NCERT Solutions for Class 11 Maths – Chapter 13 Statistics Exercise 13.2

NCERT Solutions for Class 11 Maths Chapter 13 Exercise 13.2 – Statistics covers the concepts of Mean, Variance, and Standard Deviation. This exercise includes problems on calculating variance and standard deviation for both ungrouped data (individual observations) and grouped data (discrete and continuous frequency distributions).

Prepared according to the latest CBSE Class 11 Maths syllabus 2025-26, Exercise 13.2 helps students understand how to measure the spread of data using variance and standard deviation. These are key statistical tools used in data analysis and are important for board exams and competitive exams like JEE.

NCERT Solutions for Class 11 Maths – Chapter 13 Statistics Exercise 13.2

Statistics
Medium
Q.
Find the mean deviation about the mean for the data in Exercises 1 and 2.
Q.1: 4, 7, 8, 9, 10, 12, 13, 17
Q.2: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Statistics
Difficult
Q.
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
Statistics
Medium
Q.
 Given that x‾ is the mean and σ2 is the variance of n observations x1, x2, ...,xn.Prove that the mean and variance of the observations ax1, ax2, ax3, ...., axn areax‾ and a2σ2, respectively, (a10). \begin{array}{l}\mathbf{{Given that }}\overline{\mathbf{{x}}}{\mathbf{{ is the mean and σ}}}^{\mathbf{{2}}}\mathbf{{ is the variance of n }}\\ {\mathbf{{observations x}}}_{\mathbf{{1}}}{\mathbf{{, x}}}_{\mathbf{{2}}}\mathbf{{, }}\mathbf{.}\mathbf{..}{\mathbf{{,x}}}_{\mathbf{{n}}}\mathbf{{.\hspace{0.17em}Prove that the mean and }}\\ {\mathbf{{variance of the observations ax}}}_{\mathbf{{1}}}{\mathbf{{, ax}}}_{\mathbf{{2}}}{\mathbf{{, ax}}}_{\mathbf{{3}}}\mathbf{{, }}\mathbf{.}\mathbf{...}{\mathbf{{, ax}}}_{\mathbf{{n}}}\mathbf{{ are\hspace{0.17em}a}}\overline{\mathbf{{x}}}\mathbf{{ }}\\ {\mathbf{{and a}}}^{\mathbf{{2}}}{\mathbf{{σ}}}^{\mathbf{{2}}}\mathbf{{, respectively, (a¹0).}}\end{array} 
Statistics
Difficult
Q.
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
 ∑i=150xi=212,∑i=150xi2=902.8,∑i=150yi=261,∑i=150yi2=1457.6Whichismorevarying,thelengthorweight? \begin{array}{l}\sum _{i=1}^{50}{x}_{i}=212,{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{x}_{i}}^{2}=902.8,{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{y}_{i}=261,{\hspace{0.17em}\hspace{0.17em}}\sum _{i=1}^{50}{{y}_{i}}^{2}=1457.6\\ \mathbf{Which}\mathbf{\quad }\mathbf{is}\mathbf{\quad }\mathbf{more}\mathbf{\quad }\mathbf{varying}\mathbf{,}\mathbf{\quad }\mathbf{the}\mathbf{\quad }\mathbf{length}\mathbf{\quad }\mathbf{or}\mathbf{\quad }\mathbf{weight}\mathbf{?}\end{array} 
Statistics
Difficult
Q.
The following is the record of goals scored by team A in a football session:
No. of goals scored
0
1
2
3
4
No. of matched
1
9
7
5
3
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Statistics
Difficult
Q.
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

                                                                  Firm A              Firm B
      No. of wage earners                              586                   648
      Mean of monthly wages                     Rs. 5253            Rs. 5253
      Variance of distribution of wages          100                   121

1. Which firm A or B pays larger amount as monthly wages?
2. Which firm, A or B, shows greater variability in individual wages?
Statistics
Difficult
Q.
From the prices of shares X and Y below, find out which is more stable in value:
X
35
54
52
53
56
58
52
50
51
49
Y
108
107
105
105
106
107
104
103
104
101
Statistics
Difficult
Q.
From the data given below state which group is more variable, A or B?
Marks
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Group A
9
17
32
33
40
10
9
Group B
10
20
30
25
43
15
7
Statistics
Difficult
Q.
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.
Statistics
Difficult
Q.
Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Q7.
Classes
0-30
30-60
60-90
90-120
120-150
150-180
180-210
Frequencies
2
3
5
10
3
5
2

Q8.
Classes
0-10
10-20
20-30
30-40
40-50
Frequencies
5
8
15
16
6
Statistics
Easy
Q.
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
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Statistics
Medium
Q.
Find the mean and standard deviation using short-cut method.
Xi
60
61
62
63
64
65
66
67
68
fi
2
1
12
29
25
12
10
4
5
Statistics
Difficult
Q.
Find the mean and variance for each of the data in Exercies 1 to 5.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
Q2.   First n natural numbers.
Q3.  First 10 multiples of 3.
Q4.  
Xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3
 
Q5.
Xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3
Statistics
Medium
Q.
Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age
16-20
21-25
36-30
31-35
36-40
41-45
46-50
51-55
Number
5
6
12
14
26
12
16
9
Statistics
Medium
Q.
Find the mean deviation about median for the  following data :
Marks
0-10
Oct-20
20-30
30-40
40-50
50-60
Number of Girls
6
8
14
16
4
2
Statistics
Medium
Q.
Find the mean deviation about the mean for the data in Exercises 9 and 10.

9.
Income per day 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800
Number of persons 4 8 9 10 7 5 4 3
 
10.
Height in cms 95-105 105-115 115-125 125-135 135-145 145-155
Number of boys 9 13 26 30 12 10
Statistics
Medium
Q.
Find the mean deviation about the median for the data in Exercises 7 and 8.
7.
Xi
5
7
9
10
12
15
fi
8
6
2
2
2
6
 
8.
Xi
15
21
27
30
35
fi
3
5
6
7
8
Statistics
Medium
Q.
5. Find the mean deviation about the mean for the data in Exercises 5 and 6.
Xi
5
10
15
20
25
fi
7
4
6
3
5
6. 
Xi
10
30
50
70
90
fi
4
24
28
16
8
Statistics
Difficult
Q.
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
        Subject      Mathematics       Physics       Chemistry
        Mean              42                     32               40.9
        Standard         12                     15                20
        Deviation
Which of the three subjects shows the highest variability in marks and which shows the lowest?

NCERT Solutions for Class 11 Maths – Chapter 13 Statistics Exercise 13.2

The solutions are explained in a clear, step-by-step format to help students understand each calculation methodically.

Q1. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.

Ans:

Mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9

xi xi − x̄ (xi − x̄)²
6 −3 9
7 −2 4
10 1 1
12 3 9
13 4 16
4 −5 25
8 −1 1
12 3 9
Total 74

Variance (σ²) = Σ(xi − x̄)²/n = 74/8 = 9.25

Q2. Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12.

(Note: This is the first six natural numbers problem in the textbook)

Q2. Find the mean and variance for first n natural numbers.

Ans:

Mean of first n natural numbers = (n+1)/2

Variance = (n²−1)/12

For first 6 natural numbers (1, 2, 3, 4, 5, 6):

Mean = (1+2+3+4+5+6)/6 = 21/6 = 3.5

xi xi − x̄ (xi − x̄)²
1 −2.5 6.25
2 −1.5 2.25
3 −0.5 0.25
4 0.5 0.25
5 1.5 2.25
6 2.5 6.25
Total 17.5

Variance σ² = 17.5/6 = 35/12 ≈ 2.92

Q3. Find the mean and variance for the data:

xi 6 10 14 18 24 28 30
fi 2 4 7 12 8 4 3

Ans:

n = Σfi = 2+4+7+12+8+4+3 = 40

Σfi·xi = 12+40+98+216+192+112+90 = 760

Mean = 760/40 = 19

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
6 2 −13 169 338
10 4 −9 81 324
14 7 −5 25 175
18 12 −1 1 12
24 8 5 25 200
28 4 9 81 324
30 3 11 121 363
Total 40 1736

Variance σ² = 1736/40 = 43.4

Standard Deviation σ = √43.4 ≈ 6.59

Q4. Find the mean and variance for the data:

xi 92 93 97 98 102 104 109
fi 3 2 3 2 6 3 3

Ans:

n = Σfi = 3+2+3+2+6+3+3 = 22 (actually n=22... let us verify: total = 22)

Σfi·xi = 276+186+291+196+612+312+327 = 2200

Mean = 2200/22 = 100

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
92 3 −8 64 192
93 2 −7 49 98
97 3 −3 9 27
98 2 −2 4 8
102 6 2 4 24
104 3 4 16 48
109 3 9 81 243
Total 22 640

Variance σ² = 640/22 = 29.09 (approx)

Standard Deviation σ = √29.09 ≈ 5.39

Q5. Find the mean and variance for the data:

Class 30–40 40–50 50–60 60–70 70–80 80–90 90–100
fi 3 7 12 15 8 3 2

Ans:

Mid-values: 35, 45, 55, 65, 75, 85, 95

n = Σfi = 3+7+12+15+8+3+2 = 50

Σfi·xi = 105+315+660+975+600+255+190 = 3100

Mean = 3100/50 = 62

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
35 3 −27 729 2187
45 7 −17 289 2023
55 12 −7 49 588
65 15 3 9 135
75 8 13 169 1352
85 3 23 529 1587
95 2 33 1089 2178
Total 50 10050

Variance σ² = 10050/50 = 201

Standard Deviation σ = √201 ≈ 14.18

Q6. Find the mean and standard deviation for the data:

Class 0–30 30–60 60–90 90–120 120–150 150–180 180–210
fi 2 3 5 10 3 5 2

Ans:

Mid-values: 15, 45, 75, 105, 135, 165, 195

n = Σfi = 2+3+5+10+3+5+2 = 30

Σfi·xi = 30+135+375+1050+405+825+390 = 3210

Mean = 3210/30 = 107

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
15 2 −92 8464 16928
45 3 −62 3844 11532
75 5 −32 1024 5120
105 10 −2 4 40
135 3 28 784 2352
165 5 58 3364 16820
195 2 88 7744 15488
Total 30 68280

Variance σ² = 68280/30 = 2276

Standard Deviation σ = √2276 ≈ 47.71

Q7. Find the mean and variance for the following frequency distribution:

Class 0–10 10–20 20–30 30–40 40–50
fi 5 8 15 16 6

Ans:

Mid-values: 5, 15, 25, 35, 45

n = Σfi = 5+8+15+16+6 = 50

Σfi·xi = 25+120+375+560+270 = 1350

Mean = 1350/50 = 27

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
5 5 −22 484 2420
15 8 −12 144 1152
25 15 −2 4 60
35 16 8 64 1024
45 6 18 324 1944
Total 50 6600

Variance σ² = 6600/50 = 132

Standard Deviation σ = √132 ≈ 11.49

Q8. Find the mean and variance for the following frequency distribution:

Class 0–10 10–20 20–30 30–40 40–50
fi 2 3 4 3 2

Ans:

Mid-values: 5, 15, 25, 35, 45

n = Σfi = 2+3+4+3+2 = 14 (actually = 14)

Σfi·xi = 10+45+100+105+90 = 350

Mean = 350/14 = 25

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
5 2 −20 400 800
15 3 −10 100 300
25 4 0 0 0
35 3 10 100 300
45 2 20 400 800
Total 14 2200

Variance σ² = 2200/14 = 157.14 (approx)

Standard Deviation σ = √157.14 ≈ 12.54

Q9. Find the mean, variance and standard deviation for the data:

Height (cm) 70–75 75–80 80–85 85–90 90–95 95–100 100–105 105–110 110–115
fi 3 4 7 7 15 9 6 6 3

Ans:

Mid-values: 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5

n = Σfi = 3+4+7+7+15+9+6+6+3 = 60

Σfi·xi = 217.5+310+577.5+612.5+1387.5+877.5+615+645+337.5 = 5580

Mean = 5580/60 = 93

Σfi·(xi−x̄)² = 3(420.75)+4(240.25)+7(110.25)+7(30.25)+15(0.25)+9(20.25)+6(90.25)+6(210.25)+3(380.25)

= 1262.25+961+771.75+211.75+3.75+182.25+541.5+1261.5+1140.75 = 6336.5

Variance σ² = 6336.5/60 ≈ 105.58

Standard Deviation σ = √105.58 ≈ 10.27

Q10. The diameters of circles (in mm) drawn in a design are given below:

Diameters 33–36 37–40 41–44 45–48 49–52
fi 15 17 21 22 25

Ans:

Mid-values: 34.5, 38.5, 42.5, 46.5, 50.5

n = Σfi = 15+17+21+22+25 = 100

Σfi·xi = 517.5+654.5+892.5+1023+1262.5 = 4350

Mean = 4350/100 = 43.5

xi fi xi−x̄ (xi−x̄)² fi(xi−x̄)²
34.5 15 −9 81 1215
38.5 17 −5 25 425
42.5 21 −1 1 21
46.5 22 3 9 198
50.5 25 7 49 1225
Total 100 3084

Variance σ² = 3084/100 = 30.84

Standard Deviation σ = √30.84 ≈ 5.55

FAQs – Class 11 Maths Chapter 13 Exercise 13.2 Statistics

Q1. What is variance?

Variance (σ²) is the average of the squared deviations of each data value from the mean. It measures how spread out the data is. Formula: σ² = Σ(xi − x̄)²/n for ungrouped data.

Q2. What is standard deviation?

Standard deviation (σ) is the square root of variance. It is expressed in the same units as the data, making it easier to interpret than variance. σ = √(variance).

Q3. What is the formula for variance in grouped data?

For grouped/frequency data: σ² = Σfi(xi − x̄)² / Σfi, where xi are mid-values, fi are frequencies, and x̄ is the mean.

Q4. What is the difference between variance and standard deviation?

Variance is the squared average deviation and has squared units. Standard deviation is its square root and has the same units as the original data. Standard deviation is more commonly used in practice.

Q5. Why is Exercise 13.2 important for board exams?

Variance and standard deviation are directly tested in CBSE board exams. Questions on both ungrouped and grouped data appear regularly. Understanding these measures of dispersion is also essential for Class 12 statistics and probability.

Q6. How can students prepare effectively for Exercise 13.2?

Students should practise constructing the table with columns xi, fi, fi·xi, (xi−x̄), (xi−x̄)², and fi·(xi−x̄)² for every problem. Accuracy in arithmetic is key — always verify the mean before computing variance.