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NCERT Solutions · Class 11 Maths · Chapter 13

Chapter 13: Statistics (Statistics and Probability)

Step-by-step solutions to all exercises of Chapter 13, Statistics (NCERT Class 11 Maths, 2026-27): mean deviation about mean and median, variance and standard deviation of raw, discrete and grouped data, and the Miscellaneous Exercise. All 28 questions are answered, with the key answer highlighted.

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Mean deviation $= \frac{\sum f _{i} ∣ x _{i} - a ∣}{N}$ (a = mean or median). Variance $σ^{2} = \frac{\sum f _{i} ( x _{i} - x ˉ ) ^{2}}{N} = \frac{\sum f _{i} x _{i}^{2}}{N} - \overset{x}{ˉ}^{2}$ ; σ = standard deviation. Coefficient of variation $= \frac{σ}{x ˉ} \times 100$ .

Exercise 13.1

M.D. about the mean: 4, 7, 8, 9, 10, 12, 13, 17.

Solution

x̄ = 80/8 = 10. |x − 10| = 6, 3, 2, 1, 0, 2, 3, 7; sum 24. M.D. = 24/8 = 3.

M.D. = 24/8 = 3

M.D. about the mean: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.

Solution

x̄ = 500/10 = 50. |x − 50| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6; sum 84. M.D. = 8.4.

M.D. = 8.4

M.D. about the median: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

Solution

Sorted: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 (n = 12). M = (6th + 7th)/2 = (13 + 14)/2 = 13.5. |x − 13.5| = 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5; sum 28. M.D. = 28/12 = 2.33.

M.D. = 28/12 = 2.33

M.D. about the median: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Solution

Sorted: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72. M = (46 + 49)/2 = 47.5. |x − 47.5| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5; sum 70. M.D. = 7.

M.D. = 7

M.D. about the mean: x 5, 10, 15, 20, 25; f 7, 4, 6, 3, 5.

Solution

N = 25, Σfx = 35 + 40 + 90 + 60 + 125 = 350, x̄ = 14. |x − 14| = 9, 4, 1, 6, 11; f|x − 14| = 63, 16, 6, 18, 55; sum 158. M.D. = 158/25 = 6.32.

M.D. = 158/25 = 6.32

M.D. about the mean: x 10, 30, 50, 70, 90; f 4, 24, 28, 16, 8.

Solution

N = 80, Σfx = 40 + 720 + 1400 + 1120 + 720 = 4000, x̄ = 50. |x − 50| = 40, 20, 0, 20, 40; f|x − 50| = 160, 480, 0, 320, 320; sum 1280. M.D. = 16.

M.D. = 16

M.D. about the median: x 5, 7, 9, 10, 12, 15; f 8, 6, 2, 2, 2, 6.

Solution

N = 26; c.f. 8, 14, 16, 18, 20, 26. The 13th and 14th values are both 7: M = 7. |x − 7| = 2, 0, 2, 3, 5, 8; f|x − 7| = 16, 0, 4, 6, 10, 48; sum 84. M.D. = 84/26 = 3.23.

M.D. = 84/26 = 3.23

M.D. about the median: x 15, 21, 27, 30, 35; f 3, 5, 6, 7, 8.

Solution

N = 29; c.f. 3, 8, 14, 21, 29. The 15th value is 30: M = 30. |x − 30| = 15, 9, 3, 0, 5; f|x − 30| = 45, 45, 18, 0, 40; sum 148. M.D. = 148/29 = 5.1.

M.D. = 148/29 = 5.1

M.D. about the mean: income 0–100, …, 700–800; persons 4, 8, 9, 10, 7, 5, 4, 3.

Solution

Mid-points 50, 150, …, 750; N = 50, Σfx = 17900, x̄ = 358. |x − 358| = 308, 208, 108, 8, 92, 192, 292, 392; f|x − x̄| = 1232, 1664, 972, 80, 644, 960, 1168, 1176; sum 7896. M.D. = 157.92.

M.D. = 157.92

M.D. about the mean: height 95–105, …, 145–155; boys 9, 13, 26, 30, 12, 10.

Solution

Mid-points 100, 110, …, 150; N = 100, Σfx = 12530, x̄ = 125.3. |x − x̄| = 25.3, 15.3, 5.3, 4.7, 14.7, 24.7; f|x − x̄| = 227.7, 198.9, 137.8, 141, 176.4, 247; sum 1128.8. M.D. = 11.28.

M.D. = 11.28

M.D. about the median: marks 0–10, …, 50–60; girls 6, 8, 14, 16, 4, 2.

Solution

N = 50, c.f. 6, 14, 28, 44, 48, 50; N/2 = 25 lies in 20–30: l = 20, C = 14, f = 14, h = 10. M = 20 + (25 − 14)/14 × 10 = 27.86. |x − M| for mid-points 5, 15, …, 55: 22.86, 12.86, 2.86, 7.14, 17.14, 27.14; f|x − M| = 137.14, 102.86, 40, 114.29, 68.57, 54.29; sum 517.14. M.D. = 10.34.

M.D. = 10.34

M.D. about the median age: 16–20, 21–25, …, 51–55; numbers 5, 6, 12, 14, 26, 12, 16, 9.

Solution

Make the classes continuous: 15.5–20.5, …, 50.5–55.5 (h = 5); mid-points 18, 23, …, 53. N = 100, c.f. 5, 11, 23, 37, 63, …; median class 35.5–40.5: M = 35.5 + (50 − 37)/26 × 5 = 38. |x − 38| = 20, 15, 10, 5, 0, 5, 10, 15; f|x − 38| = 100, 90, 120, 70, 0, 60, 160, 135; sum 735. M.D. = 7.35.

M.D. = 7.35

Exercise 13.2

Mean and variance: 6, 7, 10, 12, 13, 4, 8, 12.

Solution

x̄ = 72/8 = 9. (x − 9)² = 9, 4, 1, 9, 16, 25, 1, 9; sum 74. Mean 9, variance 74/8 = 9.25.

Mean 9, variance 74/8 = 9.25

Mean and variance of the first n natural numbers.

Solution

x̄ = [n(n + 1)/2]/n = (n + 1)/2. Σx² = n(n + 1)(2n + 1)/6, so σ² = (n + 1)(2n + 1)/6 − (n + 1)²/4 = (n + 1)[2(2n + 1) − 3(n + 1)]/12 = (n² − 1)/12.

(n + 1)/2; (n² − 1)/12

Mean and variance of the first 10 multiples of 3.

Solution

3, 6, …, 30: x̄ = 165/10 = 16.5. Σ(x − 16.5)² = 2(13.5² + 10.5² + 7.5² + 4.5² + 1.5²) = 742.5. Mean 16.5, variance 74.25.

Mean 16.5, variance 74.25

Find the mean and variance for the data: x 6, 10, 14, 18, 24, 28, 30; f 2, 4, 7, 12, 8, 4, 3.

Solution

N = 40, Σfx = 760, x̄ = 19. (x − 19)² = 169, 81, 25, 1, 25, 81, 121; f(x − 19)² = 338, 324, 175, 12, 200, 324, 363; sum 1736. Mean 19, variance 43.4.

Mean 19, variance 43.4

Find the mean and variance for the data: x 92, 93, 97, 98, 102, 104, 109; f 3, 2, 3, 2, 6, 3, 3.

Solution

N = 22, Σfx = 2200, x̄ = 100. (x − 100)² = 64, 49, 9, 4, 4, 16, 81; f(x − 100)² = 192, 98, 27, 8, 24, 48, 243; sum 640. Mean 100, variance 640/22 = 29.09.

Mean 100, variance 640/22 = 29.09

Mean and SD by the short-cut method: x 60, 61, …, 68; f 2, 1, 12, 29, 25, 12, 10, 4, 5.

Solution

A = 64, h = 1: y = −4, …, 4. Σfy = −8 − 3 − 24 − 29 + 0 + 12 + 20 + 12 + 20 = 0; Σfy² = 32 + 9 + 48 + 29 + 0 + 12 + 40 + 36 + 80 = 286; N = 100. x̄ = 64 + 0 = 64; σ² = 286/100 − 0 = 2.86; σ = 1.69.

64; σ = 1.69

Classes 0–30, …, 180–210; f 2, 3, 5, 10, 3, 5, 2.

Solution

Mid-points 15, 45, …, 195; A = 105, h = 30, y = −3, …, 3. Σfy = −6 − 6 − 5 + 0 + 3 + 10 + 6 = 2; Σfy² = 18 + 12 + 5 + 0 + 3 + 20 + 18 = 76; N = 30. x̄ = 105 + 30 × 2/30 = 107; σ² = (900/900)[30 × 76 − 4] = 2276.

107; 2276

Classes 0–10, …, 40–50; f 5, 8, 15, 16, 6.

Solution

Mid-points 5, …, 45; A = 25, h = 10, y = −2, …, 2. Σfy = −10 − 8 + 0 + 16 + 12 = 10; Σfy² = 20 + 8 + 0 + 16 + 24 = 68; N = 50. x̄ = 25 + 100/50 = 27; σ² = (100/2500)[50 × 68 − 100] = 132.

27; 132

Mean, variance, SD (short-cut): height 70–75, …, 110–115; children 3, 4, 7, 7, 15, 9, 6, 6, 3.

Solution

Mid-points 72.5, …, 112.5; A = 92.5, h = 5, y = −4, …, 4. Σfy = −12 − 12 − 14 − 7 + 0 + 9 + 12 + 18 + 12 = 6; Σfy² = 48 + 36 + 28 + 7 + 0 + 9 + 24 + 54 + 48 = 254; N = 60. x̄ = 92.5 + 30/60 = 93; σ² = (25/3600)[60 × 254 − 36] = 105.58; σ = 10.27.

93; 105.58; σ = 10.27

Diameters 33–36, …, 49–52; circles 15, 17, 21, 22, 25. Find the SD and mean diameter.

Solution

Continuous classes 32.5–36.5, …, 48.5–52.5; mid-points 34.5, 38.5, 42.5, 46.5, 50.5; A = 42.5, h = 4, y = −2, …, 2. Σfy = −30 − 17 + 0 + 22 + 50 = 25; Σfy² = 60 + 17 + 0 + 22 + 100 = 199; N = 100. x̄ = 42.5 + 4 × 25/100 = 43.5 mm; σ² = (16/10000)[100 × 199 − 625] = 30.84; σ = 5.55 mm.

43.5 mm; σ = 5.55 mm

Miscellaneous Exercise on Chapter 13

Mean and variance of 8 observations are 9 and 9.25; six are 6, 7, 10, 12, 12, 13. Find the other two.

Solution

Σx = 72 ⇒ x + y = 72 − 60 = 12. Σx² = 8(9.25 + 81) = 722; the six squares add to 642 ⇒ x² + y² = 80. 2xy = 144 − 80 = 64 ⇒ xy = 32: 4 and 8.

4 and 8

Mean and variance of 7 observations are 8 and 16; five are 2, 4, 10, 12, 14. Find the other two.

Solution

Σx = 56 ⇒ x + y = 14. Σx² = 7(16 + 64) = 560; five squares = 460 ⇒ x² + y² = 100. xy = (196 − 100)/2 = 48: 6 and 8.

6 and 8

Mean and SD of six observations are 8 and 4. Each is multiplied by 3. Find the new mean and SD.

Solution

Mean × 3 = 24; SD × |3| = 12.

24; 12

Prove that ax₁, …, axₙ have mean ax̄ and variance a²σ².

Solution

Mean = (Σax_(i))/n = a(Σx_(i)/n) = ax̄. Variance = (1/n)Σ(ax_(i) − ax̄)² = a²(1/n)Σ(x_(i) − x̄)² = a²σ².

(i) a²σ²

Mean and SD of 20 observations are 10 and 2; one observation 8 was incorrect. Find the correct mean and SD if (i) it is omitted (ii) it is replaced by 12.

Solution

Σx = 200, Σx² = 20(4 + 100) = 2080. (i) n = 19, Σx = 192, Σx² = 2016: mean = 192/19 = 10.1; σ² = 2016/19 − (192/19)² = 106.105 − 102.116 = 3.989 ⇒ σ ≈ 2.00 (the NCERT key prints 1.99). (ii) Σx = 204, Σx² = 2080 − 64 + 144 = 2160: mean = 10.2; σ² = 108 − 104.04 = 3.96 ⇒ σ ≈ 1.99 (key: 1.98).

(i) 10.1, σ ≈ 2.00; (ii) 10.2, σ ≈ 1.99

Mean and SD of 100 observations are 20 and 3; three observations 21, 21, 18 were incorrect. Find the mean and SD if they are omitted.

Solution

Σx = 2000, Σx² = 100(9 + 400) = 40900. Remove: Σx = 2000 − 60 = 1940, Σx² = 40900 − 1206 = 39694, n = 97. Mean = 1940/97 = 20; σ² = 39694/97 − 400 = 9.216 ⇒ σ = 3.036. Quick-reference sheet ItemKey fact rangemax − min M.D. (ungrouped)Σ|xᵢ − a|/n, a = x̄ or M M.D. (grouped)Σfᵢ|xᵢ − a|/N median (classes)l + [(N/2 − C)/f] h varianceΣfᵢ(xᵢ − x̄)²/N = Σfx²/N − x̄² short-cuth²[Σfy²/N − (Σfy/N)²], x̄ = A + hΣfy/N SDσ = √variance x + avariance, SD unchanged kxvariance × k², SD × |k| first n naturalsmean (n + 1)/2, variance (n² − 1)/12 For inclusive classes (16–20, 21–25) subtract 0.5 and add 0.5 to make them continuous. Case study (practice): weekly temperatures. The noon temperatures (°C) on five days were 20, 22, 25, 23, 20. (i) Find the range. (ii) Find the mean and the mean deviation about the mean. (iii) Find the variance and the standard deviation. (iv) If each reading is converted to F = 1.8C + 32, what is the new SD?

20; σ = 3.036; Case study (practice): weekly temperatures.

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