RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 11 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 11. Students can also read RBSE Class 11 Maths Important Questions for exam preparation. Students can also go through RBSE Class 11 Maths Notes to understand and remember the concepts easily.

RBSE Class 11 Maths Solutions Chapter 15 Statistics Ex 15.1

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

Question 1.
4, 7, 8, 9, 10, 12, 13, 17
Answer:
Given terms are: 4, 7, 8, 9, 10, 12, 13, 17

Given terms xi

Deviation from mean x̄, di = (xi - x̄)

Absolute values of deviations |xi - x̄|

(1)

(2)

(3)

4

4 – 10 = - 6

|-6| = 6

7

7 – 10 = - 3

| - 3| = 3

8

8 – 10 = - 2

|- 2| = 2

9

9 – 10 = - 1

|- 1| = 1

10

10 – 10 = 0

|0| = 0

12

12 – 10 = + 2

|2| = 2

13            '

13 – 10 = + 3

|3| = 3

17

17 – 10 = + 7

|7| =  7

\(\sum_{i=1}^8\) xi = 80

 

\(\sum_{=1}^8\) |xi - x̄| = 24

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 1
Thus, mean deviation about mean of given data = 3

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

Question 2.
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:
Mean x̄ = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}\)
⇒ x̄ = \(\frac{500}{10}\) = 50

Table for computing mean deviation

Given terms xi

Deviation from mean di = (xi - x̄); x̄ = 50  

Absolute value of deviation |xi - |

(1)

(2)

(3)

38

38 – 50 = - 12

|- 12 | = 12

70

70 – 50 = + 20

|20| = 20

48

48 – 50 = - 2

|- 2| = 2

40

42 – 50 = - 8

|- 10| = 10

42

55 – 50 = + 5

|- 8| = 8

55

63 – 50 = + 3

|5| = 5

63

46 – 50 = - 4

|13| = 13

46

54 – 50 = + 4

|- 4| = 4

54

44- 50 = - 6

|4| = 4

44

 

|- 6| = 6

\(\sum_{i=1}^{10}\) xi = 500

 

\(\sum_{i=1}^{10}\) |xi - | = 84

Mean Deviation (M.D.) = \(\frac{\sum_{i=1}^{10}\left|\left(x_i-\bar{x}\right)\right|}{N}\) = \(\frac{84}{10}\) = 8.4
Thus, mean deviation about mean of given data = 8.4

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

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

Question 3.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer:
Writing the given data in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 2
Median(M) = \(\frac{N+1}{2}\) th term = \(\frac{12+1}{2}\) th term = \(\frac{13}{2}\)th term = 6\(\frac{1}{2}\)th term
= Mean of 6th and 7th terms = \(\frac{13+14}{2}\) = \(\frac{27}{2}\) = 13.5
Mean Deviation (M.D.) = \(\frac{\sum_{i=1}^n\left|x_i-\mathrm{M}\right|}{N}=\frac{28}{12}\) = 2.33
Thus, mean deviation about median for given data = 2.33

Question 4.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 3
Median (M) = Value of \(\frac{N+1}{2}\)th term.= Value of \(\frac{10+1}{2}\)th term
= Value of \(\frac{11}{2}\)th term = Value of 5 \(\frac{1}{2}\)th term
= Mean of 5th and 6th term = \(\frac{46+49}{2}\) = 47.5
Mean Deviation (M.D.) = \(\frac{\sum_{i=1}^{10}\left|x_i-\mathbf{M}\right|}{N}=\frac{70}{10}\) = 7
Thus, mean deviation about median for given data = 7

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

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

Question 5.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 4Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 5
Thus, mean deviation about mean for given data = 6.32

Question 6.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 6Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 7Thus, mean deviation about mean for given data = 16

Find mean deviation about median for the data In ExercIses 7 and 8.

Question 7.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 8Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 9Thus, mean deviation about median for given data = 3.23

Question 8.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 10Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 11Median term = \(\left(\frac{N+1}{2}\right)\)th term = \(\frac{29+1}{2}\) th term = \(\frac{30}{2}\) th term = 15 th term
15th term is under of cumulative frequency 21.
Thus, for cumulative frequency 21, xi = 30
∴ Median = 30
Then, Mean Deviation (M.D.) = \(\frac{\sum_{i=1}^{29} f_i\left|x_i-\mathbf{M}\right|}{N}\) = \(\frac{148}{29}\) = 5.10
Thus, mean deviation about median of given data = 5.10

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

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

Question 9.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 12Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 13Thus, mean deviation about median of given frequency distribution = 157.92

Question 10.
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 14Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 15RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 16
Thus, mean deviation about mean of given frequency distribution = 11.288.

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1

Question 11.
Find mean deviation about median for the following data:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 17Answer:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 18(1) Median term = \(\frac{N+1}{2}\)th term = \(\frac{50+1}{2}\)th term = 25\(\frac{1}{2}\)th term
(2) Median class = The class which includes median term (25\(\frac{1}{2}\)th term)
= 20 - 30 (from table)
Then, li = 20, li + 1 - li = 30 - 20 = 10
N = 50, ci - 1 = 14 (c.f. of class just before the median class)
f = 14 (frequency of median class)
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 19
Thus, mean deviation about median of given frequency distribution = 10.342

Question 12.
Find mean deviation about median age for the age distribution of 100 persons given below:
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 20Answer:
Here given frequency distribution is discontinuous. To make it continuous frequency distribution, subtract 0.5 from lower limit of each class interval and add 0.5 in upper limit.

Discontinuous C.I.

Continuous C.I.

 

16-20

16 – 0.5 – 20 + 0.5

15.5 – 20.5

21-25

21 – 0.5 – 25 + 0.5

20.5 – 25.5

26-30

26 – 0.5 – 30 + 0.5

25.5 – 30.5

31-35

31 – 0.5 – 35 + 0.5

30.5 – 35.5

36-40

36 – 0.5 – 40 + 0.5

35.5 – 40.5

41-45

41 – 0.5 – 45 + 0.5

40.5 – 45.5

46-50

46 – 0.5 – 50 + 0.5

45.5 – 50.5

51-55

51 – 0.5 – 55 + 0.5

50.5 – 55.5

RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 21
Middle terms for Median = \(\left(\frac{N+1}{2}\right)\)th term = \(\frac{100+1}{2}\) th term = \(\frac{101}{2}\)th term = 50\(\frac{1}{2}\)th term
Median Class = Class which includes 50\(\frac{1}{2}\)th term
= 35.5 - 40.5 (from table)
Then, li = 35.5, li + 1 = 40.5, li + 1 - li = 5, fi = 26 (frequency of median class) N = 100
ci - 1 = 37 (cumulative frequency of class just before the median class)
RBSE Solutions for Class 11 Maths Chapter 15 Statistics Ex 15.1 22
Thus, mean deviation about median for given frequency distribution = 7.35

Bhagya
Last Updated on Nov. 18, 2023, 5:07 p.m.
Published Nov. 17, 2023