ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.3

 Exercise 20.3

Question 1

Sol :
The required bar group as shown below

Scale on y-axis=1 unit=30 cm






















Question 2

Sol :
The bar graph required is shown below

Scale on y-axis 1 unit = 50 cm




















Question 3

Sol :
The required bar graph is shown below
Scale on x-axis 1 cm=1 unit
Scale on y-axis 10 cm=1 unit



















Question 4

Sol :
The required horizontal bar graph as shown below
Scale on y-axis 1 unit = 5 cm






















Question 5

Sol :
The required bar graph as shown below
Scale on y-axis 1 unit = 10 cm






















Question 6

Sol :
The histogram is shown below

Scale on x-axis 1 unit=10 cm
Scale on y-axis 1 unit=1 cm





















Question 7

Sol :
The required histogram is shown below
Scale on x-axis 1 unit = 100 cm
Scale on y-axis 1 unit = 5 cm





















Question 8

Sol :
The required histogram as shown below
Scale on x-axis 1 unit = 5 cm
Scale on y-axis 1 unit = 5 cm























Question 9

Sol :
The required histogram as shown below
Scale on x-axis 1 unit = 2 cm
Scale on y-axis 1 unit = 5 cm























Question 10

Sol :
The required histogram as shown below
∴The given frequency distribution is discontinuous , to convert into continuous frequency distribution convert into continuous frequency distribution

Adjustment factor =lower limit of one class - upper limit of previous class / 2
$=\frac{66-65}{2}=\frac{1}{2}$
=0.5

Continuous frequency distribution for the given data is 

Class before adjustment Class after adjustment Frequency
59-65 59.5-65.5 10
66-72 65.5-72.5 5
73-79 72.5-79.5 25
80-86 78.5-86.5 15
87-93 86.5-93.5 30
94-100 93.5-100.5 10

Scale on x-axis 1 unit=6 cm
Scale on y-axis 1 unit=5 cm


















Question 11

Sol :
The required frequency polygon as shown below
Scale on x-axis 1 unit = 10 cm
Scale on y-axis 1 unit = 5 cm




















Question 12

Sol :
The required frequency polygon as shown below
Scale on x-axis 1 unit = 10 cm
Scale on y-axis 1 unit = 5 cm





























Question 13

Sol :
The given frequency distribution is discontinuous, to convert it into a continuous distribution.

Adjustment factor =lower limit of one class - upper limit of previous class / 2

$=\frac{21-20}{2}=\frac{1}{2}$
=0.5

Class before adjustment Class after adjustment Frequency
16-20 16.5-20.5 4
21-25 20.5-25.5 12
26-30 25.5-30.5 18
31-35 30.5-35.5 26
36-40 35.5-40.5 14
41-45 40.5-45.5 10
46-50 45.5-50.5 6

Scale on x-axis 1 unit=4 cm
Scale on y-axis 1 unit=5 cm





















Question 14

Sol :

The required frequency polygon as shown below

Scale on x-axis 1 unit=10 cm
Scale on y-axis 1 unit=5 cm



















Question 15

Sol :
The required histogram and frequency polygon as shown below
Scale on x-axis 1 unit=15 cm
Scale on y-axis 1 unit=5 cm




















Question 16

Sol :

The required histogram and frequency polygon as shown below 

Scale on x-axis 1 unit=10 cm
Scale on y-axis 1 unit=5 cm























Question 17

Sol :
Given water bills of 32 houses in a locality are
30, 48, 52, 78 , 103, 85 ,  37, 94 , 72, 73 ,66, 52 , 92 , 65 ,78 ,81 ,64 ,60 ,75 ,78 , 108 ,63 ,71 ,54 ,59 ,75 ,100 ,103 ,35 ,89 ,95 ,73

Class interval 30-40 40-50 50-60 60-70 70-80 80-90 90-100 100-110
Frequency 3 1 4 5 9 3 3 3

∴The required histogram and frequency polygon as shown below
Scale on x-axis 1 unit = 10 cm
Scale on y-axis 1 unit = 1 cm

















Question 18

Sol :
Class Interval 40-42 42-44 44-46 46-48 48-50 50-52 52-54
Frequency 2 12 7 6 7 4 2

Scale on x-axis 1 unit = 2 cm
Scale on y-axis 1 unit = 1 cm























Question 19

Sol :
(i) 18
(ii) 475-500
(iii) 34
(iv) 
ClassesFrequencyCumulative Frequency
375-40066
400-1251824
425-4501034
450-4752054
475-500458

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

Image
 Exercise 15.1 Question 1 Using the given information, find the value of x in each of the following figures : Sol : (i) ∠ADB and ∠ACB are in the same segment. ∠ADB = ∠ACB = 50° Now in ∆ADB, ∠DAB + X + ∠ADB = 180° (Angles of a triangles) $\Rightarrow 42^{\circ}+x+50^{\circ}=180^{\circ}$ $\Rightarrow x+92^{\circ}=180^{\circ}$ $\Rightarrow x=180^{\circ}-92^{\circ}=88^{\circ}$ (ii) ∠ABD=∠ACD (Angles in the same segment) But ∠ACD=32° ∴∠ABD=32° Now in ΔABD ∠ABD+∠ADB+∠DAB=180° ⇒32°+45°+x=180° ⇒77°+x=180° ⇒x=180°-77°=103° (iii) ∠BAD=∠BCD (Angles in the same segement) But ∠BAD=20° $\therefore \angle B C D=20^{\circ}$ $\because \angle \mathrm{CEA}=90^{\circ}$ $\therefore \angle \mathrm{CED}=90^{\circ}$ Now in $\Delta$ CED, $\angle C E D+\angle B C D+\angle C D E=180^{\circ}$ $\Rightarrow 90^{\circ}+20^{\circ}+x=180^{\circ}$ $\Rightarrow 110^{\circ}+x=180^{\circ}$ $\Rightarrow x=180^{\circ}-110^{\circ}=70^{\circ}$ (iv) In ΔABC ∠ABC+∠ACB+∠BAC=180° (∵Sum of angles of a triangle) $\Rightarrow 69...
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference be...
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S...
Read more