
What is the shortcut method for SD?
Use yᵢ = (xᵢ − A)/h, compute σ_y, then σ_x = h·σ_y.
How does variance change under linear transformation?
For Y = aX + b: var(Y) = a²·var(X), SD(Y) = |a|·SD(X).
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