NCERT Solutions for Class 11 Maths Chapter 13: Statistics

These Class 11 Maths Chapter 13 solutions cover Statistics — the measures of dispersion. Every question in Exercise 13.1, Exercise 13.2 and the Miscellaneous Exercise is reproduced verbatim from the NCERT textbook and solved step by step, with full working tables for range, mean deviation, variance and standard deviation, all verified against the book’s answer key for the 2026–27 session.

Class: 11 Subject: Mathematics Chapter: 13 – Statistics Exercises: 13.1, 13.2, Miscellaneous Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Important formulas
Exercise 13.1 solutions
Exercise 13.2 solutions
Miscellaneous Exercise solutions
Common mistakes to avoid
Practice MCQs & Assertion–Reason
Quick revision summary
FAQs

Chapter 13 Overview

Chapter 13 of Class 11 Maths, Statistics, builds on the measures of central tendency (mean, median, mode) learned earlier by introducing measures of dispersion — single numbers that describe how scattered or spread out a data set is. The chapter studies the range, mean deviation (about the mean and about the median, for ungrouped, discrete and continuous data), and the variance and standard deviation, including the shortcut (step-deviation) method. It closes by comparing two distributions and analysing the effect of changing each observation by a constant. These Class 11 Maths Chapter 13 solutions work every exercise question with neat tables and verified final answers.

Key Concepts & Definitions

Dispersion: the scatter or spread of data about a central value. Common measures: range, quartile deviation, mean deviation, variance, standard deviation.

Range: the difference between the maximum and minimum values of the data — a quick but crude measure that ignores the central tendency.

Mean deviation, M.D.(a): the mean of the absolute deviations of the observations from a central value a (usually mean or median). It always uses |x_i − a| so that positive and negative deviations do not cancel.

Variance (σ²): the mean of the squared deviations from the mean. Squaring removes the sign problem and allows further algebraic treatment.

Standard deviation (σ): the positive square root of the variance; it has the same unit as the observations, so it is the most reliable measure of dispersion.

Median (grouped): for a continuous distribution, the median class is the class whose cumulative frequency first reaches N/2, and the median is found by interpolation inside that class.

Important Formulas (Chapter 13)

Range = Maximum value − Minimum value.

Mean deviation (ungrouped): M.D.(x̄) = (1/n)Σ|x_i − x̄|; M.D.(M) = (1/n)Σ|x_i − M|.

Mean deviation (grouped): M.D.(x̄) = (1/N)Σf_i|x_i − x̄|; M.D.(M) = (1/N)Σf_i|x_i − M|, N = Σf_i.

Median of grouped data: M = l + [(N/2 − C)/f] × h, where l = lower limit, C = c.f. before median class, f = frequency, h = width.

Variance & SD (ungrouped): σ² = (1/n)Σ(x_i − x̄)²; σ = √[(1/n)Σ(x_i − x̄)²].

Variance & SD (frequency): σ² = (1/N)Σf_i(x_i − x̄)²; σ = (1/N)√[NΣf_ix_i² − (Σf_ix_i)²].

Shortcut (step-deviation), y_i = (x_i − A)/h: σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²]; x̄ = A + h·(Σf_iy_i/N).

Effect of a constant: adding a constant to every observation leaves the variance unchanged; multiplying every observation by k multiplies the variance by k² and the SD by |k|.

Exercise 13.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

Find the mean deviation about the mean for the data in Exercises 1 and 2.

1. 4, 7, 8, 9, 10, 12, 13, 17

SOLUTION Mean x̄ = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/8 = 80/8 = 10. Absolute deviations |x_i − 10|: 6, 3, 2, 1, 0, 2, 3, 7. Their sum = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24. M.D.(x̄) = 24/8 = 3.

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

SOLUTION Sum = 38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44 = 500, n = 10, so x̄ = 500/10 = 50. Absolute deviations |x_i − 50|: 12, 20, 2, 10, 8, 5, 13, 4, 4, 6. Sum = 84. M.D.(x̄) = 84/10 = 8.4.

Find the mean deviation about the median for the data in Exercises 3 and 4.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

SOLUTION Arrange in ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18 (n = 12, even). Median = mean of 6th and 7th observations = (13 + 14)/2 = 13.5. Absolute deviations |x_i − 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.(M) = 28/12 = 2.333… ≈ 2.33.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

SOLUTION Arrange: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 (n = 10, even). Median = mean of 5th and 6th observations = (46 + 49)/2 = 47.5. Absolute deviations |x_i − 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.(M) = 70/10 = 7.

Find the mean deviation about the mean for the data in Exercises 5 and 6.

5. x_i: 5, 10, 15, 20, 25 f_i: 7, 4, 6, 3, 5

SOLUTION

x_i	f_i	f_ix_i	\|x_i − x̄\|	f_i\|x_i − x̄\|
5	7	35	9	63
10	4	40	4	16
15	6	90	1	6
20	3	60	6	18
25	5	125	11	55
Total	25	350		158

N = 25, Σf_ix_i = 350, so x̄ = 350/25 = 14. M.D.(x̄) = (1/N)Σf_i|x_i − x̄| = 158/25 = 6.32.

6. x_i: 10, 30, 50, 70, 90 f_i: 4, 24, 28, 16, 8

SOLUTION

x_i	f_i	f_ix_i	\|x_i − x̄\|	f_i\|x_i − x̄\|
10	4	40	40	160
30	24	720	20	480
50	28	1400	0	0
70	16	1120	20	320
90	8	720	40	320
Total	80	4000		1280

N = 80, Σf_ix_i = 4000, so x̄ = 4000/80 = 50. M.D.(x̄) = 1280/80 = 16.

Find the mean deviation about the median for the data in Exercises 7 and 8.

7. x_i: 5, 7, 9, 10, 12, 15 f_i: 8, 6, 2, 2, 2, 6

SOLUTION

x_i	f_i	c.f.	\|x_i − M\|	f_i\|x_i − M\|
5	8	8	2	16
7	6	14	0	0
9	2	16	2	4
10	2	18	3	6
12	2	20	5	10
15	6	26	8	48
Total	26			84

N = 26 (even); N/2 = 13. The 13th and 14th observations both fall in the c.f. 14, which corresponds to x = 7, so Median M = 7. M.D.(M) = (1/N)Σf_i|x_i − M| = 84/26 = 3.23 (approx.) = 3.23.

8. x_i: 15, 21, 27, 30, 35 f_i: 3, 5, 6, 7, 8

SOLUTION

x_i	f_i	c.f.	\|x_i − M\|	f_i\|x_i − M\|
15	3	3	15	45
21	5	8	9	45
27	6	14	3	18
30	7	21	0	0
35	8	29	5	40
Total	29			148

N = 29 (odd); the median is the (29 + 1)/2 = 15th observation. The c.f. just reaching 15 is 21, corresponding to x = 30, so Median M = 30. M.D.(M) = 148/29 = 5.103… ≈ 5.1.

Find the mean deviation about the mean for the data in Exercises 9 and 10.

9. Income per day (in ₹): 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800Number of persons: 4, 8, 9, 10, 7, 5, 4, 3

SOLUTION

Income	Mid x_i	f_i	f_ix_i	\|x_i − x̄\|	f_i\|x_i − x̄\|
0-100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	2250	392	1176
Total		50	17900		7896

N = 50, Σf_ix_i = 17900, so x̄ = 17900/50 = 358. M.D.(x̄) = 7896/50 = 157.92.

10. Height in cms: 95-105, 105-115, 115-125, 125-135, 135-145, 145-155Number of boys: 9, 13, 26, 30, 12, 10

SOLUTION

Height	Mid x_i	f_i	f_ix_i	\|x_i − x̄\|	f_i\|x_i − x̄\|
95-105	100	9	900	25.3	227.7
105-115	110	13	1430	15.3	198.9
115-125	120	26	3120	5.3	137.8
125-135	130	30	3900	4.7	141.0
135-145	140	12	1680	14.7	176.4
145-155	150	10	1500	24.7	247.0
Total		100	12530		1128.8

N = 100, Σf_ix_i = 12530, so x̄ = 12530/100 = 125.3. M.D.(x̄) = 1128.8/100 = 11.288 ≈ 11.28.

11. Find the mean deviation about median for the following data: Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60 Number of Girls: 6, 8, 14, 16, 4, 2

SOLUTION

Marks	f_i	c.f.	Mid x_i	\|x_i − M\|	f_i\|x_i − M\|
0-10	6	6	5	22.86	137.14
10-20	8	14	15	12.86	102.86
20-30	14	28	25	2.86	40.0
30-40	16	44	35	7.14	114.29
40-50	4	48	45	17.14	68.57
50-60	2	50	55	27.14	54.29
Total	50				517.14

N = 50, N/2 = 25, which lies in the class 20-30 (c.f. 28), so median class = 20-30 with l = 20, C = 14, f = 14, h = 10. Median M = 20 + [(25 − 14)/14] × 10 = 20 + 110/14 = 20 + 7.857 = 27.857. M.D.(M) = 517.14/50 = 10.343 ≈ 10.34.

12. Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age (in years): 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55 Number: 5, 6, 12, 14, 26, 12, 16, 9 [Hint: Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

SOLUTION Convert to continuous classes (subtract 0.5 from each lower limit, add 0.5 to each upper limit):

Age	f_i	c.f.	Mid x_i	\|x_i − M\|	f_i\|x_i − M\|
15.5-20.5	5	5	18	20	100
20.5-25.5	6	11	23	15	90
25.5-30.5	12	23	28	10	120
30.5-35.5	14	37	33	5	70
35.5-40.5	26	63	38	0	0
40.5-45.5	12	75	43	5	60
45.5-50.5	16	91	48	10	160
50.5-55.5	9	100	53	15	135
Total	100				735

N = 100, N/2 = 50, which lies in the class 35.5-40.5 (c.f. 63), so l = 35.5, C = 37, f = 26, h = 5. Median M = 35.5 + [(50 − 37)/26] × 5 = 35.5 + 65/26 = 35.5 + 2.5 = 38. M.D.(M) = 735/100 = 7.35.

Exercise 13.2 Solutions

Find the mean and variance for each of the data in Exercises 1 to 5.

1. 6, 7, 10, 12, 13, 4, 8, 12

SOLUTION Sum = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72, n = 8, so mean x̄ = 72/8 = 9. Squared deviations (x_i − 9)²: 9, 4, 1, 9, 16, 25, 1, 9. Sum = 74. Variance σ² = (1/n)Σ(x_i − x̄)² = 74/8 = 9.25. ∴ Mean = 9, Variance = 9.25.

2. First n natural numbers

SOLUTION Mean = (sum of first n natural numbers)/n = [n(n + 1)/2]/n = (n + 1)/2. Σx_i² = n(n + 1)(2n + 1)/6, so variance = (1/n)Σx_i² − (x̄)² = (n + 1)(2n + 1)/6 − (n + 1)²/4. = (n + 1)[2(2n + 1) − 3(n + 1)]/12 = (n + 1)(4n + 2 − 3n − 3)/12 = (n + 1)(n − 1)/12. ∴ Mean = (n + 1)/2, Variance = (n² − 1)/12.

3. First 10 multiples of 3

SOLUTION Data: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Sum = 165, n = 10, so mean = 165/10 = 16.5. These are 3 × (first 10 natural numbers). Variance of first 10 natural numbers = (10² − 1)/12 = 99/12 = 8.25. Multiplying each observation by 3 multiplies the variance by 3² = 9: Variance = 9 × 8.25 = 74.25. ∴ Mean = 16.5, Variance = 74.25.

4. x_i: 6, 10, 14, 18, 24, 28, 30 f_i: 2, 4, 7, 12, 8, 4, 3

SOLUTION

x_i	f_i	f_ix_i	x_i − x̄	(x_i − x̄)²	f_i(x_i − x̄)²
6	2	12	−13	169	338
10	4	40	−9	81	324
14	7	98	−5	25	175
18	12	216	−1	1	12
24	8	192	5	25	200
28	4	112	9	81	324
30	3	90	11	121	363
Total	40	760			1736

N = 40, Σf_ix_i = 760, so mean x̄ = 760/40 = 19. Variance σ² = (1/N)Σf_i(x_i − x̄)² = 1736/40 = 43.4. ∴ Mean = 19, Variance = 43.4.

5. x_i: 92, 93, 97, 98, 102, 104, 109 f_i: 3, 2, 3, 2, 6, 3, 3

SOLUTION

x_i	f_i	f_ix_i	x_i − x̄	(x_i − x̄)²	f_i(x_i − x̄)²
92	3	276	−8	64	192
93	2	186	−7	49	98
97	3	291	−3	9	27
98	2	196	−2	4	8
102	6	612	2	4	24
104	3	312	4	16	48
109	3	327	9	81	243
Total	22	2200			640

N = 22, Σf_ix_i = 2200, so mean x̄ = 2200/22 = 100. Variance σ² = 640/22 = 29.09 (approx.). ∴ Mean = 100, Variance = 29.09.

6. Find the mean and standard deviation using short-cut method. x_i: 60, 61, 62, 63, 64, 65, 66, 67, 68 f_i: 2, 1, 12, 29, 25, 12, 10, 4, 5

SOLUTION Take assumed mean A = 64, h = 1, y_i = x_i − 64.

x_i	f_i	y_i	f_iy_i	f_iy_i²
60	2	−4	−8	32
61	1	−3	−3	9
62	12	−2	−24	48
63	29	−1	−29	29
64	25	0	0	0
65	12	1	12	12
66	10	2	20	40
67	4	3	12	36
68	5	4	20	80
Total	100		0	286

N = 100, Σf_iy_i = 0, Σf_iy_i² = 286. Mean x̄ = A + h·(Σf_iy_i/N) = 64 + 1·(0/100) = 64. σ = (h/N)√[NΣf_iy_i² − (Σf_iy_i)²] = (1/100)√[100×286 − 0] = (1/100)√28600 = 169.11/100 = 1.69. ∴ Mean = 64, Standard deviation = 1.69.

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

7. Classes: 0-30, 30-60, 60-90, 90-120, 120-150, 150-180, 180-210 Frequencies: 2, 3, 5, 10, 3, 5, 2

SOLUTION Take assumed mean A = 105, h = 30, y_i = (x_i − 105)/30.

Class	Mid x_i	f_i	y_i	f_iy_i	f_iy_i²
0-30	15	2	−3	−6	18
30-60	45	3	−2	−6	12
60-90	75	5	−1	−5	5
90-120	105	10	0	0	0
120-150	135	3	1	3	3
150-180	165	5	2	10	20
180-210	195	2	3	6	18
Total		30		2	76

N = 30, Σf_iy_i = 2, Σf_iy_i² = 76. Mean x̄ = A + h·(Σf_iy_i/N) = 105 + 30·(2/30) = 105 + 2 = 107. σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²] = (900/900)[30×76 − 4] = (2280 − 4)/1 = 2276. ∴ Mean = 107, Variance = 2276.

8. Classes: 0-10, 10-20, 20-30, 30-40, 40-50 Frequencies: 5, 8, 15, 16, 6

SOLUTION Take assumed mean A = 25, h = 10, y_i = (x_i − 25)/10.

Class	Mid x_i	f_i	y_i	f_iy_i	f_iy_i²
0-10	5	5	−2	−10	20
10-20	15	8	−1	−8	8
20-30	25	15	0	0	0
30-40	35	16	1	16	16
40-50	45	6	2	12	24
Total		50		10	68

N = 50, Σf_iy_i = 10, Σf_iy_i² = 68. Mean x̄ = A + h·(Σf_iy_i/N) = 25 + 10·(10/50) = 25 + 2 = 27. σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²] = (100/2500)[50×68 − 100] = (1/25)[3400 − 100] = 3300/25 = 132. ∴ Mean = 27, Variance = 132.

9. Find the mean, variance and standard deviation using short-cut method Height in cms: 70-75, 75-80, 80-85, 85-90, 90-95, 95-100, 100-105, 105-110, 110-115 No. of children: 3, 4, 7, 7, 15, 9, 6, 6, 3

SOLUTION Take assumed mean A = 92.5, h = 5, y_i = (x_i − 92.5)/5.

Height	Mid x_i	f_i	y_i	f_iy_i	f_iy_i²
70-75	72.5	3	−4	−12	48
75-80	77.5	4	−3	−12	36
80-85	82.5	7	−2	−14	28
85-90	87.5	7	−1	−7	7
90-95	92.5	15	0	0	0
95-100	97.5	9	1	9	9
100-105	102.5	6	2	12	24
105-110	107.5	6	3	18	54
110-115	112.5	3	4	12	48
Total		60		6	254

N = 60, Σf_iy_i = 6, Σf_iy_i² = 254. Mean x̄ = A + h·(Σf_iy_i/N) = 92.5 + 5·(6/60) = 92.5 + 0.5 = 93. σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²] = (25/3600)[60×254 − 36] = (25/3600)(15240 − 36) = (25×15204)/3600 = 380100/3600 = 105.58. σ = √105.58 = 10.27. ∴ Mean = 93, Variance = 105.58, Standard deviation = 10.27.

10. The diameters of circles (in mm) drawn in a design are given below: Diameters: 33-36, 37-40, 41-44, 45-48, 49-52 No. of circles: 15, 17, 21, 22, 25 Calculate the standard deviation and mean diameter of the circles. [Hint: First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5-48.5, 48.5-52.5 and then proceed.]

SOLUTION Make continuous classes; mid-points 34.5, 38.5, 42.5, 46.5, 50.5. Take A = 42.5, h = 4, y_i = (x_i − 42.5)/4.

Class	Mid x_i	f_i	y_i	f_iy_i	f_iy_i²
32.5-36.5	34.5	15	−2	−30	60
36.5-40.5	38.5	17	−1	−17	17
40.5-44.5	42.5	21	0	0	0
44.5-48.5	46.5	22	1	22	22
48.5-52.5	50.5	25	2	50	100
Total		100		25	199

N = 100, Σf_iy_i = 25, Σf_iy_i² = 199. Mean x̄ = A + h·(Σf_iy_i/N) = 42.5 + 4·(25/100) = 42.5 + 1 = 43.5. σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²] = (16/10000)[100×199 − 625] = (16/10000)(19900 − 625) = (16×19275)/10000 = 308400/10000 = 30.84. σ = √30.84 = 5.55. ∴ Mean diameter = 43.5 mm, Standard deviation = 5.55 mm.

Miscellaneous Exercise on Chapter 13 Solutions

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

SOLUTION Let the two unknown observations be x and y. Mean = 9 with n = 8, so total = 72. 6 + 7 + 10 + 12 + 12 + 13 + x + y = 72 ⇒ 60 + x + y = 72 ⇒ x + y = 12 … (1) Variance = 9.25 = (1/8)Σx_i² − (9)² ⇒ (1/8)Σx_i² = 9.25 + 81 = 90.25 ⇒ Σx_i² = 722. Sum of squares of the six known values = 36 + 49 + 100 + 144 + 144 + 169 = 642. So x² + y² = 722 − 642 = 80 … (2) From (1): (x + y)² = 144 ⇒ x² + y² + 2xy = 144 ⇒ 2xy = 144 − 80 = 64. Then (x − y)² = 80 − 64 = 16 ⇒ x − y = ±4. Solving x + y = 12 and x − y = 4: x = 8, y = 4. ∴ The remaining two observations are 4 and 8.

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

SOLUTION Let the two unknown observations be x and y. Mean = 8 with n = 7, so total = 56. 2 + 4 + 10 + 12 + 14 + x + y = 56 ⇒ 42 + x + y = 56 ⇒ x + y = 14 … (1) Variance = 16 = (1/7)Σx_i² − (8)² ⇒ (1/7)Σx_i² = 16 + 64 = 80 ⇒ Σx_i² = 560. Sum of squares of the five known values = 4 + 16 + 100 + 144 + 196 = 460. So x² + y² = 560 − 460 = 100 … (2) From (1): (x + y)² = 196 ⇒ 2xy = 196 − 100 = 96. Then (x − y)² = 100 − 96 = 4 ⇒ x − y = ±2. Solving x + y = 14 and x − y = 2: x = 8, y = 6. ∴ The remaining two observations are 6 and 8.

3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

SOLUTION When every observation is multiplied by a constant k, the new mean = k × (old mean) and the new standard deviation = |k| × (old SD). Here k = 3: New mean = 3 × 8 = 24; New standard deviation = 3 × 4 = 12. ∴ New mean = 24, New standard deviation = 12.

4. Given that x̄ is the mean and σ² is the variance of n observations x₁, x₂, …,x_n. Prove that the mean and variance of the observations ax₁, ax₂, ax₃, …., ax_n are a x̄ and a² σ², respectively, (a ≠ 0).

SOLUTION Let the new observations be y_i = ax_i for i = 1, 2, …, n. Mean: ȳ = (1/n)Σy_i = (1/n)Σax_i = a·(1/n)Σx_i = a x̄. Hence the new mean is a x̄. Variance: Var(y) = (1/n)Σ(y_i − ȳ)² = (1/n)Σ(ax_i − a x̄)² = (1/n)Σa²(x_i − x̄)². = a²·(1/n)Σ(x_i − x̄)² = a²σ². Hence the new variance is a²σ². Proved.

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12.

SOLUTION Given n = 20, mean = 10 ⇒ Σx_i = 200. SD = 2 ⇒ variance = 4 = (1/20)Σx_i² − 100 ⇒ Σx_i² = 20×104 = 2080. (i) Item 8 omitted. New n = 19. Correct Σx_i = 200 − 8 = 192 ⇒ Correct mean = 192/19 = 10.1 (approx.). Correct Σx_i² = 2080 − 8² = 2080 − 64 = 2016. Variance = 2016/19 − (10.1)² = 106.1 − 102.01 = 4.09 (approx.). SD = √4.09 ≈ 1.99 (approx., taking mean ≈ 10.105). ∴ (i) Correct mean = 10.1, Correct SD = 1.99. (ii) Item 8 replaced by 12. New n = 20. Correct Σx_i = 200 − 8 + 12 = 204 ⇒ Correct mean = 204/20 = 10.2. Correct Σx_i² = 2080 − 64 + 144 = 2160. Variance = 2160/20 − (10.2)² = 108 − 104.04 = 3.96. SD = √3.96 = 1.98 (approx.). ∴ (ii) Correct mean = 10.2, Correct SD = 1.98.

6. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

SOLUTION Given n = 100, mean = 20 ⇒ Σx_i = 2000. SD = 3 ⇒ variance = 9 = (1/100)Σx_i² − 400 ⇒ Σx_i² = 100×409 = 40900. Omit the three wrong values 21, 21, 18 (sum = 60; sum of squares = 441 + 441 + 324 = 1206). New n = 97. Correct Σx_i = 2000 − 60 = 1940 ⇒ Correct mean = 1940/97 = 20. Correct Σx_i² = 40900 − 1206 = 39694. Variance = 39694/97 − (20)² = 409.216 − 400 = 9.216. SD = √9.216 = 3.036 (approx.). ∴ Mean = 20, Standard deviation = 3.036.

Common Mistakes to Avoid

Watch out for these

Forgetting to use the absolute value in mean deviation — the signed deviations from the mean always sum to zero, so |x_i − a| is essential.
Using N (total frequency) instead of n (number of distinct values) when dividing in grouped data, or vice versa — always divide by the sum of frequencies N for grouped data.
Picking the wrong median class: the median class is the first whose cumulative frequency reaches N/2, not the class with the largest frequency.
In Exercise 13.1 Q12, forgetting the hint to make the data continuous (subtract/add 0.5) before finding the median class.
Mixing up variance and standard deviation — SD is the square root of variance and carries the unit of the data.
Confusing the rules for a constant: adding a constant leaves variance unchanged, while multiplying by k scales variance by k² (SD by |k|).
In step-deviation, forgetting the factor h² in the variance formula σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²].

Practice MCQs & Assertion–Reason

1. The range of the data 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 is:

(a) 80 (b) 112 (c) 117 (d) 71

2. The mean deviation about the mean for the data 4, 7, 8, 9, 10, 12, 13, 17 is:

(a) 2.75 (b) 3 (c) 3.5 (d) 10

3. The standard deviation is the:

(a) square of the variance (b) positive square root of the variance (c) mean of the deviations (d) range ÷ 2

4. The variance of the first n natural numbers is:

(a) (n + 1)/2 (b) (n² − 1)/12 (c) (n² + 1)/12 (d) n(n + 1)/2

5. If each observation of a data is increased by 5, the variance:

(a) increases by 5 (b) increases by 25 (c) remains unchanged (d) becomes 25 times

6. If each observation is multiplied by 3, the new variance becomes the old variance multiplied by:

(a) 3 (b) 6 (c) 9 (d) 1/3

7. The variance of the data 6, 7, 10, 12, 13, 4, 8, 12 is:

(a) 9 (b) 9.25 (c) 2.75 (d) 74

8. In a grouped frequency distribution, the median class is the class whose cumulative frequency is just greater than or equal to:

(a) N (b) N/2 (c) N/4 (d) 2N

9. The mean and variance of 8 observations are 9 and 9.25. The sum of squares of all the observations is:

(a) 648 (b) 722 (c) 74 (d) 81

10. Which of the following is NOT a measure of dispersion?

(a) Range (b) Mean deviation (c) Median (d) Standard deviation

Answer key: 1-(c), 2-(b), 3-(b), 4-(b), 5-(c), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The sum of the deviations of observations from their mean is always zero.

Reason: This is why mean deviation uses the absolute values of the deviations.

A-R 2. Assertion: Adding a constant to every observation does not change the variance.

Reason: Adding a constant shifts the mean by the same constant, so each deviation (x_i − x̄) is unchanged.

A-R 3. Assertion: Standard deviation can never be negative.

Reason: Standard deviation is defined as the positive square root of the variance.

A-R 4. Assertion: If each observation is multiplied by 3, the standard deviation is multiplied by 9.

Reason: Multiplying each observation by k multiplies the variance by k².

A-R 5. Assertion: The range gives complete information about the dispersion of a data set.

Reason: The range depends only on the maximum and minimum values and ignores how the other observations are spread.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(D).

Quick Revision Summary

Measures of dispersion: range, quartile deviation, mean deviation, variance and standard deviation. Range = Max − Min.
Mean deviation = mean of the absolute deviations from a central value; computed about the mean or about the median.
For grouped data divide the weighted sum f_i|x_i − a| by N = Σf_i; locate the median class by N/2.
Variance σ² = mean of squared deviations; standard deviation σ = √(variance), carrying the unit of the data.
Shortcut formula: σ² = (h²/N²)[NΣf_iy_i² − (Σf_iy_i)²] with y_i = (x_i − A)/h.
Adding a constant: variance unchanged. Multiplying by k: variance × k², SD × |k|, mean × k.
Variance of the first n natural numbers = (n² − 1)/12; mean = (n + 1)/2.

How to score full marks in this chapter

Always show a neat working table with clearly labelled columns (f_ix_i, deviations, f_iy_i, f_iy_i²) — method marks come from the table even if a final number slips. Use the step-deviation method whenever mid-points are large to keep arithmetic simple, and remember the h² factor in the variance formula. State the formula before substituting, and finish standard-deviation answers with the unit (cm, mm, etc.). For “correct mean/SD” problems, work with Σx_i and Σx_i² and adjust both before recomputing.

Frequently Asked Questions

What is Class 11 Maths Chapter 13 Statistics about?

Chapter 13 is about measures of dispersion — range, mean deviation (about mean and median), variance and standard deviation — for ungrouped, discrete and continuous data, including the shortcut (step-deviation) method and the effect of changing each observation by a constant.

How many exercises are there in Class 11 Maths Chapter 13?

There are two numbered exercises, Exercise 13.1 (mean deviation) and Exercise 13.2 (mean, variance and standard deviation), plus a Miscellaneous Exercise on Chapter 13. Every question of all three is solved on this page.

What is the difference between variance and standard deviation?

Variance is the mean of the squared deviations from the mean, so its unit is the square of the data’s unit. Standard deviation is the positive square root of the variance and has the same unit as the data, which makes it the most useful measure of dispersion.

Are these Class 11 Maths Chapter 13 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every final answer verified against the book’s answer key.

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