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

10. From the given data we can insulate the following. Take the assumed mean a=120 and h=10 Heigts in cm. No. of boys fi Mid-pointsxi fidi |xi - x? | fi |xi - x? | 95-105 9 100 -2 -18 25.3 227.7 105-115 13 110 -1 -13 15.3 198.9 115-125 26 120 0 5.3 137.8 125-135 30 130 1 30 4.7 141 135-145 12 140 2 24 14.7 176.4 145-155 10 150 3 30 24.7 247 Total 100 53 1128.8

Statistics Class 11 NCERT Solutions – Step-by-Step Answers

Q: 22. 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.]

22. The given data is converted into continuous frequency duration by subtracting and adding 0.5 from lower and upper limit respectively. Lit the assumed mean be A=42.5 and h=4

Q: 33. 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?

33. C.V in mathematics = 1242×100=28.57. C.V in Physics = 1532×100=46.87. C.V in chemistry = 2040.9×100=48.89 ∴ Chemistry has the highest variability and mathematics has the lowest variability.

Q: Class 11 Maths Statistics exercise 13.2 Solution 13. Find the mean and variance for each of the data in Exercies 13 to 17. 6, 7, 10, 12, 13, 4, 8, 12

13. The given data can be tabulated as. we have, mean, x¯=∑i=1nxin=6+7+10+12+13+4+8+128=728=9. So, variance, a2=18∑i=1n (xi−x¯)2 =18×74=9.25

Q: 14. First n natural numbers

14. We know that, Sum of first'n ' natural no =n(n+1)2 So, mean, x¯= sumoffirst(n) naturalno.=n (n +1)/2 nno of observations =n+12 So, Variance, a2=1n∑i=1n(xi−x¯)2=1n∑i=1n(xi−(n+12))2 a2=1n[∑i=1nxi2−∑i=1n2xi(n+12)+∑i=1n(n+12)2] _______(1) So, ∑i=1nxi2=(1)2+(2)2+(3)2+…+(n)2=n(n+1)(2n+1)6 _____(2). And ∑i=1n(n+12)2=(n+1)24∑i=1n1.=n(n+1)24⋅ ____(4). Putting (2), (3) and (4) in (1) we get, a2=1n[n(n+1)(2n+1)6−n(n+1)22+n(n+1)24] =(n+1)(2n+1)6−(n+1)22+(n+1)24 =(n+1)(2n+1)6−(n+1)24. =(n+1)[2n+16−(n+1)4] =(n+1)[4n+2−3n−312] =(n+1)(n−1)12=n2−112.

Updated on Jul 25, 2025 14:55 IST

By Pallavi Pathak, Assistant Manager Content

Class 11 Statistics NCERT Solutions deals with the concepts of statistics, which include the data collected for specific purposes. By analyzing and interpreting the data, we make decisions out on the data. To make a better interpretation from the data, we must have an idea of how much the data is bunched around a measure of central tendency and how the data are scattered. It also covers the concepts of variability and measures of central tendency.
NCERT Solutions for Class 11 Statistics covers key topics like Mean, Deviation, range, measures of dispersion, and variance and standard deviation. The well-structured solutions are created by the subject matter experts at Shiksha. It will help the students in improving their problem-solving skills, and in scoring good marks in the Class 11 exams, CBSE Board, and entrance exams like the JEE Main exam.
If you are looking for a go-to place for the NCERT solutions of Maths, Physics, and Chemistry of Class 11 and Class 12, you must check here. The solutions are in a step-by-step format to help students understand the concepts better.

Table of content

Quick Summary of Chapter 13 Statistics – Class 11 Maths
Class 11 Math Statistics: Key Topics, Weightage
Important Formulas of Class 11 Statistics NCERT Solutions
Statistics Class 11 NCERT Solution PDF: Download Free PDF
Class 11 Math Statistics Exercise 13.1 Solutions
Class 11 Math Statistics Exercise 13.2 Solutions
Class 11 Math Statistics Exercise 13.3 Solutions
Class 11 Math Statistics Miscellaneous Exercise Solutions

Quick Summary of Chapter 13 Statistics – Class 11 Maths

Find below the highlights of the Statistics Class 11 Maths:

The Statistics Class 11 NCERT Solutions cover the concepts of measures of dispersion, mean deviation for ungrouped data, mean deviation for grouped data, variance, and standard deviation for ungrouped data.
It also covers the variance and standard deviation of a discrete frequency distribution, the variance and standard deviation of a continuous frequency distribution, and a shortcut method to find variance and standard deviation.
The chapter includes the mathematical representations for these concepts.

Students must find the free PDFs of all NCERT solutions of all chapters of Class 11 Maths here. They will also get the weightage information and the important topics through this link.

Class 11 Math Statistics: Key Topics, Weightage

In the NCERT Solutions Class 11 Statistics, students should focus on the mode, median, mean, variance, and standard deviation to have a better understanding of probability and statistics problems. See here the topics covered in this chapter:

Exercise	Topics Covered
13.1	Introduction
13.2	Measures of Dispersion
13.3	Range
13.4	Mean Deviation
13.5	Variance and Standard Deviation

Statistics Class 11 Weightage in JEE Mains

The Class 11 Statistics doesn't have a stand-alone and direct weightage in the JEE Mains exam. However, the concepts of this chapter are foundational for probability and statistics, which are part of the JEE Mains. To prepare well for JEE Mains exam, the students should also focus on the probability distributions, random variables, Bayes' theorem, and conditional probability.

Important Formulas of Class 11 Statistics NCERT Solutions

Important Formulae for Class 11 Math Statistics for CBSE and Competitive Exams

Measures of Dispersion

Range $\text{Range} = \text{Maximum Value} - \text{Minimum Value}$
Mean Deviation
- About Mean $\text{Mean Deviation} = \frac{\sum |X - \bar{X}|}{n}$
- About Median $\text{Mean Deviation} = \frac{\sum |X - M|}{n}$
Variance & Standard Deviation
- Variance: $\sigma^2 = \frac{\sum (X - \bar{X})^2}{n}$
- Standard Deviation: $\sigma = \sqrt{\sigma^2}$
- Shortcut Formula: $\sigma^2 = \left( \frac{\sum X^2}{n} \right) - \bar{X}^2$
Coefficient of Variation (C.V.) $C.V. = \frac{\sigma}{\bar{X}} \times 100\%$

Statistics Class 11 NCERT Solution PDF: Download Free PDF

Students also get the link to the free Statistics Class 11 PDF below, which they must download to practice the NCERT exercises related to this chapter. The PDF is created by the experts in Shiksha for students to deepen their understanding of concepts. It can help students to score well in the CBSE Board exams and competitive exams like JEE Main.

Chapter 5 Statistics Class 11 NCERT Solutions: Download PDF for Free

Class 11 Math Statistics Exercise 13.1 Solutions

Class 11 Math Statistics Exercise 13.1 focuses on fundamental concepts such as measures of dispersion, mean deviation for grouped and ungrouped data, variance, and standard deviation. Ex 13.1 Statistics solutions will contain well-explained answers to all 12 questions of the exercise. Students can check the solutions below.

Class 11 Maths Statistics exercise 15.1 Solution

Q1. Find the mean deviation about the mean for the data in Exercises 1 and 2.

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

A.1. Mean of the given observation is.

$\bar{x} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8}$

$= \frac{80}{8} = 10 .$

Deviation of the respective observation about the mean $\bar{x}$ i.e., $x_{i} - \bar{x}$ are 4–10,7–10,8–10,9–10,10–10,12–10,13–10,17–10

=6,-3,-2,-1,0,2,3,7

The absolute value of the deviation i.e., $| x_{i} - \bar{x} |$ are 6,3,2,1,0,2,3,7.

Therefore, the required mean deviation about the mean is

$M.D = (\bar{x}) = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} | = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}$

$= \frac{24}{8}$

= 3.

Q2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

A.2. Mean of the given observation is.

$\bar{x} = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}$

$= \frac{500}{10} = 50 .$

So,

x_i	38	10	48	40	42	55	63	46	54	44
\|x_i- 50\|	12	20	2	10	8	5	13	4	4	6

Therefore, the required mean deviation about the mean is

$M.D. (\bar{x}) = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} |$

$= \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}$

$= \frac{84}{10}$

= 8.4

Q3. Find the mean deviation about the median for the data in Exercises 3 and 4.

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

A.3. Arranging the data in ascending order we get,

10,11,1112,13,13,14,16,16,17,17,18

As n=12, even

So, median is the mean of ${(\frac{M}{2})th}^{}$ and ${(\frac{M}{2} + 1)th}^{}$ observation.

$\Rightarrow = \frac{{6thobservation}^{} + {7thobservation}^{}}{2}$

$M= \frac{13 + 14}{2} = \frac{27}{2} = 13.5 .$

So, deviation of respective observation about the median. $M, | x_{i} - M |$ are

x_i	10	11	11	12	13	13	14	16	16	17	17	18
\|x_i- M\|	3.5	2.5	2.5	1.5	0.5	0.5	0.5	2.5	2.5	3.5	3.5	4.5

Therefore the mean deviation about the mean is

$M.D. (M) = \frac{1}{n} \times \sum_{i = 1}^{n} | x_{i} - M |$

$= \frac{1}{12} \times (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5)$

$= \frac{28}{12} = 2.33 .$

Q4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

A.4. Arranging the given data in ascending order we get,

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

As n = 10 (even)

$= \frac{(\frac{n}{2}) {thObservation}^{} + {(\frac{n}{2} + 1)thObservation}^{}}{2}$

$\Rightarrow = \frac{{5thobservation}^{} + {6thobservation}^{}}{2}$

$= \frac{46 + 49}{2}$

$= \frac{95}{2}$

= 47.5

x_i	36	42	45	46	46	49	51	53	60	72
\|x_i- M\|	11.5	5.5	2.5	1.5	1.5	1.5	3.5	5.5	12.5	24.5

$∴$ M.D. (M) $= \frac{1}{n} \times \sum_{i = 1}^{n} | x_{i} - M |$

$= \frac{1}{10} \times (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)$

$= \frac{70}{10} = 7 .$

Q5. Find the mean deviation about the mean for the data in Exercises 5 and 6.

x_i5 10 15 20 25

f_i7 4 6 3 5

A.5. From the given data we have,

x_i	f_i	x_i f_i	\|x_i - 14\|	f_i\|x_i - 14\|
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

Q6. x_i10 30 50 70 90

f_i4 24 28 16 8

A.6. From the given data we tabulate the following.

x_i	f_i	x_if_i	\|x_i - x̄\|	f_i\|x_i - x̄\|
10	4	4	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

We have,

Q7. Find the mean deviation about the median for the data in Exercises 7 and 8.

x_i5 7 9 10 12 15

f_i 8 6 2 2 2 6

A.7. From the given data we cantabulate the following.

x_i	f_i	Cumulative frequency 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

Now, N=26 which is even.

So, Median is the mean of 13^th and 14^th observation. Both of these observations lie in the cumulative frequency 14 for which corresponding observation is 7.

Q8. x_i15 21 27 30 35

f_i3 5 6 7 8

A.8. From the given data we can tabulate the following.

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

Here N = 29 which is odd.

Q9. Find the mean deviation about the mean for the data in Exercises 9 and 10.

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

A.9. From the given data we cantabulate the following.

Income per day in `	Number of person f_i	Mid points x_i	f_i x_i	\|x_i - x̄\|	f_i\|x_i - x̄\|
01-00	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
700-800	3	750	2250	392	1176
Total	50		17900		7896

Q10.

Height in cms	95-105	105-115	115-125	125-135	135-145	145-155
Number of persons	9	13	26	30	12	10

A.10. From the given data we can insulate the following.

Take the assumed meana=120 and h=10

Heigts in cm.	No. of boys f_i	Mid-pointsx_i		f_id_i	\|x_i - x̄\|	f_i\|x_i - x̄\|
95-105	9	100	-2	-18	25.3	227.7
105-115	13	110	-1	-13	15.3	198.9
115-125	26	120	0	0	5.3	137.8
125-135	30	130	1	30	4.7	141
135-145	12	140	2	24	14.7	176.4

145-155	10	150	3	30	24.7	247
Total	100			53		1128.8

Q11.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

A.11. From the given data we can tabulate the following.

Marks	No. of girls (f_i)	c.f.	mid-pointsx_i	\|x_i - M\|	f_i\|x_i - M\|
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
20-30	14	28	35	2.85	39.9
30-40	16	44	35	7.15	114.4
40-50	4	48	45	17.15	68.6
50-60	2	50	55	27.15	54.3
Total	50				517.1

Q12.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

[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]

A.12. The given data is made'continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. So, we cam tabulate as.

Age	number f_i	c.f.	mid-point 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

Commonly asked questions

22. 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.]

22. The given data is converted into continuous frequency duration by subtracting and adding 0.5 from lower and upper limit respectively. Lit the assumed mean be A=42.5 and h=4

33. 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?

33. C.V in mathematics = $\frac{12}{42} \times 100 = 28.57 .$

C.V in Physics = $\frac{15}{32} \times 100 = 46.87 .$

C.V in chemistry = $\frac{20}{40.9} \times 100 = 48.89$

$∴$ Chemistry has the highest variability and mathematics has the lowest variability.

Class 11 Maths Statistics exercise 13.2 Solution

13. Find the mean and variance for each of the data in Exercies 13 to 17.

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

13. The given data can be tabulated as.

we have,

mean, $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9 .$

So, variance, $\begin{matrix} a^{2} = \frac{1}{8} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \end{matrix}$

$= \frac{1}{8} \times 74 = 9.25$

14. First n natural numbers

14. We know that,

Sum of first'n ' natural no $= \frac{n (n + 1)}{2}$

So, mean, $\begin{matrix} \bar{x} = \frac{sumoffirst (n) naturalno .=}{} \frac{n (n +1)/2}{n} \\ no of observations \end{matrix}$

$= \frac{n + 1}{2}$

So, Variance, $a^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - (\frac{n + 1}{2}))}^{2}$

$\begin{array}{l} a^{2} = \frac{1}{n} [\sum_{i = 1}^{n} x_{i}^{2} - \sum_{i = 1}^{n} 2 x_{i} (\frac{n + 1}{2}) + \sum_{i = 1}^{n} {(\frac{n + 1}{2})}^{2}] \\ _______(1) \end{array}$

So, $\begin{matrix} \sum_{i = 1}^{n} x_{i}^{2} = {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + \dots + {(n)}^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6}_____(2) . \end{matrix}$

And $\sum_{i = 1}^{n} {(\frac{n + 1}{2})}^{2} = \frac{{(n + 1)}^{2}}{4} \sum_{i = 1}^{n} 1 . = \frac{n {(n + 1)}^{2}}{4 \cdot}____(4) .$

Putting (2), (3) and (4) in (1) we get,

$\begin{matrix} a^{2} = \frac{1}{n} [\frac{n (n + 1) (2 n + 1)}{6} - \frac{n {(n + 1)}^{2}}{2} + \frac{n {(n + 1)}^{2}}{4}] \end{matrix}$

$\begin{matrix} = \frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{2} + \frac{{(n + 1)}^{2}}{4} \end{matrix}$

$\begin{matrix} = \frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{4} . \end{matrix}$

$\begin{matrix} = (n + 1) [\frac{2 n + 1}{6} - \frac{(n + 1)}{4}] \end{matrix}$

$= (n + 1) [\frac{4 n + 2 - 3 n - 3}{12}]$

$= \frac{(n + 1) (n - 1)}{12} = \frac{n^{2} - 1}{12} .$

10.

Height in cms	95-105	105-115	115-125	125-135	135-145	145-155
Number of persons	9	13	26	30	12	10

10. From the given data we can insulate the following.

Take the assumed mean a=120 and h=10

Heigts in cm.	No. of boys f_i	Mid-pointsx_i		f_id_i	\|x_i - x? \|	f_i\|x_i - x? \|
95-105	9	100	-2	-18	25.3	227.7
105-115	13	110	-1	-13	15.3	198.9
115-125	26	120	0	5.3	137.8
125-135	30	130	1	30	4.7	141
135-145	12	140	2	24	14.7	176.4

145-155	10	150	3	30	24.7	247
Total	100		53		1128.8

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

[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]

12. The given data is made'continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. So, we cam tabulate as.

Age	number f_i	c.f.	mid-point 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

21. 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	100-125	105-110	110-115
Frequencies	3	4	7	7	15	9	6	6	3

21. Let the assumed mean be A=92.5 and h=5

Class 11 Maths Statistics exercise 15.1 Solution

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

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

1. Mean of the given observation is.

$\bar{x} = \frac{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}{8}$

$= \frac{80}{8} = 10 .$

Deviation of the respective observation about the mean $\bar{x}$ i.e., $x_{i} - \bar{x}$ are 4–10,7–10,8–10,9–10,10–10,12–10,13–10,17–10

=6, -3, -2, -1,0,2,3,7

The absolute value of the deviation i.e., $| x_{i} - \bar{x} |$ are 6,3,2,1,0,2,3,7.

Therefore, the required mean deviation about the mean is

$M.D = (\bar{x}) = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} | = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}$

$= \frac{24}{8}$

= 3.

18.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

18. Let the assumed mean be A=64 and it the width, h=1.

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

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

3. Arranging the data in ascending order we get,

10,11,1112,13,13,14,16,16,17,17,18

As n=12, even

So, median is the mean of ${(\frac{M}{2})th}^{}$ and ${(\frac{M}{2} + 1)th}^{}$ observation.

$\Rightarrow = \frac{{6thobservation}^{} + {7thobservation}^{}}{2}$

$M= \frac{13 + 14}{2} = \frac{27}{2} = 13.5 .$

So, deviation of respective observation about the median. $M, | x_{i} - M |$ are

x_i	10	11	11	12	13	13	14	16	16	17	17	18
\|x_i- M\|	3.5	2.5	2.5	1.5	0.5	0.5	0.5	2.5	2.5	3.5	3.5	4.5

Therefore the mean deviation about the mean is

$M.D. (M) = \frac{1}{n} \times \sum_{i = 1}^{n} | x_{i} - M |$

$= \frac{1}{12} \times (3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5)$

$= \frac{28}{12} = 2.33 .$

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

x_i5 10 15 20 25

f_i7 4 6 3 5

5. From the given data we have,

x_i	f_i	x_i f_i	\|x_i - 14\|	f_i\|x_i - 14\|
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

8. x_i15 21 27 30 35

f_i3 5 6 7 8

8. From the given data we can tabulate the following.

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

Here N = 29 which is odd.

= 1/29 * 148

= 5. 10

15. First 10 multiples of 3

15. We have, first 10 multiples of 3=3,6,9,12,15,18,21,24,27,30.

So, $\bar{x} = \frac{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}{10} = \frac{165}{10} = 16.5$

We can now tabulate the given data as following.

Therefore, variance, $a^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}$

$\begin{matrix} = \frac{1}{10} \times 742.5 \end{matrix}$

= 74.25.

Class 11 Maths Statistics exercise 13.3 Solution

23. 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

23. Let the assumed mean be A=45 and h=10. Then we can tabulate the given data as following.

$\begin{matrix} = \frac{- 6}{150} \times 10 + 45 . \end{matrix}$

= 44.6

25. 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 ? 5253 ?5253

Variance of the distribution 100 121

of wages

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

(i) The number of wage earner in firm A, n_A=586. Mean monthly wages of firm

Total no of were canner in firm B.

Total monthly mages in firm. B = ?5253 × n_B

=? 5253 × 586

=? 34, 03, 944

Firm B pays larger amount of monthly wages.

(ii) Since both the firm A and B has same mean monthly wages the firm with greater standard duration i.e, greater variance will have more variability in individual ways. Therefore, firm B will have more variability in individual wages.

27. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

$\sum_{i = 1}^{50} = x_{i} = 212, \sum_{i = 1}^{50} x_{i}^{2} = 902.8, \sum_{i = 1}^{50} y_{i} = 261, \sum_{i = 1}^{50} {y_{i}}^{2} = 1457.6$

Which is more varying, the length or weight?

27. Given, n=50.

$\begin{array}{l} \sum_{i = 1}^{n} x i = 212 \sum_{i = 1}^{50} x i^{2} = 902.8 \end{array}$

$\sum_{i = 1}^{50} y i = 261 \sum_{i = 1}^{50} y i^{2} = 1457.6$

So, $\bar{x} = \frac{\sum_{i = 1}^{50} x_{i}}{n} = \frac{212}{50} = 4.24$

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

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

9. From the given data we cantabulate the following.

Income per day in '	Number of person f_i	Mid points x_i	f_i x_i	\|x_i - x? \|	f_i\|x_i - x? \|
01-00	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
700-800	3	750	2250	392	1176
Total	50		17900		7896

32. 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.

32. (i) Given, n = 20.

Incorrect mean $(\bar{x}) = 10$

Incorrect standard deviation $(σ) = 2$

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

$\Rightarrow 10 = \frac{1}{20} \sum_{i = 1}^{20} x_{i}$

$\Rightarrow \sum_{i = 1}^{20} x_{i} = 200$

So, incorrect sum of observation = 200.

correct sum of observation =200 – 8 = 192

And correct mean $= \frac{192}{19} = 10.1$

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

2. Mean of the given observation is.

$\bar{x} = \frac{38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44}{10}$

$= \frac{500}{10} = 50 .$

So,

x_i	38	10	48	40	42	55	63	46	54	44
\|x_i- 50\|	12	20	2	10	8	5	13	4	6

Therefore, the required mean deviation about the mean is

$M.D. (\bar{x}) = \frac{1}{n} \sum_{i = 1}^{n} | x_{i} - \bar{x} |$

$= \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}$

$= \frac{84}{10}$

= 8.4

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

4. Arranging the given data in ascending order we get,

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

As n = 10 (even)

$= \frac{(\frac{n}{2}) {thObservation}^{} + {(\frac{n}{2} + 1)thObservation}^{}}{2}$

$\Rightarrow = \frac{{5thobservation}^{} + {6thobservation}^{}}{2}$

$= \frac{46 + 49}{2}$

$= \frac{95}{2}$

= 47.5

x_i	36	42	45	46	46	49	51	53	60	72
\|x_i- M\|	11.5	5.5	2.5	1.5	1.5	1.5	3.5	5.5	12.5	24.5

$∴$ M.D. (M) $= \frac{1}{n} \times \sum_{i = 1}^{n} | x_{i} - M |$

$= \frac{1}{10} \times (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)$

$= \frac{70}{10} = 7 .$

6. x_i10 30 50 70 90

f_i4 24 28 16 8

6. From the given data we tabulate the following.

x_i	f_i	x_if_i	\|x_i - x? \|	f_i\|x_i - x? \|
10	4	4	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

We have,

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

x_i5 7 9 10 12 15

f_i 8 6 2 2 2 6

7. From the given data we cantabulate the following.

x_i	f_i	Cumulative frequency 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

Now, N=26 which is even.

So, Median is the mean of 13^th and 14^th observation. Both of these observations lie in the cumulative frequency 14 for which corresponding observation is 7.

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

11. From the given data we can tabulate the following.

Marks	No. of girls (f_i)	c.f.	mid-pointsx_i	\|x_i - M\|	f_i\|x_i - M\|
0-10	6	6	5	22.85	137.1
10-20	8	14	15	12.85	102.8
20-30	14	28	35	2.85	39.9
30-40	16	44	35	7.15	114.4
40-50	4	48	45	17.15	68.6
50-60	2	50	55	27.15	54.3
Total	50				517.1

16.

*x_i*	6	10	14	18	24	28	30
*f_i*	2	4	7	12	8	4	3

16. The given data can be tabulated as follow.

17.

*x_i*	92	93	97	98	102	104	109
*f_i*	3	2	3	2	6	3	3

17. The given data can be tabulated as follow

19. Find the mean and variance for the following frequency distributions in Exercises 19 and 20.

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

19. Let the assumed mean be A=105 and class width, h=30. The given data can be tabulated as

= 2280 - 4

= 2276.

20.

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

20.

We have, $N = \sum_{i = 1}^{n} f i = 50$ .

So, mean, $\bar{x} = \frac{1}{N} \times \sum_{i = 1}^{n} f i x i = \frac{1350}{5} = 27 .$

$\begin{matrix} = \frac{1}{50} \times 6600 \end{matrix}$

= 132.

24. 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

24. We can tabulate the given data as follows.

Total =

Mean of $x, \bar{x} = \frac{\sum_{}^{} x i}{n} = \frac{510}{10} = 51 .$

Mean of $Y, \bar{Y} = \frac{\sum_{}^{} Y i}{n} = \frac{1050}{10} = 105$ .

26. 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 matches	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?

26. For team A.

Hence, we conclude that team A is more consistent.

30. 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.

30. Let 'x' be the given observations with n = 6.

31. Given that $\bar{x}$ 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, ...., ax_n are a $\bar{x}$ and a₂ σ₂, respectively, (a ≠ 0).

31. For n observations x₁, x₂,……., x_n .

We have mean = $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n} (1)$

and variance = $σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} (2)$

Let y_i be the new observations with same n.

So, y_i = ax_i (3)

Now mean, $\bar{y} = \frac{\sum_{i = 1}^{n} y_{i}}{n} = \frac{\sum_{i = 1}^{n} a x_{i}}{n} = \frac{a \sum_{i = 1}^{n} y_{i}}{n} = a \bar{x} [(1)]$

So $\bar{y} = a \bar{x} (4)$

And, putting (3) and (4) in (2) we get,

$σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(\frac{y_{i}}{a} - \frac{\bar{y}}{a})}^{2}$

$\Rightarrow σ^{2} = \frac{1}{a^{2}} [\frac{1}{n} \sum_{i = 1}^{n} {(y_{i} - \bar{y})}^{2}]$

$\Rightarrow {(σ^{'})}^{2} = σ^{2} α^{2} .$

Hence, the mean and variance of ax₁, ax₂, ……, ax_n are $a \bar{x}$ and a² σ² .

34. 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.

34. Given, n = 100.

incorrect mean ( $\bar{x}$ ) = 20.

incorrect standard deviation (σ) = 3

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow 20 = \frac{1}{100} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow \sum_{i = 1}^{n} n_{i} = 2000 .$

So, incorrect sum of observation = 2000

Correct sum of observation $= 2000 - 21 - 21 - 18$

= 1940

Class 11 Math Statistics Exercise 13.2 Solutions

Class 11 Math Statistics Exercise 13.2 deals with more advanced concepts, such as measures of dispersion, variance, and standard deviation, which helps students understand how data is distributed around the mean and median. Statistics Exercise 13.2 contains a total of 10 short and long answer questions. Students can check the solution of all the question below;

Class 11 Maths Statistics exercise 13.2 Solution

Q1. Find the mean and variance for each of the data in Exercies 1 to 5.

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

A.1. The given data can be tabulated as.

we have,

mean, $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9 .$

So, variance, $\begin{matrix} a^{2} = \frac{1}{8} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} \end{matrix}$

$= \frac{1}{8} \times 74 = 9.25$

Q2. First n natural numbers

A.2. We know that,

Sum of first'n ' natural no $= \frac{n (n + 1)}{2}$

So, mean, $\begin{matrix} \bar{x} = \frac{sumoffirst (n) naturalno .=}{} \frac{n (n +1)/2}{n} \\ no of observations \end{matrix}$

$= \frac{n + 1}{2}$

So, Variance, $a^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - (\frac{n + 1}{2}))}^{2}$

$\begin{array}{l} a^{2} = \frac{1}{n} [\sum_{i = 1}^{n} x_{i}^{2} - \sum_{i = 1}^{n} 2 x_{i} (\frac{n + 1}{2}) + \sum_{i = 1}^{n} {(\frac{n + 1}{2})}^{2}] \\ _______(1) \end{array}$

So, $\begin{matrix} \sum_{i = 1}^{n} x_{i}^{2} = {(1)}^{2} + {(2)}^{2} + {(3)}^{2} + \dots + {(n)}^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6}_____(2) . \end{matrix}$

And $\sum_{i = 1}^{n} {(\frac{n + 1}{2})}^{2} = \frac{{(n + 1)}^{2}}{4} \sum_{i = 1}^{n} 1 . = \frac{n {(n + 1)}^{2}}{4 \cdot}____(4) .$

Putting (2), (3) and (4) in (1) we get,

$\begin{matrix} a^{2} = \frac{1}{n} [\frac{n (n + 1) (2 n + 1)}{6} - \frac{n {(n + 1)}^{2}}{2} + \frac{n {(n + 1)}^{2}}{4}] \end{matrix}$

$\begin{matrix} = \frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{2} + \frac{{(n + 1)}^{2}}{4} \end{matrix}$

$\begin{matrix} = \frac{(n + 1) (2 n + 1)}{6} - \frac{{(n + 1)}^{2}}{4} . \end{matrix}$

$\begin{matrix} = (n + 1) [\frac{2 n + 1}{6} - \frac{(n + 1)}{4}] \end{matrix}$

$= (n + 1) [\frac{4 n + 2 - 3 n - 3}{12}]$

$= \frac{(n + 1) (n - 1)}{12} = \frac{n^{2} - 1}{12} .$

Q3. First 10 multiples of 3

A.3. We have, first 10 multiples of 3=3,6,9,12,15,18,21,24,27,30.

So, $\bar{x} = \frac{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}{10} = \frac{165}{10} = 16.5$

We can now tabulate the given data as following.

Therefore, variance, $a^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2}$

$\begin{matrix} = \frac{1}{10} \times 742.5 \end{matrix}$

= 74.25.

Q4.

*x_i*	6	10	14	18	24	28	30
*f_i*	2	4	7	12	8	4	3

Q5.

*x_i*	92	93	97	98	102	104	109
*f_i*	3	2	3	2	6	3	3

A.5. The given data can be tabulated as follow

Q6.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

Q7. Find the mean and variance for the following frequency distributions in Exercises7 and 8.

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

A8.

We have, $N = \sum_{i = 1}^{n} f i = 50$ .

So, mean, $\bar{x} = \frac{1}{N} \times \sum_{i = 1}^{n} f i x i = \frac{1350}{5} = 27 .$

$\begin{matrix} = \frac{1}{50} \times 6600 \end{matrix}$

= 132.

Q9. 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	100-125	105-110	110-115
Frequencies	3	4	7	7	15	9	6	6	3

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
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.]

A.10. The given data is converted into continuous frequency duration by subtracting and adding 0.5 from lower and upper limit respectively. Lit the assumed mean be A=42.5 and h=4

Class 11 Math Statistics Exercise 13.3 Solutions

Class 11 Math Exercise 13.3 of Statistics focuses on problems related to big or grouped data set, such as finding the mean deviation with mode or median, finding standard deviation with mean or median, and more. There are 5 questions in Exercise 13.3, which is deleted in the latest CBSE syllabus. Students can find the solution below;

Class 11 Maths Statistics exercise 13.3 Solution

Q1. 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

A.1. Let the assumed mean be A=45 and h=10. Then we can tabulate the given data as following.

$\begin{matrix} = \frac{- 6}{150} \times 10 + 45 . \end{matrix}$

= 44.6

Q2. 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

A.2. We can tabulate the given data as follows.

Total =

Mean of $x, \bar{x} = \frac{\sum_{}^{} x i}{n} = \frac{510}{10} = 51 .$

Mean of $Y, \bar{Y} = \frac{\sum_{}^{} Y i}{n} = \frac{1050}{10} = 105$ .

Q3. 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 ₹ 5253 ₹5253

Variance of the distribution 100 121

of wages

(i) Which firm A or B pays larger amount as monthly wages?

(ii) Which firm, A or B, shows greater variability in individual wages?

A.3. (i) The number of wage earner in firm A, n_A=586. Mean monthly wages of firm

Total no of were canner in firm B.

Total monthly mages in firm. B = ₹5253 × n_B

= ₹ 5253 × 586

₹ 34, 03, 944

Firm B pays larger amount of monthly wages.

(ii) Since both the firm A and B has same mean monthly wages the firm with greater standard duration i.e, greater variance will have more variability in individual ways. Therefore, firm B will have more variability in individual wages.

Q4. 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 matches	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?

Q5. The sum and sum of squares corresponding to length x (in cm) and weight y (ingm) of 50 plant products are given below:

$\sum_{i = 1}^{50} = x_{i} = 212, \sum_{i = 1}^{50} x_{i}^{2} = 902.8, \sum_{i = 1}^{50} y_{i} = 261, \sum_{i = 1}^{50} {y_{i}}^{2} = 1457.6$

Which is more varying, the length or weight?

A.5. Give, n=50.

$\begin{array}{l} \sum_{i = 1}^{n} x i = 212 \sum_{i = 1}^{50} x i^{2} = 902.8 \end{array}$

$\sum_{i = 1}^{50} y i = 261 \sum_{i = 1}^{50} y i^{2} = 1457.6$

So, $\bar{x} = \frac{\sum_{i = 1}^{50} x_{i}}{n} = \frac{212}{50} = 4.24$

Class 11 Math Statistics Miscellaneous Exercise Solutions

Class 11 Statistics Miscellaneous Exercise covers all the topics covered in the chapter. Students can use this exercise as a summary of the problem-solving skills needed for this chapter. This exercise includes all the topics such as mean,median, mode, standard deviation, mean deviation, and others. There are 7 long-answer type questions in the miscellaneous exercise. Students can find below solutions to all the questions in this exercise.

Class 11 Maths Statistics Miscellaneous exercise Solution

Q1. 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.

A.1. Let the other two observation be x and y.

Therefore, the series is 6, 7, 10, 12, 12, 13, x, y.

So, Mean $\bar{x} = 9$

$\Rightarrow \frac{6 + 7 + 10 + 12 + 12 + 13 + x + y}{8} = 9$

$\Rightarrow x + y = 72 - 60$

$\Rightarrow x + y = 12$ ….. (1)

And, variance = 9.25

$\Rightarrow \frac{1}{8} \sum_{i = 1}^{8} {(x_{i} - \bar{x})}^{2} = 9.25$

$\Rightarrow {(- 3)}^{2} + {(- 2)}^{2} + {(1)}^{2} + {(3)}^{2} + {(3)}^{2} + {(4)}^{2} + {(x - 9)}^{2} + {(y - 9)}^{2} = 9.25 \times 8$

$\Rightarrow$ 9 + 4 + 1 + 9 + 9 + 16 + x2 + 81 - 18x + y2 + 81 - 18y = 74

$\Rightarrow 210 + x^{2} + y^{2} - 18 (x + y) = 74$

$\Rightarrow x^{2} + y^{2} = 74 - 210 + 18 (12)$ [ $∴$ equation (1)]

$\Rightarrow x^{2} + y^{2} = 80$ ……(2)

Squaring Equation (1) we get,

$x^{2} + y^{2} + 2 x y = 144$

$\Rightarrow 80 + 2 x y = 144$

$\Rightarrow 2 x y = 64$ (3)

Subtracting (3) from (2) we get,

$x^{2} + y^{2} - 2 x y = 16$

$\Rightarrow {(x - y)}^{2} = 4^{2}$

$\Rightarrow x - y = \pm 4$ ……(4)

From (1) and (4)

Case I, x + y =12 and x - y = 4

$\Rightarrow$ x = 8 and y = 4

Case II,x + y =12 and x – y = -4

$\Rightarrow$ x = 4 and y = 8.

Hence, the remaining observations are 8 and 4.

Q2. 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.

A.2. Let the other two observation be x and y. Then, the series is 2, 4, 10, 12, 14, x, y.

So, mean, $\bar{x} = \frac{2 + 4 + 10 + 12 + 14 + x + y}{7} = 8 .$

$\Rightarrow 8 \times 7 = 42 + x + y .$

$\Rightarrow x + y = 14$ ….(1)

And variance = $\frac{1}{n} \sum_{i = 1}^{7} {(x_{i} - \bar{x})}^{2}$

$\Rightarrow 16 = \frac{1}{7} [{(- 6)}^{2} + {(- 4)}^{2} + {(- 2)}^{2} + {(4)}^{2} + {(6)}^{2} + {(x - 8)}^{2} + {(y - 8)}^{2}]$

$\Rightarrow 36 + 16 + 4 + 36 + x^{2} 64 - 16 x + y^{2} + 64 - 16 y = 112 .$

$\Rightarrow 236 + x^{2} y^{2} - 16 (x + y) = 112$

$\Rightarrow x^{2} + y^{2} = 112 - 236 + 16 (14) [(1)]$

$\Rightarrow x^{2} + y^{2} = 100$ ….(2)

Squaring eqn (1) we get,

$x^{2} + y^{2} + 2 x y = 196$

$\Rightarrow 2 x y = 96 - 100$

$\Rightarrow 2 x y = 96$ …..(3)

Subtracting eqn (3) from (2) we get,

$x^{2} + y^{2} - 2 x y = 4$

$\Rightarrow {(x - y)}^{2} = 2^{2}$

$\Rightarrow x - y = \pm 2 .$ ….(4)

Using eqn (1) and (2) we get,

Case I. x + y = 14 and x – y = 2

$\Rightarrow$ x = 8 and y = 6

Case II. x + y = 14 and x – y = -2

$\Rightarrow$ x = 6 and y = 8

Hence, the remaining observations are 6 and 8.

Q3. 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.

A.3. Let ‘x’ be the given observations with n = 6.

Q4. Given that $\bar{x}$ 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, ...., ax_n are a $\bar{x}$ and a₂ σ₂, respectively, (a ≠ 0).

A.4. For n observations x₁, x₂,……., x_n .

We have mean = $\bar{x} = \frac{\sum_{i = 1}^{n} x_{i}}{n} (1)$

and variance = $σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(x_{i} - \bar{x})}^{2} (2)$

Let y_i be the new observations with same n.

So, y_i = ax_i (3)

Now mean, $\bar{y} = \frac{\sum_{i = 1}^{n} y_{i}}{n} = \frac{\sum_{i = 1}^{n} a x_{i}}{n} = \frac{a \sum_{i = 1}^{n} y_{i}}{n} = a \bar{x} [(1)]$

So $\bar{y} = a \bar{x} (4)$

And, putting (3) and (4) in (2) we get,

$σ^{2} = \frac{1}{n} \sum_{i = 1}^{n} {(\frac{y_{i}}{a} - \frac{\bar{y}}{a})}^{2}$

$\Rightarrow σ^{2} = \frac{1}{a^{2}} [\frac{1}{n} \sum_{i = 1}^{n} {(y_{i} - \bar{y})}^{2}]$

$\Rightarrow {(σ^{'})}^{2} = σ^{2} α^{2} .$

Hence, the mean and variance of ax₁, ax₂, ……, ax_n are $a \bar{x}$ and a² σ² .

Q5. 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.

A.5. (i) Given, n = 20.

Incorrect mean $(\bar{x}) = 10$

Incorrect standard deviation $(σ) = 2$

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} x_{i}$

$\Rightarrow 10 = \frac{1}{20} \sum_{i = 1}^{20} x_{i}$

$\Rightarrow \sum_{i = 1}^{20} x_{i} = 200$

So, incorrect sum of observation = 200.

correct sum of observation =200 – 8 = 192

And correct mean $= \frac{192}{19} = 10.1$

Q6. 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?

A.6. C.V in mathematics = $\frac{12}{42} \times 100 = 28.57 .$

C.V in Physics = $\frac{15}{32} \times 100 = 46.87 .$

C.V in chemistry = $\frac{20}{40.9} \times 100 = 48.89$

$∴$ Chemistry has highest variability and mathematics has lowest variability.

Q7. 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.

A.7. Given, n = 100.

incorrect mean ( $\bar{x}$ ) = 20.

incorrect standard deviation (σ) = 3

We know that,

$\bar{x} = \frac{1}{n} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow 20 = \frac{1}{100} \sum_{i = 1}^{n} n_{i}$

$\Rightarrow \sum_{i = 1}^{n} n_{i} = 2000 .$

So, incorrect sum of observation = 2000

Correct sum of observation $= 2000 - 21 - 21 - 18$

= 1940

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